A 1-m3 rigid tank contains 10 kg of water (in any phase or phases) at 160 0C. The pressure in the tank is: a)-370 kPa b)- 618 kP
1 answer:
Answer:
Option C is the correct answer.
Explanation:
We have ideal gas equation, PV = nRT
Volume of tank, V = 1 m³
Mass of water = 10 kg
Mass of 1 mole of water = 18 g = 0.018 kg
Number of moles,
Temperature, T = 160 °C = 160 + 273 = 433 K
Ideal gas constant, R = 8.314 JK⁻¹mol⁻¹
Substituting
P x 1 = 555 x 8.314 x 433
P = 2000000 Pa = 2000 kPa
The pressure in the tank is 2000 kPa
Option C is the correct answer.
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Answer:
5308.34 N/C
Explanation:
Given:
Surface density of each plate (σ) = 47.0 nC/m² =
Separation between the plates (d) = 2.20 cm
We know, from Gauss law for a thin sheet of plate that, the electric field at a point near the sheet of surface density 'σ' is given as:
Now, as the plates are oppositely charged, so the electric field in the region between the plates will be in same direction and thus their magnitudes gets added up. Therefore,
Now, plug in for 'σ' and
for
and solve for the electric field. This gives,
Therefore, the electric field between the plates has a magnitude of 5308.34 N/C
Answer:
Explanation:
If a particle move with time and expressed according to the formula:
f(t) = 0.01t⁴ − 0.03t³
a) Velocity is the change in motion of the particle with respect to time and it is expressed as;
Hence the velocity of the particle at time t is
b) To calculate the velocity after 1 second, we will substitute t = 1 into the function v(t) in (a) as shown:
Hence the velocity after 1second is -0.05
c) The particle is at rest when when the time is zero.
Initially, the body is not moving and the time during this time is 0. Hence the particle is at rest when t = 0second