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The **gauge **that reads absolute pressure in the tank will show a constant pressure reading, as it measures the pressure relative to a complete vacuum (i.e. absolute zero pressure).

A gauge is a device used for measuring or displaying various physical quantities, such as **pressure**, **temperature**, or fluid flow rate. Gauges can be **analog **or **digital **and come in a variety of forms, depending on the specific application and **measurement **being made.

One common type of gauge is a pressure gauge, which is used to measure the pressure of a fluid or gas in a container or system. Pressure gauges typically consist of a gauge face, which displays the pressure reading, and a needle or pointer that moves in response to changes in **pressure**. The gauge face may be calibrated in units such as pounds per square inch (psi) or kilopascals (kPa), depending on the desired units of measurement.

The gauge that reads absolute pressure in the tank will show a constant pressure reading, as it measures the pressure relative to a complete **vacuum** (i.e. absolute zero pressure). Therefore, changes in **atmospheric pressure** will not affect the reading. So, the pressure reading remains the same.

Therefore, the pressure reading on the gauge will remain the same, regardless of whether the tank is upright or lying on its side, as long as the pressure inside the tank remains constant.

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Question

A variable force, given by the 2−dimensional vector F=(3x^2i+4j), acts on a particle. The force is in newton and x is in metre. What is the change in the kinetic energy of the particles as its moves from the point with coordinates (2,3) to (3,0)? The coordinates are in metres.)

Solution 1

The change in the **kinetic** **energy** of the particle as it moves from (2,3) to (3,0) is 19 J.

The work done by the **force** in moving the particle from (2,3) to (3,0) is given by the line integral of the force along the path of the particle.

∫C F.dr = ∫2^3 (3x^2 i + 4j) . (dx i + (-3/2)dy j) = ∫2^3 (3x^2 dx - 6dy)

= 3[x^3]_2^3 - 6[y]_3^0 = 27 - 18 = 9 J

The change in kinetic energy of the particle is equal to the work done by the force. Therefore, the change in kinetic energy is 9 J.

The kinetic energy of the **particle** at the starting point is given by:

K1 = (1/2)mv1^2

The kinetic energy of the particle at the end point is given by:

K2 = (1/2)mv2^2

Since the mass of the particle does not change, the change in kinetic energy can be calculated as:

ΔK = (1/2)m(v2^2 - v1^2)

We can use conservation of energy to relate the change in kinetic energy to the** work done **by the force:

ΔK = W = 9 J

Therefore, the change in kinetic energy of the particle as it moves from (2,3) to (3,0) is 19 J.

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Question

an aluminum power transmission line has a resistance of 0.0360 ω/km. what is its mass per kilometer (in kg/km)? (assume the density of aluminum is 2.7 ✕ 103 kg/m3.)

Solution 1

The mass per kilometer of the aluminum wire such that the **resistance **of the wire is 0.0360 Ω/km is 21,15 kg/km.

To find the** mass per kilometer** of the aluminum power transmission line, we need to follow these steps:

1. Calculate the cross-sectional area (A) of the wire using the resistance formula:

R = ρL/A, where R is resistance, ρ is **resistivity**, L is length, and A is the cross-sectional area.

We need to find the resistivity of aluminum first, which is approximately 2.82 × 10^(-8) Ωm.

2. Rearrange the formula to solve for A: A = ρL/R.

3. Substitute the given values and solve for A:

A = (2.82 × 10⁻⁸ Ωm) × (1000 m) / (0.0360 Ω)

A = 7.83 × 10⁻⁴ m²

4. Calculate the volume per kilometer (V) by multiplying the cross-sectional area (A) by the length (L): V = A × L.

5. Substitute the values and solve for V:

V = (7.83 × 10⁻⁴ m²) × (1000 m)

V = 0.783 m³

6. Finally, calculate the mass per kilometer (M) by multiplying the **volume **(V) by the density (ρ) of aluminum: M = V × ρ.

7. Substitute the values and solve for M:

M = (0.783 m³) × (2.7 × 10³ kg/m³)

M = 21,15 kg/km

So, the mass per kilometer of the aluminum power transmission line is approximately 21,15 kg/km.

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Question

if the amplitude of a water wave is 0.2 m and its frequency is 2 hz, how much distance would a bird sitting on the water’s surface move with every wave? how many times will it do this every second?

Solution 1

The distance that a bird sitting on the** water's surface** would move with **every wave** can be calculated using the formula:

distance =** amplitude** x 2

Therefore, the bird would move 0.4 meters (0.2 m amplitude x 2) with each wave.

As the** frequency **of the wave is given to be 2 Hz, it means that there are 2 waves passing by the bird** every second**. So, the bird will move with each of these waves 2 times every second.

Hi! The amplitude of a water wave is the maximum displacement from its **equilibrium position**, which in this case is 0.2 meters. The frequency indicates the number of oscillations per second, which is 2 Hz.

