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2 years ago

We cab calculate work because we know the force and distance.what else would we need to know to calculate power

Physics

1 answer:

san4es73 [151]2 years ago

6 0

We would need to know how long it took to do the work.

Power = (work) / (time).

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The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s. (a) What

Neko [114]

Complete Question

The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.

(a) What is its angular acceleration in revolutions per minute-squared

(b) How many revolutions does the engine make during this 20 s interval?

rev

Answer:

$\alpha = 6261 \ rev/minutes^2$

$\theta = 613 \ revolutions$

Explanation:

From the question we are told that

The initial angular speed is $w_i = 1120 \ rev/minutes$

The angular speed after $t = 13.8 s = \frac{13.8}{60 } = 0.23 \ minutes$ is $w_f = 2560 \ rev/minutes$

The time for revolution considered is $t_r = 20 \ s = \frac{20}{60} = 0.333 \ minutes$

Generally the angular acceleration is mathematically represented as

$\alpha = \frac{w_f - w_i }{t}$

=> $\alpha = \frac{2560 - 1120 }{0.23}$

=> $\alpha = 6261 \ rev/minutes^2$

Generally the number of revolution made is $t_r = 20 \ s = \frac{20}{60} = 0.333 \ minutes$ is mathematically represented as

$\theta = \frac{1}{2} * (w_i + w_f)* t$

=> $\theta = \frac{1}{2} * (1120+ 2560 )* 0.333$

=> $\theta = 613 \ revolutions$

5 0

2 years ago

If a pizza delivery guy declares himself the leader over the other pizza delivery people, he probably won't have many people obe

NemiM [27]

Answer:

the last one i think...

4 0

2 years ago

Two students, Student X and Student Y, stand on a long skateboard that is at rest on a flat, horizontal surface, as shown. In or

OleMash [197]

Answer:

the answer is B.

Explanation:

The claim is correct because Student Y can apply a force that is greater in magnitude than the frictional forces that are exerted on the student-student-skateboard system

6 0

2 years ago

Read 2 more answers

the total positive charge is QQQ = 1.62×10−6 CC , what is the magnitude of the electric field caused by this charge at point P,

balu736 [363]

Answer:

6.1 × 10^9 Nm-1

Explanation:

The electric field is given by

E= Kq/d^2

Where;

K= Coulombs constant = 9.0 × 10^9 C

q = magnitude of charge = 1.62×10−6 C

d = distance of separation = 1.53 mm = 1.55 × 10^-3 m

E= 9.0 × 10^9 × 1.62×10−6/(1.55 × 10^-3 )^2

E= 14.58 × 10^3/2.4 × 10^-6

E= 6.1 × 10^9 Nm-1

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2 years ago

Write the nuclear equation for the release of the beta particles by Pb-210.

Liula [17]

210 Pb ---> -ie + 210 B:
84 8.3

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3 years ago