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zlopas [31]
3 years ago
15

If a cart of 4 kg mass has a force of 8 newtons exerted on it, what is its acceleration

Physics
1 answer:
Vsevolod [243]3 years ago
4 0
4 because you have to subtract the newtons from the mass
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A girl (mass M) standing on the edge of a frictionless merry-go-round (radius R, rotational inertia I) that is not moving. She t
vladimir1956 [14]

a) \omega=\frac{-mvR}{I+MR^2}

b) v=\frac{-mvR^2}{I+MR^2}

Explanation:

a)

Since there are no external torques acting on the system, the total angular momentum must remain constant.

At the beginning, the merry-go-round and the girl are at rest, so the initial angular momentum is zero:

L_1=0

Later, after the girl throws the rock, the angular momentum will be:

L_2=(I_M+I_g)\omega +L_r

where:

I is the moment of inertia of the merry-go-round

I_g=MR^2 is the moment of inertia of the girl, where

M is the mass of the girl

R is the distance of the girl from the axis of rotation

\omega is the angular speed of the merry-go-round and the girl

L_r=mvR is the angular momentum of the rock, where

m is the mass of the rock

v is its velocity

Since the total angular momentum is conserved,

L_1=L_2

So we find:

0=(I+I_g)\omega +mvR\\\omega=\frac{-mvR}{I+MR^2}

And the negative sign indicates that the disk rotates in the direction opposite to the motion of the rock.

b)

The linear speed of a body in rotational motion is given by

v=\omega r

where

\omega is the angular speed

r is the distance of the body from the axis of rotation

In this problem, for the girl, we have:

\omega=\frac{-mvR}{I+MR^2} is the angular speed

r=R is the distance of the girl from the axis of rotation

Therefore, her linear speed is:

v=\omega R=\frac{-mvR^2}{I+MR^2}

5 0
3 years ago
An object weighs 32 newtons. What is its mass if a gravitometer indicates that g = 8.25 m/s?
Elan Coil [88]

Explanation:

weight=mass×g

32=mass×8.25

mass=

\frac{128}{33}  = 3.878kg

8 0
3 years ago
I need help answering 30A and B. Help a guy out?
dmitriy555 [2]

Uh so I'm no master at this subject, but all stuffs accelerate at 9.8 m/s squared. So you multiply the 9.8 and the 0.20 it's given for reasons unknown other than that's what I see in my notes... and that gives you 1.96 m/s squared.

As for B, I have no idea. I think you may multiply the 1.96 by 4. Tell me your thoughts and maybe we can work it out together

5 0
3 years ago
Calculate how much work is required to launch a spacecraft of mass m from the surface of the earth (mass mE, radius RE) and plac
mylen [45]

To solve this problem it is necessary to apply the concepts related to the conservation of energy, through the balance between the work done and its respective transformation from the gravitational potential energy.

Mathematically the conservation of these two energies can be given through

W = U_f - U_i

Where,

W = Work

U_f = Final gravitational Potential energy

U_i = Initial gravitational Potential energy

When the spacecraft of mass m is on the surface of the earth then the energy possessed by it

U_i = \frac{-GMm}{R}

Where

M = mass of earth

m = Mass of spacecraft

R = Radius of earth

Let the spacecraft is now in an orbit whose attitude is R_{orbit} \approx R then the energy possessed by the spacecraft is

U_f = \frac{-GMm}{2R}

Work needed to put it in orbit is the difference between the above two

W = U_f - U_i

W = -GMm (\frac{1}{2R}-\frac{1}{R})

Therefore the work required to launch a spacecraft from the surface of the Eart andplace it ina circularlow earth orbit is

W = \frac{GMm}{2R}

3 0
3 years ago
Earth pulls on the Moon with a gravitational force, keeping it in its nearly circular, 27‑day orbit. According to Newton's third
arlik [135]

Answer:

Centripetal Force

Explanation:

The reaction to the earth's pull on the moon is the centripetal force

4 0
3 years ago
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