When a liquid is cooled, the kinetic energy of the particles ?

Gravitational force exist between you and the building why are you not pulled towards the building?

Nataly_w [17]

Answer:

You are pulled towards that building. At the same time, that building is pulled towards you. Neither object creates enough gravitational force to really do anything. That is why you never notice any affect by either body, (you and a building).

Explanation:

You will surely get attracted towards the building.But it takes a lot of time depending on their masses.

This happens only when you are away from earth with that building.

Both of you will get attracted to it

if a third party with mass more than you or building is with you.

If it is on the earth.. Then the gravity between you and the building is negligible compared to the earth.Hence you will not get attracted towards the building in this case.

3 0

2 years ago

A 0.01-kg object is initially sliding at 9.0 m/s. It goes up a ramp (increasing its elevation by 1.5 m), and then moves horizont

barxatty [35]

Answer:

During this motion, 0.133 J of heat energy was created

Explanation:

Hi there!

Let´s calculate the energy of the object in each phase of the motion.

At first, the object has only kinetic energy (KE):

KE = 1/2 · m · v²

Where:

m = mass of the object.

v = velocity.

KE = 1/2 · 0.01 kg · (9 m/s)²

KE = 0.405 J

When the object goes up the ramp, it gains some gravitational potential energy (PE). Due to the conservation of energy, the object must convert some of its kinetic energy to obtain potential energy. By calculating the potential energy that the object acquires, we can know the loss of kinetic energy:

PE = m · g · h

Where:

m = mass of the object.

g = acceleration due to gravity (9.81 m/s²)

h = height.

PE = 0.01 kg · 9.81 m/s² · 1.5 m

PE = 0.147 J

The object "gives up" 0.147 J of kinetic energy to be converted into potential energy.

Then, after going up the ramp, the kinetic energy of the object will be:

0.405 J - 0.147 J = 0.258 J

When the object reaches the spring, kinetic energy is used to compress the spring and the object obtains elastic potential energy (EPE). Let´s calculate the EPE obtained by the object:

EPE = 1/2 · k · x²

Where:

k = spring constant.

x = compression of the spring

EPE = 1/2 · 100 N/m · (0.05 m)² = 0.125 J

Then, only 0.125 J of kinetic energy was converted into elastic potential energy. The object is at rest at the end of the motion, i.e., the object does not have kinetic energy when it compresses the spring by 5.0 cm. Since energy can´t be lost, the rest of the kinetic energy, that was not used to compress the spring, had to be converted into heat energy:

Heat energy = initial kinetic energy - obtained elastic potential energy

Heat energy = 0.258 J - 0.125 J = 0.133 J

During this motion, 0.133 J of heat energy was created.

7 0

3 years ago

Consider the Uniform Circular Motion Gizmo configured as shown. Notice that, under the current settings, |a|=0.50m/s2. What chan

Eddi Din [679]

To increase the centripetal acceleration to $2.00 m/s^2$ , you can double the speed or decrease the radius by 1/4

Explanation:

An object is said to be in uniform circular motion when it is moving at a constant speed in a circular path.

The acceleration of an object in uniform circular motion is called centripetal acceleration, and it is given by

$a=\frac{v^2}{r}$

where

v is the speed of the object

r is the radius of the circular path

In the problem, the original centripetal acceleration is

$a=0.50 m/s^2$

We want to increase it by a factor of 4, i.e. to

$a'=2.00 m/s^2$

We notice that the centripetal acceleration is proportional to the square of the speed and inversely proportional to the radius, so we can do as follows:

- We can double the speed:

v' = 2v

This way, the new acceleration is

$a'=\frac{(2v)^2}{r}=4(\frac{v^2}{r})=4a$