Answer:
a) The net force on the ball is instantaneously equal to zero newtons at the top of the flight path.
Explanation:
At an instantenous time at the top of the flight path, the upward force due to the Canon explosion on the ball is just equal to the weight of the ball, this will equate the net force on the ball to zero. At this point the velocity of the ball is zero before it decends down to earth under its own weight.
Answer:
BYEEEEEEEEEEEE3EEEEEEEEEE
Explanation:
dawg
Answer:
Given:
high temperature reservoir ![T_{H} =1000k](https://tex.z-dn.net/?f=T_%7BH%7D%20%3D1000k)
low temperature reservoir ![T_{L} =400k](https://tex.z-dn.net/?f=T_%7BL%7D%20%3D400k)
thermal efficiency ![n_{1}= n_{2}](https://tex.z-dn.net/?f=n_%7B1%7D%3D%20n_%7B2%7D)
The engines are said to operate on Carnot cycle which is totally reversible.
To find the intermediate temperature between the two engines, The thermal efficiency of the first heat engine can be defined as
![n_{1} =1-\frac{T}{T_{H} }](https://tex.z-dn.net/?f=n_%7B1%7D%20%3D1-%5Cfrac%7BT%7D%7BT_%7BH%7D%20%7D)
The thermal efficiency of second heat engine can be written as
![n_{2} =1-\frac{T_{L} }{T}](https://tex.z-dn.net/?f=n_%7B2%7D%20%3D1-%5Cfrac%7BT_%7BL%7D%20%7D%7BT%7D)
The temperature of intermediate reservoir can be defined as
![1-\frac{T}{T_{H} } =1-\frac{T_{L} }{T} \\T^2=T_{L} T_{H} \\T=\sqrt{T_{L} T_{H} }\\T=\sqrt{400*1000} =632k](https://tex.z-dn.net/?f=1-%5Cfrac%7BT%7D%7BT_%7BH%7D%20%7D%20%3D1-%5Cfrac%7BT_%7BL%7D%20%7D%7BT%7D%20%5C%5CT%5E2%3DT_%7BL%7D%20T_%7BH%7D%20%5C%5CT%3D%5Csqrt%7BT_%7BL%7D%20T_%7BH%7D%20%7D%5C%5CT%3D%5Csqrt%7B400%2A1000%7D%20%3D632k)
Explanation:
इसिसिसिसैस्स्स्स्स्स्स्स्स्स्सूस्सोस्स्स्स्स्स
Answer:
V = 0.30787 m³/s
m = 2.6963 kg/s
v2 = 0.3705 m³/s
v2 = 6.017 m/s
Explanation:
given data
diameter = 28 cm
steadily =200 kPa
temperature = 20°C
velocity = 5 m/s
solution
we know mass flow rate is
m = ρ A v
floe rate V = Av
m = ρ V
flow rate = V =
V = Av = ![\frac{\pi}{4} * d^2 * v1](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%7D%7B4%7D%20%2A%20d%5E2%20%2A%20v1)
V = ![\frac{\pi}{4} * 0.28^2 * 5](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cpi%7D%7B4%7D%20%2A%200.28%5E2%20%2A%205)
V = 0.30787 m³/s
and
mass flow rate of the refrigerant is
m = ρ A v
m = ρ V
m =
= ![\frac{0.30787}{0.11418}](https://tex.z-dn.net/?f=%5Cfrac%7B0.30787%7D%7B0.11418%7D)
m = 2.6963 kg/s
and
velocity and volume flow rate at exit
velocity = mass × v
v2 = 2.6963 × 0.13741 = 0.3705 m³/s
and
v2 = A2×v2
v2 = ![\frac{v2}{A2}](https://tex.z-dn.net/?f=%5Cfrac%7Bv2%7D%7BA2%7D)
v2 = ![\frac{0.3705}{\frac{\pi}{4} * 0.28^2}](https://tex.z-dn.net/?f=%5Cfrac%7B0.3705%7D%7B%5Cfrac%7B%5Cpi%7D%7B4%7D%20%2A%200.28%5E2%7D)
v2 = 6.017 m/s