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Alinara [238K]
3 years ago
10

A child carries a 3N book at a constant velocity of 4 meters across a horizontal floor. What is the net work done?

Physics
1 answer:
s2008m [1.1K]3 years ago
6 0
This problem is getting you used to the definition of 'work' = (force) x (distance).

It's hard to really imagine, but technically, when an object moves horizontally,
no work is done, because there's no lifting against the force of gravity.

Another way to look at it:  The potential energy of the object never changes,
because its height never changes, so no energy ever goes into it ot gets
taken out of it.

Also, if the object moves at a constant speed, then it never accelerated,
and that means that no force ever acted on it.

So technically, when an object moves horizontally at a constant speed,
the work done on it is zero.
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If points a and b are connected by a wire with negligible resistance, find the magnitude of the current in the 12.0 v battery.
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Why is vehicle systems forensics useful today? Cars are less computerized than before More people own cars Devices like smartpho
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A simple and common technique for accelerating electrons is shown in the figure, where there is a uniform electric field between
Sunny_sXe [5.5K]

Answer : 4.483\times 10^{15}\ m/s^2.

Explanation:

It is given that,

Electric field strength, E=2.55\times 10^{4}\ N/C

We know that,

Charge of electron, q=1.6\times 10^{-19}\ C

Mass of electron, m=9.1\times 10^{-31}\ kg

From the definition of electric field, F=qE...............(1)

According to Newton's second law, F = ma..........(2)

From equation (1) and (2)

ma=qE

a=\dfrac{qE}{m}

a=\dfrac{1.6\times 10^{-19}\ C\times 2.55\times 10^{4}\ N/C }{9.1\times 10^{-31}\ kg}

a=0.4483\times 10^{16}\ m/s^2

or

a=4.483\times 10^{15}\ m/s^2

So, the horizontal component of acceleration of an electron is 4.483\times 10^{15}\ m/s^2.

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7 0
3 years ago
Hey help me plzzzzz i will mark brainliest​
vodomira [7]

Answer:

The answer to your question is given below.

Explanation:

Mechanical advantage (MA) = Load (L)/Effort (E)

MA = L/E

Velocity ratio (VR) = Distance moved by load (l) / Distance moved by effort (e)

VR = l/e

Efficiency = work done by machine (Wd) /work put into the machine (Wp) x 100

Efficiency = Wd/Wp x100

Recall:

Work = Force x distance

Therefore,

Work done by machine (wd) = load (L) x distance (l)

Wd = L x l

Work put into the machine (Wp) = effort (E) x distance (e)

Wp = E x e

Note: the load and effort are measured in Newton (N), while the distance is measured in metre (m)

Efficiency = Wd/Wp x100

Efficiency = (L x l) / (E x e) x 100

Rearrange

Efficiency = L/E ÷ l/e x 100

But:

MA = L/E

VR = l/e

Therefore,

Efficiency = L/E ÷ l/e x 100

Efficiency = MA ÷ VR x 100

Efficiency = MA / VR x 100

7 0
3 years ago
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