Charles Law
Explanation:
Step 1:
It is given that the original volume of the gas is 250 ml at 300 K temperature and 1 atmosphere pressure. We need to find the volume of the same gas when the temperature is 350 K and 1 atmosphere pressure.
Step 2:
We observe that the gas pressure is the same in both the cases while the temperature is different. So we need a law that explains the volume change of a gas when temperature is changed, without any change to the pressure.
Step 3:
Charles law provides the relationship between the gas volume and temperature, at a given pressure
Step 4:
Hence we conclude that Charles law can be used.
Answer:
v= 449.8 m/s
Explanation:
Given data
Frequency= 346Hz
Wave length= 1.4m
The expression below is used to find the speed

substitute

Hence the speed is v= 449.8 m/s
Answer: Tension = 47.8N, Δx = 11.5×
m.
Tension = 95.6N, Δx = 15.4×
m
Explanation: A speed of wave on a string under a tension force can be calculated as:

is tension force (N)
μ is linear density (kg/m)
Determining velocity:


0.0935 m/s
The displacement a pulse traveled in 1.23ms:


Δx = 11.5×
With tension of 47.8N, a pulse will travel Δx = 11.5×
m.
Doubling Tension:



|v| = 0.1252 m/s
Displacement for same time:


15.4×
With doubled tension, it travels
15.4×
m
Answer:
The answer is below
Explanation:
The length of the rope is equal to the radius of the circle formed by the complete rotation of the rope. Therefore the radius = 1.50 m.
a) The distance covered by the rope when completing one rotation is the same as the perimeter of the circle. Hence:
Distance covered in one rotation = 2π * radius = 2π * 1.5 = 3π meters
The velocity of the ball = Distance / time = 3π meters / 3.4 seconds = 2.77 m/s
b) The initial velocity (u) is 0 m/s, the final velocity is 2.77 m/s during time (t) = 3.4 s. Hence acceleration (a):
v = u + at
2.77 = 3.4a
a = 0.82 m/s²
c) Force on ball = mass * acceleration = 4 * 0.82 = 3.28 N
Answer:
Explanation:
Total Resistance, Rt = (0.35 + 0.002 + 5200 + 0.5 + 0.008 + 1100) = 6300 ohms
a) Capacity of transformer, Pt = 150KVA = 150,000 W
Input Voltage, Vp = 2.8 KV = 2800 V
Current, Ip = 150000/2800 = 53.57 A
Input impedance, Zp = Vp/Ip = 2800/5357 = 52.27 ohms
b) i) Input current = 53.57 A
ii) Voltage, V = Ip * Rt = 53.57 X 6300 = 337.5 KV
iii) Power, P = I² * Rt = (53.57)² X 6300 = 18.08 MW
iv) Power factor = 0.83