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natita [175]
3 years ago
6

Some engineers have suggested that we can simulate gravity in outer space by having a circular rotating space station where pers

ons feel an outward – directed fictitious force due to the rotation of the station. The reason they feel such a force is because
Physics
1 answer:
Thepotemich [5.8K]3 years ago
8 0

Answer:

Centrifugal force

Explanation:

In Space, It is in a way possible create an " artificial gravity" just by spinning the aircraft or space station. On spinning the space station the inhabitants feel an outward force or the centrifugal force, this outward force when equivalent to gravitational force is able to stimulate gravity.

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The force of replusion between two like charged particles will increase if​
Alex_Xolod [135]

Answer:

The distance of separation is decreased

Explanation:

From Cuolomb's law, we know that the strength of charge is inversely proportional to the distance of separation between the charges. To mean that increasing the distance let's say from 2m to 3 m would mean initial strength getting form 1/4 to 1/9 which is a decrease. The vice versa is true hence the force of repulsion can increase only when we decrease the distance of separation.

7 0
3 years ago
a dragster gets to a peed of 112m/s over a distance of 300m from rest. Determine the acceleration of the dragster.
madreJ [45]

A dragster gets to a speed of 112 m/s over a distance of 300 m from the rest, then the acceleration will be 20.9 m/s².

<h3>What is Acceleration?</h3>

The rate of change in an object's velocity with respect to time is known as acceleration in mechanics. The vector quantity of accelerations. The direction of the net force that is acting on an object determines its acceleration.

Since acceleration has both a magnitude and a direction, it is a vector quantity. Velocity is a vector quantity as well. The definition of acceleration is the change in velocity vector over a time interval divided by the time interval.

According to the question, the given values are :

Final velocity, v = 112 m/s.

Initial velocity, u = 0 m/s

Distance, s = 300 m.

Use the equation of motion,

v² - u² = 2as

112 ² - 0 = 2 (a)(300)

a = 12544/600

a = 20.9 m/s².

Hence, the acceleration of the dragster will be 20.9 m/s².

To get more information about Acceleration :

brainly.com/question/20382454

#SPJ1

3 0
1 year ago
At the start of a basketball game, a referee tosses a basketball straight into the air by giving it some initial speed. After be
max2010maxim [7]

Answer:

0.32 m.

Explanation:

To solve this problem, we must recognise that:

1. At the maximum height, the velocity of the ball is zero.

2. When the velocity of the ball is 2.5 m/s above the ground, it is assumed that the potential energy and kinetic energy of the ball are the same.

With the above information in mind, we shall determine the height of the ball when it has a speed of 2.5 m/s. This can be obtained as follow:

Mass (m) = constant

Acceleration due to gravity (g) = 9.8 m/s²

Velocity (v) = 2.5 m/s

Height (h) =?

PE = KE

Recall:

PE = mgh

KE = ½mv²

Thus,

PE = KE

mgh = ½mv²

Cancel m from both side

gh = ½v²

9.8 × h = ½ × 2.5²

9.8 × h = ½ × 6.25

9.8 × h = 3.125

Divide both side by 9.8

h = 3.125 / 9.8

h = 0.32 m

Thus, the height of the ball when it has a speed of 2.5 m/s is 0.32 m.

3 0
3 years ago
F(x) = 3x^2+5x-14<br> Find f(-9)
Llana [10]

Answer:

f(-9) = 184

Explanation:

f(x)=3x²+5x-14

f(-9)= 3(-9)² +5(-9)-14 Order of Operations : Exponents

     =  3(81)+5(-9)-14

     = 243+5(-9)-14

     =  243-45-14

     = 198-14

f(-9)= 184

Hope this helps :)

8 0
3 years ago
A _____ line on a typical Speed vs. Time graph means an object is experiencing a constant speed.
stepladder [879]
D.horizontal has better speed
7 0
4 years ago
Read 2 more answers
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