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2 years ago

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A record turntable is rotating at 33 rev/min. A watermelon seed is on the turntable 9.6 cm from the axis of rotation. (a) Calcul

ate the acceleration of the seed, assuming that it does not slip. (b) What is the minimum value of the coefficient of static friction between the seed and the turntable if the seed is not to slip

1 answer:

Vanyuwa [196]2 years ago

6 0

Answer:

(A) Acceleration will be $1.145rad/sec^2$

(B) Coefficient of static friction will be 0.116

Explanation:

We have given angular speed

$\omega =33rpm=\frac{2\pi \times 33}{60}=3.454rad/sec$

Distance from the axis r = 9.6 cm = 0.096 m

(a) Acceleration is equal to

$a_c=\omega ^2r=3.454^2\times 0.096=1.145rad/sec^2$

(b) For seed is not to slip

$ma=\mu mg$

$\mu =\frac{a}{g}$

$\mu =\frac{1.145}{9.8}=0.116$

So coefficient of static friction will be 0.116

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b. A string is wrapped around a pulley of radius 0.05 m and moment of inertia 0.2 kg  m2. If the string is pulled with a force

Oduvanchick [21]

Answer:

f = 8 N

Explanation:

Data provided in the question

Radius of the pulley = r = 0.05 m

Moment of inertia = (I) = 0.2 kg.m^{2}

Angular acceleration = ∝ = 2 rad/sec

Based on the above information

As we know that

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$= force \times radius$

$= f \times r$

And,

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$= I \times \alpha$

So,

We can say that

$f \times r = I \times \alpha$

$f \times 0.05 = 0.2 \times 2$

0.05f = 0.4

f = 8 N

We simply applied the above formulas

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3 years ago

A less massive moving object has an elastic collision with a more massive object that is not moving. Compare the initial velocit

stepladder [879]

Assume that the small-massed particle is $m$ and the heavier mass particle is $M$ .

Now, by momentum conservation and energy conservation:

$mv = mv_{m} + Mv_{M}$

$mv^{2} = mv^{2}_{m} + Mv^{2}_{M}$

Now, there are 2 solutions but, one of them is useless to this question's main point so I excluded that point. Ask me in the comments if you want the excluded solution too.

$v_{m} = v\frac{m - M}{m + M}\\\\\\v_{M} = v\frac{m + M}{m + M}\\$

So now, we see that $v_{m} < 0$ and $v_{M} > 0$ . So therefore, the smaller mass recoils out.

Hope this helps you!

Bye!

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3 years ago

20 characters or more

andreev551 [17]

Answer:

yes 20 characters or more

Explanation:

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2 years ago

According to Newton's third law, if you push on a wall, the wall ?

Lady bird [3.3K]

The wall will push back, in exactly the opposite direction, and with
exactly the same size force.

That's why the net force on the palm of your hand is zero, and that
in turn is the reason that your hand doesn't accelerate.

If you keep increasing the strength of your push, then eventually you
exceed the force that the wall is capable of delivering. Then the wall
crumbles and falls, your hand accelerates in the direction you're pushing,
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3 years ago

Read 2 more answers

On a horizontal frictionless surface, a small block with mass 0.200 kg has a collision with a block of mass 0.400 kg. Immediatel

Akimi4 [234]

Answer:48.2 Joules

Explanation:

Given

two masses of 0.2 kg and 0.4 kg collide with each other

after collision 0.2 kg deflect 30 north of east and 0.4 kg deflects 53.1 south of east

Velocity of 0.2 kg mass is

$v_{0.2}=12\cos (30)\hat{i}+12\sin (30)\hat{j}$

$|v_{0.2}|=11.99 m/s\approx 12 m/s$

Velocity of 0.4 kg mass

$v_{0.4}=13\cos (53.1)\hat{i}-12\sin (53.1)\hat{j}$

$|v_{0.4}|=12.99 m/s\approx 13 m/s$

Thus total Kinetic energy $=\frac{0.2\times 12^2}{2}+\frac{0.4\times 13^2}{2}$

Kinetic energy=48.2 J

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3 years ago