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NemiM [27]
3 years ago
11

What happens to the force of gravitational pull if we tripled the distance and one of the masses is doubled

Physics
2 answers:
Svetllana [295]3 years ago
5 0
Triple the distance ... The force drops to 1/9 of the original force.

Double one mass ... The original force doubles.

Do both ... The original force becomes 2/9 as strong, = about 22.2% of what it was originally.
kicyunya [14]3 years ago
4 0
This depends on the original mass of the object having its mass doubled and the the original distance before the distance was tripled.
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A weight lifter applies an upward force of 1100 N while lowering a dumbbell
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Answer:

A

Explanation:

work = force \times distance

work = 1100 \times 0.5

= 550 \: j

hope it helped a lot

pls mark brainliest with due respect .

6 0
3 years ago
Pls help I will mark brainliest
earnstyle [38]
It will be unstable system because it will not be able to recover from the disturbance
7 0
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1.imagine you were able to throw a ball in a frictionless environment such as outer space.once you let go of the ball,what will
Scorpion4ik [409]
Imagine you were able to throw a ball in a frictionless environment
such as outer space.  Once you let go of the ball, it will travel forever
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5 0
3 years ago
Nitrogen at 100 kPa and 25oC in a rigid vessel is heated until its pressure is 300 kPa. Calculate (a) the work done and (b) the
nignag [31]

Answer:

A. The work done during the process is W = 0

B. The value of heat transfer during the process Q = 442.83 \frac{KJ}{kg}

Explanation:

Given Data

Initial pressure P_{1} = 100 k pa

Initial temperature T_{1} = 25 degree Celsius = 298 Kelvin

Final pressure P_{2} = 300 k pa

Vessel is rigid so change in volume of the gas is zero. so that initial volume is equal to final volume.

⇒ V_{1} = V_{2} ------------- (1)

Since volume of the gas is constant so pressure of the gas is directly proportional to the temperature of the gas.

⇒ P ∝ T

⇒ \frac{P_{2} }{P_{1}} = \frac{T_{2} }{T_{1}}

⇒ Put all the values in the above formula we get the final temperature

⇒ T_{2} = \frac{300}{100} × 298

⇒ T_{2} = 894 Kelvin

(A). Work done during the process is given by W = P × (V_{2} -V _{1})

From equation (1), V_{1} = V_{2} so work done W = P × 0 = 0

⇒ W = 0

Therefore the work done during the process is zero.

Heat transfer during the process is given by the formula Q = m C_{v} ( T_{2} -T_{1} )

Where m = mass of the gas = 1 kg

C_{v} = specific heat at constant volume of nitrogen = 0.743 \frac{KJ}{kg k}

Thus the heat transfer Q = 1 × 0.743 × ( 894- 298 )

⇒ Q = 442.83 \frac{KJ}{kg}

Therefore the value of heat transfer during the process Q = 442.83 \frac{KJ}{kg}

6 0
3 years ago
You need to put a metal rod
Lubov Fominskaja [6]
-- Put the rod into the freezer for a while.  As it cools,
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-- Bring the rod and the cylinder togther quickly, before the
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-- I bet it'll fit now.

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8 0
3 years ago
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