HELP ASAP TIMED TEST

A snowball is dropped from a height of H and reaches the ground with a speed of V. If you want to double the speed that the snowball has when it reaches the ground, at what height should you drop it from?

a
2H
b
4H
c
√2H

d
8H
e
16H

Answers

Answer 1

Answer:

Correct choice: b 4H

Explanation:

Conservation of the mechanical energy

The mechanical energy is the sum of the gravitational potential energy GPE (U) and the kinetic energy KE (K):

E = U + K

The GPE is calculated as:

U = mgh

And the kinetic energy is:

[tex]\displaystyle K=\frac{1}{2}mv^2[/tex]

Where:

m = mass of the object

g = gravitational acceleration

h = height of the object

v = speed at which the object moves

When the snowball is dropped from a height H, it has zero speed and therefore zero kinetic energy, thus the mechanical energy is:

[tex]U_1 = mgH[/tex]

When the snowball reaches the ground, the height is zero and the GPE is also zero, thus the mechanical energy is:

[tex]\displaystyle U_2=\frac{1}{2}mv^2[/tex]

Since the energy is conserved, U1=U2

[tex]\displaystyle mgH=\frac{1}{2}mv^2 \qquad\qquad [1][/tex]

For the speed to be double, we need to drop the snowball from a height H', and:

[tex]\displaystyle mgH'=\frac{1}{2}m(2v)^2[/tex]

Operating:

[tex]\displaystyle mgH'=4\frac{1}{2}m(v)^2 \qquad\qquad [2][/tex]

Dividing [2] by [1]

[tex]\displaystyle \frac{mgH'}{mgH}=\frac{4\frac{1}{2}m(v)^2}{\frac{1}{2}m(v)^2}[/tex]

Simplifying:

[tex]\displaystyle \frac{H'}{H}=4[/tex]

Thus:

H' = 4H

Correct choice: b 4H


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Answers

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What is amplitude ?

Amplitude of a wave is the distance from the midpoint of a crest to its top or the distance from the midpoint of a trough to its bottom. For a wave the peak to the top is called crest and peak to the bottom or decline is called troughs.

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Answer:

a. i. 188.49 m/s. ii. 235.62 m/s. b. i. 117.81 m/s² ii. 4.71 m/s²

Explanation:

a. Since the linear velocity v = rω where r = radius of path = length of horizontal bar = 1.5 m and the initial angular speed ω₀ = 1200 r.P.M = 1200 × 2π/60 = 125.66 rad/s and final angular speed after the 5 seconds ω₁ = 1500 r.P.M = 1500 × 2π/60 = 157.08 rad/s

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ii. The final linear velocity v₁ = rω₁ = 1.5 m × 157.08 rad/s = 235.62 m/s.

b. The mid-point of the bar is at a distance of one-half from the end of the bar. So its radius r' = r/2 = 1.5 m/2 = 0.75 m

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Answer:

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Answers

Answer:

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Explanation:

Since the block reads 24.5 N before the block starts to move, this is its weight. Now, when the block starts to move at a constant velocity, it experiences a frictional force which is equal to the force with which the student pulls.

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Answers

Answer:

0.905

Explanation:

right answer

The radius of the swing is 0.905 m

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From the question, we are to determine the radius of the swing

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