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3 years ago

12

two object gravitationally attract with a force of 20N.if the distance between the two object centers is doubled, then what is t

he new force?

1 answer:

Usimov [2.4K]3 years ago

8 0

Gravity decreases with the square of the distance, so the new force is (20)/(2*2) = 5N.

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A sled slides down a hill with friction between the sled and hill but negligible alr resistance. Which of the following must be

Andreyy89

Answer:

A) The sum of the kinetic energy and the gravitational potential energy changes by an amount equal to the energy dissipated by friction,

Explanation:

The kinetic energy is the energy that the object has and is defied by the work that is needed to accelerate the body.

The gravitational potential is a mechanism by which an equal amount of energy is being transferred per unit mass that is needed for the object to move from the specific location.

Hence when the sled moves down the hill with the force of gravity it has negligible resistance as an equal amount of energy is dissipated.

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2 years ago

Visible light includes all of the

krek1111 [17]

Answer:Visible light is a small part of the electromagnetic spectrum. The spectrum covers everything from gamma rays, x-rays, visible light, infrared, microwave and radio waves. Each part of the spectrum, including the different colors of visible light, have different wavelengths (the space between each wave).

Explanation:

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2 years ago

What the electron and outermost energy level of periods 2and3,4,5 and 6

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3 years ago

When you whirl a can at the end of a string in a circular path, what is the direction of the force that acts on the can

andreyandreev [35.5K]

Answer:

Toward the centre of the circular path

Explanation:

The can is moved in a circular path: this means that it is moving by circular motion (uniform circular motion if its tangential speed is constant).

In order to keep a circular motion, an object must have a force that pushes it towards the centre of the circular trajectory: this force is called centripetal force, and its magnitude is given by

$F=m\frac{v^2}{r}$

where m is the mass of the object, v its tangential speed, r the radius of the trajectory. This force always points towards the centre of the circular path.

3 0

3 years ago

A 392 N wheel comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at 24

alex41 [277]

Answer:

h=12.41m

Explanation:

N=392

r=0.6m

w=24 rad/s

$I=0.8*m*r^{2}$

So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion

$N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}$

$m=40kg$

moment of inertia

$M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}$

Kinetic energy of the rotation motion

$K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J$

Kinetic energy translational

$K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J$

Total kinetic energy

$K=3317.79J+4147.2J\\K=7464.99J$

Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height

$K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m$

6 0

3 years ago