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MAVERICK [17]
3 years ago
14

Is this true or false the kinetic energy of a freely falling object decreases as it begins to fall

Physics
2 answers:
Naddika [18.5K]3 years ago
8 0
The kinetic energy of an object is the energy of the object as it moves. It has more kinetic energy the faster it moves. Because a falling object is increasing in speed, it would also increase in kinetic energy. Hope this helps! :)
rjkz [21]3 years ago
7 0
IT INCREASES AS IT STARTED TO FALL, 
The kinetic energy = K.E =  1/2 * m *V^2  
FROM THAT EQUATION,  IT SAYS THAT THE K.E INCREASES AS THE VELOCITY INCREASE , SINCE IT FALLS SO IT WILL ACCELERATES, THAT MEANS THE VELOLCITY INCREASES, SO THE K.E INCREASE
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Answer:

i) the minimum acceleration to take off is 22500 km/h²

ii) the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) required force that the engine must exert to attain acceleration is 625 kN

Explanation:

Given the data in the question;

mass of plane m = 360,000 kg

take of speed v = 300 km/hr = 83.33 m/s

i)

What should be the minimum acceleration to take off if the length of the runway is 2.00 km

from Newton's equation of motion;

v² = u² + 2as

we know that a plane starts from rest, so; u = 0

given that distance S = 2 km

we substitute

(300)² = 0² + ( 2 × a × 2 )

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a = 90000 / 4

a = 22500 km/h²

Therefore,  the minimum acceleration to take off is 22500 km/h²

ii) At this acceleration, how much time would the plane need from starting to takeoff.

from Newton's equation of motion;

v = u + at

we substitute

300 = 0 + 22500 × t

t = 300 / 22500

t = 0.0133 hrs

Therefore, the required time needed by the plane from starting to takeoff is 0.0133 hrs

iii) What force must the engines exert to attain this acceleration

we know that;

F = ma

acceleration a = 22500 km/hr² = 1.736 m/s²

so we substitute

F = 360,000 kg × 1.736 m/s²

F =  624960 N

F = 625 kN

Therefore, required force that the engine must exert to attain acceleration is 625 kN

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A 2 kg, frictionless block is attached to a horizontal, ideal spring with spring constant 300 N/m. At t = 0 the spring is neithe
schepotkina [342]

Answer:

Explanation:

Given that,

Mass of block

M = 2kg

Spring constant k = 300N/m

Velocity v = 12m/s

At t = 0, the spring is neither stretched nor compressed. Then, it amplitude is zero at t=0

xo = 0

It velocity is 12m/s at t=0

Then, it initial velocity is

Vo = 12m/s

Then, amplitude is given as

A = √[xo + (Vo²/ω²)]

Where

xo is the initial amplitude =0

Vo is the initial velocity =12m/s

ω is the angular frequency and it can be determine using

ω = √(k/m)

Where

k is spring constant = 300N/m

m is the mass of object = 2kg

Then,

ω = √300/2 = √150

ω = 12.25 rad/s²

Then,

A = √[xo + (Vo²/ω²)]

A = √[0 + (12²/12.5²)]

A = √[0 + 0.96]

A = √0.96

A = 0.98m

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