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romanna [79]
4 years ago
10

A boy who exerts a 300-N force on the ice of a skating rink is pulled by his friend with a force of 75 N, causing the boy to acc

elerate across the ice. If drag and the friction from the ice apply a force of 5 N on the boy, what is the magnitude of the net force acting on him?
Physics
1 answer:
Bezzdna [24]4 years ago
7 0

Answer:

70 N

Explanation:

Draw a free body diagram of the boy.  There are four forces:

Weight force mg pulling down,

A 300 N normal force pushing up,

A 75 N applied force pulling right,

and a 5 N friction force pushing left.

The boy's acceleration in the y direction is 0, so the net force in the y direction is 0.

The net force in the x direction is 75 N − 5 N = 70 N.

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Vector has a magnitude of 6.0 m and points 30° north of east. Vector has a magnitude of 4.0 m and points 30° east of north. The
Brut [27]

Answer:

The resultant vector is \vec R = \vec A + \vec B = 7.196\,i + 6.464\,j.

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\vec A = (6\,m)\cdot (\cos30^{\circ} \,i + \sin 30^{\circ}\,j)

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4-meter vector with direction: 30º east of north.

\vec B = (4\,m)\cdot (\cos 60^{\circ}\,i + \sin 60^{\circ}\,j)

\vec B = 2\,i + 3.464\,j

The resultant vector is obtaining by sum of components:

\vec R = \vec A + \vec B = 7.196\,i + 6.464\,j

The resultant vector is \vec R = \vec A + \vec B = 7.196\,i + 6.464\,j.

5 0
4 years ago
A 4.0-kilogram ball moving at 8.0 m/s to the right collides with a 1.0-kilogram ball at rest. After the collision, the 4.0-kilog
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Its about momentum. Momentum (p)=mass(m)xvelocity(v)
So for the first ball P=4x8=32kgm/s
For the second the momentum is zero as it is still.
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Momentum has to be conserved
After the collision the momentum of the 4kg ball is 4x4.8=19.2kgm/s
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3 0
3 years ago
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A 60.0-kg man stands at one end of a 20.0-kg uniform 10.0-m long board. How far from the man is the center of mass of the man-bo
timama [110]

Answer:

x=1.25m

Explanation:

The <em>Center of mass </em>of the system is defined as the point where whole mass of the body is appeared to be  concentrated.

The center of mass of the system is given by

         x=  \frac{m1x1+m2x2}{m1+m2}      

where m1 is mass of man =60 kg

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by substituting in above formula

x =\frac{(60*0)+(20*5)}{60+20} = \frac{100}{80} =1.25 m

x=1.25m

So the center of mass of the system is at 1.25 m from man.

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4 years ago
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Power = work /time

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6 0
4 years ago
Two children fight over a 200g stuffed bear. The 25kg boy pulls to the right with a 15N force and the 20kg girl pulls to the lef
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The masses of the children may not be taken into account since these are towards the ground and directly affecting the bear. The net force is the difference of the forces since they are acting on opposite sides. The magnitude of this net force is equal to 2N and the sign is to where the greater force is from. Therefore, the force is directed towards the left. 
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3 years ago
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