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const2013 [10]
3 years ago
14

How much work is needed to push a 5kg bookshelf 4.52 m with the force of 15.3N

Physics
1 answer:
larisa86 [58]3 years ago
4 0

The work done is 69.2 J

Explanation:

The work done by a force when pushing an object is given by

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the direction of the force and of the displacement

In this problem, we have:

F = 15.3 N is the force applied

d = 4.52 m is the displacement of the bookshelf

\theta=0 (assuming the force is applied in the direction of motion)

Therefore, the work done is:

W=(15.3)(4.52)(cos 0)=69.2 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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Suppose the battery in a clock wears out after moving thousand coulombs of charge through the clock at a rate of 0.5 Ma how long
Ksivusya [100]

Answer:

Hello your question is poorly written below is the complete question

Suppose the battery in a clock wears out after moving Ten thousand coulombs of charge through the clock at a rate of 0.5 Ma how long did the clock run on does battery and how many electrons per second slowed?

answer :

a) 231.48 days

b) n = 3.125 * 10^15

Explanation:

Battery moved 10,000 coulombs

current rate = 0.5 mA

<u>A) Determine how long the clock run on the battery. use the relation below</u>

q = i * t ----- ( 1 )

q = charge , i = current , t = time

10000 = 0.5 * 10^-3 * t

hence  t = 2 * 10^7 secs

hence the time = 231.48  days

<u>B) Determine how many electrons per second flowed </u>

q = n*e ------ ( 2 )

n = number of electrons

e = 1.6 * 10^-19

q = 0.5 * 10^-3 coulomb ( charge flowing per electron )

back to equation 2

n ( number of electrons ) = q / e = ( 0.5 * 10^-3 ) / ( 1.6 * 10^-19 )

hence : n = 3.125 * 10^15

8 0
2 years ago
An object moving with a speed of 35 m/a and has a kinetic energy of 1500j, what is the mass of the object
EleoNora [17]

Explanation:

Speed or velocity (V) = 35 m/s

Kinetic energy (K. E) = 1500 Joule

mass (m) = ?

We know

K.E = 1/2 * m * v²

1500 = 1/2 * m * 35²

1500 * 2 = 1225m

m = 3000 / 1225

m = 2.45 kg

The mass of the object is 2.45 kg

Hope it will help :)

5 0
3 years ago
Energy is transferred by the process of convection from the hot water at the bottom of the tank to the cooler water at the top.
iragen [17]
The transfer of energy means, in convention process, transport of matter. In this case, hot water has lower density than cool water. The water with less density ascends and leaves gaps that are occupied with cooler water "packages".
7 0
3 years ago
An old-fashioned LP record rotates at 33 1/3 RPM
Ksenya-84 [330]

Answer:

Part a) \frac{5}{9}\ \frac{rev}{sec}

Part b) \frac{9}{5}\ \frac{sec}{rev}

Explanation:

Part a) what is its frequency, in rev/s

we have that

An old-fashioned LP record rotates at 33 1/3 RPM

so

33\frac{1}{3}\ \frac{rev}{min}

Convert mixed number to an improper fraction

33\frac{1}{3}\ \frac{rev}{min}=\frac{33*3+1}{3}=\frac{100}{3}\ \frac{rev}{min}

Remember that

1\ min=60\ sec

Convert rev/min to rev/sec

\frac{100}{3}\ \frac{rev}{min}=\frac{100}{3}(\frac{1}{60})=\frac{100}{180}\ \frac{rev}{sec}

Simplify

\frac{5}{9}\ \frac{rev}{sec}

Part b) what is it period, in seconds

we know that

The period is the reciprocal of the frequency

therefore

the frequency is

\frac{9}{5}\ \frac{sec}{rev}

4 0
3 years ago
Why are noise considerations important in optical fiber communications? 3. Describe the principle of "population inversion".
pochemuha

Answer and Explanation:

the electronic devices always have some noises present in the signal

there are some important considerations in optical fiber communications these are.

  • the noise which is contributed by transmitter are electronic random noise, low frequency noise
  • noise which is contributed by laser are relative intensity noise, mode partition noise, conversion of phase noise to amplitude noise.
  • noise contributed by photo detector are quantum shot noise, shot noise from dark current, avalanche multiplication noise.

PRINCIPLE OF POPULATION INVERSION :

The principle of population inversion is defined as for production of high percentage of simulated emission for a laser beam the number of atoms in higher state should be greater than lower energy state

7 0
3 years ago
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