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Digiron [165]
3 years ago
11

The gauge pressure in your car tires is 2.20 x 10^5 N/m^2 at a temperature of 35.0°C when you drive it onto a ferry boat to Alas

ka. What is their gauge pressure (in atm) later, when their temperature has dropped to −38.0°C? (Assume that their volume has not changed.)
Physics
1 answer:
Elanso [62]3 years ago
3 0

Answer:

1.425 atm

Explanation:

Gauge pressure, P1' = 2.20 x 10^5 N/m^2 = 2.178 atm

Absolute pressure P1 = P1' + Po

where, Po be the atmospheric pressure

P1 = 2.178 + 1 = 3.178 atm

T1 = 35 degree C = 35 + 273 = 308 k

T2 = - 38 degree c = - 38 + 273 = 235 k

let final absolute pressure is P2.  Use

P1 / T1 = P2 / T2

3.178 / 308 = P2 / 235

P2 = 2.425 atm

So, the final gauge pressure = 2.425 - 1 = 1.425 atm

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In-s [12.5K]

The correct option is (b) Sea breeze

During the day, susan notices that the wind is blowing onshore at the beach called sea breeze.

What is sea breeze?

  • Any wind that flows from a big body of water onto or onto a landmass is called a sea breeze or an onshore breeze.
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8 0
2 years ago
Two facing surfaces of two large parallel conducting plates separated by 8.5 cm have uniform surface charge densities such that
elena-s [515]

Answer:

positive plate

E = 5.764 KV / m

W = 490eV or 7.85 * 10^-17 J

E_p = 4.74 *10^(-12) eV

E_k = 490 eV

Explanation:

part a

The potential difference between two plates = 490 V

Distance between two plates = 8.5 cm

Answer: The positive plate is at higher potential because of convention.

part b

Electric Field between the plates

E = V / d

E = 490 / 0.085 = 5.764 KV / m

Answer: Electric Field between the plates E = 5.764 KV / m

part c

Work done by electric field

W = V*q

W = 490 * 1.602*10^-19

W = 7.85 * 10^-17 J

or W = 490 eV

Answer: Work done by electric field W = 490eV or 7.85 * 10^-17 J

part d

Potential Energy of an electron gained:

E_p = m_e * g * d / (1.602*10^-19)

E_p =  9.109*10^-31* 9.81 * 0.085 / (1.602*10^-19)

E_p = 4.74 *10^(-12) eV

Very very small E_p approximately 0

Answer: Potential Energy of an electron gained E_p = 4.74 *10^(-12) eV or 0.

part e

Kinetic Energy of an electron gained:

W - E_p = E_k

E_k = 490eV - 4.74*10^(-12)eV

E_k = 490 eV

Answer: Kinetic Energy of an electron gained E_k = 490 eV

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What happens to the brightness of bulb a when the switch is closed and bulb b lights up?
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When only "bulb a" is connected to the battery then more current is flowing to "bulb a" causing it to be bright.

Closing the switch would mean that "bulb b" is already included in the circuit and the battery will push small current to flow around the whole circuit. The more bulbs are connected, the harder for the current to flow because the resistance will be very high.

So the light of "bulb a" will be dimmer.
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