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ivanzaharov [21]
3 years ago
12

Imagine a raindrop starting from rest in a cloud 2 km in the air. If it fell with no air friction at all, it would accelerate to

ward the ground with gravitational acceleration of 9.8m/s2. What would the speed of the drop be when it reached the ground? Convert your answer into miles per hour.
1) 212 mph
2) 433 mph
3) 774 mph
4) 72.5 mph
Physics
1 answer:
LenKa [72]3 years ago
5 0

Answer:

2) 433 mph

Explanation:

The final velocity of the raindrop as it reaches the ground can be found by using the equation for a uniformly accelerated motion:

v^2 = u^2 + 2ad

where

v is the final velocity

u = 0 is the initial velocity (the raindrop starts from rest)

a = g = 9.8 m/s^2 is the acceleration due to gravity

d = 2 km = 2000 m is the distance covered

Solving for v,

v=\sqrt{u^2 +2gd}=\sqrt{0^2+2(9.8 m/s^2)(2000)}=198 m/s

And keeping in mind that

1 mile = 1609 metres

1 hour = 3600 s

The speed converted into miles per hour is

v=198 \frac{m}{s}\cdot \frac{3600 s/h}{1609 m/mi}=433 mph

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We need to find the force the ball exert on the bat.

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2 years ago
An oscillator consists of a block of mass 0.628 kg connected to a spring. When set into oscillation with amplitude 27 cm, the os
oksian1 [2.3K]

Answer:

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Explanation:

a.)

Period: It is already given in the question "oscillator repeats its motion every 0.372 s".

So T=0.372 s

b)

frequency= f = 1/ T

f = 1/ 0.372

f=2.7 Hz

c).

Angular frequency= w= 2πf

w= 2*π*2.7

w=16.9 rad/s

d)

Spring Constant:

As w=\sqrt{k/m}

⇒w²= k/m

⇒k= m*w²

⇒k= 0.628 * 16.9² N/m

⇒k=179.2 N/m

e)

The mass will have maximum speed when it passes through the mean position.

At mean position

Maximum elastic potential energy = Maximum kinetic energy

1/2 k A² = 1/2 m v²    ( A is amplitude of oscillation)

⇒ v=\sqrt{k A^2/m}

⇒ v= \sqrt{179.2 * 0.27/ 0.628}\

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f)

Maximum force will be exerted on the block when it is at maximum distance.

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F= 179.2 * 0.27 N

F= 48.4 N

5 0
3 years ago
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