1 and 4 are tire.
2 and 3 are not.
Answer:
A= 203 KJ
B= 54 Kg
Explanation:
The initial specific volumes and internal energies are obtained from A-12 for a given pressure and state. The enthalpy of the refrigerant in the supply line is determined using the saturated liquid approximation for the given temperature with data from A-11. The mass that has entered the tank is:
Δm = m₂ – m₁
= V(1/α₂ – 1/α₁)
= 0.05 (1/0.0008935 – 1/ 0.025645)Kg
= 54Kg
The heat transfer is obtained from the energy balance:
ΔU=
+ 
m₂u₂ – m₁u₂ = 
+ 
= m₂u₂ – m₁u₁ –
= V/α₂u₂ - V/α₁u₁ –
=(0.05/0.0008935 . 116.72 – 0.05/0.025645 . 246.82 – 54.108.28) Kj
= 203 KJ
the soviet union was the first country.
Explanation:
- Initial velocity (u) = 10 m/s
- Final velocity (v) = 30 m/s
- Mass (m) = 2400 kg
- Force (F) = 12000 N
Let us find the time taken first.
→ F = ma
- Acceleration (a) = (v – u)/t
→ 12000 = 2400 × (30 – 10)/t
→ 12000 ÷ 2400 = (20)/t
→ 5 = 20/t
→ 5t = 20
→ t = 20 ÷ 5
→ <u>t</u><u> </u><u>=</u><u> </u><u>4</u><u> </u><u>seconds</u>
Now, find the acceleration.
→ a = (v – u)/t
→ a = (30 – 10)/4
→ a = 20/4
→ <u>a</u><u> </u><u>=</u><u> </u><u>5</u><u> </u><u>m</u><u>/</u><u>s²</u>
Now, by using the third equation of motion,
→ v² – u² = 2as
→ (30)² – (10)² = 2 × 5 × s
→ 900 – 100 = 10s
→ 800 = 10s
→ 800 ÷ 10 = s
→ <u>8</u><u>0</u><u> </u><u>m</u><u> </u><u>=</u><u> </u><u>s</u>
Therefore, distance travelled is 80 m.
When two charged plates are separated by a distance, the potential difference between the plates is 54 Volts.
<h3>What is uniform electric field?</h3>
The electric field intensity which does not increase or decrease with time or remain constant is called the uniform electric field.
Given is the plates separated by distance d = 15 cm =0.15 m, electric field E= 360 N/C.
The relation between the electric field and voltage is
E=V/d
Substitute the values into the expression to get the potential difference V,
V = 360 x 0.15
V = 54 Volts
Thus, the potential difference between the plates is 54 Volts.
Learn more about uniform electric field.
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