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3 years ago

5

A paper airplane is thrown horizontally with a velocity of 20 mph. The plane is in the air for 7.63 s before coming to a standst

ill on the ground. What is the acceleration of the plane?

1 answer:

Inessa05 [86]3 years ago

7 0

Answer:

-1.17 m/s²

Explanation:

Given:

v₀ = 20 mph = 8.94 m/s

v = 0 m/s

t = 7.63 s

Find: a

v = at + v₀

0 m/s = a (7.63 s) + 8.94 m/s

a = -1.17 m/s²

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The frequency of sound wave A is 250 Hz. Find the time period.

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3 years ago

Banked Curves: A 600-kg car is going around a banked curve with a radius of 110 m at a steady speed of 24.5 m/s. What is the app

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The angle of baking from the calculation is obtained as 30°.

<h3>What is banking?</h3>

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2 years ago

a bullet with a mass of 4.0g and a speed of 650m/s is fired at a block of wood with a mass of 0.095kg. the block rests on a fric

n200080 [17]

Part a)

Here in this we can use momentum conservation as there is no external force on it

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} +m_2v_{2f}$

here we know that

$m_1 = 0.004 kg$

$v_{1i} = 650 m/s$

$m_2 = 0.095$

$v_{2i} = 0$

$v_{2f} = 23 m/s$

now by above equation

$0.004*650 + 0.095* 0 = 0.004*v + 0.095*23$

$2.6 + 0 = 0.004*v + 2.185$

$v = 103.75 m/s$

Part b)

Final kinetic energy of the system

$KE_f = \frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2$

$KE_f = \frac{1}{2}*0.004*(103.75^2) + \frac{1}{2}*(0.095)*23^2$

$KE_f = 46.65 J$

Initial Kinetic energy of the system will be

$KE_f = \frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2$

$KE_f = \frac{1}{2}*0.004*(650^2) + \frac{1}{2}*(0.095)*0^2$

$KE_f = 845 J$

So here kinetic energy is decreased for this system

final energy is less than initial energy

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3 years ago