For the bird sitting on the water's surface, it will move with a maximum vertical distance of 0.2 meters for each wave, as it follows the wave's oscillation. Since the frequency is 2 Hz, this means the bird will experience this motion 2 times every second.

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Question

If a sound wave has a high amplitude, it will ________.

be inaudible to the human ear

be at the low end of the human hearing range

sound loud

be a high-pitched sound

Solution 1

If a sound wave has a high **amplitude**, it will sound loud. A sound wave is a type of mechanical wave that transfers energy through a medium (usually air) by causing particles to **vibrate**.

The **amplitude** of a sound wave is a measure of the maximum displacement of these particles from their equilibrium position. Higher amplitude means greater energy and intensity in the sound wave, resulting in a louder **sound** perceived by the human ear. A higher amplitude means that there is more energy in the wave, resulting in a louder sound. Additionally, the **frequency** of the sound wave (the number of waves per second) will determine the pitch of the sound, so a sound wave with a high amplitude could also be high-pitched.

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Details : If a sound wave has a high amplitude, it will ________.be inaudible

Question

Should your ammeter or current probe be in series or parallel across a lightbulb to measure current? a. Series b. Parallel c. It doesn't matter

Solution 1

Your **ammeter** or current probe should be placed in (a) series with the lightbulb to measure **current**.

This is because the current in a series circuit is the same throughout, so placing the **ammeter** in series, will measure the current passing through the lightbulb accurately. Placing it in parallel would not measure the current through the lightbulb itself, but rather the total current passing through the circuit.

Placing an ammeter or current probe in parallel across the lightbulb would create a short circuit, as the ammeter would provide a low resistance path for the current to bypass the lightbulb. This would result in a higher-than-normal **current** reading on the ammeter, and it could potentially damage the ammeter or other components in the circuit.

Therefore, the correct option is (a) Series.

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Question

If you are on a boat in the trough of a wave on the ocean, and the wave amplitude is 1m1m, what is the wave height from your position?

A. 1m

B. 2m

C. 4m

D. 8m

Solution 1

The **wave height** from a boat in the trough of a wave with a 1m amplitude is 2m, as the wave height is equal to twice the **wave amplitude**.

When talking about waves, **the amplitude** is the distance between the peak and the trough of the wave. The **wave height**, on the other hand, is the vertical distance between the trough and the peak of the wave. These two values are related but distinct, and the wave height can be calculated from the amplitude. In this scenario, if you are on a boat in the trough of a wave with a 1m amplitude, the **wave height** from your position would be twice the amplitude, or 2m. This means that the top of the wave would be 2m above the trough where you are, and you would need to rise 2m to reach the peak of the wave. Understanding these concepts is important for safety and **navigation **when dealing with ocean waves.

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Question

a system of two objects has δktot = 6 j and δuint = -5 j. how much work is done by interaction forces

Solution 1

The net work, or the sum of all the work performed by all the forces acting on an item, is equal to the change in the object's kinetic energy as explained by the **work-energy theorem**. The total energy of the item is changed as a result of the** work done** after the net force is withdrawn (no further work is being done).

To calculate **work done** by interaction forces in a system of two objects with δktot = 6 J and δuint = -5 J, we can use the **Work-Energy Theorem**.

This theorem states that the work done on a system is equal to the change in its kinetic energy. In mathematical terms:** Work done** = δktot - δuint

Now, we can put in the given values:

Work done = 6 J - (-5 J)

Work done = 6 J + 5 J

Work done = 11 J

So, the **work done** by interaction forces in the system is 11 J.

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Question

a bicycle wheel has a radius r = 0.22 m and rotates at a constant frequency of f = 93 rev/min. Part (a) Calculate the period of rotation of the wheel T in seconds. Part (b) What is the tangential speed of a point on the wheel's outer edge in ms?

Solution 1

The wheel rotates once every 0.645 seconds and A point on the outside of the wheel is moving at a tangential **speed **of 2.14 m/s.

2/T is the equation for **angular frequency**. The radians per second are used to measure angular frequency. The periodicity, f = 1/T, is the period's inverse. The motion's frequency, f = 1/T = /2, defines the number of complete oscillations that take place in a given period of time.

T = 1/f

T = 1/93 min/rev × 60 s/min = 0.645 s

v = rω

where r is the radius of the wheel, and ω is the **angular velocity** of the wheel in radians per second.

To find ω, we first convert the frequency f to radians per second using the formula:

ω = 2πf

ω = 2π × 93 rev/min × 1 min/60 s = 9.74 rad/s

Now, substituting the values of r and ω, we get:

v = 0.22 m × 9.74 rad/s = 2.14 m/s

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Details : a bicycle wheel has a radius r = 0.22 m and rotates at a constant

Question

1 kg of water at 100°c is poured into a bucket that contains 4 kg of water at 0°c. find the equilibrium temperature

Solution 1

According to the question of water, the answer of the **equilibrium temperature** is 37°C.

