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kari74 [83]
3 years ago
9

Hold a pencil in front of your eye at a position where its blunt end just blocks out the Moon. Make appropriate measurements to

estimate the diameter of the Moon, given that the Earth–Moon distance is 3.8 × 105 km. Assume that the pencil has a diameter of about 0.7 cm, and it just blocks out the Moon if it is held about 0.75 m from the eye.Express your answer using two significant figures.
Physics
1 answer:
kenny6666 [7]3 years ago
4 0

Answer:

D = 3.6 \times 10^6 m

Explanation:

Since the diameter of the moon position and the diameter of pencil position are in same line

so the angle made by the rays coming from the sun will be same as the angle made by the pencil lead at eye position

so we will have

\frac{D}{L} = \frac{d}{x}

here we have

D = diameter of moon

d = diameter of pencil lead = 0.7 cm

L = 3.8 \times 10^8 m

x = 0.75 m

so we have

\frac{0.007}{0.75} = \frac{D}{3.8 \times 10^8}

D = 3.6 \times 10^6 m

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A) The answer is 11.53 m/s

The final kinetic energy (KEf) is the sum of initial kinetic energy (KEi) and initial potential energy (PEi).
KEf = KEi + PEi

Kinetic energy depends on mass (m) and velocity (v)
KEf = 1/2 m * vf²
KEi = 1/2 m * vi²

Potential energy depends on mass (m), acceleration (a), and height (h):
PEi = m * a * h

So:
KEf = KEi + <span>PEi
</span>1/2 m * vf² =  1/2 m * vi² + m * a * h
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Divide all sides by m:
1/2 vf² =  1/2 vi² + a * h

We know:
vi = 9.87 m/s
a = 9.8 m/s²
h = 1.81 m

1/2 vf² =  1/2 * 9.87² + 9.8 * 1.81
1/2 vf² = 48.71 + 17.74
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b) The answer is 6.78 m

The kinetic energy at the bottom (KE) is equal to the potential energy at the highest point (PE)
KE = PE

Kinetic energy depends on mass (m) and velocity (v)
KE = 1/2 m * v²

Potential energy depends on mass (m), acceleration (a), and height (h):
PE = m * a * h

KE = PE
1/2 m * v² = m * a * h

Divide both sides by m:
1/2 * v² = a * h
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a = 9.8 m/s² 
h = ?

1/2 * 11.53² = 9.8 * h
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3 0
3 years ago
A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)
kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

a)

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

W=Fd cos \theta

where

F is the magnitude of the force applied

d is the displacement of the crate

\theta is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

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In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

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W=(1300)(4.0)(cos 0^{\circ})=5200 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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