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Gre4nikov [31]
3 years ago
10

a force of 25 newtons moves a box a distance of 4 meeters in 5 seconds.the work done on the box is ? NM and the power is. ? nm/

s​
Physics
1 answer:
miskamm [114]3 years ago
6 0

Answer:

The work done on the box is 100 Nm

The power is 20 Nm/s

Explanation:

There is a force 25 newtons moves a box a distance of 4 meters in

5 seconds

The work done on the box is the product of the force and the distance

that the box moves ⇒ <em>work = force × distance</em>

The force = 25 newtons

the distance = 4 meters

Work = 25 × 4 = 100 NM

<em>The work done on the box is 100 Nm</em>

<em></em>

The force moves the box 4 meters in 5 seconds

The power is the rate of work

<em>The power = work ÷ time</em>

The work = 100 Nm

The time = 5 seconds

The power = 100 ÷ 5 = 20 Nm/s

<em>The power is 20 Nm/s</em>

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Give a specific example of a vector quantity. why this type of quantity would be used to describe this event ?
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now by energy equation above when photon of 303 nm incident on the surface

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6 0
3 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

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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

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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
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