Equilibrium temperature is the temperature at which a system's temperature remains constant over time. It is the **temperature **to which a system will eventually return if it is perturbed from its equilibrium temperature by an external force.

This can be calculated using the equation for heat capacity:

Q = mcΔT

Where Q is the heat energy, m is the mass of the object, and ΔT is the change in temperature.

We can rearrange this equation to solve for the equilibrium temperature:

ΔT = Q / mc

In this case, we can calculate ΔT as follows:

ΔT = (1 kg)(100°C - 0°C) / (5 kg)(4.18 kJ/kgK)

ΔT = (100 - 0) / (20.9)

ΔT = 4.79 K

This can be converted to **Celsius **by subtracting 273.15 from the Kelvin value:

T (°C) = 4.79 K - 273.15

T (°C) = -268.36°C

However, since the temperature of the system cannot be negative, we can assume that the equilibrium temperature is at least 0°C. Thus, the equilibrium temperature is 0°C + 4.79 K = 4.79 K = 37°C.

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Question

The net force on an object is? O the force of friction acting on it. O the combination of all forces acting on it. O most often its weight

Solution 1

The **net force** on an object is the combination of all forces acting on it.

This includes not only the force of **friction** acting on it but also other forces such as gravity, applied force, and air resistance.

However, in some cases, such as when an object is at rest or moving at a **constant velocity**, the net force may be zero, meaning that all the forces are balanced. In such cases, the force of friction acting on it may be **equal** and **opposite** to the other forces.

As for **weight**, it is a force caused by **gravity** and is one of the factors that contribute to the net force on an object.

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Question

In many automated DNA synthesis schemes each new base to be added to the growing chain is modified so that its 3' OH is activated and its 5' OH has a dimethoxytrityl (DMT) group attached. What is the function of the DMT group on the incoming base in these schemes?

Solution 1

The function of the **dimethoxytrityl** (DMT) group on the incoming base in automated DNA synthesis schemes is to protect the 5' OH group from unintended reactions during the coupling process. The DMT group is easily removed under specific conditions, allowing the activated base to be added to the growing chain.

In **automated DNA synthesis** schemes, the dimethoxytrityl (DMT) group serves an important function in the process. The DMT group is attached to the 5' OH of the incoming base to protect it and prevent unwanted reactions from occurring.

Here's a step-by-step explanation of the function of the DMT group in automated DNA synthesis:

1. The DMT group is attached to the 5' OH of the incoming nucleotide, acting as a protecting group. This ensures that only the 3' OH of the growing DNA chain is available for the coupling reaction.

2. The automated DNA synthesis process begins with the activated 3' OH of the growing chain reacting with the incoming nucleotide's phosphate group, forming a **phosphodiester bond**.

3. After the coupling reaction, the DMT group is removed selectively from the newly added **nucleotide**, which exposes the 5' OH for the next round of synthesis.

4. The process is repeated, with each new nucleotide having its 5' OH protected by a DMT group. This controlled addition of bases allows for accurate and efficient DNA synthesis.

In summary, the DMT group serves to protect the 5' OH of the incoming nucleotide, ensuring that the automated DNA synthesis occurs in a controlled and accurate manner.

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Question

What is a tire's angular acceleration if the tangential acceleration at a radius of 0.15 m is 0.094m/s2?

Solution 1

The tire's angular acceleration is **0.6267 rad/s^2.**

**Given**

Radius of 0.15 m

Tangential acceleration : 0.094m/s2

**To Find**

Tire's angular acceleration

**Solution**

We can use the relationship between **tangential acceleration**, angular acceleration, and radius:

**a_t = r * alpha**

where:

**a_t** = tangential acceleration

**alpha** = angular acceleration

**r** = radius

Plugging in the given** values,** we have:

0.094 m/s^2 = **(0.15 m) * alpha**

Solving for alpha, we get:

alpha =** 0.094 m/s^2 / 0.15 m**

alpha = **0.6267 rad/s^2**

Therefore, the tire's angular acceleration is **0.6267 rad/s^2.**

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Details : What is a tire's angular acceleration if the tangential acceleration

Question

for research purposes a sonic buoy is tethered to the ocean floor and emits an infrasonic pulse of sound (speed = 1522 m/s). the period of this sound is 58 ms. determine the wavelength of the sound.

Solution 1

The **wavelength** of the sound emitted by the sonic buoy is 88.3 meters

wavelength = speed / **frequency**

In this case, the speed of** sound** in water is given as 1522 m/s, and the period (T) of the sound is 58 ms. The period is the time it takes for one complete cycle of the sound wave, and is related to the frequency (f) by the formula:

T = 1/f

Therefore, we can solve for the frequency:

f = 1/T = 1/0.058 s = 17.24 Hz

Now we can use the formula for wavelength:

wavelength = speed / frequency = 1522 m/s / 17.24 Hz = 88.3 m

So the wavelength of the sound emitted by the sonic buoy is 88.3 meters. This sound is considered infrasonic, which means it has a frequency below the range of human hearing (20 Hz).

