What force would be required ....

What is the minimum tangential velocity a space station would need to simulate earth gravity if the radius is 50 meters ?

Aneli [31]

Explanation:

angular velocity is given by

$w = \sqrt{ \frac{g}{r} }$

$w = \sqrt{ \frac{9.8}{25} }$

w = 0.626

now tangential velocity is

V = rw

= 25 x 0.626

= 15.65 m/s

5 0

3 years ago

When a 100-N force acts horizontally on an 8.0-kg chair, the chair moves at a constant speed across the level floor. Which state

TiliK225 [7]

Answer:

The applied force is greater than the frictional force.

Explanation:

the chair moves at a constant speed therefore, the answer is not A or C.

if there is no friction then the chair would accelerate and it would not be at a constant speed.

hence, the only possible answer is B.

7 0

3 years ago

A moving curling stone, A, collides head on with stationary stone, B. Stone B has a larger mass than stone A. If friction is neg

Kitty [74]

Answer:

The correct answer is option 'c': Smaller stone rebounds while as larger stone remains stationary.

Explanation:

Let the velocity and the mass of the smaller stone be 'm' and 'v' respectively

and the mass of big rock be 'M'

Initial momentum of the system equals

$p_i=mv+0=mv$

Now let after the collision the small stone move with a velocity v' and the big roch move with a velocity V'

Thus the final momentum of the system is

$p_f=mv'+MV'$

Equating initial and the final momenta we get

$mv=mv'+MV'\\\\m(v-v')=MV'.....i$

Now since the surface is frictionless thus the energy is also conserved thus

$E_i=\frac{1}{2}mv^2$

Similarly the final energy becomes

$E_f=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2$ \

Equating initial and final energies we get

$\frac{1}{2}mv^2=\frac{1}{2}mv'^2+\frac{1}{2}MV'^2\\\\mv^2=mv'^2+MV'^2\\\\m(v^2-v'^2)=MV'^2\\\\m(v-v')(v+v')=MV'^2......(ii)$

Solving i and ii we get

$v+v'=V'$

Using this in equation i we get

$v'=\frac{v(m-M)}{(M-m)}=-v$

Thus putting v = -v' in equation i we get V' = 0

This implies Smaller stone rebounds while as larger stone remains stationary.

4 0

3 years ago

34. (Double points) You help your friend construct a soap box derby car for the All-American Soap Box derby's Stock Division. He

Elden [556K]

To accelerate a 34.01 kg-car at 0.55 m/s², a force of 19 N will be required, according to Newton's Second Law of Motion.

<h3>What does Newton's Second Law of Motion state?</h3>

Newton's Second Law of Motion states that acceleration (a) happens when a force (F) acts on a mass (m).

We want a car of mass 34.01 kg to have an acceleration of 0.55 m/s². We can calculate the required force using Newton's Second Law of Motion.

F = m × a = 34.01 kg × 0.55 m/s² = 19 N

To accelerate a 34.01 kg-car at 0.55 m/s², a force of 19 N will be required, according to Newton's Second Law of Motion.

Learn more about Newton's Second Law of Motion here: brainly.com/question/25545050

#SPJ1

4 0

2 years ago

Find the deceleration of a cyclist who comes to rest from 10m/s in 2 seconds.

I am Lyosha [343]

Answer:

-5 m/s^2

Explanation:

10/2=5 and is is a stop so it is negitive

5 0

3 years ago