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4 years ago

To have momentum, an object must be ______.

Physics

1 answer:

Basile [38]4 years ago

7 0

It must be massive because momentum is the product of mass and velocity.

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4 years ago

What do you mean by kinetic energy and potential energy ?

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Kinetic Energy of an object is the measure of the work an object can do by virtue of its motions..

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2 years ago

Which body is in equilibrium?

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3 years ago

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A 7 kg sled is initially at rest on a horizontal road. the sled is pulled a distance of 2.9 m by a force of 42 n applied to the

Artemon [7]

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4 years ago

Suppose an object is launched from a point 320 feet above the earth with an initial velocity of 128 ft/sec upward, and the only

Ne4ueva [31]

Answer:

(a)Therefore the highest altitude attained by the object is =576 ft .

(b)Therefore the object takes 6 sec to fall to the ground.

Explanation:

Initial velocity: Initial velocity is a velocity from which an object starts to move.

u is usually used for notation of initial notation.

Final velocity: Final velocity is a velocity of an object after certain second from starting.

The final velocity is denoted by v.

Acceleration: The difference of final velocity and initial velocity per unit time

The S.I unit of acceleration is m/s².

(a)

Given that u= 128 ft\sec and g = 32 ft/sec².

At highest point the velocity of the object is 0 i.e v=0

Since the displacement is opposite to the gravity.

Therefore acceleration( a)= -g = -32 ft/sec².

To find the time this to happen we use the following formula

$v=u+at$

Here v=0

⇒0=128+(-32) t

⇒32t=128

⇒t = 4 sec

To determine the height we use the following formula

$s=ut+\frac{1}{2} at^2$

$\Rightarrow s= (128\times4)+\frac{1}{2}\times (-32) \times4^2$

⇒s= 256 ft

Therefore the highest altitude attained by the object is =(320+256)ft=576 ft .

(b)

At the highest point the velocity of the object is 0.

so u=0. a=g= 32 ft/sec² [ since the direction of gravity and the displacement are same] s= 576 ft

To determine the time to fall we use the following formula

$s=ut+\frac{1}{2} at^2$

$\Rightarrow 576 = (0\times t)+\frac{1}{2} \times 32 \times t^2$

$\Rightarrow 16\times t^2=576$

$\Rightarrow t^2=\frac{576}{16}$

$\Rightarrow t^2=36$

⇒t=6 sec

Therefore the object takes 6 sec to fall to the ground.

8 0

3 years ago