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Novay_Z [31]
4 years ago
5

What is the efficiency of a 0.60 kg basketball that, one dropped from a 2.0 m height (from basket rim) , rebounds to 1.2 m in he

ight (G= 10 N/KG)
Physics
1 answer:
Zinaida [17]4 years ago
4 0
<h2>Let us find the efficiency : Ans = 0.6</h2>

Explanation:

we know :

efficiency = output/input

We also know that :

output = m x g x h

where :

m = mass of body

g = acceleration due to gravity

h = height of body from floor

Thus, output = 0.6 x 10 x 1.2 = 7.2J

Similarly ,input = 0.6 x 10 x 2 = 12J

Thus efficiency = 7.2/12 = 0.6

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A lightning rod effectively protects a building by transferring the electrical current to the ____________.
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It transfers the electrical current from the point of contact on the lightning rod into the ground.
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Which two factors affect the amount of thermal energy an object has?
leonid [27]

Answer:

B

Explanation:

When the average kinetic energy increases, the thermal energy increases

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An experiment invilves three charges objects: A, B, and C. Object A repels object B and attracts onject C. object C ir repelled
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Answer:

A and B is positive charge

C_negative

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because when an ebonite is rubbed with fur produce negative charge due to law of electrostatic like charge repel and unlike attract

7 0
3 years ago
The higher the pressure on a liquid, the higher its boiling point temperature. true false
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3 years ago
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An unmanned spacecraft is in a circular orbit around the moon, observing the lunar surface from an altitude of 43.0 km . To the
GalinKa [24]

Answer: v₂ = 5962 km

the spacecraft  will crash into the lunar surface at a speed of 5962 km if nothing is done to correct its orbits

Explanation:

Given that;

Lunar surface is in an altitude h = 43.0 km =  43 × 10³ m

we know; Radius of moon R₁ = 1.74 × 10⁶, mass of moon = 7.35 × 10²²

speed of the space craft when it crashes into the lunar surface , v

decreasing speed of the space craft = 23 m/s

Now since the space craft travels in a circular orbit, we use centrifugal expression Fe = mv²/r

but the forces is due to gravitational forces between space craft and lunar surface Fg = GMn/r²

HERE r = Rm + h

we substitute

r = 1.74 × 10⁶ m + 43 × 10³ m

= 1.783 × 10⁶ m

On equating these, we have

G is gravitational force ( 6.673 × 10⁻¹¹ Nm²/kg²)

v²/r = GM/r²

v = √ ( GM/r)

v = √ ( 6.673 × 10⁻¹¹ Nm²/kg² × 7.35 × 10²² / 1.783 × 10⁶ )

v = √ (2750787.9978)

v = 1658.55 m/s

Now since speed is decreasing by 23 m/s

the speed of the space craft into the lunar face is,

v₁ = 1658.55 m/s - 23 m/s

v₁ = 1635.55 m/s

Now applying conversation of energy, we say

1/2mv₂² = 1/2mv₁² + GMem (1/Rm - 1/r)

v₂ =  √ [ v₁² + GMe (1/Rm - 1/r)]

v₂ =   √ [ 1635.55²  + ( 6.673 × 10⁻¹¹ Nm²/kg² × 7.35 × 10²²) (1/ 1.74 × 10⁶ - 1 / 1.783 × 10⁶)]

v₂ =  √ (2675023.8025 + 67979.24)

v₂ = √(2743003.046)

v₂ = 1656.2 m/s

now convert

v₂ = 1656.2 × 1km/1000m × 3600s/1hrs

v₂ = 5962 km

Therefore the spacecraft  will crash into the lunar surface at a speed of 5962 km if nothing is done to correct its orbits

8 0
3 years ago
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