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3 years ago

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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Explanation:

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2 years ago

The angular velocity of a flywheel obeys the equa tion w(1) A Br2, where t is in seconds and A and B are con stants having numer

makkiz [27]

Answer:

$A \to rad/s$

$B \to rad/s^3$

Explanation:

$\omega_z(t)=A + Bt^2$

Required

The units of A and B

From the question, we understand that:

$\omega_z(t) \to rad/s$

This implies that each of $A$ and $Bt^2$ will have the same unit as $\omega_z(t)$

So, we have:

$A \to rad/s$

$Bt^2 \to rad/s$

The unit of t is (s); So, the expression becomes

$B * s^2 \to rad/s$

Divide both sides by $s^2$

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5 0

2 years ago

To calibrate the calorimeter electrically, a constant voltage of 3.6 V is applied and a current of 2.6 A flows for a period of 3

iren [92.7K]

Answer:

$372.3 J/^{\circ}C$

Explanation:

First of all, we need to calculate the total energy supplied to the calorimeter.

We know that:

V = 3.6 V is the voltage applied

I = 2.6 A is the current

So, the power delivered is

$P=VI=(3.6)(2.6)=9.36 W$

Then, this power is delivered for a time of

t = 350 s

Therefore, the energy supplied is

$E=Pt=(9.36)(350)=3276 J$

Finally, the change in temperature of an object is related to the energy supplied by

$E=C\Delta T$

where in this problem:

E = 3276 J is the energy supplied

C is the heat capacity of the object

$\Delta T =29.1^{\circ}-20.3^{\circ}=8.8^{\circ}C$ is the change in temperature

Solving for C, we find:

$C=\frac{E}{\Delta T}=\frac{3276}{8.8}=372.3 J/^{\circ}C$

5 0

2 years ago

A 55 newton force applied on an object moves the object 10 meters in the same direction as the force. What is the value of work

kifflom [539]

Answer: Option D: 5.5×10²Joules

Explanation:

Work done is the product of applied force and displacement of the object in the direction of force.

W = F.s = F s cosθ

It is given that the force applied is, F = 55 N

The displacement in the direction of force, s = 10 m

The angle between force and displacement, θ = 0°

Thus, work done on the object:

W = 55 N × 10 m × cos 0° = 550 J = 5.5 × 10² J

Hence, the correct option is D.

3 0

2 years ago