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Yanka [14]
3 years ago
7

What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?Use 1.67×

10−27 kg for the mass of a proton, 1.60×10−19 C for the magnitude of the charge on an electron, and 9.81 m/s2 for the magnitude of the acceleration due to gravity.
Physics
1 answer:
vazorg [7]3 years ago
5 0

The electric force on the proton is:

F = Eq

F = electric force, E = electric field strength, q = proton charge

The gravitational force on the proton is:

F = mg

F = gravitational force, m = proton mass, g = gravitational acceleration

Since the electric force and gravitational force balance each other out, set their magnitudes equal to each other:

Eq = mg

Given values:

q = 1.60×10⁻¹⁹C, m = 1.67×10⁻²⁷kg, g = 9.81m/s²

Plug in and solve for E:

E(1.60×10⁻¹⁹) = 1.67×10⁻²⁷(9.81)

E = 1.02×10⁻⁷N/C

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Annette [7]

The car tyre contains air initially at a pressure of 195 kPa after travelling several km the temperature of the air inside a car tyre rises from 30 to 70°C if the tyre is rigid and does not expand then the new pressure inside the tyre would be 220.74 kPa.

<h3>What is pressure?</h3>

The total applied force per unit of area is known as the pressure.

The pressure depends both on externally applied force as well the area on which it is applied.

The mathematical expression for the pressure

Pressure = Force /Area

the pressure is expressed by the unit pascal or N /m²

By using the Charles law for gases which states that the volume of the gas remains constant then the pressure of the gas is directly proportional to the temperature.

As given in the problem the tyre is rigid and does not expand this means the volume of the tyre remains constant.

The mathematical expression for Charles's law is as follows

P₁/P₂ = T₁/T₂

First, we have to change the temperature from degree Celcius to the kelvin temperature scale

K = 273 + C

where k is the temperature in kelvin and the C is degrees of Celcius

Initially, the temperature was 30° C

T₁ = 273 + 30

T₁ = 303 K

Then after travelling the temperature of the air inside a car tyre rises from 30 to 70°C

T₂= 273+ 70

T₂ =343 K

The car tyre contains air initially at a pressure of 195 KPa

P₁ = 195 kPa

Lets us take the final pressure of the air would be P₂

By substituting the values in the formula

P₁/P₂ = T₁/T₂

195/P₂ = 303/343

P₂ = 220.74 kPa

Thus, the new pressure inside the tyre would be  220.74 kPa.

Learn more about pressure learn more

brainly.com/question/28012687

#SPJ1

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