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3 years ago

An object is moving from north to south what is the direction of the force of friction of the object

Physics

2 answers:

aev [14]3 years ago

6 0

Answer:

North

Explanation:

Friction is a reaction force against the direction of movement. So since the direction of movement is south the friction would be opposite and move north.

Murljashka [212]3 years ago

3 0

Answer:

South To North

Explanation:

Frictional force acts in the direction opposite to the direction of motion of a body. Because the object is moving from north to south, the direction of frictional force is from south to north

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Name three bright Saturn ring features, and explain why they are so bright.

Leviafan [203]

The F Ring, the Cassini Division, and the C Ring are bright ring features. They are bright due to the low concentration of materials within them, which allows sunlight to shine through.

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3 years ago

Why are north and south poles important to magnets

worty [1.4K]

Answer:

All magnets have north and south poles. Opposite poles are attracted to each other, while the same poles repel each other. When you rub a piece of iron along a magnet, the north-seeking poles of the atoms in the iron line up in the same direction. The force generated by the aligned atoms creates a magnetic field.

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3 years ago

Read 2 more answers

A ball is dropped from rest from the top of a building of height h. At the same instant, a second ball is projected vertically u

uranmaximum [27]

Answer:

a) $t = \sqrt{\frac{h}{2g}}$

b) Ball 1 has a greater speed than ball 2 when they are passing.

c) The height is the same for both balls = 3h/4.

Explanation:

a) We can find the time when the two balls meet by equating the distances as follows:

$y = y_{0_{1}} + v_{0_{1}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{1}}$ : is the initial height = h

$v_{0_{1}}$ : is the initial speed of ball 1 = 0 (it is dropped from rest)

$y = h - \frac{1}{2}gt^{2}$ (1)

Now, for ball 2 we have:

$y = y_{0_{2}} + v_{0_{2}}t - \frac{1}{2}gt^{2}$

Where:

$y_{0_{2}}$ : is the initial height of ball 2 = 0

$y = v_{0_{2}}t - \frac{1}{2}gt^{2}$ (2)

By equating equation (1) and (2) we have:

$h - \frac{1}{2}gt^{2} = v_{0_{2}}t - \frac{1}{2}gt^{2}$

$t=\frac{h}{v_{0_{2}}}$

Where the initial velocity of the ball 2 is:

$v_{f_{2}}^{2} = v_{0_{2}}^{2} - 2gh$

Since $v_{f_{2}}^{2}$ = 0 at the maximum height (h):

$v_{0_{2}} = \sqrt{2gh}$

Hence, the time when they pass each other is:

$t = \frac{h}{\sqrt{2gh}} = \sqrt{\frac{h}{2g}}$

b) When they are passing the speed of each one is:

For ball 1:

$v_{f_{1}} = - gt = -g*\sqrt{\frac{h}{2g}} = - 0.71\sqrt{gh}$

The minus sign is because ball 1 is going down.

For ball 2:

$v_{f_{2}} = v_{0_{2}} - gt = \sqrt{2gh} - g*\sqrt{\frac{h}{2g}} = (\sqrt{1} - \frac{1}{\sqrt{2}})*\sqrt{gh} = 0.41\sqrt{gh}$

Therefore, taking the magnitude of ball 1 we can see that it has a greater speed than ball 2 when they are passing.

c) The height of the ball is:

For ball 1:

$y_{1} = h - \frac{1}{2}gt^{2} = h - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

For ball 2:

$y_{2} = v_{0_{2}}t - \frac{1}{2}gt^{2} = \sqrt{2gh}*\sqrt{\frac{h}{2g}} - \frac{1}{2}g(\sqrt{\frac{h}{2g}})^{2} = \frac{3}{4}h$

Then, when they are passing the height is the same for both = 3h/4.

I hope it helps you!

7 0

3 years ago

PLEASE HELP! A river current has a velocity of 5km/h relative to the shore, and a boat moves in the same direction as the curren

Usimov [2.4K]

add the river's velocity to the boat's velocity

5+ 5 = 10 km/h

Hope this helps!

5 0

4 years ago

Luis is trying to push a box of new soccer balls across the floor. In the illustration, the arrow on the box is a vector represe

solong [7]

Answer:

Complete question

Luis is trying to push a box of new soccer balls across the floor.

If the box is not moving, which of the following must be true?

A. The box is exerting a larger force on Luis than he is exerting on the box

B. There is another force acting on the box that balances Luis's force.

C. Luis is applying a force that acts at a distance.

D. There is no force of friction acting on the box.

Explanation:

Using newton second law of motion

ΣF = ma

Now, for a body not to move when a force is acting on it means that the body is in equilibrium and acceleration a = 0

Therefore,

Fnet = 0

So, if he is applying a force F to push the box and the box is not moving then, there is an external force that is pushing the force back opposite the direction he his pushing and this force counterbalance is own force.

F—F' = 0

F' = F

So, F' is the counter balance force and it is equal to the force applied by Luis

Or it might be frictional force, because if the static friction is not overcome, then, the body will not leave it's state of rest. So if the fictional force is very high, then the box will not leave it rest position and we also know that frictional force opposes motion,

F—Fr=0

F = Fr

So using this explanation,.

The answer is B

B. There is another force acting on the box that balances Luis's force.

3 0

3 years ago