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4 years ago

Which of the following particle is responsible for the transfer of static charge?

Physics

1 answer:

djverab [1.8K]4 years ago

4 0

Answer:

B. Electron

Static charge is produced by electron transfer.

Explanation:

Answer A: The neutron does not possess a charge and is said to be neutral.

Answer D: Protons and neutrons never move from object to object.

Only negative charges can move freely from one object to another. The energy that comes from these charges particles is called electrical energy.

....

Hope this answer can help you.

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Consider a system consisting of an ideal gas confined within a container, one wall of which is a movable piston. Energy can be a

Helen [10]

Complete Question

Consider a system consisting of an ideal gas confined within a container, one wall of which is a movable piston. Energy can be added to the gas in the form of heat by applying a flame to the outside of the container. Conversely, energy can also be removed from the gas in the form of heat by immersing the container in ice water. Energy can be added to the system in the form of work by pushing the piston in, thereby compressing the gas. Conversely, if the gas pushes the piston out, thereby pushing some atmosphere aside, the internal energy of the gas is reduced by the amount of work done.

$pV=nRT$

so the absolute temperature T is directly proportional to the product of the absolute pressure p and the volume V,Here n denotes the amount of gas moles,which is a constant because the gas is confined and R is the universal constant

What is the $\triangle U$ as the system of ideal gas goes from point A to point B on the graph recall u is proportional to T

Answer:

$\triangle T=0$

$\triangle V=0$

The gas A and B have same internal energy

Explanation:

From the question we are told that

$Pa=u atm\\Va=1m^3\\Pb=1 atm\\Vb=4m^3$

Generally the equation of temperature is mathematically given as

$Ta=\frac{Pv}{nR}$

$Ta=\frac{u*1}{nR}$

And

$Tb=\frac{PbVb}{nR}$

$Tb=\frac{u*1}{nR}$

Generally the change in temperature $\triangle T$ is mathematically given as

$\triangle T=Tb-Ta=Tb=\frac{u*1}{nR}-\frac{u*1}{nR}$

$\triangle T=0$

Generally the change in internal energy $\triangle V$

$\triangle V=nC_v \triangle T\\$

$\triangle V=0$

Therefore with

$\triangle T=0$

$\triangle v=0$

The gas A and B have same internal energy

6 0

3 years ago

A cheetah bites into its prey. One tooth exerts a force of 320 N. The area of the point of the tooth is 0.5 cm². The pressure of

mote1985 [20]

Answer:

Answer D is correct

Explanation:

$pressure = \frac{force}{area} \\ = \frac{320}{0.5} \\ = 640$

hope this helps

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6 0

3 years ago

The heat released by burning candle is an example of thermal energy

yawa3891 [41]

False the correct answer is chemical bonds instead of thermal energy

4 0

3 years ago

Read 2 more answers

What is the specific heat of a substance that absorbs 2500 joules of heat when a sample of 1.200 kg of the substance increases i

Anarel [89]

Answer:

0.035 J/g°C

Explanation:

From the question given above, the following data were obtained:

Heat (Q) absorbed = 2500 J

Mass (M) = 1.2 Kg

Initial Temperature (T₁) = 10 °C

Final Temperature (T₂) = 70 °C

Specific heat capacity (C) =?

Next, we shall determine the change in temperature. This can be obtained as follow:

Initial Temperature (T₁) = 10 °C

Final Temperature (T₂) = 70 °C

Change in temperature (ΔT) =

ΔT = T₂ – T₁

ΔT = 70 – 10

ΔT = 60 °C

Thus, the change in the temperature of the substance is 60 °C

Next, we shall convert 1.2 Kg to grams (g). This can be obtained as follow:

1 Kg = 1000 g

Therefore,

1.2 Kg = 1.2 Kg × 1000 g / 1 Kg

1.2 Kg = 1200 g

Thus, 1.2 Kg is equivalent to 1200 g.

Finally, we shall determine the specific heat capacity of substance. This can be obtained as follow:

Heat (Q) absorbed = 2500 J

Mass (M) = 1200 g

Change in temperature (ΔT) = 60 °C

Specific heat capacity (C) =?

Q = MCΔT

2500 = 1200 × C × 60

2500 = 72000 × C

Divide both side by 72000

C = 2500 / 72000

C = 0.035 J/g°C

Therefore, the specific heat capacity of the substance is 0.035 J/g°C

3 0

3 years ago

15) A 328-kg car moving at 19.1 m/s in the +x direction hits from behind a second car moving at 13 m/s in the same direction. If

kolbaska11 [484]

Answer:

Explanation:

Given that,

Mass of first car

M1= 328kg

The car is moving in positive direction of x axis with velocity

U1 = 19.1m/s

Velocity of second car

U2 = 13m/s, in the same direction as the first car..

Mass of second car

M2 = 790kg

Velocity of second car after collision

V2 = 15.1 m/s

Velocity of first car after collision

V1 =?

This is an elastic collision,

And using the conservation of momentum principle

Momentum before collision is equal to momentum after collision

P(before) = P(after)

M1•U1 + M2•U2 = M1•V1 + M2•V2

328 × 19.1 + 790 × 13 = 328 × V1 + 790 × 15.1

16534.8 = 328•V1 + 11929

328•V1 = 16534.8—11929

328•V1 = 4605.8

V1 = 4605.8/328

V1 = 14.04 m/s

The velocity of the first car after collision is 14.04 m/s

5 0

3 years ago