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beks73 [17]
3 years ago
7

A bicycle travels at a velocity of 5.33 m/s for distance of 215 m. How much time did it take?

Physics
1 answer:
valina [46]3 years ago
4 0

Answer:

40.34 s

Explanation:

Given,

Velocity (v) = 5.33 m/s

Distance (s) = 215 m

Time (t) = ?

We know that,

Speed (v) = distance (s)/time (t)  

⇒ time(t) = distance (s)/speed (v)

= 215/5.33 = 40.34 sec

∴ time(t) = 40.34 sec.

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Conservation of momentum: total momentum before = total momentum after

Momentum = mass x velocity

So before the collision:
4kg x 8m/s = 32
1kg x 0m/s = 0
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Therefore after the collision
4kg x 4.8m/s = 19.2
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Answer:

Clockwise and counter clockwises, depands.

Explanation:

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What is the pooled variance for the following two samples?
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I think it is B as 168/20
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3 years ago
A ball is projected at an immovable wall with a speed vi and bounces back the wall in such a manner that it only has 1/3 of its
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The fraction of the kinetic energy of the ball lost during the collision is \frac{1}{3}.

The given parameters;

  • <em>initial speed of the ball, = vi</em>
  • <em>final momentum of the ball, Pf = ¹/₃Pi</em>

The initial and final momentum of the ball is calculated as;

P_i = m_ivi

P_f = m_fv_f = \frac{1}{3} m_iv_i

The initial and final kinetic energy of the ball is calculated as;

K.E_i = \frac{1}{2} m_iv_i^2 = \frac{1}{2} P_iv_i\\\\K.E_f = \frac{1}{2} m_fv_f^2= \frac{1}{2} (\frac{1}{3} P_iv_i)= \frac{1}{6} P_iv_i

The change in the kinetic energy is calculated as;

\Delta K.E = K.E_f - K.E_i \\\\\Delta K.E = \frac{1}{6} P_iv_i - \frac{1}{2} P_iv_i = \frac{1}{3} P_iv_i

Thus, the fraction of the kinetic energy of the ball lost during the collision is \frac{1}{3}.

Learn more here:brainly.com/question/18566218

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3 years ago
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