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2 years ago

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Two cylindrical space stations, the second four times the diameter of the first, rotate so as to provide the same amount of arti

ficial gravity. If the first station makes one rotation in the time T, then the second station makes one rotation in time A. T/4 B. 2T C. 4T D. 16T

1 answer:

dybincka [34]2 years ago

5 0

Answer:

B. T1 = 2T2

Explanation:

Given

r2 = 4r1

In this case, the centripetal forces must be equal (a1 = a2) for a person on the inside of the outer wall to feel the same amount of artificial gravity.

For a uniform circular motion, v = 2πr/T

Since a1 = a2 and a = v²/r

Then

(v1)²/r1 = (V2)²/r2

(2πr1/T)²/r1 = (2πr2/T)²/r2

4π(r1)²/(r1T1²) = 4π(r2)²/(r2T2²) ---- divide both sides by 4π

(r1)²/r1T1² = (r2)²/r2T2²

r1/T1² = r2/T2²----- make T2 the subject of formula

T2² = r2 * T1²/r1 ---- Square root both sides

T2 = T1√(r2/r1)

Substitute 4r1 for r2

T2 = T1√(4r1/r1)

T2 = T1√4

T2 = 2T1

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A uniform Rectangular Parallelepiped of mass m and edges a, b, and c is rotating with the constant angular velocity ω around an

Sonbull [250]

Answer:

(a) k = $\frac{Mw^{2} }{6} (a^{2} +b^{2} )$

(b) τ = $\frac{M}{3} (a^{2} +b^{2} )$ ∝

Explanation:

The moment of parallel pipe rotating about it's axis is given by the formula;

I = $\frac{M}{3} (a^{2} +b^{2} )$ ---------------------------------1

(a) The kinetic energy of a parallel pipe is also given as;

k = $\frac{1}{2} Iw^{2}$ --------------------------------2

Putting equation 1 into equation 2, we have;

k = $\frac{M}{6} (a^{2} +b^{2} )w^{2}$

k = $\frac{Mw^{2} }{6} (a^{2} +b^{2} )$

(b) The angular momentum is given by the formula;

τ = Iw -----------------------3

Putting equation 1 into equation 3, we have

τ = $\frac{Mw}{3} (a^{2} +b^{2} )$

But

τ = dτ/dt = $\frac{M}{3} (a^{2} +b^{2} )\frac{dw}{dt}$ ------------------4

where

dw/dt = angular acceleration =∝

Equation 4 becomes;

τ = $\frac{M}{3} (a^{2} +b^{2} )$ ∝

8 0

2 years ago

49. A vertically hung 0.50-meter- long spring is stretched from its equilibrium position to a length of 1.00 meter by a weight a

Natali5045456 [20]

Answer:

$K=120$

Explanation:

From the question we are told that

Length of spring $l_1=0.5m$

Length of stretched $l_s=1m$

Potential energy of spring $E=15J$

Generally equation for energy stored is mathematically given as

$U=1/2K \triangle x^2$

$K=\frac{2U}{\triangle x^2}$

$K=\frac{2*15}{ 0.5^2}$

Therefore value of the spring constant in N/m? is given as

$K=120$

4 0

1 year ago

Multiplying or dividing vectors by scalars results in a. vectors. b. scalars. c. vectors if multiplied or scalars if divided. d.

klemol [59]

Let A = i+j+k be a vector and B = 3 be any scalar,
Multiplying A and B,
AB = (i+j+k)3 = 3i+3j+3k

Which is a new vector whose direction is same as the old but it's 3 times greater in length than the old vector(i+j+k).

Now, dividing A and B,
A/B = (i+j+k)/3 = $\frac{1}{3}i + \frac{1}{3}j + \frac{1}{3}k$

Which is again a new vector whose direction is same as the old but now it's 1/3 times small in length than the old vector.

Direction is same because we multiplied by positive scalar. If we multiply A by suppose -1, -4, -1000000 or any negative number, it's direction will reverse.

Thus, if we multiply a vector with scalar, it's length increases. If we divide, it shrinks.

8 0

2 years ago

Read 2 more answers

A brick is resting on a rough incline as shown in the figure. The friction force acting on the brick, along the incline, is

Tasya [4]

B) equal to the gravitational force of the brick

6 0

2 years ago

Read 2 more answers

A wooden disk of mass m and radius r has a string of negligible mass is wrapped around it. If the disk is allowed to fall and th

Tju [1.3M]

Answer:

$a = \frac{2}{3}g$

$T = \frac{mg}{3}$

Explanation:

As the disc is unrolling from the thread then at any moment of the time

We have force equation as

$mg - T = ma$

also by torque equation we can say

$TR = I\alpha$

$TR = \frac{1}{2}mR^2(\frac{a}{R})$

$T = \frac{1}{2}ma$

Now we have

$mg - \frac{1}{2}ma = ma$

$mg = \frac{3}{2}ma$

$a = \frac{2}{3}g$

Also from above equation the tension force in the string is

$T = \frac{1}{2}ma$

$T = \frac{mg}{3}$

7 0

2 years ago