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3 years ago

Explain the centripetal force or Newton's second law...

2 answers:

8 0

Answer: a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. According to Newton's second law of motion, net force is mass times acceleration: net . For uniform circular motion, the acceleration is the centripetal acceleration

Explanation:

stepan [7]3 years ago

6 0

Newton’s second law states that the acceleration of an object is directly related to the net force and inversely related to its mass. Acceleration of an object depends on two things, force and mass.

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3 years ago

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Which terrestrial planet experiences the shortest year?

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3 years ago

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Vlad [161]

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3 years ago

If a proton were released from rest at the sphere's surface, what would be its speed far from the sphere?

balu736 [363]

Let the sphere is having charge Q and radius R

Now if the proton is released from rest

By energy conservation we can say

$U = K$

$\frac{kQe}{R} = \frac{1}{2}mv^2$

$\frac{2kQe}{mR} = v^2$

now take square root of both sides

$v =\sqrt{\frac{2kQe}{mR}}$

so the proton will move by above speed and

here Q = charge on the sphere

R = radius of sphere

$k = 9 * 10^9$

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3 years ago

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A 2-lb slider is propelled upward at A along the fixed curved bar which lies in a vertical plane. If the slider is observed to h

timofeeve [1]

To develop this problem it is necessary to apply the concepts given in the balance of forces for the tangential force and the centripetal force. An easy way to detail this problem is through a free body diagram that describes the behavior of the body and the forces to which it is subject.

PART A) Normal Force.

$F_n = \frac{mv^2}{r}$

$N+mgcos\theta = \frac{mv^2}{r}$

Here,

Normal reaction of the ring is N and velocity of the ring is v

$N+mgcos\theta = \frac{mv^2}{r}$

$N+Wcos\theta = \frac{W}{g} (\frac{v^2}{r})$

$N+2cos30\° = \frac{2}{32.2}*\frac{10^2}{2}$

$N = 1.374lb$

PART B) Acceleration

$F_t = ma_t$

$-mgsin\theta = ma_t$

$-W sin\theta = \frac{W}{g} a_t$

$-2Sin30\° = (\frac{2}{32.2})a_t$

$a_T = -16.10ft/s^2$

Negative symbol indicates deceleration.

NOTE: For the problem, the graph in which the turning radius and the angle of suspension was specified was not supplied. A graphic that matches the description given by the problem is attached.

8 0

3 years ago