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AnnyKZ [126]
3 years ago
11

One car is sitting still the other is moving at a velocity of 4m/s. If two cars have a mass of 0.04 kg run into each other, what

is the momentum of the pared cars?
Physics
1 answer:
Leni [432]3 years ago
7 0

Answer:

p = 0.16 kgm/s

Explanation:

the initial momentum combined of the two cars and the final momentum of the paired cares are the same, so we just need to find the initial momentum

p = m1v1 + m2v2

p = 0.04*4 + 0.04*0

p = 0.04*4

p = 0.16 kgm/s

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The maximum displacement in an oscillatory motion is A = 0.49 m. Determine the position x at which the kinetic energy of the par
amid [387]

Answer:

x = 0.40 m

Explanation:

  • When the displacement is maximum, the particle is momentarily at rest, which means that at this point (assuming no friction present) all the mechanical energy is elastic potential, which can be written as follows:

      E_{tot} = U_{o} = \frac{1}{2} *k*A^{2}  (1)

  • Since in absence of friction, total mechanical energy must keep constant, this means that at any time, the sum of the kinetic and potential energy, must be equal to (1), as follows:

       E_{tot} = U_{o} = \frac{1}{2} *k*A^{2}  = (KE)_{f} + U_{f}  (2)

  • If KEf = U/2f, replacing in (2), we get:

      E_{tot} = U_{o} = \frac{1}{2} *k*A^{2}  = (U/2)_{f} + U_{f} =  \frac{3}{2} *U_{f}  (3)

  • At any point, the elastic potential energy is given by the following expression:

       U_{f} = \frac{1}{2} *k*x^{2}   (4)

      where k= spring constant (N/m) and x is the displacement from the

      equilibrium position.

  • Replacing (4) in (3), simplifying and rearranging, we get:

       E_{tot} = U_{o} = \frac{1}{2} *A^{2}  =  \frac{3}{4} *x^{2}   (5)

  • Finally, solving for x, we get:

        x = \sqrt{\frac{2}{3} } * A =  \sqrt{\frac{2}{3} } * 0.49m = 0.40 m  (6)

8 0
3 years ago
What is the frequency of a clock waveform whose period is 750 microseconds?
Allushta [10]
Use this formula to find your answer...

Determine the frequency of a clock waveform whose period is 2us or (micro) and 0.75ms

frequency (f)=1/( Time period).

Frequency of 2 us clock =1/2*10^-6 =10^6/2 =500000Hz =500 kHz.

Frequency of 0..75ms clock =1/0.75*10^-3 =10^3/0.75 =1333.33Hz =1.33kHz.

6 0
3 years ago
1. What was the electromagnetic spectrum ?
Maurinko [17]
The answer is C. I hope this helps you.
4 0
4 years ago
How high does Pete lift his sledge hammer if he used a force of 25N to lift the hammer while doing 50J of work?
sasho [114]

Answer:He lifts 2 meters

Explanation:We are trying to find the distance. The formula for distance is W/Force. 50 is our amount of work. 25N is how much force was used. Divide the work bye the force. 50/25=2M

8 0
3 years ago
a student pushes a 40-N block across the floor for a distance 0f 10 m how much work was done to move the block?
slava [35]
W=f*d
= 40N * 10m
work= 400 joules
6 0
3 years ago
Read 2 more answers
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