Q1: We drop this 0.1 kg apple 1 m. What speed does it hit the ground

1 answer:

5 0

Potential energy is measured using formula Ep=mgh
m=mass (kg)
g= acceleration due to gravity (which is 9.8 on earth)
h= height in metres above ground

For this question
m=0.1
g=9.8
h=1
So Ep=0.1(9.8)(1)
Ep=0.98 Joules

When it is dropped all of this potential energy is converted into kinetic energy which can be measured using formula
Ek=1/2m(v^2) (v=final velocity)

Since all potential energy in this q is converted to kinetic we know Ek=0.98Joules and our mass is the same (0.1kg)

So when we sub everything in we get
0.98=1/2(0.1)(v^2)
0.98=0.05(v^2)||divide both side by 0.05
19.6=v^2 ||square root both sides
v=4.4 m/s

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Answer: 259.2 KJ

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Given : Current : $I=3A$

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A person stands at the base of a hill that is a straight incline making an angle φ with the horizontal. For a given initial spee

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Answer:

θ = sin⁻¹ $\sqrt{2gd}$

Explanation:

From one of the equations of motion, v² = u² + 2as.......... equation 1

Since the object thrown was moving against gravity, then the acceleration, a would change to -g and the initial velocity u would change to V₀ sin θ because the object is travelling at angle of θ to the horizontal. By inputting all these parameter into equation 1, we would arrive at:

v² = (u sin θ)² - 2gd

(u sin θ)² = 2gd

d = (u sin θ)²/2g

sin² θ = 2gd

sin θ = $\sqrt{2gd}$

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