Given:
m = 555 g, the mass of water in the calorimeter
ΔT = 39.5 - 20.5 = 19 °C, temperature change
c = 4.18 J/(°C-g), specific heat of water
Assume that all generated heat goes into heating the water.
Then the energy released is
Q = mcΔT
= (555 g)*(4.18 J/(°C-g)*(19 °C)
= 44,078.1 J
= 44,100 J (approximately)
Answer: 44,100 J
A).
It would decrease because the speed of sound and temperature are proportional.
The applicable equation:
P = F/A
P = pressure
F = Force or weight
A = surface area
Pressure on each cylinder = (W/n)/A
Where n = number of cylinders. Additionally, pressure in the reservoir is equivalent to the pressure in each cylinder.
Net pressure = 75 - 14.7 = 60.3 psi
Therefore,
60.3 = (W/n)/A = (450/n)/(πD^2/4) = (450/n)/(π*1.5^2/4) = (450/n)/(1.7671)
60.3*1.7671 = 450/n
106.03 = 450/n
n = 450/106.3 = 4.244 ≈ 5
The number of cylinders is 5.
Answer:
The magnitude will be "353.5 N". A further solution is given below.
Explanation:
The given values is:
F = 500 N
According to the question,
In ΔABC,
⇒ 
⇒ 
then,
⇒ 
⇒ 
Now,
The corresponding angle will be:
⇒ 
⇒ 
⇒ 
Aspect of F across the AC arm will be:
= 
On putting the values of F, we get
= 
= 
Component F along the AC (in magnitude) will be:
= 
= 
= 