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weeeeeb [17]
3 years ago
7

If you are given force and distance, you can determine power if you know (2 points)

Physics
1 answer:
dangina [55]3 years ago
4 0

Answer:

The answer is time

Explanation:

So when youre given force and distance, you can determine work done

Work Done = Force × Distance travelled in the

direction of the force

Since Power = Work Done/ Time

when you know work done, and you want to find power, you will need time.

Because you have work done already, you dont need energy. Though you can use energy and time to find work too. The alternative formula for Power would be:

Power = Energy Converted/Time

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Reflection off of a smooth surface like a mirror is an example of diffuse reflection.
Akimi4 [234]

Answer:

true

Explanation:

Reflection is when light bounces off an object. If the surface is smooth and shiny, like glass, water or polished metal, the light will reflect at the same angle as it hit the surface. ... This is called diffuse reflection. Diffuse reflection is when light hits an object and reflects in lots of different directions.

8 0
3 years ago
Read 2 more answers
A 1425 kg truck driving at 13.0 m/s collides elastically with a stationary 1175 kg car. If the car is traveling 14.25 m/s just a
TEA [102]

Answer:

1.25 m/s

Explanation:

m1v1+m2v2=m1v1f+m2v2f

(1425*13)+(1175*0)=(1425*v1f)+(1175*14.25)

18525+0=1425(v1f)+16743.75

1781.25=1425(v1f)

v1f=1.25 m/s

4 0
2 years ago
___is found in fruits and honey. *<br> 1.Maltose<br> 2.Sucrose<br> 3.Fructose <br> 4.Galactose
Oksi-84 [34.3K]

Answer:

3. Fructose

Explanation:

Fructose is a sugar found naturally in fruits, fruit juices, some vegetables and honey.

8 0
3 years ago
Based on the data given, in what direction will the car accelerate?
skad [1K]
Vertical forces:
There is a force of 579N acting upward, and a force of 579N
acting downward.
The vertical forces are balanced ... they add up to zero ...
so there's no vertical acceleration. 
Not up, not down.

Horizontal forces:
There is a force of 487N acting to the left, and a force of 632N
acting to the right.
The net horizontal force is

        (487-left + 632-right)  -  (632-right - 487-right) =  145N to the right.

The net force on the car is all to the right.
The car accelerates to the right.
7 0
3 years ago
A disk rotates about its central axis starting from rest and accelerates with constant angular acceleration. At one time it is r
atroni [7]

(a) 2.79 rev/s^2

The angular acceleration can be calculated by using the following equation:

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha is the angular acceleration

\theta=50.0 rev is the number of revolutions made by the disk while accelerating

Solving the equation for \alpha, we find

\alpha=\frac{\omega_f^2-\omega_i^2}{2d}=\frac{(20.0 rev/s)^2-(11.0 rev/s)^2}{2(50.0 rev)}=2.79 rev/s^2

(b) 3.23 s

The time needed to complete the 50.0 revolutions can be found by using the equation:

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 20.0 rev/s is the final angular speed

\omega_i = 11.0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{20.0 rev/s-11.0 rev/s}{2.79 rev/s^2}=3.23 s

(c) 3.94 s

Assuming the disk always kept the same acceleration, then the time required to reach the 11.0 rev/s angular speed can be found again by using

\alpha = \frac{\omega_f-\omega_i}{t}

where

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

t is the time

Solving for t, we find

t=\frac{\omega_f-\omega_i}{\alpha}=\frac{11.0 rev/s-0 rev/s}{2.79 rev/s^2}=3.94 s

(d) 21.7 revolutions

The number of revolutions made by the disk to reach the 11.0 rev/s angular speed can be found by using

\omega_f^2 - \omega_i^2 = 2 \alpha \theta

where:

\omega_f = 11.0 rev/s is the final angular speed

\omega_i = 0 rev/s is the initial angular speed

\alpha=2.79 rev/s^2 is the angular acceleration

\theta=? is the number of revolutions made by the disk while accelerating

Solving the equation for \theta, we find

\theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}=\frac{(11.0 rev/s)^2-0^2}{2(2.79 rev/s^2)}=21.7 rev

4 0
3 years ago
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