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Citrus2011 [14]
3 years ago
14

You are pushing a wooden crate across the floor at constant speed. You decide to turn the crate on end, reducing by half the sur

face area in contact with the floor. In the new orientation, to push the same crate across the same floor with the same speed, the force that you apply must be aboutA. Four times as greatB. Twice as greatC. Equally greatD. Half as greatE. One-fourth as greatas the force required before you changed the crate's orientation.
Physics
1 answer:
borishaifa [10]3 years ago
3 0

Answer:

D. Half as great

Explanation:

Since we know that the friction force between the surface of crate and ground is given as

F_f = \mu F_n

so here we know that

\mu = friction coefficient between two surfaces which depends on the effective contact area between two surfaces

F_n = normal force due to the object

So when we turn the object on another side such that the surface area is half then the friction coefficient will become also half

So here the friction force will also reduce to half

so correct answer will be

D. Half as great

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water flows horizontally from a hose that is 3m above the ground. the water lands 7.2m to the right of the hose . how long does
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Answer:

0.782 s

Explanation:

The water flows horizontally from the hose, so its initial vertical velocity is 0.

Given:

y₀ = 3 m

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v₀ = 0 m/s

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8 0
3 years ago
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If the electron has half the speed needed to reach the negative plate, it will turn around and go towards the positive plate. Wh
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Answer:

 v = -v₀ / 2

Explanation:

For this exercise let's use kinematics relations.

Let's use the initial conditions to find the acceleration of the electron

            v² = v₀² - 2a y

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now they tell us that the initial velocity is half

          v’² = v₀’² - 2 a y’

          v₀ ’= v₀ / 2

at the point where turn v = 0              

          0 = v₀² /4  - 2 a y '

          v₀² /4 = 2 (v₀² / 2y)  y’

          y = 4 y'

          y ’= y / 4

We can see that when the velocity is half, advance only ¼ of the distance between the plates, now let's calculate the velocity if it leaves this position with zero velocity.

         v² = v₀² -2a y’

         v² = 0 - 2 (v₀² / 2y) y / 4

         v² = -v₀² / 4

         v = -v₀ / 2

We can see that as the system has no friction, the arrival speed is the same as the exit speed, but with the opposite direction.

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The forces of attraction and repulsion in liquids are comparable. Compared to the solid state, they move a little bit more. They then assume the shape of the container while still having a fixed capacity.

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