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Oduvanchick [21]
3 years ago
5

A 4kg watermelon is dropped from a height of 45m. What is the velocity of the watermelon just before it hits the ground?

Physics
1 answer:
Makovka662 [10]3 years ago
3 0

Answer:

v=30 m/s

Explanation:

h - height

g - acceleration due to gravity=10

t - time

v- velocity

h =  \frac{1}{2}  \times g \times t {}^{2}

45 = 5t²

t² = 9

t=3 seconds

v=g×t

v=10×3

v=30 m/s

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Answer:

Identify the object to be analyzed. For some systems in equilibrium, it may be necessary to consider more than one object. Identify all forces acting on the object. Identify the questions you need to answer. Identify the information given in the problem. In realistic problems, some key information may be implicit in the situation rather than provided explicitly.

Explanation:

Identify the object to be analyzed. For some systems in equilibrium, it may be necessary to consider more than one object. Identify all forces acting on the object. Identify the questions you need to answer. Identify the information given in the problem. In realistic problems, some key information may be implicit in the situation rather than provided explicitly.

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3 years ago
Rachel and Sarah are on a bus travelling at 5 mph past John who is standing on the sidewalk. Rachel then throws a ball
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I think C

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{C - V}.

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2 years ago
Why does immersion oil improve resolution?
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Answer:

d. It increases numerical aperture and maintains a uniform light speed.

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7 0
3 years ago
Convert 1,265,341 mm to m.
Thepotemich [5.8K]
The answer in Meters is going to to 1265.341
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3 years ago
Read 2 more answers
It has been suggested that rotating cylinders about 10 mi long and 5.9 mi in diameter be placed in space and used as colonies. T
tekilochka [14]

Answer:

ω = 0.05 rad/s

Explanation:

We consider the centripetal force acting as the weight force on the surface of the cylinder. Therefore,

Centripetal Force = Weight\\\frac{mv^{2}}{r} = mg\\\\here,\\v = linear\ speed = r\omega \\therefore,\\\frac{(r\omega)^{2}}{r} = g\\\\\omega^{2} = \frac{g}{r}\\\\\omega = \sqrt{\frac{g}{r}}\\

where,

ω = angular velocity of cylinder = ?

g = required acceleration = 9.8 m/s²

r = radius of cylinder = diameter/2 = 5.9 mi/2 = 2.95 mi = 4023.36 m

Therefore,

\omega = \sqrt{\frac{9.8\ m/s^{2}}{4023.36\ m}}\\\\

<u>ω = 0.05 rad/s</u>

7 0
3 years ago
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