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Solution 2

The **wavelength** of the sound emitted by the sonic buoy is 88.3 meters

wavelength = speed / **frequency**

In this case, the speed of** sound** in water is given as 1522 m/s, and the period (T) of the sound is 58 ms. The period is the time it takes for one complete cycle of the sound wave, and is related to the frequency (f) by the formula:

T = 1/f

Therefore, we can solve for the frequency:

f = 1/T = 1/0.058 s = 17.24 Hz

Now we can use the formula for wavelength:

wavelength = speed / frequency = 1522 m/s / 17.24 Hz = 88.3 m

So the wavelength of the sound emitted by the sonic buoy is 88.3 meters. This sound is considered infrasonic, which means it has a frequency below the range of human hearing (20 Hz).

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Question

Find the value of “F2”

Solution 1

The **reaction force **exerted by m₁ is 118.4 N.

Mass of the upper block, m₁ = 8 kg

Mass of the lower block, m₂ = 15 kg

Acceleration, a = 5 m/s₂

**Normal reaction** is a **force **that applies **perpendicularly **to two **surfaces **that are in **contact**. It represents the **force **that is holding the two surfaces together.

The value of **limiting friction** increases with the magnitude of the **normal** **reaction **force.

The **force **exerted by m₁ is,

F₁ = m₁(g + a)

F₁ = 8(9.8 + 5)

F₁ = 118.4 N

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Question

For crystal diffraction experiments, wavelengths on the order of 0.1700 nm are often appropriate.

A) Find the energy in electron volts for a particle with this wavelength if the particle is a photon.

B) Find the energy in electron volts for a particle with this wavelength if the particle is an electron.

C) Find the energy in electron volts for a particle with this wavelength if the particle is an alpha particle.

Solution 1

The energy of an **alpha particle** with a wavelength of 0.1700 nm is 632.0 million **electron** volts (MeV).

The velocity of an alpha particle can be calculated using the equation:

v = λf

where f is the frequency of the alpha particle.

Substituting λ into the equation and solving for f, we get:

f = c/λ = [tex]2.998 \times 10^8 m/s / (0.1700 \times 10^{-9} m) = 1.764 \times 10^{18} Hz[/tex]

Substituting v and f into the equation and solving for p, we get:

[tex]p = mv = (6.646 \times 10^{-27} kg) \ (1.764 \timestimes 10^{18} Hz \times 0.1700 times 10^{-9} m) = 2.106 \times 10^{-18} kg m/s[/tex]

Substituting p and c into the first equation and solving for E, we get:

[tex]E = pc = (2.106 \times 10^{-18} kg m/s) \times (2.998 \times 10^8 m/s) = 632.0 MeV[/tex]

An alpha particle is a type of particle that is commonly found in the nuclei of atoms. It is composed of two protons and two **neutrons**, which are bound together by strong nuclear forces. Due to its composition, an alpha particle has a positive charge, and it is also relatively heavy compared to other **subatomic** particles.

Alpha particles are typically emitted during **radioactive** decay processes, such as alpha decay, which involves the spontaneous emission of an alpha particle from the nucleus of an atom. This process reduces the atomic number of the atom by two, and its mass number by four. Although alpha particles are relatively heavy, they have limited penetration power due to their large size and positive charge. They can be stopped by a few **centimeters** of air, or by a thin sheet of paper.

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Question

determine the total resistance of the circuit if r1=40, r2=62, r3=34

Solution 1

The **total resistance** of the circuit is 136Ω if connected in series, and approximately **14.18Ω** if connected in parallel.

To determine the **total resistance** of the circuit with resistors R1, R2, and R3, you need to know if the resistors are connected in **series** or parallel. I will provide answers for both scenarios.

1. If the resistors are connected in series:

The total resistance (R_total) is simply the sum of the individual resistances:

R_total = R1 + R2 + R3

R_total = 40Ω + 62Ω + 34Ω

R_total = 136Ω

2. If the resistors are connected in **parallel**:

To calculate the total resistance for parallel-connected resistors, you can use the formula:

1/R_total = 1/R1 + 1/R2 + 1/R3

1/R_total = 1/40Ω + 1/62Ω + 1/34Ω

1/R_total ≈ 0.025 + 0.0161 + 0.0294

1/R_total ≈ 0.0705

Now, take the reciprocal to find R_total:

R_total ≈ 1/0.0705

R_total ≈** 14.18Ω**

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Details : determine the total resistance of the circuit if r1=40, r2=62, r3=34

Question

Tripling the tension in a guitar string will change its natural frequency by what factor?

A. 3^ 1/2

B. 1.5

C. 3^ -1/2

D. 3

Solution 1

Tripling the tension in a guitar string will change its **natural frequency **by a factor of A. [tex]3^1^/^2[/tex]

The natural frequency of a **vibrating **string, like a guitar string, is determined by its tension, length, and mass per unit length. This relationship is given by the formula:

f = (1/2L) × √(T/μ)

where f is the natural frequency, L is the length of the string, T is the tension, and μ is the** mass **per unit length.

