A gas at 5.00 atm pressure was stored in a tank during the winter at 5.0 °C. During the summer, the temperature in the storage a
1 answer:
Answer:
a) 5.63 atm
Explanation:
We can use combined gas law
<em>The combined gas law</em> combines the three gas laws:
- Boyle's Law, (P₁V₁ =P₂V₂)
- Charles' Law (V₁/T₁ =V₂/T₂)
- Gay-Lussac's Law. (P₁/T₁ =P₂/T₂)
It states that the ratio of the product of pressure and volume and the absolute temperature of a gas is equal to a constant.
P₁V₁/T₁ =P₂V₂/T₂
where P = Pressure, T = Absolute temperature, V = Volume occupied
The volume of the system remains constant,
So, P₁/T₁ =P₂/T₂
a)
You might be interested in
Given Information:
Potential difference = V = 100 V
Capacitance C₁ = 10 mF
Capacitance C₂ = 5 mF
Capacitance C₃ = 4 mF
Required Information:
a. Charge q₃
b. Potential difference V₃
c. Stored energy U₃
d. Charge q ₁
e. Potential difference V₁
f. Stored energy U₁
g. Charge q ₂
h. Potential difference V₂
i. Stored energy U₂
Answer:
a. Charge q₃ = 0.4 C
b. Potential difference V₃ = 100 V
c. Stored energy U₃ = 20 J
d. Charge q ₁ = 0.33 C
e. Potential difference V₁ = 33 V
f. Stored energy U₁ = 5.445 J
g. Charge q ₂ = 0.33 C
h. Potential difference V₂ = 66 V
i. Stored energy U₂ = 10.89 J
Explanation:
Please refer to the circuit attached in the diagram
a. Charge q₃
As we know charge in a capacitor is given by
q₃ = C₃V₃
q₃ = 4x10⁻³*100
q₃ = 0.4 C
b. Potential difference V₃
The potential difference V₃ is same as V
V₃ = 100 V
c. Stored energy U₃
Energy stored in a capacitor is given by
U₃ = ½C₃V₃²
U₃ = ½*4x10⁻³*100²
U₃ = 20 J
d. Charge q ₁
Since capacitor C₁ and C₂ are in series their equivalent capacitance is
Ceq = C₁*C₂/C₁ + C₂
Ceq = 10x10⁻³*5x10⁻³/10x10⁻³ + 5x10⁻³
Ceq = 3.33x10⁻³ F
q ₁ = Ceq*V
q ₁ = 3.33x10⁻³*100
q ₁ = 0.33 C
e. Potential difference V₁
V₁ = q ₁/C₁
V₁ = 0.33/10x10⁻³
V₁ = 33 V
f. Stored energy U₁
U₁ = ½C₁V₁²
U₁ = ½*10x10⁻³*(33)²
U₁ = 5.445 J
g. Charge q ₂
q₂ = Ceq*V
q₂ = 3.33x10⁻³*100
q₂ = 0.33 C
h. Potential difference V₂
V₂ = q ₂/C₂
V₂ = 0.33/5x10⁻³
V₂ = 66 V
i. Stored energy U₂
U₂ = ½C₂V₂²
U₂ = ½*5x10⁻³*(66)²
U₂ = 10.89 J
Answer:
The rocket above the ground is in 44 sec.
Explanation:
Given that,
Initial velocity = 92 m/s
Acceleration = 4 m/s²
Altitude = 1200 m
Suppose, How long was the rocket above the ground?
We need to calculate the time
Using equation of motion
Put the value into the formula
We need to calculate the velocity
Using equation of motion
Put the value into the formula
When the rocket hits the ground,
Then, h'=0
We need to calculate the time
Using equation of motion
Put the value into the formula
When the rocket is in the air it is the sum of the time when it reaches 1000 m and the time when it hits the ground
So, the total time will be
Hence, The rocket above the ground is in 44 sec.