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finlep [7]
3 years ago
15

Suppose several particles are in concentric circular orbits under the influence of some centripetal component of force, the magn

itude of which depends on distance from the center of the motion. Each particle has a different orbital radius r and a corresponding period T.How does the force depend on r if T is proportional to sqrt(r) ?
How does the force depend on r if T is proportional to r^3/2 ?
How does the force depend on r if T is independent of r ?
Physics
1 answer:
olga55 [171]3 years ago
6 0

Explanation:

The centripetal force acting on a particle is given by :

F=\dfrac{mv^2}{r}

Since, v=r\omega

F=mr\omega^2

F\propto r\omega^2

Case 1.

If T\propto \sqrt{r}

Since, T=\dfrac{2\pi}{\omega}

So, \omega\propto \dfrac{1}{\sqrt{r} }

\omega^2\propto \dfrac{1}{r}

So, the force becomes,

F\propto r\times \dfrac{1}{r}

F is independent of r

Case 2.

If T\propto r^{3/2}

\omega\propto \dfrac{1}{r^{3/2}}

\omega^2 \propto \dfrac{1}{r^3}

So, F\propto r\times \dfrac{1}{r^2}

So, the force is inversely proportional to the square of radius.

Case 3.

If T is independent of r, the force will be directly proportional to the radius of orbit.

Hence, this is the required solution.

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Can someone check my answers and tell me if their correct?
Otrada [13]

Seven

The magnitude is pointing towards the origin and is at - 20 degrees. The combination makes 160 with the x axis: C answer

Eight

They keep doing this. They use distance where they should use displacement but they use distance to try and fool you. It's a mighty poor practice.

The distance between the start and end points is the displacement. That "distance" is 180*sqrt(25) = 900 . The actual distance should be 180*4 + 180*3 = 720 + 540 = 1260. That's what a car's odometer or a bicycle odometer would read.  the difference is 360.

I really do object to the wording, but what can I do?

Nine

Nine is the same thing as 8.

Displacement = sqrt(400^2 + 80^2)= sqrt(166400) = 408

The actual distance is 400 + 80 = 480

The difference is the answer = 480 - 408 = 72 <<<< Answer

Ten

This is just the displacement magnitude.

dis = sqrt(30^2 + 80^2)

dis = sqrt(900 + 6400)

dis = sqrt(7300)

dis = 85.44 <<<< Answer D

Twelve

Vi =  2.15*Sin(30) = 1.075 m/s

vf = 0

a = - 9.81

t = ?

<u>Formula</u>

a = (vf - vi)/t

<u>Solve</u>

-9.81 =  (0 - 1.075)/t

- 9.81 * t = -1.075

t = 0.11 seconds

Thirteen

I'm leaving this last one to you. You need the initial height xo to answer it properly. Judging by the other questions, this one is right.

Edit

That is a surprise! Really quickly

d = 3.2 m

a = - 9.82

vf = 0

vi = ?

vf^2 = vi^2 - 2*a*d

0 = vi^2 - 2*9.81*3.2

vi = sqrt(19.62*3.2)

vi = 8.0  m/s   But that is the vertical component of the speed

v = vi/sin(25)

v = 8.0/sin(25) = 11


6 0
3 years ago
For a top player, a tennis ball may leave the racket on the serve with a speed of 55 m/s (about 120 mi/h). If the ball has a mas
inn [45]

Answer:

Yes is large enough

Explanation:

We need to apply the second Newton's Law to find the solution.

We know that,

F= ma

And we know as well that

a= \frac{v}{t}

Replacing the aceleration value in the equation force we have,

F= \frac{mv}{t}

Substituting our values we have,

F= \frac{(0.060)(55)}{4*10^{-3}}

F=825N

The weight of the person is then,

W = mg

W = (60)(9.8) = 558N

<em>We can conclude that force on the ball is large to lift the ball</em>

6 0
3 years ago
Two 0.2304 cm x 0.2304 cm square aluminum electrodes, spaced 0.5974 mm apart are connected to a 61 V battery. What is the charge
Arisa [49]

Answer:

The charge on each plate is 0.0048 nC

Explanation:

for the distance between the plates d and given the area of plates, A, and ε = 8.85×10^-12 C^2/N.m^2, the capacitance of the plates is given by:

C = (A×ε)/d

   =[(0.2304×10^-2)(0.2304×10^-2)×(8.85×10^-12))/(0.5974×10^-3)

   = 7.86×10^-14 F

then if the plates are connected to a battery of voltage V = 61 V, the charge on the plates is given by:

q = C×V

   = (7.86×10^-14)×(61)

   = 4.80×10^-14 C

   ≈ 0.0048 nC

Therefore, the charge on each plate is 0.0048 nC.

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