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max2010maxim [7]
2 years ago
9

A body of mass 5kg is ejected vertically from the ground when a force of 600N acts on it for 0.1s.Calculate the velocity with wh

ich the body leaves the ground.​
Physics
1 answer:
Reil [10]2 years ago
7 0

1200

Explanation:

F = m x (v / t)

v / t = F / m

v = (F / m) / t

v = (600 / 5) / 0.1

v = 120 / 0.1

v = 1200 m / s

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2.0

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What causes reactants to gain potential energy?
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The St. Louis Arch has a height of 192 m. Suppose that a stunt woman of mass 84 kg jumps off the top of the arch with an elastic
Hitman42 [59]

To solve this problem it is necessary to apply the kinematic equations of motion for speed and distance, as well as the concepts related to kinetic energy.

The change in the height of a body subject to gravity is given by

h = \frac{1}{2} gt^2 \rightarrow t = \sqrt{\frac{2h}{g}}

Where

h = Height

g =Gravity

t = time

Replacing with our values we have that the time is

t = \sqrt{\frac{2h}{g}}

t = \sqrt{\frac{2(192)}{9.8}}

t = 6.25s

From speed as a function of change between acceleration and time we have then that after 2.6 seconds the speed would be

g = \frac{v}{t} \rightarrow v = g*t

v = 9.8*2.6

v = 25.48m/s

The kinetic energy would be given by

KE = \frac{1}{2} mv^2

KE = \frac{1}{2} (84)(25.48)

KE = 1070.16J

Therefore the kinetic energy after 2.6s is 1070.16J

6 0
3 years ago
riders on the tower of doom, an amusement park ride, experience 2.0 s of free fall, after which they are slowed to a stop in 0.5
Rainbow [258]

By Newton's second law, the apparent weight of rider coming at rest is <u>3548 N</u>.

According to formula, v = u + at

a = 39.2 m/s^{2}

Force, F = ma

F = m(a+g)

F = apparent weight

m = mass, given = 65 kg

a = acceleration, given = 39.2 m/s^{2}

g = 9.8 m/s^{2}

Put these values in formula, F = m(a+g)

F = 65(39.2+9.8)

F = 3548 N

Therefore, the apparent weight of rider coming at rest is <u>3548 N</u>.

Learn more about apparent weight here:- brainly.com/question/24897276

#SPJ4

6 0
2 years ago
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