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astraxan [27]
4 years ago
8

A car (mass = 1200 kg) is traveling at 31.1 m/s when it collides head-on with a sport utility vehicle (mass = 2830 kg) traveling

in the opposite direction. In the collision, the two vehicles come to a halt. At what speed was the sport utility vehicle traveling?
Physics
1 answer:
IrinaVladis [17]4 years ago
4 0

Answer:

13.18 m/s

Explanation:

Let the velocity of sports utility car is

-u as it is moving in opposite direction.

mc = 1200 kg, uc = 31.1 m/s

ms = 2830 kg, us = - u = ?

Using conservation of momentum

mc × uc + ms × us = 0

1200 × 31.1 - 2830 × u = 0

u = 13.18 m/s

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4 years ago
A mass of 150 g stretches a spring 1.568 cm. If the mass is set in motion from its equilibrium position with a downward velocity
nadezda [96]

Answer:

u(t)=\frac{1}{5} sin\ (25t)

Explanation:

Given:

  • mass of the body stretching the spring, m=150\ g
  • extension in spring, \Delta x=1.568\ cm
  • velocity of oscillation, u'(0)=20\ cm.s^{-1}
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<u>According to given:</u>

m.g=k.\Delta x

150\times 980=k\times 1.568

k=93750\ dyne.cm^{-1}

<u>we know frequency:</u>

\omega=\sqrt{\frac{k}{m} }

\omega=\sqrt{\frac{93750}{150} }

\omega=25

Now, for position of mass in oscillation:

u= A.sin\ (\omega.t)+B.cos\ (\omega.t)

u= A.sin\ (25.t)+B.cos\ (25.t)

at t=0;\ u(0)=0\ \Rightarrow A=0

∴u(t)=B.sin\ (25.t)

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u(t)=\frac{1}{5} sin\ (25t)

7 0
3 years ago
block is attached to an oscillating spring. The function below shows its position (cm) vs. time (s). What is the angular frequen
faltersainse [42]

Answer:

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Explanation:

Given that,

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The general equation of oscillating particle is given by :

x(t)=A\cos(\omega t).......(2)

Compare equation (1) and (2) we get :

\omega=20\ rad/s

So, the angular frequency of the oscillation is 20 rad/s.

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3 years ago
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