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3 years ago

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A gas expands from an initial volume of 0.040 m^3 and an initial pressure of 210 kPa to a final volume of 0.065 m^3 while its te

mperature is kept constant. How much work is done by the system?

1 answer:

USPshnik [31]3 years ago

7 0

Answer:

286

Explanation:

p1v1/T1=p2v2/T2

then subtitude your values

T2=760*0.65*273/210*0.040

T2=135/8.4=16+273=289

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An experiment to measure the speed of light uses an apparatus similar to Fizeau's. The distance between the light source and the

Marina86 [1]

Answer:

$2.88*10^{8} m/s$

Explanation:

The speed of light is given by

$c=\frac {2d}{t}$ and $t=\frac {\theta}{\omega}$ hence

$c=\omega\frac {2d}{\theta}$

Speed of light is given by

$c=900\times 2\pi(\frac {2\times 10}{\frac {2\pi}{2*800}}})=2.88*10^{8} m/s$

4 0

3 years ago

The combination of an applied force and a friction force produces a constant total torque of 35.5 N · m on a wheel rotating abou

Cerrena [4.2K]

Answer:

a) $I = 19.799\,kg\cdot m^{2}$ , b) $T = -3.405\,N\cdot m$ , c) $n_{T} \approx 54.842\,rev$

Explanation:

a) The net torque is:

$T = I\cdot \alpha$

Let assume a constant angular acceleration, which is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{10.4\,\frac{rad}{s} - 0\,\frac{rad}{s} }{5.80\,s}$

$\alpha = 1.793\,\frac{rad}{s^{2}}$

The moment of inertia of the wheel is:

$I = \frac{T}{\alpha}$

$I = \frac{35.5\,N\cdot m}{1.793\,\frac{rad}{s^{2}} }$

$I = 19.799\,kg\cdot m^{2}$

b) The deceleration of the wheel is due to the friction force. The deceleration is:

$\alpha = \frac{\omega-\omega_{o}}{t}$

$\alpha = \frac{0\,\frac{rad}{s} - 10.4\,\frac{rad}{s}}{60.4\,s}$

$\alpha = - 0.172\,\frac{rad}{s^{2}}$

The magnitude of the torque due to friction:

$T = (19.799\,kg\cdot m^{2})\cdot (-0.172\,\frac{rad}{s^{2}} )$

$T = -3.405\,N\cdot m$

c) The total angular displacement is:

$\theta_{T} = \theta_{1} + \theta_{2}$

$\theta_{T} = \frac{(10.4\,\frac{rad}{s} )^{2}-(0\,\frac{rad}{s} )^{2}}{2\cdot (1.793\,\frac{rad}{s^{2}} )} + \frac{(0\,\frac{rad}{s} )^{2}-(10.4\,\frac{rad}{s} )^{2}}{2\cdot (-0.172\,\frac{rad}{s^{2}} )}$

$\theta_{T} = 344.580\,rad$

The total number of revolutions of the wheel is:

$n_{T} = \frac{\theta_{T}}{2\pi}$

$n_{T} = \frac{344.580\,rad}{2\pi}$

$n_{T} \approx 54.842\,rev$

5 0

3 years ago

What is the purpose of a tuner in a radio?

Crank

Answer:

A tuner is a subsystem that receives radio frequency (RF) transmissions and converts the selected carrier frequency and its associated bandwidth into a fixed frequency that is suitable for further processing, usually because a lower frequency is used on the output.

hope it helps ya mate.

7 0

3 years ago

You are designing a new piece of fishing equipment that will allow you to catch a fish from the surface. You know that water ref

OleMash [197]

Answer:

Law of refraction

Explanation:

An experiment to analyze the refraction of light in water can easily be performed with a laser pointer and protractor.

We throw the fishing rod line into the water, place the protractor at the point where the line touches the water and use the direction of the line for the direction of the laser pointer (on), the laser is visible by the reflection on the particles in the air.

The vertical line is called Normal and all angles must be measured with respect to this reference in optics.

Having these angles and the refractive index of water we can use the law of refraction

n₁ sin θ₁ = n₂ sin θ₂

θ₂ = $sin^{-1} ( \frac{n_1}{n_2} \ sin \ \theta_1 )$

we can repeat several times to analyze several different input points (different angles) and to decrease the errors in the measurements.

the refractive index of air is n1 = 1 and n2= 1.33 (water)

4 0

3 years ago

Calculate the maximum deceleration (in m/s2) of a car that is heading down a 14° slope (one that makes an angle of 14° with the

lesya [120]

The question is incomplete. Here is the complete question.

Calculate the maximum deceleration of a car that is heading down a 14° slope (one that makes an anlge of 14° with the horizontal) under the following road conditions. You may assum that the weight of the car is evenlydistributed on all four tires and that the sttic coefficient of friction is involved - that is, the tires are not allowed to slip during the deceleration. (Ignore rolling) Calculate for a car: (a) On a dry concrete. (b) On a wet concrete. (c) On ice, assuming that μs = 0.100, the same as for shoes on ice.

Answer: (a) a = - 11.05 m/s²; (b) a = - 10.64 m/s²; (c) a = - 9.84m/s²

Explanation: The image in the attachment describe the forces acting on the car. Observing that, we know that:

$F_{net}$ = - $W_x$ - $f_s$

The $W_x$ is a x-component of force due to gravity (W) and, in this case, is given by: $W_x$ = W.sin(14)

W is described as: W = m.g

Force due to friction ( $f_s$ ) is given by: $f_s$ = μs.N

N is the normal force and, in the system, is equivalent of $W_y$ , so:

$W_y$ = m.g.cos(14)

Therefore, the formula will be:

$F_{net}$ = - $W_x$ - $f_s$

m.a = - (m.g.sin14) - (μs.mg.cos14)

a = - g (sin14 + μscos 14)

a) For dry concrete, μs = 1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 1.cos14)

a = - 11.05 m/s²

b) For wet concrete, μs = 0.7:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin 14 + 0.7.cos14)

a = - 10.64 m/s²

c) For ice, μs = 0.1:

a = - g (sin14 + μscos 14)

a = - 9.8 (sin14 + 0.1cos14)

a = - 9.84 m/s²

3 0

3 years ago