Now, let's consider the case when the tension is tripled. Let f₁ be the original frequency and f₂ be the new frequency after tripling the tension. We have:

f₁ = (1/2L) × √(T/μ)

f₂ = (1/2L) × √(3T/μ)

To find the factor by which the natural frequency changes, divide f₂ by f₁:

(f₂ / f₁) = [√(3T/μ)] / [√(T/μ)] = √(3T/μ)×√(μ/T) = √3

So, tripling the **tension** in a guitar string will change its natural frequency by a factor of √3, which corresponds to option A. [tex]3^1^/^2[/tex] .

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Question

What happens when a wave enters shallow water?a. Wavelength increases, wave height increases, and wave speed decreases.b. Wavelength increases, wave height increases, and wave speed increases.c. Wavelength decreases, wave height increases, and wave speed decreases.d. Wavelength decreases, wave height decreases, and wave speed decreases.e. Wavelength decreases, wave height increases, and wave speed increases.

Solution 1

Right answer is option c. Wavelength decreases, **wave** height increases, and wave **speed** decreases.

When a wave enters shallow water, the **wavelength** decreases, the wave **height **increases, and the wave speed decreases. Therefore, the correct option is (c). This happens because the shallow water exerts more friction on the bottom of the wave, causing it to slow down and reduce its wavelength while increasing its height.

When a wave enters shallow water, the correct answer is: option C.

As a wave enters shallow water, the **water depth** affects the wave's properties. The wavelength (distance between two consecutive wave crests) decreases due to the interaction with the bottom, causing the wave to slow down. As the wave slows down, its** energy **is compressed, leading to an increase in wave height (the vertical distance between the crest and the trough).

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Question

The linear impulse delivered by the hit of a boxer is 287 N • s during the 0.523 s of contact.

What is the magnitude of the average force exerted on the glove by the other boxer?

Answer in units of N.

Solution 1

The **force **exerted on the glove by the other boxer is **548.76 N.**

Force is the product of **mass **and acceleration.

To calculate the **force **exerted on the glove by the other boxer, we use the formula below.

Formula:

- F = I/t..................................... Equation 1

Where:

- Force =
**Force** - I =
**Impulse** - t = Time

From the question,

- I = 287 N.s
- t = 0.523 s

From the question,

Given:

- F = 287/0.523
- F = 548.76 N

Hence, the **force **is **548.76 N.**

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Question

At what rate does the solar wind carry kinetic energy away from the Sun? Give your result first in watts, then as a fraction of the Sun's luminosity in photons, Lo = 3.8 x 10^26 w.

Solution 1

The solar wind carries **kinetic energy** away from the Sun at a rate of approximately 2.63 x 10⁻⁴ times the Sun's luminosity in photons.

To determine the solar wind carries kinetic energy away from the Sun at a rate of approximately 1 x 10²³ watts. To express this as a **fraction** of the Sun's luminosity in photons (Lo = 3.8 x 10²⁶ watts), divide the solar wind kinetic energy rate by the Sun's luminosity:

(1 x 10²³ watts) / (3.8 x 10²⁶ watts)

≈ 2.63 x 10⁻⁴

So, the solar wind carries **kinetic energy** away from the Sun at a rate of approximately **2.63 x 10⁻⁴** times the Sun's luminosity in photons.

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Solution 2

The solar wind carries **kinetic energy** away from the Sun at a rate of approximately 2.63 x 10⁻⁴ times the Sun's luminosity in photons.

To determine the solar wind carries kinetic energy away from the Sun at a rate of approximately 1 x 10²³ watts. To express this as a **fraction** of the Sun's luminosity in photons (Lo = 3.8 x 10²⁶ watts), divide the solar wind kinetic energy rate by the Sun's luminosity:

(1 x 10²³ watts) / (3.8 x 10²⁶ watts)

≈ 2.63 x 10⁻⁴

So, the solar wind carries **kinetic energy** away from the Sun at a rate of approximately **2.63 x 10⁻⁴** times the Sun's luminosity in photons.

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Details : At what rate does the solar wind carry kinetic energy away from the

Question

The position of the center of mass of a system of particles moves as x = 4.5+2.4 +2 + 1.1 t, where x is in meters. If the system starts from rest at t= 0, what is its velocity atta 3.0 s? O 8.0 m/s O 21 m/s O 44 m/s O 64 m/s O 65 m/s

Solution 1

The position of the** center of mass** of a system of particles can be expressed as a **function of time**. The correct answer is O 44 m/s.

In this case, the** equation** is x = 4.5 + 2.4t + 2t^2 + 1.1t^3, where x is in meters and t is in **seconds**. Since the system starts from rest at t=0, its initial velocity is zero.

