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Nimfa-mama [501]
2 years ago
8

Q1) An aeroplane covers a certain distance at a speed of 240 kmph in 5 hours.To cover the same distance in 1(2/3) hours, it must

travel at a speed of;
A)300kmph
B)360kmph
C)600kmph
D)720kmph

Explain how you got the answer

Physics
1 answer:
Pepsi [2]2 years ago
8 0

Answer:

D) 720 kmph

Explanation:

First, let's find what distance this aeroplane covered. Distance (d) is the product of speed and time - here, we have a speed of 240 kmph and a time of 5 hours, gives us

d=240\cdot5=1200 km

Using that same fact, we can set up a new equation to solve for speed (s) when we have a distance of 1200 km and a time of 1 2/3 hours. For the sake of cleanliness, I'm gonna rewrite 1 2/3 as the improper fraction 5/3:

1200=(5/3)s

Multiplying both sides of the equation by 3/5:

s=(3/5)(1200)=720 kmph

So our answer is D.

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A ball of mass 0.150 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.665 m. W
Liula [17]

Answer:

Impulse is 1.239 kg.m/s in upward direction

Explanation:

Taking upward motion as positive and downward motion as negative.

Downward motion:

Given:

Mass of ball (m) = 0.150 kg

Displacement of ball (S) = -1.25 m

Initial velocity (u) = 0 m/s

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

v_d^2=u^2+2aS\\\\v=\pm\sqrt{u^2+2aS}\\\\v_d=\pm\sqrt{0+2\times -9.8\times -1.25}\\\\v_d=\pm\sqrt{24.5}=\pm4.95\ m/s

Since, the motion is downward, final velocity must be negative. So,

v_d=-4.95\ m/s

Upward motion:

Given:

Displacement of ball (S) = 0.665 m

Initial velocity (v_d) = 4.95 m/s(Upward direction)

Acceleration is due to gravity (g) = -9.8 m/s²

Using equation of motion, we have:

v_{up}^2=v_d^2+2aS\\\\v_{up}=\pm\sqrt{v_d^2+2aS}\\\\v_{up}=\pm\sqrt{24.5+2\times -9.8\times 0.665}\\\\v_{up}=\pm\sqrt{10.966}=\pm3.31\ m/s

Since, the motion is upward, final velocity must be positive. So,

v_{up}=3.31\ m/s

Now, impulse is equal to change in momentum. So,

Impulse = Final momentum - Initial momentum

J=m(v_{up}-v_d)\\\\J=(0.150\ kg)(3.31-(-4.95))\ m/s\\\\J=0.150\ kg\times 8.26\ m/s\\\\J=1.239\ Ns

Therefore, the impulse given to the ball by the floor is 1.239 kg.m/s in upward direction.

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3 years ago
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Explanation:

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