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12345 [234]
3 years ago
5

When a 12 N horizontal force is applied to a box on a horizontal tabletop, the box remains at rest. The force of friction acting

on the box is
1) 0N
2) Between 0N and 12N
3) Exactly 12 N
4) Greater than 12N​
Physics
1 answer:
Mrrafil [7]3 years ago
3 0

Answer:

12N

Explanation:

when a force is applied to a body but still stays at rest or moves at a constant speed , the frictional force is equal to the force applied

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Answer:

Put water at room temperature into a vacuum chamber and begin removing the air. Eventually, the boiling temperature will fall below the water temperature and boiling will begin without heating. Or if you want to be easy but messy, add dry ice to a bowl of water and watch how the water starts to boil.

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The driver of a car slams on the brakes, causing the car to slow down at a rate of
Alona [7]

Answer:

The time taken for the car to stop is 5.43 s.

The initial velocity of the car is 108.6 ft/s

Explanation:

The following data were obtained from the question:

Acceleration (a) = –20 ft/s² (since the car is coming to rest)

Distance travalled (s) = 295 ft

Final velocity (v) = 0 ft/s

Time taken (t) =?

Initial velocity (u) =?

Next, we shall determine the initial velocity of the car as shown below:

v² = u² + 2as

0² = u² + (2 × –20 × 295)

0 = u² + (–11800)

0 = u² – 11800

Collect like terms

0 + 11800 = u²

11800 = u²

Take the square root of both side

u = √11800

u = 108.6 ft/s

Therefore, the initial velocity of the car is 108.6 ft/s.

Finally, we shall determine the time taken for the car to stop as shown below:

Acceleration (a) = –20 ft/s² (since the car is coming to rest)

Final velocity (v) = 0 ft/s

Initial velocity (u) = 108.6 ft/s

Time taken (t) =?

v = u + at

0 = 108.6 + (–20 × t)

0 = 108.6 + (–20t)

0 = 108.6 – 20t

Collect like terms

0 – 108.6 = – 20t

– 108.6 = – 20t

Divide both side by –20

t = – 108.6 / –20

t = 5.43 s

Therefore, the time taken for the car to stop is 5.43 s.

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3 years ago
The gas in a cylinder has a volume of 3 liters at a pressure of 117 kpa. the pressure of the gas is increased to 219 kpa. assumi
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V = (1 / p) × p0×V0 = (1 / 219kPa) × 117kPa × 3L = 1.06L
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