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NNADVOKAT [17]
3 years ago
11

Future space stations will create an artificial gravity by rotating. Consider a cylindrical space station 100 m diameter rotatin

g about its central axis. Astronauts walk on the inside surface of the space station. What rotation period will provide "normal" gravity?
Physics
1 answer:
Zielflug [23.3K]3 years ago
3 0

Given Information:

Diameter = d = 100 m

Required Information:

Period = T = ?

Answer:

Period = T = 14.2 seconds

Explanation:

We know that a station revolving  at an angular velocity ω,  have an  acceleration given by

α = ω²r

Where ω is angular velocity and r is the radius of cylindrical space station.

Normal gravity means α = g = 9.8 m/s²

ω² = α/r

ω = √(α/r)

The radius is given by

r = d/2

r = 100/2

r = 50 m

ω = √(9.8/50)

ω = 0.4427 rad/sec

We also know that

ω = 2πf

f = ω/2π

f = 0.4427/2π

f = 0.0704 rev/sec

Finally time period is given by

T = 1/f

T = 1/0.0704

T = 14.2  sec

Therefore, the rotation period is 14.2 seconds.

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Weight is influenced by which of the following
nignag [31]

Out of the given options, weight is influenced by mass and gravity

Answer: Option A

<u>Explanation: </u>

The object's mass is defined as the quantity of a matter with which the object is formed. It can change its state of matter but the quantity will remain the same. However, the weight is defined as how much force gravity exerts on the object's mass to pull it.

The mass is always same irrespective the location but the weight may vary from one place to the other while talking for the bigger picture. For example, the object's weight may be 60 kg on Earth but when it is measured on the moon, it will be lesser.

The weight of an object generally has nothing doing with the volume and it doesn't depend solely on the gravitational pull. The mass plays a crucial role.

                                  W=F=m \times g

6 0
3 years ago
A block of weight mg sits on an inclined plane as shown in (Figure 1) . A force of magnitude F1 is applied to pull the block up
svlad2 [7]

Answer:

W = (F1 - mg sin θ) L,   W = -μ  mg cos θ L

Explanation:

Let's use Newton's second law to find the friction force. In these problems the x axis is taken parallel to the plane and the y axis perpendicular to the plane

Y Axis  

       N - W_{y} =

       N = W_{y}

X axis

       F1 - fr - Wₓ = 0

       fr = F1 - Wₓ

Let's use trigonometry to find the components of the weight

     sin θ = Wₓ / W

     cos θ = W_{y} / W

      Wₓ = W sin θ

      W_{y} = W cos θ

We substitute

      fr = F1 - W sin θ

Work is defined by

        W = F .dx

        W = F dx cos θ

The friction force is parallel to the plane in the negative direction and the displacement is positive along the plane, so the Angle is 180º and the cos θ= -1

         

        W = -fr x

        W = (F1 - mg sin θ) L

Another way to calculate is

         fr = μ N

         fr = μ W cos θ

the work is

         W = -μ  mg cos θ L

4 0
3 years ago
A block on a horizontal frictionless plane is attached to a spring, as shown below. The block oscillates along the x-axis with s
AleksandrR [38]

The question is about unclear since no picture provided. But from the question, it could be guessed that the box is moving back and forth on the frictionless plane at the amplitude of A in simple harmonic motion.

Answer:

D. At x=0, it's acceleration is at a maximum

Explanation:

As the box move forward, it reaches point A and than move backward. Theoretically, the box will move backwards, through its origin, to point -A and then going forward.

Point A is the maximum displacement of the box in this case. At this point, the box instantaneously stop to go backward. Therefore the velocity at that moment is zero.

From point -A, the box travel forward and keep building up speed due to the release in potential energy of the spring. And at point x=0, the velocity become maximum. After point x=0, the velocity of the box slows down due to the conversion of kinetic energy to potential energy of the spring. And as it reaches point A, it reaches zero velocity.

The same can be said as the box travels backward from point A to -A

8 0
4 years ago
The Steamboat Geyser in Yellowstone National Park shoots water into the air at 48.0 m/s. How
qwelly [4]

Answer:

The maximum height reached by the water is 117.55 m.

Explanation:

Given;

initial velocity of the water, u = 48 m/s

at maximum height the final velocity will be zero, v = 0

the water is going upwards, i.e in the negative direction of gravity, g = -9.8 m/s².

The maximum height reached by the water is calculated as follows;

v² = u² + 2gh

where;

h is the maximum height reached by the water

0 = u² + 2gh

0 = (48)² + ( 2 x -9.8 x h)

0 = 2304 - 19.6h

19.6h = 2304

h = 2304 / 19.6

h = 117.55 m

Therefore, the maximum height reached by the water is 117.55 m.

7 0
3 years ago
Read 2 more answers
A solution is prepared by dissolving 49.3 g of KBr in enough water to form 473 mL of solution. Calculate the mass % of KBr in th
son4ous [18]

Answer:

9.31%

Explanation:

We are given that

Mass of KBr=49.3 g

Volume of solution=473 mL

Density of solution =1.12g/mL

We have to find the mass% of KBr.

Mass =volume\times density

Using the formula

Mass of solution=1.12\times 473=529.76 g

Mass % of KBr=\frac{mass\;of\;KBr}{Total\;mass\;of\;solution}\times 100

Mass % of KBr=\frac{49.3}{529.76}\times 100

Mass % of KBr=9.31%

Hence, the mass% of KBr=9.31%

7 0
3 years ago
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