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lubasha [3.4K]
2 years ago
6

A ball is gently dropped from a height of 20 m. If its velocity Increases uniformly at the rate of 10 m s2, with what velocity w

ill it strike the ground? After what time will it strike the ground?​
Physics
2 answers:
Pie2 years ago
5 0

Answer:

20 m/s

Explanation:

Height (s)= 20m

acceleration (a) = 10 m/s^2

v=?

we know,

V^2= U^2 + 2as. (U is initial velocity)

or,. V^2 = 0 + 2 × 10 × 20

or, V^2 = 400

or,. V = 20 m/s

belka [17]2 years ago
5 0

Explanation:

Initial Velocity of the ball (u) = 0

Distance or height of fall (s) = 20 m

Downward acceleration (a) = 10 m s-2

As we know

2as = v2 – u2

v2 = 2as + u2

v2 = (2 x 10 x 20 ) + 0

v2 = 400

Final velocity of ball (v) = 20 ms-1

t = (v – u)/a

Time taken by the ball to strike (t) = (20 – 0)/10

t = 20/10

t = 2 seconds

The final velocity with which the ball will strike the ground is (v) = 20 ms-1

The time it takes to strike the ground (t) = 2 seconds

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In Anchorage, collisions of a vehicle with a moose are so common that they are referred to with the abbreviationMVC. Suppose a 1
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Answer:

Part a)

f = \frac{8}{9}

Part b)

f = \frac{120}{169}

Part c)

So from above discussion we have the result that energy loss will be more if the collision occurs with animal with more mass

Explanation:

Part a)

Let say the collision between Moose and the car is elastic collision

So here we can use momentum conservation

m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

1000 v_o = 1000 v_{1f} + 500 v_{2f}

also by elastic collision condition we know that

v_{2f} - v_{1f} = v_o

now we have

2v_o = 2v_{1f} + v_o + v_{1f}

now we have

v_{1f} = \frac{v_o}{3}

Now loss in kinetic energy of the car is given as

\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)

\Delta K = \frac{1}{2}m(v_o^2 - \frac{v_o^2}{9})

so fractional loss in energy is given as

f = \frac{\Delta K}{K}

f = \frac{\frac{4}{9}mv_o^2}{\frac{1}{2}mv_o^2}

f = \frac{8}{9}

Part b)

Let say the collision between Camel and the car is elastic collision

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m_1v_{1i} = m_1v_{1f} + m_2v_{2f}

1000 v_o = 1000 v_{1f} + 300 v_{2f}

also by elastic collision condition we know that

v_{2f} - v_{1f} = v_o

now we have

10v_o = 10v_{1f} + 3(v_o + v_{1f})

now we have

v_{1f} = \frac{7v_o}{13}

Now loss in kinetic energy of the car is given as

\Delta K = \frac{1}{2}m(v_o^2 - v_{1f}^2)

\Delta K = \frac{1}{2}m(v_o^2 - \frac{49v_o^2}{169})

so fractional loss in energy is given as

f = \frac{\Delta K}{K}

f = \frac{\frac{60}{169}mv_o^2}{\frac{1}{2}mv_o^2}

f = \frac{120}{169}

Part c)

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8 0
3 years ago
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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