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OLga [1]
3 years ago
13

A box has a 20 N force applied to it to move it 5 m. What is the work done on the box? 4 J 4 N 25 J 100 J

Physics
2 answers:
Sunny_sXe [5.5K]3 years ago
7 0
100 J

Explanation:
multiply the force by the distance
20 N x 5 meters = 100 J
please mark brainliest
Rzqust [24]3 years ago
3 0
To solve for work you would need to take your force and multiply it by your displacement which would be 20F*5m=100J
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a condenser of capacitor 50 micro coulomb is charged to 10 volt the energy stored is can you do derivation please
Sedbober [7]

Answer:

2.5x10^-3 J

Explanation:

C=50mC

V=10V

U=1/2 C (V)^2

=1/2(50)(10)^2 x (10)^-6

=25(100)/1000000

=2.5x10^-3

8 0
2 years ago
Determine the angular velocity of the merry-go-round if a jumps off horizontally in the −n direction with a speed of 2 m/s , mea
lapo4ka [179]

by angular momentum conservation we will have

angular momentum of child + angular momentum of merry go round = 0

angular momentum of child = mvR

m = mass of child

R = radius of child

v = speed = 2 m/s

now let's say moment of inertia of merry go round is I

so we will have

m*2*R + Iw = 0

w = -\frac{2mR}{I}

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6 0
3 years ago
Find the magnitude of the sum of two vectors; A is 5 km , and B is 7 , when the angle btween them is 120
Dimas [21]

Answer: 6.24 km

Explanation:

Given

The magnitude of the first vector(say) \left |  a\right |=5\ km

the magnitude of the second vector(say) \left |  b\right |=7\ km

the angle between them is 120^{\circ}

The resultant vector magnitude is given by

\left |  \vec{R}\right |=\sqrt{a^2+b^2+2ab\cos \theta}

\left |  \vec{R}\right |=\sqrt{5^2+7^2+2\times 5\times 7\cdot \cos 120^{\circ}}\\\left |  \vec{R}\right |=\sqrt{74-35}=\sqrt{39}\\\left |  \vec{R}\right |=6.24\ km

4 0
2 years ago
The cricket player while catches the ball wears gloves and why​
gregori [183]

Answer:

the ball is travelling very fast and the player can get injured if he doesn't wear gloves

Explanation:

3 0
2 years ago
How many joules of heat are needed to raise the temperature of 50.0 g of aluminum from 10°C to 110°C, if the specific heat of al
dusya [7]

Answer:

Heat required to raise the temperature of the aluminium is 4750 J

Explanation:

As we know that the heat energy required to raise the temperature of the aluminium is given as

Q = ms\Delta T

here we know that

m = 50 g

\Delta T = 110 - 10

\Delta T = 100 ^oC

so we have

Q = 50(0.95)(100)

Q = 4750 J

5 0
3 years ago
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