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2 years ago

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A 2.0kg cart moving right at 5.0 m/s on a frictionless track collided and sticks to a 3.0 kg cart moving left at 3.0 m/s. What i

s the final speed of the carts?

1 answer:

Mrac [35]2 years ago

8 0

Answer:

the final speed is 0.20

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A car is travelling at a constant speed of 26.5 m/s. Its tires have a radius of 72 cm. If the car slows down at a constant rate

maksim [4K]

Answer:

Magnitude of angular acceleration = -3.95 rad/s²

Explanation:

Angular acceleration is the ratio of linear acceleration and radius.

That is

$\texttt{Angular acceleration}=\frac{\texttt{Linear acceleration}}{\texttt{Radius}}\\\\\alpha =\frac{a}{r}$

Radius = 72 cm = 0.72 m

Linear acceleration is rate of change of velocity.

$a=\frac{11.7-26.5}{5.2}=-2.85m/s^2$

Angular acceleration

$\alpha =\frac{a}{r}=\frac{-2.85}{0.72}=-3.95rad/s^2$

Angular acceleration = -3.95 rad/s²

Magnitude = 3.95 rad/s²

4 0

3 years ago

I need to find 1).a,b,c

Aleksandr [31]

Let's cut through the weeds and the trash
and get down to the real situation:

A stone is tossed straight up at 5.89 m/s .
      Ignore air resistance.

Gravity slows down the speed of any rising object by 9.8 m/s every second.
So the stone (aka Billy-Bob-Joe) continues to rise for

   (5.89 m/s / 9.8 m/s²) = 0.6 seconds.

At that timer, he has run out of upward gas. He is at the top
of his rise, he stops rising, and begins to fall.

His average speed on the way up is (1/2) (5.89 + 0) = 2.945 m/s .

Moving for 0.6 seconds at an average speed of 2.945 m/s,
he topped out at

(2.945 m/s) (0.6 s) = 1.767 meters above the trampoline.

With no other forces other than gravity acting on him, it takes him
the same time to come down from the peak as it took to rise to it.

   (0.6 sec up) + (0.6 sec down) = 1.2 seconds until he hits rubber again.

5 0

3 years ago

Helium gas is compressed by an adiabatic compressor from an initial state of 14 psia and 50°F to a final temperature of 320°F in

iVinArrow [24]

Answer:

The value is $P_2 = 40.54 \ psla$

Explanation:

From the question we are told that

The initial pressure is $P_1 = 14\ psla$

The initial temperature is $T_1 = 50 \ F = (50 - 32) * [\frac{5}{9} ] + 273 = 283 \ K$

The final temperature is $T_2 = 320 \ F = (320 - 32) * [\frac{5}{9} ] + 273 =433 \ K$

Generally the equation for adiabatic process is mathematically represented as

$PT^{\frac{\gamma}{1- \gamma} } = Constant$

=> $P_1T_1^{\frac{\gamma}{1- \gamma} } = P_2T_2^{\frac{\gamma}{1- \gamma} }$

Generally for a monoatomic gas $\gamma = \frac{5}{3}$

So

$14 * 283^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} } =P_2 * 433^{\frac{\frac{5}{3} }{1- [\frac{5}{3} ]} }$

=> $14 * 283^{-2.5} =P_2 * 433^{-2.5}$

=> $P_2 = 40.54 \ psla$

8 0

2 years ago

Caroline conducts research on how the amount of fiber in a student's breakfast affects their grades in school. How should she re

Evgesh-ka [11]

Answer:

data table and line graph ( first choice)

8 0

3 years ago

Read 2 more answers

What is the most likely outcome of increasing the number of slits per unit distance on a diffraction grating?1) lines become nar

Nina [5.8K]

It's important to know that diffraction gratings can be identified by the number of lines they have per centimeter. Often, more lines per centimeter is more useful because the images separation is greater when this happens. That is, the distance between lines increases.

<h2>Therefore, the answer is 2.</h2>

8 0

1 year ago