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3 years ago

13

A 20,000 kg railroad car is traveling at 5 m/s when it collides and couples with a second, identical car at rest. What is the re

sulting speed of the combined cars?

1 answer:

PilotLPTM [1.2K]3 years ago

7 0

Ok I don’t know the answer I’m in 5th grade but 2431 is wrong

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The collision between a hammer and a nail can be considered to be approximately elastic. Estimate the kinetic energy acquired by

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4 years ago

If a book is knocked off a desk that is .75 m tall at a rate of 2.0 m/s, how far away from the desk does it fall

ANEK [815]

Answer: 0.8 m

Explanation:

In the vertical direction, the speed is zero, u = 0.

Distance covered in the vertical direction, s = 0.75 m.

The book would fall with acceleration due to gravity in the vertical direction, a = g = 9.8 m/s²

From the equation of motion,

s = u t + 0.5 a t²

Substituting the above values, we will find out the time taken for the book to hit the ground.

⇒0.75 m=0+0.5×9.8 m/s²×t²

⇒t = √0.153 = 0.39 s ≈ 0.40 s

Now, the horizontal distance covered,

d = v×t ⇒d= 2.0 m/s × 0.40 s =0.8 m

Hence, the book falls 0.8 m away from the desk.

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3 years ago

You're sitting in the back of a taxi and place your bag on the seat next to you. The taxi enters a traffic circle and travels th

svp [43]

Answer:

(d) III only

Explanation:

We have to observe the motion of the bag with respect to taxi , considering taxi as stationary or inertial frame . Since bag is not moving with respect to taxi , the inertial frame that means , net force on it is zero .So option i and ii are ruled out .

Now how to explain motion of the bag ie why it is stationary ie what are the balancing force acting on it. We know that on a body on circular path , a force called centripetal force is acting on it . So that force must be acting on it . The balancing force is the frictional force which is keeping it stationary with respect to taxi . Hence the third option is correct.

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3 years ago

You are traveling in a car toward a hill at a speed of 36.4 mph. The car's horn emits sound waves of frequency 231 Hz, which mov

Marina CMI [18]

Answer:

<em>a. The frequency with which the waves strike the hill is 242.61 Hz</em>

<em>b. The frequency of the reflected sound wave is 254.23 Hz</em>

<em>c. The beat frequency produced by the direct and reflected sound is </em>

<em> 11.62 Hz</em>

Explanation:

Part A

The car is the source of our sound, and the frequency of the sound wave it emits is given as 231 Hz. The speed of sound given can be used to determine the other frequencies, as expressed below;

$f_{1} = f[\frac{v_{s} }{v_{s} -v} ]$ ..............................1

where $f_{1}$ is the frequency of the wave as it strikes the hill;

f is the frequency of the produced by the horn of the car = 231 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 36.4 mph

Converting the speed of the car from mph to m/s we have ;

hint (1 mile = 1609 m, 1 hr = 3600 secs)

v = $36.4 mph *\frac{1609 m}{1 mile} *\frac{1 hr}{3600 secs}$

v = 16.27 m/s

Substituting into equation 1 we have

$f_{1}$ = $231 Hz (\frac{340 m/s}{340 m/s - 16.27 m/s})$

$f_{1}$ = 242.61 Hz.

Therefore, the frequency which the wave strikes the hill is 242.61 Hz.

Part B

At this point, the hill is the stationary point while the driver is the observer moving towards the hill that is stationary. The frequency of the sound waves reflecting the driver can be obtained using equation 2;

$f_{2} = f_{1} [\frac{v_{s}+v }{v_{s} } ]$

where $f_{2}$ is the frequency of the reflected sound;

$f_{1}$ is the frequency which the wave strikes the hill = 242.61 Hz;

$v_{s}$ is the speed of sound = 340 m/s;

v is the speed of the car = 16.27 m/s.

Substituting our values into equation 1 we have;

$f_{2} = 242.61 Hz [\frac{340 m/s+16.27 m/s }{340 m/s } ]$

$f_{2}$ = 254.23 Hz.

Therefore, the frequency of the reflected sound is 254.23 Hz.

Part C

The beat frequency is the change in frequency between the frequency of the direct sound and the reflected sound. This can be obtained as follows;

Δf = $f_{2}$ - $f_{1}$

The parameters as specified in Part A and B;

Δf = 254.23 Hz - 242.61 Hz

Δf = 11.62 Hz

Therefore the beat frequency produced by the direct and reflected sound is 11.62 Hz

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3 years ago

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3 years ago