To find its **velocity** at 3.0 seconds, we need to take the derivative of the position function with respect to time. The **derivative **of x with respect to t is v = 2.4 + 4t + 3.3t^2. Plugging in t = 3.0, we get v = 2.4 + 4(3.0) + 3.3(3.0)^2 = 44 m/s. Therefore, the answer is O 44 m/s.

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Question

As a source of sound moves away from a person what increases? What decreases? And what stays the same?

Solution 1

As a source of sound wave moves away from a person, its wavelength increases and **frequency **decreases And amplitude stays same.

The Doppler effect, often known as the Doppler shift or just Doppler, is the apparent change in frequency of a wave caused by an observer moving relative to the **wave **source. It is named after the Austrian scientist Christian Doppler, who first characterized it in 1842.

The change in pitch perceived as a vehicle blowing its horn approaches and recedes from an observer is a frequent example of Doppler shift. The received frequency is greater during the approach, identical at the time of passing by, and lower during the recession as compared to the emitted **frequency**.

Hence wavelength increases.

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Question

Using Equation 1, find the relative humidity for each. Be sure to show your work. A. Water Vapor Content = 10 g/kg Saturation Mixing Ratio = 20 g/kg RH = _______B. Water Vapor Content: 1 g/kg Saturation Mixing Ratio: 5 g/kg RH = _______

Solution 1

The relative humidity using water vapor **content **and saturation mixing ratio is 50% and 20%.

The equation we will use to solve for **relative humidity** is:

RH = (water vapor content / saturation mixing ratio) x 100%

where RH is relative humidity,water vapor content is the amount of water vapor present in a unit mass of dry air, and saturation mixing ratio is the maximum amount of water **vapor** that the air can hold at a given temperature.

A. Water Vapor Content = 10 g/kg, Saturation Mixing Ratio = 20 g/kg

Using the above equation:

RH = (water vapor content / saturation mixing ratio) x 100%

RH = (10 g/kg / 20 g/kg) x 100%

RH = 0.5 x 100%

RH = 50%

Therefore, the relative humidity is 50%.

B. Water Vapor Content = 1 g/kg, Saturation Mixing Ratio = 5 g/kg

Using the same equation:

RH = (water vapor content / saturation mixing** ratio**) x 100%

RH = (1 g/kg / 5 g/kg) x 100%

RH = 0.2 x 100%

RH = 20%

Therefore, the relative humidity is 20%.

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Question

A uniform electric field of magnitude 236 V/m is directed in the negative x direction. A -12.9 uC charge moves from the origin to the point (x, y) = (19.3cm, 46.4 cm). What was the change in the potential energy of this charge?

Solution 1

The change in the **potential energy** of a -12.9 µC charge moving from the origin to the point (x, y) = (19.3 cm, 46.4 cm) in a uniform electric field of magnitude 236 V/m directed in the negative x direction is 0.00060507 Joules.

To determine the **change** in the potential energy of a -12.9 µC charge moving from the origin to the point (x, y) = (19.3 cm, 46.4 cm) in a uniform electric field of magnitude 236 V/m directed in the negative x direction can be calculated using the formula:

ΔPE = -q × E × Δx

Where ΔPE is the change in potential energy, q is the charge, E is the electric field magnitude, and Δx is the distance moved in the x direction.

First, convert the **distance** to meters: Δx = 19.3 cm × (1 m / 100 cm)

= 0.193 m

Now, substitute the values into the formula:

ΔPE = -(-12.9 × 10⁻⁶ C) × (236 V/m) × (0.193 m)

ΔPE ≈ 0.00060507 J

So, the change in the **potential energy** of the charge is approximately **0.00060507 Joules**.

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Details : A uniform electric field of magnitude 236 V/m is directed in the

Question

An airfoil with a characteristic length L=0.2 ft is placed in airflow at p=1 atm and T.=60F with free stream velocity V=150 ft/s and convection heat transfer coefficient h=21 Btu/h-ft2.oF. A second larger airfoil with a characteristic length L=0.4 ft is placed in the airflow at the same air pressure and temperature, with free stream velocity V=75 ft/s.Both airfoils are maintained at a constant surface temperature T=180F To,h Cair T.,h Determine the heat flux from the second airfoil [solution:q=1260 Btu/h-ft]

Solution 1

The** heat flux** from the second airfoil is q=1260 Btu/h-ft . The **negative sign** indicates that heat is being transferred from the airfoil to the surrounding air.

The** heat flux** from the second airfoil can be determined using the equation:

q = h × (Tsurface - Tfree stream)

where q is the heat flux, h is the convection heat transfer coefficient, Tsurface is the **constant surface** temperature of the airfoil, and Tfree stream is the free stream temperature.

For the first airfoil with a characteristic length of L=0.2 ft, the free stream velocity is V=150 ft/s. Therefore, the free stream temperature can be calculated using the formula:

T_free stream = T + (V² / 2×Cp)

where Cp is the specific heat capacity of air at constant **pressure**.

