Science: Which statement is true about star A

(a) In the baggage claim area of an airport, a particular baggage carousel is shaped like a section of a large cone, steadily ro

Tom [10]

Answer:

Part a)

$F_f = 107.8 N$

Part b)

$\mu = 0.415$

Explanation:

Part a)

Time period of one revolution is given as

T = 42 s

now the angular speed of the belt is given as

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{42}$

$\omega = 0.15 rad/s$

Now at rest position the force along the surface of carousel must be constant

so we will have

$mgsin\theta + m\omega^2 r cos\theta = F_f$

$30(9.81)sin20.5 + 30(0.15^2)(7.46)cos20.5 = F_f$

$F_f = 107.8 N$

Part b)

New Time period of one revolution is given as

T = 30 s

now the angular speed of the belt is given as

$\omega = \frac{2\pi}{T}$

$\omega = \frac{2\pi}{30}$

$\omega = 0.21 rad/s$

Now at rest position the force along the surface of carousel must be constant

so we will have

$mgsin\theta + m\omega^2 r cos\theta = F_f$

$30(9.81)sin20.5 + 30(0.21^2)(7.94)cos20.5 = F_f$

$F_f = 112.9 N$

Also we know that in perpendicular direction also force is balanced

$F_n + m\omega^2 r sin\theta = mgcos\theta$

$F_n = mgcos\theta - m\omega^2 r sin\theta$

$F_n = 30(9.81)cos20.5 - 30(0.21)^2(7.94)sin20.5$

$F_n = 272 N$

now for friction coefficient we will have

$F_f = \mu F_n$

$112.9 = \mu 272$

$\mu = 0.415$

8 0

3 years ago

Two identical conducting spheres A and B carry equal charge. They are separated by a distance much larger than their diameters.

VashaNatasha [74]

What a delightful little problem !
Here's how I see it:

When 'C' is touched to 'A', charge flows to 'C' until the two of them are equally charged. So now, 'A' has half of its original charge, and 'C' has the other half.

Then, when 'C' is touched to 'B', charge flows to it until the two of them are equally charged. How much is that ? Well, just before they touch, 'C' has half of an original charge, and 'B' has a full one, so 1/4 of an original charge flows from 'B' to 'C', and then each of them has 3/4 of an original charge.

To review what we have now: 'A' has 1/2 of its original charge, and 'B' has 3/4 of it.

The force between any two charges is:

F = (a constant) x (one charge) x (the other one) / (the distance between them)².

For 'A' and 'B', the distance doesn't change, so we can leave that out of our formula.

The original force between them was 3 = (some constant) x (1 charge) x (1 charge).

The new force between them is F = (the same constant) x (1/2) x (3/4) .

Divide the first equation by the second one, and you have a proportion:

3 / F = 1 / ( 1/2 x 3/4 )

Cross-multiply this proportion:

3 (1/2 x 3/4) = F

F = 3/2 x 3/4 = 9/8 = 1.125 newton.

That's my story, and I'm sticking to it.

5 0

3 years ago

Naphthalene combustion can be used to calibrate the heat capacity of a bomb calorimeter. the heat of combustion of naphthalene i

barxatty [35]

My namer js gpvjt ahsdhhsdbdb

6 0

3 years ago

Question 1 of 10

Neko [114]

Answer:

D: A mathematical model

Hope its right if not so sorry :)

3 0

3 years ago

What unit is used to express work?

MAXImum [283]

The unit to express work is called Joule.

4 0

3 years ago