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Ugo [173]
2 years ago
14

A 63.0kg sprinter starts a race with an acceleration of 24.0m/s/s . what is the net force of him

Physics
1 answer:
shusha [124]2 years ago
8 0

Answer:

1512N

Explanation:

f = ma

here, m=63kg

a=24ms^-²

so, f=ma=63×24=1512N (Newton)

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A )voltage
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a rocket, initially at rest, is fired vertically with a net upward acceleration of 12 m/s2 . at an altitude of 0.50 km, the engi
kobusy [5.1K]

The rocket travelled a maximum height at 1.0102 km.

Given,

The acceleration of a rocket (a) = 12 m/s²

The altitude of the rocket (s) =  0.50 km = 0.5×10³m

The maximum height of the rocket (h) = ?

Solution,

A rocket is a spacecraft, aircraft, vehicle or projectile that obtains thrust from a rocket engine.

The rate of change of the velocity of an object with respect to time is known as acceleration. It is denoted by (a).i.e. unit is m/s²

(a) = Δv/Δt

Where , Δv is change in velocity and Δt is change in time.

The rate of change in position with respect to time is known as velocity. i.e. Its unit is m/s.

(v)= Δx/Δt

Where,Δx is the change in position and Δt is change in time & v is velocity.

Therefore we know the equation of motion is written as,

v² = u² +2as

Where, v  is final velocity , u is initial velocity , a is acceleration and s is altitude of the rocket.

Then putting the value ,

v² = 0 + ( 2× 10 × 0.5×10³)m/s

v² = \sqrt{10000} m/s

v = 100 m/s

Therefore, at altitude of 0.50 km the initial velocity of rocket (u) will be 100 m/s, final velocity v become zero and under free falling the acceleration will be taken (-g) then equation of motion can be given as ,

v² = u² - 2(g)h

h = (v²- u² ) / 2g

h = 10,000/2×9.8

h = 510.2 m

So that the rocket travelled the maximum height ,

(h)= (0.5 km + 510.2m)

(h) = 1.0102 km

Hence, the rocket travelled at the maximum height h is 1.0102 km

To know more about acceleration

brainly.com/question/15135960

#SPJ4

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1 year ago
Which of two factors influence the weight of an object due to gravitational pull?
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The masses of the object and the planet it's on, and the distance between their centers.
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3 years ago
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Suppose that during any period of ¼ second there is one instant at which the crests or troughs of component waves are exactly in
kolezko [41]
<h3><u>Answer;</u></h3>

A. 4

<h3><u>Explanation;</u></h3>
  • <em><u>The period of a wave or periodic time is the time taken for a complete oscillation to occur. </u></em>For example its is the time taken between two successive crests or troughs.
  • <em><u>The beats or oscillation that occur in one second represents the frequency. Frequency is the number of complete oscillations or beats in one second in a wave.</u></em>
  • Frequency, measured in Hertz is given by the reciprocal of the periodic time.
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6 0
3 years ago
Suppose the entire population of the world gathers in ONE spot and everyone jumps at the sound of a prearranged signal. While ev
puteri [66]

Answer:

(b) Yes, the earth gains momentum but the change in momentum of the earth is much lesser compared to that of everyone in the air. The resistance to motion (inertia of the earth), which is a function of its mass is so great that the earth's acceleration is small in the given time frame.

Explanation:

From Newton's second law which can be stated mathematically as

F = m(v-u)/t = ma.

By Newton's law of gravitation, there is a force between the earth and everyone in the air. This force is responsible for the change in momentum of everyone in the air and this force gives them an acceleration equal to g = 9.80m/s². By Newton's law of gravitation and Newton's third law of motion, this force is also equal to the force exerted by everyone on the earth.

For this to be true,

F = M (everyone) ×a (everyone) = M(earth) × a (earth).

And

a (earth) = {M (everyone) ×a (everyone) }/M (earth)

Then

a (earth) must be lesser than a (everyone) since M(earth) >> M(everyone).

a = change in momentum/ time

Therefore the earth will have a much lesser change in momentum which is the reason we won't notice the earth's movement.

Thank you for reading.

6 0
3 years ago
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