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olya-2409 [2.1K]
3 years ago
15

In order to simulate weightlessness for astronauts in training, they are flown in a vertical circle. if the passengers are to ex

perience weightlessness, how fast should an airplane be moving at the top of a vertical circle with a radius of 2.50 km
Physics
1 answer:
sleet_krkn [62]3 years ago
8 0
The answer is "156.6 m/s".

This is how we calculate this;

-N + mg = ma = mv²/r

For "weightlessness" N = 0, so

0 = mg - mv²/r 

g - v²/r = 0 

v =√( gr)
g = 9.8 and r = 2.5km = 2500 m

v = √(9.8 x 2500)

= 156.6 m/s
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Explanation: Cluster 3 is the oldest and it could be the source from which cluster 2 and 3 got their spectral type from.

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Can an object have increasing speed while its acceleration is decreasing?
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The best option is C. This is due to friction.
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3 years ago
Classify the following situations into contact and non-contact forces.
NARA [144]

Answer: Contact force

a. Applying break in a vehicle.

d. The speed of ball rolling on ground is reduced

Non contact force

b. A coconut falling from a coconut tree.

c. The planets revolving around the sun.

Explanation:

The contact force is the force which exerts when one object or entity comes in contact with other object or entity. For example, on application of break the vehicle stops, the force is applied on the breaks to stop the vehicle. The ball rolling on the ground the speed reduces so the application of force on the ground also reduces.

The non contact force is the force one object exerts on the other without coming in direct contact with the other object. The force exerted by one object on other due to gravity is a non contact force. The coconut falling on the ground and planets revolving around the sun are examples of non contact force due to gravity.

8 0
3 years ago
A projectile is launched from ground level with an initial speed of 47 m/s at an angle of 0.6 radians above the horizontal. It s
Zarrin [17]

Answer:

30.67m

Explanation:

Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;

s = ut ± \frac{1}{2} at^2               ------------(i)

Where;

s = horizontal/vertical displacement

u = initial horizontal/vertical component of the velocity

a = acceleration of the projectile

t = time taken for the projectile to reach a certain horizontal or vertical position.

Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;

h = vt ± \frac{1}{2} at^2               ------------(ii)

Where;

h = vertical displacement

v = initial vertical component of the velocity = usinθ

a = acceleration due to gravity (since vertical motion is considered)

t = time taken for the projectile to hit the target

<em>From the question;</em>

u = 47m/s, θ = 0.6rads

=> usinθ = 47 sin 0.6

=> usinθ = 47 x 0.5646 = 26.54m/s

t = 1.7s

Take a = -g = -10.0m/s   (since motion is upwards against gravity)

Substitute these values into equation (ii) as follows;

h = vt - \frac{1}{2} at^2

h = 26.54(1.7) - \frac{1}{2} (10)(1.7)^2

h = 45.118 - 14.45

h = 30.67m

Therefore, the vertical distance is 30.67m        

7 0
4 years ago
If a rock has a speed of 12 m/s as it hits the ground, from what height did it
grin007 [14]

Answer:

To find the height the following formula should be used:  

v 2 = u 2 + 2aH

Explanation:

Assuming this occurs on earth,  a= 9.8 ms -2

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12=0+2 x9.8 x H

144

_______ =H

2 x 9.8

H= 7.35m

6 0
3 years ago
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