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mojhsa [17]
3 years ago
12

| A T-ball with a mass of 0.6 kg travels in the

Physics
1 answer:
r-ruslan [8.4K]3 years ago
7 0

Answer:

|I|=6\ Kg.m/s

F=120\ N

Explanation:

Impulse and Momentum

They are similar concepts since they deal with the dynamics of objects having their status of motion changed by the sudden application of a force. The momentum at a given initial time is computed as

p_o=m.v_o

When a force is applied, the speed changes to v_1 and the new momentum is

p_1=m.v_1

The change of momentum is

\Delta p=p_1-p_0=m(v_1-v_o)

The impulse is equal to the change of momentum of an object and it's defined as the average net force applied times the time it takes to change the object's motion

I=F.t=\Delta p

Part 1

The T-ball initially travels at 10 m/s and then suddenly it's stopped by the glove. The final speed is zero, so

\Delta p=0.6\ Kg(0-10\ m/s)=-6\ Kg.m/s

The impulse is

I=\Delta p

I=-6\ Kg.m/s

The magnitude is

|I|=6\ Kg.m/s

Part 2

The force can be computed from the formula

I=F.t

The direction of the impulse the T-ball receives is opposite to the direction of the force exerted by the ball on the glove, thus I_b=6\ kg.m/s

\displaystyle F=\frac{I}{t}=\frac{6\ kg.m/s}{0.05\ s}

\boxed{F=120\ N}

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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

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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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Suppose you are pushing a 3 kg box with a force of 25 N (directed parallel to the ground) over a distance of 15 m. Afterward, th
alex41 [277]

Answer: 321 J

Explanation:

Given

Mass of the box m=3\ kg

Force applied is F=25\ N

Displacement of the box is s=15\ m

Velocity acquired by the box is v=6\ m/s

acceleration associated with it is a=\dfrac{F}{m}

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Work done by force is W=F\cdot s

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Therefore, the magnitude of work done by friction is 321\ J

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