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Vinvika [58]
3 years ago
15

15 PTS! HELP! BRAINLIEST!

Physics
1 answer:
vodka [1.7K]3 years ago
3 0
I cant do all but I can do number 4
      When force moves ur leg u start skatin and when that happenes ur using a lot of force to move
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A classmate is attempting to draw a free body diagram for a box being pushed across the floor at a constant speed. The image bel
aivan3 [116]

Answer:

A. The applied force should be the same size as the friction force

Explanation:

Whenever we apply a force to an object it moves if the force applied to that object is unbalanced and there is no force or a lesser force to counter it. According to Newton's Second Law of motion, when an unbalanced force is applied to an object it produces an acceleration in the object in its own direction. So, the two forces acting on this box are the frictional force and the applied force in horizontal direction. In order to move the box at constant speed, the applied force must first, overcome the frictional force, so the object can start its motion. Since, the motion has constant velocity, it means no acceleration. So, the force must be balanced in order to avoid acceleration as a consequence of Newton's Second Law of motion. Therefore, the correction in this case will be:

<u>A. The applied force should be the same size as the friction force</u>

7 0
2 years ago
Does the KE of a car change more when it accelerates from 22 km/h to 32 km/h or when it accelerates from 32 km/h to 42 km/h
mote1985 [20]

Answer:

The change in kinetic energy (KE) of the car is more in the second case.

Explanation:

Let the mass of the car = m

initial velocity of the first case, u = 22 km/h = 6.11 m/s

final velocity of the first case, v = 32 km/h = 8.89 m/s

change in kinetic energy (K.E) = ¹/₂m(v² - u²)

                                          ΔK.E = ¹/₂m(8.89² - 6.11²)

                                                    = 20.85m J

 

initial velocity of the second case, u = 32 km/h = 8.89 m/s

final velocity of the second case, v = 42 km/h = 11.67 m/s

change in kinetic energy (K.E) = ¹/₂m(v² - u²)

                                          ΔK.E = ¹/₂m(11.67² - 8.89²)

                                                    = 28.58m J

The change in kinetic energy (KE) of the car is more in the second case.

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Mercury is 35.98 million miles from the Sun.

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The answer is D for sure.

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