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charle [14.2K]
3 years ago
6

A box of weight 74 N is sliding on a horizontal surface at a constant velocity due to an external force F-> of magnitude 4.8

N. A normal force (n->) and a friction force (f->) also act on the box. What is the magnitude of the friction force?
74 N


2.4 N


69 N


4.8 N


9.6 N

Physics
1 answer:
n200080 [17]3 years ago
6 0

Answer

4.8 N

If the box is moving with a constant velocity, then we can say that the system is in equilibrium. This is because if the external force (F->) was greater than other forces the box would be accelerating. This tells us that this force (F->) is just enough to overcome friction and so it must be equal to 4.8 N.

The normal force has no effect to the horizontal velocities or forces. It is equal to -Weight. That is -74 N. The negative sign shows that the force is in opposite direction.

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Answer:

10N

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1. Every Action has an equal and opposite reAction.

2. If 10N of force is acted upon an wrench, then the wrench will react with an equal amount of force, but in the opposite direction.

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If a car increases its velocity from zero to 60 m/s in 10 seconds, its acceleration is A. 600 m/s2 B. 60 m/s2 C. 3 m/s2 D. 6 m/s
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Lamar writes several equations trying to better understand potential energy. What conclusion is best supported by Lamar’s work?
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Water falls without splashing at a rate of 0.200 L/s from a height of 3.60 m into a 0.730 kg bucket on a scale. If the bucket is
dimaraw [331]

Answer:

15.106 N

Explanation:

From the given information,

The weight of the bucket can be calculated as:

W_b = m_bg =  \\ \\  W_b = (0.730 \  kg) ( 9.80 \ m/s^2) \\ \\ W_b = 7.154 \ N

The mass of the water accumulated in the bucket after 3.20s is:

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m _w=0.64 \ kg

To determine the weight of the water accumulated in the bucket, we have:

W_w = m_w g

W_w = ( 0.64  \ kg )(9.80\  m  \  /s^2)

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For the speed of the water before hitting the bucket; we have:

v = \sqrt{2gh}

v = \sqrt{2*9.80 \ m/s^2 * 3.60 \ m}

v = 8.4 m/s

Now, the force required to stop the water later when it already hit the bucket is:

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Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 mm away horizontal
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Answer:

a. 8.96 m/s b. 1.81 m

Explanation:

Here is the complete question.

a) A long jumper leaves the ground at 45° above the horizontal and lands 8.2 m away.

What is her "takeoff" speed  v 0 ?

b) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 m away horizontally and 2.5 m, vertically below.  

If she long jumps from the edge of the left bank at 45° with the speed calculated in part a), how long, or short, of the opposite bank will she land?

a. Since she lands 8.2 m away and leaves at an angle of 45 above the horizontal, this is a case of projectile motion. We calculate the takeoff speed v₀ from R = v₀²sin2θ/g. where R = range = 8.2 m.

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b. We use R = v₀²sin2θ/g to calculate how long or short of the opposite bank she will land. With v₀ = 8.96 m/s and θ = 45

R = 8.96²sin(2 × 45)/9.8 = 80.2816/9.8 = 8.192 m.

So she land 8.192 m away from her bank. The distance away from the opposite bank she lands is 10 - 8.192 m = 1.808 m ≅ 1.81 m

8 0
3 years ago
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