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nalin [4]
2 years ago
11

A ball is thrown horizontally from the top of a building 0.10 km high. The ball strikes the ground at a point 65 m horizontally

away from and below the point of release. What is the speed of the ball just before it strikes the ground?
a) 43 m/s
b) 47 m/s
c) 39 m/s
d) 36 m/s
e) 14 m/s
Physics
1 answer:
aivan3 [116]2 years ago
5 0

Answer:

option B

Explanation:

given,

height of building = 0.1 km

ball strikes horizontally to ground at = 65 m    

speed at which the ball strike = ?

vertical velocity  = 0 m/s

time at which the ball strike

s = \dfrac{1}{2}gt^2

t = \sqrt{\dfrac{2s}{g}}

t = \sqrt{\dfrac{2\times 100}{9.8}}

t = 4.53 s

vertical velocity at the time  4.53 s = g × t = 9.8 × 4.53 = 44.39 m/s

horizontal velocity = \dfrac{65}{4.53} =14.35 m/s

speed of the ball = \sqrt{44.39^2+14.35^2}

                             = 46.65 m/s

hence, the speed of the ball just before it strike the ground = 47 m/s

The correct answer is option B

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A small 20-kg canoe is floating downriver at a speed of 2 m/s. What is the canoe’s kinetic energy? A. 40 J B. 80 J C. 18 J
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A small 20-kg canoe is floating downriver at a speed of 2 m/s. 40 J is the canoe’s kinetic energy.

Answer: Option A

<u>Explanation:</u>

The given canoe has the mass and is being given to move at a speed. Therefore the kinetic energy of the canoe can be calculated using the following method,

Given that mass of the canoe = 20 kg and its speed =1 m/s

As we know that the Kinetic energy has the formula,

\text {Kinetic energy}=\frac{1}{2} \boldsymbol{m} \boldsymbol{v}^{2}

Therefore, substituting the value into the equation, we get,  

K . E .=\frac{1}{2} \times 20 \times 2^{2} = 40 J

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The Burj Khalifa in Dubai is the world's tallest building. The structure is 828 m 828 m (2716.5 ft) and has more than 160 storie
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Answer:

h = height of the hotel room from the ground floor = 237.4m

Explanation:

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PE1 is the potential energy of tourist at the ground floor

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Given

PE1 = − 2.01 × 10⁵ J

PE2 = 0J

PE2 – PE1 = mgh

0 – (− 2.01 × 10⁵ J) = mgh

2.01 × 10⁵ J = 86.4×9.8×h

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An observer sees a full moon in the night sky. What will the Moon's phase be 10 days later?
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A reconnaissance plane flies 560 km away from its base at 602 m/s, then flies back to its base at 903 m/s.
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Answer:

Approximately 722\; \rm m\cdot s^{-1}.

Explanation:

The average speed of a vehicle is calculated as:

\displaystyle \text{average speed} = \frac{\text{total distance}}{\text{total time}}.

In this question, the total distance is 2 \times 560\; \rm km = 1120\; \rm km.

The unit of the speeds in this question is meters per second, while the unit of distance is kilometers. Convert the unit of distance to meters:

560 \; \rm km = 560 \times 10^{3} \; \rm m = 5.6 \times 10^{5}\; \rm m.

1120 \; \rm km = 1120 \times 10^{3} \; \rm m = 1.12 \times 10^{6}\; \rm m.

Time required for the first part of this trip:

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Time required for the second part of this trip:

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The time required for the entire trip would be approximately 930 + 620 = 1550\; \rm s.

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6 0
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