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katrin [286]
3 years ago
7

An object has an acceleration of 12.0 m/s/s. The mass of the object is doubled while the net force on the object is held constan

t. What will be the new acceleration?
Physics
1 answer:
Aloiza [94]3 years ago
7 0

Answer:

The net force acting on the object is doubled while the mass of the object is held constant. What will be the new acceleration? An object has an acceleration of 12.0 m/s^2. The net force acting on the object is halved (decreased to one half its original value) while the mass of the object is held constant.

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Arthur conducts a controlled experiment several times. The results of each test show that his hypothesis is not supported. Which
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Hello. You did not inform the experiment that Arthur is conducting, which makes it impossible for your question to be answered accurately. However, I will try to help you in the best possible way.

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3 years ago
A uniform solid disk with a mass of 24.3 kg and a radius of 0.364 m is free to rotate about a frictionless axle. Forces of 90.0
a_sh-v [17]

Answer:

a. -12.7 Nm

b. -7.9 rad/s^2

Explanation:

I have attached an illustration of a solid disk with the respective forces applied, as stated in this question.

Forces applied to the solid disk include:

F_1 = 90.0N\\F_2 = 125N

Other parameters given include:

Mass of solid disk, M = 24.3kg

and radius of solid disk, r = 0.364m

a.) The formula for determining torque (T), is T = r * F

Hence the net torque produced by the two forces is given as a summation of both forces:

T = T_{125} + T_{90}\\= -r(125)sin90 + r(90)sin90\\= 0.364(-125 + 90)\\= -12.7 Nm

b.)  The angular acceleration of the disk can be found thus:

using the formula for the Moment of Inertia of a solid disk;

I_{disk} = {\frac{1}{2}}Mr^2

where M = Mass of solid disk

and r = radius of solid disk

We then relate the torque and angular acceleration (\alpha) with the formula:

T = I\alpha \\-12.7 = ({\frac{1}{2}}Mr^2)\alpha \\\alpha  = -{\frac{12.7}{1.61}} = -7.9 rad/s^2

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