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Black_prince [1.1K]

2 years ago

12

A teacher pushed a 98 newton desk across a floor for a distance of 5 meters. He exerted a horizontal force of 20 N for four seco

nds. What was his power?

1 answer:

Pie2 years ago

7 0

Answer:

First let's find the work.

Work = Force*Displacement

= 20*5

= 100 Joules

Power = Work/Time

= 100/4

= 25 Watts

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You have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3

LenaWriter [7]

Answer:

0 Joules

Explanation:

The work done is given by

$W=F\times s\times cos\theta$

where,

F = Force applied

s = Displacement of the object = 0 m

$\theta$ = Angle between the force applied and the horizontal = 0

$W=F\times 0\times cos0\\\Rightarrow W=0\ J$

Work is only observed when there is a displacement.

The work done by me is 0 Joules as I was unable to move it.

6 0

3 years ago

Who is the founder of operant conditioning?

Gemiola [76]

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Hope this helps!

7 0

3 years ago

Help me please i dont understand

alexandr402 [8]

Option A is the correct answer.

5 0

3 years ago

In the steady state 1.2 ✕ 1018 electrons per second enter bulb 1. There are 6.3 ✕ 1028 mobile electrons per cubic meter in tungs

bekas [8.4K]

Answer:

E=12.2V/m

Explanation:

To solve this problem we must address the concepts of drift velocity. A drift velocity is the average velocity attained by charged particles, such as electrons, in a material due to an electric field.

The equation is given by,

$V=\frac{I}{nAq}$

Where,

V= Drift Velocity

I= Flow of current

n= number of electrons

q = charge of electron

A = cross-section area.

For this problem we know that there is a rate of 1.8*10^{18} electrons per second, that is

$\frac{I}{q} = 1.2*10^{18}$

$A= 1.3*10^{-8}m^2$

$n=6.3*10^{28} e/m^3$

$\omicron{O} = 1.2*10^{-4}(m/s)(N/c)$ Mobility

We can find the drift velocity replacing,

$V = \frac{1.2*10^{18}}{(1.3*10^{-8})(6.3*10^{28})}$

$V= 1.465*10^-3m/s$

The electric field is given by,

$E= \frac{V}{\omicron{O}}$

$E=\frac{1.465*10^-3}{1.2*10^{-4}}$

$E=12.2V/m$

7 0

3 years ago

A 1,700 kg car is being used to give a 1,400 kg car a push start by exerting a force of

Gwar [14]

Answer:

- 670 kg.m/s

Explanation:

Newton's third law states that to every action, there is equal and opposite reaction force. Since the force will be same but different in direction and acted in the same time then the impulses ( force multiply by time) of the two car be same in magnitude but different in direction - 670 kg.m/s

5 0

3 years ago