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dsp73
1 year ago
11

A flowerpot falls from a window sill 36.5 m

Physics
1 answer:
hram777 [196]1 year ago
7 0

When a flowerpot falls from a window sill 36.5 m

above the sidewalk, then the velocity of the flowerpot is 26.7 m/s.

From Newton's third equation of motion,

v^2 = u^2 + 2gh

where,

h is the height of the object or body from ground

u is the initial velocity of the body or object

v is the final velocity of the body or object

g is the acceleration due to gravity

Now, as we know that

Flowerpot is at rest. So, u = 0

g = 9.81m/s^2

h = 36.5m

By substituting all the values, we get

v^2 = 2 × 9.81 × 36.5

= 716.13

v = 26.7m/s

Thus, we concluded that when a flowerpot falls from a window sill 36.5 m

above the sidewalk, then the velocity of the flowerpot is 26.7 m/s.

learn more about Newton's equation of law of motion:

brainly.com/question/8898885

#SPJ9

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D

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3 years ago
A runner traveled 15 kilometers north then backtracked 11 kilometers south before stopping. His resultant displacement was
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<u>Answer:</u>

 Resultant displacement is 4 km north.

<u>Explanation:</u>

    Let east represents positive x- axis and north represent positive y - axis. Horizontal component is i and vertical component is j.

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3 years ago
Read 2 more answers
A 16000 kg railroad car travels along on a level frictionless track with a constant speed of 23.0 m/s. A 5400 kg additional load
Alisiya [41]

Answer:

The speed of the car when load is dropped in it is 17.19 m/s.

Explanation:

It is given that,

Mass of the railroad car, m₁ = 16000 kg

Speed of the railroad car, v₁ = 23 m/s

Mass of additional load, m₂ = 5400 kg

The additional load is dropped onto the car. Let v will be its speed. On applying the conservation of momentum as :

m_1v_1=(m_1+m_2)v

v=\dfrac{m_1v_1}{m_1+m_2}

v=\dfrac{16000\ kg\times 23\ m/s}{(16000+5400)\ kg}

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So, the speed of the car when load is dropped in it is 17.19 m/s. Hence, this is the required solution.

6 0
3 years ago
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