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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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loris [4]

The average force of the water droplets is the force given by the impact

per second of the droplets on the limestone floor.

  • The average force exerted on the limestone floor is approximately <u>1.6013 × 10⁻² N</u>

Reasons:

The given parameters are;

Volume of a droplet = 10 ml = 1 × 10⁻⁵ m³

Height from which the water falls, <em>h </em>= 5 meters

Rate at which the water falls = 10 per minute

Required:

The average force exerted on the floor by the water droplets.

Solution:

According to Newton's Second Law of motion, we have;

Force = Rate of change of momentum

Momentum = Mass × Velocity

Mass of a droplet of water = Volume × Density

Density of water = 997 kg/m³

Mass of a droplet = 1 × 10⁻⁵ m³ × 997 kg/m³ = 0.00997 kg

The velocity just before the droplet reaches the ground, v = √(2·g·h)

Where;

g = Acceleration due to gravity ≈ 9.81 m/s²

Which gives;

v = √(2 × 9.81 m/s² × 5 m) ≈ 9.905 m/s

The rate of change in momentum per minute = 1

Therefore;

\displaystyle The \ rate \ of \ change \ in \ momentum = Average \ force = \mathbf{\frac{\Delta Momentum }{\Delta Time}}

ΔMomentum = Mass × ΔVelocity

Considering the 10 drops per minute, we have;

ΔMomentum = 10 × 0.0097 kg × 9.905 m/s = 0.960785 kg·m/s

ΔTime = 1 minute = 60 seconds

Therefore;

\displaystyle Average \ force, \, F_{ave}  \frac{0.960785 \, kg\cdot m/s }{60 \, s} \approx =\mathbf{1.6013 \times 10^{-2} \, N}

  • The average force exerted on the limestone floor by the droplets of water is F_{ave} ≈ <u>1.6013 × 10⁻² N</u>

Learn more about Newton's Second Law of motion and force exerted water here:

brainly.com/question/3999427

brainly.com/question/4197598

3 0
2 years ago
The drag on a pitched baseball can be surprisingly large. Suppose a 145 g baseball with a diameter of 7.4 cm has an initial spee
kupik [55]

Answer:

<h2>Part A)</h2><h2>Acceleration of the ball is 10.1 m/s/s</h2><h2>Part B)</h2><h2>the final speed of the ball is given as</h2><h2>v_f = 35.3 m/s</h2>

Explanation:

Part a)

As we know that drag force is given as

F = \frac{C_d \rho A v^2}{2}

C_d = 0.35

A = \frac{\pi d^2}{4}

A = \frac{\pi(0.074)^2}{4}

A = 4.3 \times 10^{-3} m^2

v = 40.2 m/s

so we have

F = \frac{0.35\times 1.2 (4.3 \times 10^{-3})(40.2)^2}{2}

F = 1.46 N

So acceleration of the ball is

a = \frac{F}{m}

a = \frac{1.46}{0.145}

a = 10.1 m/s^2

Part B)

As per kinematics we know that

v_f^2 - v_i^2 = 2 a d

v_f^2 - 40.2^2 = 2(-10.1)(18.4)

v_f = 35.3 m/s

4 0
3 years ago
A youngster having a mass of 50.0 kg steps off a 1.00 m high platform. If she keeps her legs fairly rigid and comes to rest in 1
attashe74 [19]

The average act on her during the deceleration is 4.47 meters per second.                                                                

<u>Explanation</u>:

<u>Given</u>:

youngster mass m = 50.0 kg

She steps off a 1.00 m high platform that is s = 1 meter

She comes to rest in the 10-meter second

<u>To Find</u>:

The average force and momentum

<u>Formulas</u>:

p = m * v

F * Δ t = Δ p

vf^2= vi^2+2as

<u>Solution</u>:

a = 9.8 m/s

vi = 0

vf^2= 0+2(9.8)(1)

vf^2 = 19.6

vf = 4.47 m/s .

Therefore the average force is 4.47 m/s.                                          

                                 

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2 years ago
A non-_____ rock has interlocking grains with no specific pattern.
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A non <span>foliated </span>rock has interlocking grains with no specific pattern.
3 0
3 years ago
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If an electron is accelerated from rest through a potential difference of 1200V find its approximate velocity at the end of this
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Answer: 2.1 × 10^7 m/s

Explanation:

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8 0
3 years ago
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