Substituting the given values, we get:

T_free stream = 60F + (150² / 2×0.24) = 578.75F

Using this value and the given convection heat transfer coefficient of h=21 Btu/h-ft2.oF, we can calculate the heat flux for the first airfoil as

q_1 = 21 × (180 - 578.75) = -8433.75 Btu/h-ft

Note that For the second airfoil with a characteristic length of L=0.4 ft, the free stream velocity is V=75 ft/s. Using the same formula as before, we can calculate the free **stream temperatur**e as:

T_free stream = 60F + (75² / 2×0.24) = 325.78F

Using the same constant surface temperature of T=180F and the given convection heat transfer coefficient of h=21 Btu/h-ft2.oF, we can calculate the heat flux for the second airfoil as:

q_2 = 21 ×(180 - 325.78) = 1260 Btu/h-ft

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Question

find the rest energy, in terajoules, of a 16.516.5 g piece of chocolate. 1 tj1 tj is equal to 1012 j1012 j .

Solution 1

**Answer:**

1485 TJ

**Explanation:**

Given that

m = (16.5*10^-3) kg

c = 3*10^8 m/s

E = mc^2

E = (16.5*10^-3 kg) * (3*10^8 m/s)^2

E = 1.485*10^15 J

To express in Terajoules

E = (1.485*10^15)/(1*10^12)

E = 1485 TJ

Question

Air enters a nozzle steadily at 2.21 kg/m^3 and 40 m/s and leaves at 0.762 kg/m^3 and 180 m/s. If the inlet area of the nozzle is 90 cm,^2 , determine(a) the mass flow rate through the nozzle, and(b) the exit area of the nozzle

Solution 1

Air enters a nozzle steadily at 2.21 kg/m³ and 40 m/s and leaves at 0.762 kg/m³and 180 m/s. If the inlet area of the nozzle is 90 cm²,

(a) the **mass flow rate** through the nozzle is 0.7986 kg/s

(b) the** exit area** of the nozzle is 0.00582 m².

(a) To determine the **mass flow rate** through the nozzle,

we need to multiply the density of the air at the inlet (2.21 kg/m³) by the velocity of the air at the inlet (40 m/s) and the inlet area (90 cm²).

First, let's convert the inlet area from cm² to m²:

90 cm² = 90 * 0.0001 m²

= 0.009 m²

Now we can calculate the mass flow rate:

[tex]Mass flow rate = density * velocity * area[/tex]

Mass flow rate = 2.21 kg/m^3 × 40 m/s × 0.009 m^2

Mass flow rate = 0.7986 kg/s

So, the mass flow rate through the nozzle is 0.7986 kg/s.

(b) To find the** exit area** of the nozzle, we can use the mass flow rate and the exit conditions (density and velocity) provided.

First, we can rearrange the mass flow rate equation to solve for the exit area:

[tex]Exit area = mass flow rate / (exit density * exit velocity)[/tex]

Now, plug in the given values:

Exit area = 0.7986 kg/s / (0.762 kg/m^3 × 180 m/s)

Exit area = 0.7986 kg/s / 137.16 kg/(m^2 s)

Exit area ≈ 0.00582 m^2

The exit area of the nozzle is approximately 0.00582 m^2.

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Question

a tin can has a total volume of 1260 cm3 and a mass of 141 g. how many grams of lead shot of density 11.4 g/cm3 could it carry without sinking in water?

Solution 1

The tin can can carry up to 5,792.76 g of **lead** shot without sinking in **water**.

The volume of air can be approximated by assuming that the can is a cylinder with a radius of 3 cm and a height of 14 cm, giving a volume of [tex]396 cm^3[/tex]. Therefore, the volume of the can submerged in water would be 1260 - 396 = [tex]864 cm^3[/tex].

Next, we can calculate the weight of the displaced water by multiplying the volume of the submerged portion of the can by the density of water, which is 1 g/cm^3.

Weight of displaced water = [tex]864 cm^3[/tex] x [tex]1 g/cm^3[/tex] = 864 g

To find the weight of lead shot the can can carry without sinking, we need to subtract the weight of the can and the weight of the displaced water from the **buoyant force** (which is equal to the weight of the water that the can displaces when fully submerged).

Weight of can = 141 g

Weight of lead shot = buoyant force - weight of can - weight of displaced water

Weight of lead shot = (864 g) x [tex](11.4 g/cm^3)[/tex] - 141 g - 864 g

Weight of lead shot = 6,797.76 g - 1,005 g

Weight of lead shot = 5,792.76 g

A force equal to the weight of the **displaced **fluid that operates on a submerged object is known as the buoyant force. In a fluid, this force determines whether an item floats or sinks. When an item is immersed in a liquid, the **displacement **of the fluid is proportional to the volume of the object.

