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3 years ago

Can electrostatic attract both pull and pull?

Physics

2 answers:

Delvig [45]3 years ago

8 0

Answer: Yes

Explanation:

Electrostatic forces are non-contact forces; they pull or push on objects without touching them. Rubbing some materials together can result in something called 'charge' being moved from one surface to the other. Charged objects pull on other uncharged objects and may either push or pull on other charged objects

Sedaia [141]3 years ago

6 0

Yes they can attract both pulls

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A rectangular gasoline tank can hold 50.0 kg of gasoline when full. What is the depth of the tank if it is 0.500-m wide by 0.900

sweet-ann [11.9K]

Answer:

0.16 m

Explanation:

A rectangular gasoline tank can hold 50.0 kg of gasoline when full, and the density of gasoline is 6.8 × 10² kg/m³. We can find the volume occupied by the gasoline (volume of the tank).

50.0 kg × (1 m³/6.8 × 10² kg) = 0.074 m³

The volume of the rectangular tank is:

volume = width × length × depth

depth = volume / width × length

depth = 0.074 m³ / 0.500 m × 0.900 m

depth = 0.16 m

3 0

3 years ago

What does every magnet possess?

IRINA_888 [86]

Answer:

a North and South Pole :)

Explanation:

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3 years ago

How does the size of arctic foxes ears help to keep them warm in a cold environment

DochEvi [55]

artic foxes have thick white fur to keep them warm and also for camouflage in the snow they have small ears to keep the snow out of their ears so that they can hear better they also have thick fur lining and a thick pad on their feet this stops the paws from getting cold and they have short legs so then the body weight can be spread making it easier for the animal to not sink in the snow in the summer the artic fox turns brown to help make the fox invisible when people are hunting in the long grass

3 0

3 years ago

A 175-kg roller coaster car starts from rest at the top of an 18.0-m hill and rolls down the hill, then up a second hill that ha

Anni [7]

Answer:

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

Explanation:

By Principle of Energy Conservation and Work-Energy Theorem we present the equations that describe the situation of the roller coaster car on each top of the hill. Let consider that bottom has a height of zero meters.

From top of the first hill to the bottom

$m\cdot g \cdot h_{1} = \frac{1}{2}\cdot m\cdot v_{1}^{2} +W_{1, loss}$ (1)

From the bottom to the top of the second hill

$\frac{1}{2}\cdot m\cdot v_{1}^{2} = m\cdot g \cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2}+W_{2,loss}$ (2)

Where:

$m$ - Mass of the roller coaster car, in kilograms.

$v_{1}$ - Speed of the roller coaster car at the bottom between the two hills, in meters per second.

$g$ - Gravitational acceleration, in meters per square second.

$h_{1}$ - Height of the first top of the hill with respect to the bottom, in meters.

$W_{1, loss}$ - Work done by non-conservative forces on the car between the top of the first hill and the bottom, in joules.

$v_{2}$ - Speed of the roller coaster car at the top of the second hill, in meters per seconds.

$h_{2}$ - Height of the second top of the hill with respect to the bottom, in meters.

$W_{2, loss}$ - Work done by non-conservative forces on the car bewteen the bottom between the two hills and the top of the second hill, in joules.

By using (1) and (2), we reduce the system of equation into a sole expression:

$m\cdot g\cdot h_{1} = m\cdot g\cdot h_{2} + \frac{1}{2}\cdot m \cdot v_{2}^{2} + W_{loss}$ (3)

Where $W_{loss}$ is the work done by non-conservative forces on the car from the top of the first hill to the top of the second hill, in joules.

If we know that $m = 175\,kg$ , $g = 9.807\,\frac{m}{s^{2}}$ , $h_{1} = 18\,m$ , $h_{2} = 8\,m$ and $v_{2} = 11\,\frac{m}{s}$ , then the work done by non-conservative force is:

$W_{loss} = m\cdot\left[ g\cdot \left(h_{1}-h_{2}\right)-\frac{1}{2}\cdot v_{2}^{2} \right]$

$W_{loss} = 6574.75\,J$

The work done by non-conservative forces on the car from the top of the first hill to the top of the second hill is 6574.75 joules.

8 0

2 years ago

A rigid, nonconducting tank with a volume of 4 m3 is divided into two unequal parts by a thin membrane. One side of the membrane

kondor19780726 [428]

The final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.

To find the answer, we need to know about the thermodynamic processes.

<h3>How to find the final temperature of the gas?</h3>

Any processes which produce change in the thermodynamic coordinates of a system is called thermodynamic processes.
In the question, it is given that, the tank is rigid and non-conducting, thus, dQ=0.
The membrane is raptured without applying any external force, thus, dW=0.
We have the first law of thermodynamic expression as,

$dU=dQ-dW$

Here it is zero.

$dU=0$ ,

As we know that,

$dU=C_pdT=0\\\\thus, dT=0\\\\or , T=constant\\\\i.e, T_1=T_2$

Thus, the final temperature of the system will be equal to the initial temperature,

$T_1=T_2=100^0C=373K$

We found that the process is isothermal,
Thus, the work done will be,

$W=RT*ln(\frac{V_2}{V_1} )=373R*ln(\frac{4}{\frac{4}{3} })\\ \\W=409.8R J$

Where, R is the universal gas constant.

<h3>What is a reversible process?</h3>

Any process which can be made to proceed in the reverse direction is called reversible process.
During which, the system passes through exactly the same states as in the direct process.

Thus, we can conclude that, the final temperature of the system will be equal to the initial temperature, and which is 373K. The work done by the system is 409.8R Joules.

Learn more about thermodynamic processes here:

brainly.com/question/28067625

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7 0

1 year ago

Read 2 more answers