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sergij07 [2.7K]
3 years ago
15

Two equal forces are applied to a door. The first force is applied at the midpoint of the door; the second force is applied at t

he doorknob. Both forces are applied perpendicular to the door. Which force exerts the greater torque?
Physics
1 answer:
TEA [102]3 years ago
6 0

Answer: The second force applied to the doorknob

Explanation:

The formulae for torque is simple the product of the applied force and the perpendicular distance.

The greater the perpendicular force, the greater the torque assuming a constant value of force.

Applying the force at the doorknob gives for a greater distance between the force and the turning point compared to applying the force at the midpoint of the door ( which is at a shorter distance)

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The cylinder of gravity of cylinder is where
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Explanation:

In uniform gravity it is the same as the centre of mass. For regular shaped bodies it lies at the centre of the that particular body. Hence for a cylinder centre of gravity lies at the midpoint of the axis of the cylinder.

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A batter hits a pop fly straight up. (a) Is the acceleration of the ball on the way up different from its acceleration on the wa
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a) No. If air resistance is ignored the acceleration of the ball is the same at each point on its flight.

(b). No. As long as the air resistance is negligible, the acceleration of the ball at the top of its flight different from its acceleration just before it lands.

Explanation:

This constant, similar acceleration is called the acceleration due to gravity and it is the acceleration of a body due to the influence of the pull of gravity alone.

Every object on the surface of the earth regardless of its mass is pulled towards the centre of the earth, especially when in flight. Whether flying upwards, or coming downwards, the pull of attraction towards the surface of the earth is the same!

4 0
3 years ago
1. Spring force is proportional to stretch distance, F=kd. Do you expect doubling the stretch distance double the force?
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Answer:

Yes

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The spring force is given as:

          F  = kd

F is the spring force

K is the spring constant

d is the magnitude of the stretch

 Since k is a constant, therefore, doubling the stretch distance will double the force.

Both stretch distance and force applied can be said to be directly proportional to one another.

3 0
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Tarzan, in one tree, sees Jane in another tree. He grabs the end of a vine with length 20m that makes an angle of 45° with the v
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3 0
4 years ago
coin 1 is thrown upward from the top of 100m tower with a speed of 15m/s. coin 2 is dropped from the top of the tower 2.0second
kvv77 [185]

The height below the tower at which coin 1 pass coin 2 is 89.04 m.

The given parameters:

height of the tower, h = 100 m

initial velocity of coin 1, v = 15 m/s

time spent in air by coin 1 before coin 2 was dropped = 2s

To find:

  • the height below the tower at which coin 1 passes coin 2

Find the maximum height attained by coin 1 before falling to the ground:

v^2 = u^2 - 2gh\\\\where;\\\\v \ is \ the \ final \ velocity \ of \ coin \ 1 \ at \ maximum \ height, v \ = 0\\\\0 = (15^2) - 2(10)h\\\\20h = 225\\\\h = \frac{225}{20} \\\\h = 11.25 \ m

Find the time taken for coin 1 to fall to the ground:

Total height of coin 1 above the ground, H = 11.25 m + 100 m = 111.25 m

t = \sqrt{\frac{2H}{g} } \\\\t = \sqrt{\frac{2\times 111.25}{10} } \\\\t = 4.72 \ s

But the time taken for the coin 1 to reach 11.25 m above the tower:

t_1 = \sqrt{\frac{2h}{g} } \\\\t_1 = \sqrt{\frac{2\times 11.25}{10} } \\\\t_1 = 1.5 \ s

Total time spent by coin 1 before reaching ground with respect to coin 2:

time = (1.5 s + 4.72 s) - 2 s

time = 4.22 s

<u>Note:</u><em> the 2 s was subtracted to keep both coins at a fair starting time below the tower</em>.

Find the total time taken for coin 2 to fall to the ground:

Height of coin 2 above the ground = 100 m

Total time taken by coin 2 before falling to the ground is calculated as:

t_2 = \sqrt{\frac{2(100)}{10} } \\\\t_2 =  4.47s

The time  at which coin 1 will pass coin 2 is 4.22 s.

Find the height below the tower when the time is 4.22 s.

h = \frac{1}{2} (10)(4.22)^2\\\\h = 89.04 \ m

Thus, the height below the tower at which coin 1 pass coin 2 is 89.04 m.

Learn more here: brainly.com/question/18943273

6 0
3 years ago
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