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Lubov Fominskaja [6]
2 years ago
7

What is the velocity of a jet that travels 785 m [W] in 2.39 s?

Physics
1 answer:
Cerrena [4.2K]2 years ago
7 0

Answer:

v = 328.4

Explanation:

Data:

  • Distance (d) = 785 m
  • Time (t) = 2.39 s
  • Velocity (v) = ?

Use formula:

  • \boxed{v = \frac{d}{t}}

Replace:

  • \boxed{v = \frac{785m}{2.39s}}

It divides:

  • \boxed{v = 328.4\frac{m}{s}}}

What is the velocity?

The velocity is <u>328.4 meters per second.</u>

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In the physics lab, a block of mass M slides down a frictionless incline from a height of 35cm. At the bottom of the incline it
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Solution :

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M = 0.35 kg

$m=\frac{M}{2}=0.175 \ kg$

Total mechanical energy = constant

or $K.E._{top}+P.E._{top} = K.E._{bottom}+P.E._{bottom}$

But $K.E._{top} = 0$ and $P.E._{bottom} = 0$

Therefore, potential energy at the top = kinetic energy at the bottom

$\Rightarrow mgh = \frac{1}{2}mv^2$

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      = 2.62 m/s

It is the velocity of M just before collision of 'm' at the bottom.

We know that in elastic collision velocity after collision is given by :

$v_1=\frac{m_1-m_2}{m_1+m_2}v_1+ \frac{2m_2v_2}{m_1+m_2}$

here, $m_1=M, m_2 = m, v_1 = 2.62 m/s, v_2 = 0$

∴ $v_1=\frac{0.35-0.175}{0.5250}+\frac{2 \times 0.175 \times 0}{0.525}

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Therefore, velocity after the collision of mass M = 0.33 m/s

 

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