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2 years ago

If you were trying to describe the difference between power and work you could say:

Physics

2 answers:

avanturin [10]2 years ago

6 0

Power is the energy transferred or "WORK DONE" in one second

nordsb [41]2 years ago

5 0

Answer:

Explanation:

Work done is defined as the product of force in the direction of displacement and the displacement.

Work = force in the direction of displacement x displacement

It is a scalar quantity and its SI uni is Joule.

Power is defined as the rate of doing work.

Power = Work / time

It is also a scalar quantity and its SI unit is Watt.

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A fire hose held near the ground shoots water at a speed of 6.5 m/s. At what angle(s) should the nozzle point in order that the

Mariana [72]

Answer:

17.72° or 72.28°

Explanation:

u = 6.5 m/s

R = 2.5 m

Let the angle of projection is θ.

Use the formula for the horizontal range

$R=\frac{u^{2}Sin2\theta }{g}$

$2.5=\frac{6.5^{2}Sin2\theta }{9.8}$

Sin 2θ = 0.58

2θ = 35.5°

θ = 17.72°

As we know that the range is same for the two angles which are complementary to each other.

So, the other angle is 90° - 17.72° = 72.28°

Thus, the two angles of projection are 17.72° or 72.28°.

8 0

3 years ago

If we cannot see the whole society what can we see

creativ13 [48]

I believe it's People interacting. I hope i've helped you! (:

8 0

2 years ago

A mobile starts with a speed of 250m / s and begins to decelerate at a rate of 3m / s². How fast is it after 45s?

Korvikt [17]

$\large{ \underline{ \underline{ \bf{ \purple{Given}}}}}$

Speed of the mobile = 250 m/s
It starts decelerating at a rate of 3 m/s²
Time travelled = 45s

$\large{ \underline{ \underline{ \bf{ \green{To \: find}}}}}$

Velocity of mobile after 45 seconds

$\large{ \underline{ \underline{ \red{ \bf{Now, \: What \: to \: do?}}}}}$

We can solve the above question using the three equations of motion which are:-

v = u + at
s = ut + 1/2 at²
v² = u² + 2as

So, Here a is acceleration of the body, u is the initial velocity, v is the final velocity, t is the time taken and s is the displacement of the body.

$\large{ \bf{ \underline{ \underline{ \orange{Solution:}}}}}$

We are provided with,

u = 250 m/s
a = -3 m/s²
t = 45 s

By using 1st equation of motion,

⇛ v = u + at

⇛ v = 250 + (-3)45

⇛ v = 250 - 135 m/s

⇛ v = 115 m/s

✤ Final velocity of mobile = 115 m/s

━━━━━━━━━━━━━━━━━━━━

4 0

3 years ago

¿Qué cantidad de calor absorbió una masa de 4 g de cinc al pasar de 20 °C a 180 °C?

Alborosie

(amount of heat)Q = ? , (Mass) m= 4 g , ΔT = T f - T i = 180 c° - 20 °c = 160 °c ,

Ce = 0.093 cal/g. °c

Q = m C ΔT

Q = 4 g × 0.093 cal/g.c° × ( 180 °c- 20 °c )

Q= 4×0.093 × 160

Q = 59.52 cal

I hope I helped you^_^

7 0

2 years ago

A shot is fired at an angle of 60 degree horizontal with Kinetic energy E. If air resistance is ignored, the K.E at the top of t

Lapatulllka [165]

I'm not sure what "60 degree horizontal" means.

I'm going to assume that it means a direction aimed 60 degrees
above the horizon and 30 degrees below the zenith.

Now, I'll answer the question that I have invented.

When the shot is fired with speed of 'S' in that direction,
the horizontal component of its velocity is S cos(60) = 0.5 S ,
and the vertical component is S sin(60) = S√3/2 = 0.866 S . (rounded)

-- 0.75 of its kinetic energy is due to its vertical velocity.
That much of its KE gets used up by climbing against gravity.

-- 0.25 of its kinetic energy is due to its horizontal velocity.
That doesn't change.

-- So at the top of its trajectory, its KE is 0.25 of what it had originally.

That's E/4 .

3 0

2 years ago