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lesya [120]
3 years ago
15

PLEASE HELP: A 1400 kg car moving +13.7 m/s makes an elastic collision with a 3200 kg truck, initially at rest. What is the velo

city of the truck after the collision? (Unit = m/s) remember: right is +, left is -
Physics
1 answer:
iren [92.7K]3 years ago
7 0

Answer:

8.339\  m/s

Explanation:

Given :

m1= 1400 kg \\v1=13.7 m/s\\m2=3200 kg \\v2=0

We know that for the elastic collision

V2=\frac{2m1v1}{m1+m2} \ + \frac{m2-m1 * v2}{m1+m2}

Putting the respective value we get

V2=\frac{2*1400*13.7}{1400+3200} \ + 0\\ V2=8.339\  m/s

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Determine the frequency of the 2nd harmonic of a spring that has a 3rd harmonic resonance of f3=512 Hz.
Elena L [17]

Answer:

1) 341 Hz

Explanation:

When a string vibrates, it can vibrate with different frequencies, corresponding to different modes of oscillations.

The fundamental frequency is the lowest possible frequency at which the string can vibrate: this occurs when the string oscillate in one segment only.

If the string oscillates in n segments, we say that it is the n-th mode of vibration, or n-th harmonic.

The frequency of the n-th harmonic is given by

f_n = nf_1

where

n is the number of the harmonic

f_1 is the fundamental frequency

Here we have:

f_3=512 Hz is the frequency of the 3rd harmonic

So the fundamental frequency is

f_1=\frac{f_3}{3}=\frac{512}{3}=170.7 Hz

And so, the frequency of the 2nd harmonic is:

f_2=2f_1=2(170.7)=341.3 Hz

3 0
4 years ago
A 70.0-kg skier is sliding at 4 m/s when they slide down a 2m high hill. At the bottom of the hill they run into a large 2800 N/
Katarina [22]

Answer:

\Delta x=245\ mm

Explanation:

Given:

  • mass of skier, m=70\ kg
  • initial velocity of skier, u=4\ m.s^{-1}
  • height of the hill, h=2\ m
  • spring constant, k=2800\ N.m^{-1}

<u>final velocity of skier before coming in contact of spring:</u>

Using eq. of motion:

v^2=u^2+gh

v^2=4^2+9.8\times 2

v=5.9666\ m.s^{-1}

<u>Now the time taken by the skier to reach down:</u>

v=u+gt

5.9666=4+9.8\ t

t=0.2007\ s

<u>Now we calculate force using Newton's second law:</u>

F=\frac{dp}{dt}

F=\frac{m(v-u)}{t}

F=\frac{70\times(5.9666-4)}{0.2007}

F\approx686\ N

<u>∴Compression in spring before the skier momentarily comes to rest:</u>

\Delta x=\frac{F}{k}

\Delta x=\frac{686}{2800}

\Delta x=0.245\ m

\Delta x=245\ mm

4 0
3 years ago
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Nady [450]
Answer: False

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An apple has a mass of 0.136kg and a watermelon has a mass of 9.07kg. Work out the mass of the apple as a percentage of the mass
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7 the answer is 7 JK

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Listed following are locations and times atwhich different phases of the Moon are visible fromEarth’s....?
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Answer:

Explanation:

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