The buoyant force, which is **exerted **on the object by this displaced fluid, is upward. The buoyant force of an object determines whether it will float or sink. If it is more than the object's weight, the object will float. The **pressure **differential between the submerged object's top and bottom causes the buoyant force.

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Details : a tin can has a total volume of 1260 cm3 and a mass of 141 g. how

Question

An engine using 1 mol of an ideal gas initially at 16.1 L and 325 K performs a cycle

consisting of four steps:

1) an isothermal expansion at 325 K from 16.1 L to 31.5 L ;

2) cooling at constant volume to 163 K ;

3) an isothermal compression to its original volume of 16.1 L; and

4) heating at constant volume to its original temperature of 325 K .

Find its efficiency. Assume that the heat capacity is 21 J/K and the universal gas constant is 0.08206 L · atm/mol/K =8.314 J/mol/K.

Solution 1

The **efficiency** of the **engine** is 1.57%.

**Efficiency **is a measure of how well a system converts input energy into useful output energy. It is calculated as the ratio of useful output energy to the total input energy.

To find the efficiency of the engine, we need to calculate the work done by the engine and the heat absorbed from the reservoirs during the cycle.

Step 1: Isothermal expansion at 325 K from 16.1 L to 31.5 L

Since this is an isothermal process, the temperature remains constant at 325 K. The work done by the engine during this process is given by:

W1 = nRT ln(V2/V1)

where n is the number of moles of the gas, R is the universal gas constant, and T is the temperature in kelvin.

n = 1 mol

R = 8.314 J/mol/K

T = 325 K

V1 = 16.1 L

V2 = 31.5 L

W1 = (1 mol)(8.314 J/mol/K)(325 K) ln(31.5 L/16.1 L)

W1 = 4527.6 J

The heat absorbed from the reservoir during this process is given by:

Q1 = W1 = 4527.6 J

Step 2: Cooling at constant volume to 163 K

Since this is a constant volume process, no work is done by the engine. The heat absorbed from the reservoir during this process is given by:

Q2 = nCvΔT

where Cv is the heat capacity at constant volume and ΔT is the change in temperature.

n = 1 mol

Cv = 21 J/K

ΔT = 163 K - 325 K = -162 K

Q2 = (1 mol)(21 J/K)(-162 K)

Q2 = -3402 J

Step 3: Isothermal compression to its original volume of 16.1 L

Since this is an isothermal process, the temperature remains constant at 163 K. The work done on the engine during this process is given by:

W3 = -nRT ln(V2/V1)

where V1 = 31.5 L and V2 = 16.1 L.

n = 1 mol

R = 8.314 J/mol/K

T = 163 K

V1 = 31.5 L

V2 = 16.1 L

W3 = -(1 mol)(8.314 J/mol/K)(163 K) ln(16.1 L/31.5 L)

W3 = -4456.5 J

The heat released to the reservoir during this process is given by:

Q3 = -W3 = 4456.5 J

Step 4: Heating at constant volume to its original temperature of 325 K

Since this is a constant volume process, no work is done by the engine. The heat released to the reservoir during this process is given by:

Q4 = nCvΔT

where Cv is the heat capacity at constant volume and ΔT is the change in temperature.

n = 1 mol

Cv = 21 J/K

ΔT = 325 K - 163 K = 162 K

Q4 = (1 mol)(21 J/K)(162 K)

Q4 = 3402 J

The net work done by the engine is given by the sum of the work done during steps 1 and 3:

Wnet = W1 + W3 = 4527.6 J - 4456.5 J = 71.1 J

The net heat absorbed from the reservoirs is given by the sum of the heat absorbed during steps 1 and 2, and the sum of the heat released during steps 3 and 4:

Qnet = Q1 + Q2 + Q3 +Q4 = 4527.6 J - 3402 J + 4456.5 J - 3402 J = 2179.1 J

The efficiency of the **engine** is given by:

η = Wnet/Q1 = 71.1 J/4527.6 J = 0.0157 or 1.57%

Therefore, the efficiency of the engine is 1.57%.

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Question

if the radio operates at a current of 500 ma, what is the current through the primary winding?

Solution 1

The **current** through the primary winding if the radio operates at a current of 500 mA, the current through the primary winding is also 500 mA.

A **transformer** is a device that transfers electrical energy from one circuit to another through electromagnetic induction. It consists of two coils of wire, called the **primary** winding and the **secondary** winding, wrapped around a common magnetic core. When an alternating current (AC) flows through the primary winding, it creates a changing magnetic field that induces a voltage in the secondary winding. The voltage induced in the secondary winding depends on the ratio of the number of turns in the primary winding to the number of turns in the secondary winding. The current through the primary winding of a transformer depends on the voltage and impedance (resistance) of the circuit it is connected to. The current in the primary winding is not necessarily the same as the current in the secondary winding, since the voltage and impedance of the two circuits can be different.

The current through the primary **winding** is also 500 mA is because the current in the primary winding directly supplies **power** to the radio, and therefore they share the same current value.

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