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3 years ago

Does a rigid object in uniform rotation about a fixed axis satisfy the first and second conditions for equilibrium?

Physics

1 answer:

Vladimir [108]3 years ago

7 0

Answer:

Explanation:

a rigid object in uniform rotation about a fixed axis does not satisfy both the condition of equilibrium .

First condition of equilibrium is that net force on the body should be zero.

or F net = 0

A body under uniform rotation is experiencing a centripetal force all the time so F net ≠ 0

So first condition of equilibrium is not satisfied.

Second condition is that , net torque acting on the body must be zero.

In case of a rigid object in uniform rotation , centripetal force is applied towards the centre ie towards the line joining the body under rotation with the axis .

F is along r

torque = r x F

= r F sinθ

θ = 0 degree

torque = 0

Hence 2nd condition is fulfilled.

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Which has the greater acceleration, a person going from 0 m/s to 10 m/s in 10 seconds or an ant going from 0 m/s to 0.25 m/s in

monitta

Person-1

Initial velocity=u=0m/s
Final velocity=v=10m/s
Time=10s=t

$\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}$

$\\ \sf\longmapsto Acceleration=\dfrac{10-0}{10}$

$\\ \sf\longmapsto Acceleration=\dfrac{10}{10}$

$\\ \sf\longmapsto Acceleration=1m/s^2$

Person-2

initial velocity=0m/s=u
Final velocity=v=0.25m/s
Time=t=2s

$\\ \sf\longmapsto Acceleration=\dfrac{0.25-0}{2}$

$\\ \sf\longmapsto Acceleration=\dfrac{0.25}{2}$

$\\ \sf\longmapsto Acceleration=0.125m/s^2$

Person-1 is accelerating faster.

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3 years ago

What do a control group and an experiment group have in common ?

Vlad [161]

Whatever the focus of the experiment is, plus any others factors that might influence the outcome of the experiment. If you are testing a new cancer drug, the experimental group and the control group must both be people with the same type of cancer, and both be a representative distribution of the population, all races, genders, ages, etc. You want the only difference in the two groups to be what you are studying, i.e. the effects of the drug.

7 0

2 years ago

Read 2 more answers

Derive an expression for the energy needed to launch an object from the surface of Earth to a height h above the surface.Ignorin

Sergio [31]

Answer:

Explanation:

Potential energy on the surface of the earth

= - GMm/ R

Potential at height h

= - GMm/ (R+h)

Potential difference

= GMm/ R - GMm/ (R+h)

= GMm ( 1/R - 1/ R+h )

= GMmh / R (R +h)

This will be the energy needed to launch an object from the surface of Earth to a height h above the surface.

Extra energy is needed to get the same object into orbit at height h

= Kinetic energy of the orbiting object at height h

= 1/2 x potential energy at height h

= 1/2 x GMm / ( R + h)

8 0

3 years ago

We can see Objects because of

Amiraneli [1.4K]

Answer:

I believe it's A.)

Explanation:

Although light comes into our atmosphere through refraction, it reaches our eyes only through reflection from objects. So when light rays reflect off an object and enter the eyes through the cornea you can then see that object.

Hope this helps you out : )

7 0

3 years ago

A drowsy cat spots a flowerpot that sails first up and then down past an open window. the pot was in view for a total of 0.49 s,

Alika [10]

For this case, let's assume that the pot spends exactly half of its time going up, and half going down, i.e. it is visible upward for 0.245 s and downward for 0.245 s. Let us take the bottom of the window to be zero on a vertical axis pointing upward. All calculations will be made in reference to this coordinate system. 

An initial condition has been supplied by the problem:

s=1.80m when t=0.245s

This means that it takes the pot 0.245 seconds to travel upward 1.8m. Knowing that the gravitational acceleration acts downward constantly at 9.81m/s^2, and based on this information we can use the formula:

s=(v)(t)+(1/2)(a)(t^2)

to solve for v, the initial velocity of the pot as it enters the cat's view through the window. Substituting and solving (note that gravitational acceleration is negative since this is opposite our coordinate orientation):

(1.8m)=(v)(0.245s)+(1/2)(-9.81m/s^2)(0.245s)^2

v=8.549m/s

Now we know the initial velocity of the pot right when it enters the view of the window. We know that at the apex of its flight, the pot's velocity will be v=0, and using this piece of information we can use the kinematic equation:

(v final)=(v initial)+(a)(t)

to solve for the time it will take for the pot to reach the apex of its flight. Because (v final)=0, this equation will look like

0=(v)+(a)(t)

Substituting and solving for t:

0=(8.549m/s)+(-9.81m/s^2)(t)

t=0.8714s

Using this information and the kinematic equation we can find the total height of the pot’s flight:

s=(v)(t)+(1/2)(a)(t^2)

s=8.549m/s (0.8714s)-0.5(9.81m/s^2)(0.8714s)^2

s=3.725m

This distance is measured from the bottom of the window, and so we will need to subtract 1.80m from it to find the distance from the top of the window:

3.725m – 1.8m=1.925m

Answer:

1.925m

3 0

2 years ago