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topjm [15]
3 years ago
12

A force is always required to move an object from rest. true false

Physics
1 answer:
Marina CMI [18]3 years ago
8 0
The statement is true because an object will remain not moving until an external force is applied into. so in order to move an object external force should be applied. for example an object fell down, even no one pushes it pulls it, but the force due to gravity is acting on it so there is still a force acting on it.
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If the Earth rotated in the opposite direction, how would that affect the wind? *
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The wind would still blow, but it would curve and spin in the opposite direction.

Explanation:

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Help with #3 please
Marizza181 [45]

Answer:

The answer to your question is va = 8 cm/s, vb = 12.5 cm/s, a = 9 cm/s²

Explanation:

Data

Ta = 0.125 s

Tb = 0.08 s

Δtab = 0.5 s

distance = 1 cm

Process

1.- Calculate va

va = 1/0.125 = 8 cm/s

vb = 1/0.08 = 12.5 cm/s

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3 years ago
State examples of a transverse wave. ​
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A car's bumper is designed to withstand a 5.04 km/h (1.4-m/s) collision with an immovable object without damage to the body of t
emmasim [6.3K]

Answer:

the magnitude of the average force on the bumper is 3189.8 N

Explanation:

Given the data in the question;

In terms of force and displacement, work done is;

W =F^> × x^>

W = Fxcos\theta    ------- let this be equation 1

where F is force applied, x is displacement and θ is angle between force and displacement.

Now, since the displacement of the bumper and force acting on it is in the same direction,

hence, θ = 0°

we substitute into equation 1

W = Fxcos( 0° )

W = Fx ------- let this be equation 2

Now, using work energy theorem,

total work done on the system is equal to the change in kinetic energy of the system.

W_{net = ΔKE

= \frac{1}{2}mv² -  \frac{1}{2}mu² --------- let this be equation 3

where m is mass of object, v is final velocity, u is initial velocity.

from equation 2 and 3

Fx = \frac{1}{2}mv² -  \frac{1}{2}mu²

we make F, the subject of formula

F = \frac{m}{2x}( v² - u² )

given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s

so we substitute

F = \frac{830}{(2)(0.255)}( (0)² - (1.4)² )

F = 1627.45098 ( 0 - 1.96 )

F = 1627.45098 ( - 1.96 )

F = -3189.8 N

The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.

Therefore, the magnitude of the average force on the bumper is 3189.8 N

6 0
3 years ago
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