Answer:
given,
mass of copper = 100 g
latent heat of liquid (He) = 2700 J/l
a) change in energy
Q = m Cp (T₂ - T₁)
Q = 0.1 × 376.812 × (300 - 4)
Q = 11153.63 J
He required
Q = m L
11153.63 = m × 2700
m = 4.13 kg
b) Q = m Cp (T₂ - T₁)
Q = 0.1 × 376.812 × (78 - 4)
Q = 2788.41 J
He required
Q = m L
2788.41 = m × 2700
m = 1.033 kg
c) Q = m Cp (T₂ - T₁)
Q = 0.1 × 376.812 × (20 - 4)
Q = 602.90 J
He required
Q = m L
602.9 = m × 2700
m =0.23 kg
Answer:
A sample of 5.2 mg decays to .65 mg or to 1/8 of its original amount.
1/8 = 1/2 * 1/2 * 1/2 or 3 half-lives.
3 * 30.07 = 90 yrs for 5.2 mg to decay to .65 mg
You can get these other numbers similarly:
5.2 / .0102 = 510 requires about 9 half-lives which is 30 * 9 = 270 yrs
Answer:
The magnitude of the average force exerted on the water by the blade is 960 N.
Explanation:
Given that,
The mass of water per second that strikes the blade is, 
Initial speed of the oncoming stream, u = 16 m/s
Final speed of the outgoing water stream, v = -16 m/s
We need to find the magnitude of the average force exerted on the water by the blade. It can be calculated using second law of motion as :
F = -960 N
So, the magnitude of the average force exerted on the water by the blade is 960 N. Hence, this is the required solution.
Answer: 80m
Explanation:
Distance of balloon to the ground is 3150m
Let the distance of Menin's pocket to the ground be x
Let the distance between Menin's pocket to the balloon be y
Hence, x=3150-y------1
Using the equation of motion,
V^2= U^s + 2gs--------2
U= initial speed is 0m/s
g is replaced with a since the acceleration is under gravity (g) and not straight line (a), hence g is taken as 10m/s
40m/s is contant since U (the coin is at rest is 0) hence V =40m/s
Slotting our values into equation 2
40^2= 0^2 + 2 * 10* (3150-y)
1600 = 0 + 63000 - 20y
1600 - 63000 = - 20y
-61400 = - 20y minus cancel out minus on both sides of the equation
61400 = 20y
Hence y = 61400/20
3070m
Hence, recall equation 1
x = 3150 - 3070
80m
I hope this solve the problem.
Answer:
Water flowing rate= (300000kg/s) = (300000l/s)
Explanation:
First with the section of the channel, the depth of the water and the speed of the fluid we can determine the volume of fluid that circulates per second through the channel:
Volume per time= 15m × 8m × (2.5m/s)= 300 m³/s
With this volume of circulating fluid per second elapsed, we multiply it by the density of the water to determine the kilograms or liters of water that circulate through the channel per second elapsed:
Water flowing rate= (300m³/s) × (1000kg/m³)= (300000kg/s) = (300000l/s)
Taking into account that 1kg of water is approximately equal to 1 liter of water.
