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lyudmila [28]
3 years ago
10

A rotating cylinder about 16 km long and 7.8km in diameter is designed to be used as a space colony.

Physics
1 answer:
Semmy [17]3 years ago
3 0

Answer:

0.05 rad/s

Explanation:

7.8 km = 7800 m

For the residents inside the space cylinder to experience the same gravitation acceleration g = 9.81m/s2 on Earth, the centripetal acceleration must be the same as g

a = g

But centripetal acceleration is product of angular velocity squared and radius of rotation r

\omega^2r = 9.81

\omega^2 d/2 = 9.81

\omega^2 = \frac{2*9.81}{d} = \frac{19.62}{7800} = 0.0025

\omega = \sqrt{0.0025} = 0.05 rad/s

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The answer is C) A girl hangs by both hands, motionless, from a trapeze.
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Suppose your car accelerates from rest to 9 m/s in 2 s. Assume that the acceleration is constant in this time interval. A. What
Stolb23 [73]

Answer :

(a) The acceleration of the car is, 4.5m/s^2

(b) The distance covered by the car is, 9 m

Explanation :  

By the 1st equation of motion,

v=u+at ...........(1)

where,

v = final velocity = 9 m/s

u = initial velocity  = 0 m/s

t = time = 2 s

a = acceleration of the car = ?

Now put all the given values in the above equation 1, we get:

9m/s=0m/s+a\times (2s)

a=4.5m/s^2

The acceleration of the car is, 4.5m/s^2

By the 2nd equation of motion,

s=ut+\frac{1}{2}at^2 ...........(2)

where,

s = distance covered by the car = ?

u = initial velocity  = 0 m/s

t = time = 2 s

a = acceleration  of the car = 4.5m/s^2

Now put all the given values in the above equation 2, we get:

s=(0m/s)\times (2s)+\frac{1}{2}\times (4.5m/s^2)\times (2s)^2

By solving the term, we get:

s=9m

The distance covered by the car is, 9 m

3 0
3 years ago
Vector A has magnitude of 8units and makes an angle of 45° with the positive x-axis. Vector B also has the same magnitude of 8un
iragen [17]

Answer:

Explanation:

Because vectors have direction and x and y components you can't just add them and say that their length is 16 because A is 8 units and so is B. What you're actually finding is the magnitude and direction of the vector that results from this addition. The magnitude is the length of the resultant vector, which comes from the x and y components of A and B, and the direction is the angle between the resultant vector and the positive x axis. To add the vectors, then, we need to find the x and y components of each. We'll do the x components of A and B first so we can add them to get the x component of C. Since x values are directly related to cos, the formula to find the x components of vectors is

V_x=Vcos\theta which is the magnitude of the vector (its length) and the angle. Finding the x components of A:

A_x=8.0cos45 so

A_x=5.7 and for B:

B_x=8.0cos180 since the negative x axis is the 180 degree axis and

B_x=-8.0 If we add them, we get

C_x=-2.3

Now onto the y components. The formula for that is almost the same as the x components except use sin instead of cos:

A_y=8.0sin45 so

A_y=5.7 and

B_y=8.0sin180 so

B_y=0 If we add them, we get

C_y=5.7

Now for the final magnitude:

C_{mag}=\sqrt{(-2.3)^2+(5.7)^2} and

C_{mag}=6.1 units and now onto the direction.

The x component of C is positive and the y component is negative, which means that the direction has us at an angle is quadrant 2; we add 180 to whatever the angle is. Finding the angle:

tan^{-1}(\frac{C_y}{C_x})=(\frac{5.7}{-2.3}) = -68 + 180 = 112 degrees

The resultant vector of A + B has a magnitude of 6.1 and a direction of 112°

Do the same thing for subtraction, except if you're subtracting B from A, the direction that B is pointing has to go the opposite way. That means that A doesn't change anything at all, but B is now pointing towards 0.

A_x=5.7 (doesn't change from above)

B_x=8.0cos0 and

B_x=8.0 so

C_x=13.7 and

A_y=5.7 (also doesn't change from above)

B_y=8.0sin0 so

B_y=0 and

C_y=5.7 and for the magnitude:

C_{mag}=\sqrt{(13.7)^2+(5.7)^2 so

C_{mag}=15units and for the direction:

tan^{-1}(\frac{5.7}{13.7})=23 and since both x and y components of C are in Q1, we add nothing.

And you're done!!!

3 0
3 years ago
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