Explanation:
Expression for energy balance is as follows.
or, 
Therefore,
Hence, expression for exit velocity will be as follows.
= ![V^{2}_{1} + 2C_{p}(T_{1} - T_{2})]^{0.5}](https://tex.z-dn.net/?f=V%5E%7B2%7D_%7B1%7D%20%2B%202C_%7Bp%7D%28T_%7B1%7D%20-%20T_%7B2%7D%29%5D%5E%7B0.5%7D)
As 
for the given conditions is 1.007 kJ/kg K. Now, putting the given values into the above formula as follows.
![V_{2} = V^{2}_{1} + 2C_{p}(T_{1} - T_{2})]^{0.5}](https://tex.z-dn.net/?f=V_%7B2%7D%20%3D%20V%5E%7B2%7D_%7B1%7D%20%2B%202C_%7Bp%7D%28T_%7B1%7D%20-%20T_%7B2%7D%29%5D%5E%7B0.5%7D)
= ![[(350 m/s)^{2} + 2(1.007 kJ/kg K) (30 - 90) K \frac{1000 m^{2}/s^{2}}{1 kJ/kg}]^{0.5}](https://tex.z-dn.net/?f=%5B%28350%20m%2Fs%29%5E%7B2%7D%20%2B%202%281.007%20kJ%2Fkg%20K%29%20%2830%20-%2090%29%20K%20%5Cfrac%7B1000%20m%5E%7B2%7D%2Fs%5E%7B2%7D%7D%7B1%20kJ%2Fkg%7D%5D%5E%7B0.5%7D)
= 40.7 m/s
Thus, we can conclude that velocity at the exit of a diffuser under given conditions is 40.7 m/s.
Half the potential difference of the the1-µF
A circuit must have a capacitance of 2 F across a 1 kV potential difference for an electrical technician. He has access to a sizable number of 1F capacitors, each of which can sustain a potential difference of no more than 400 V. Please suggest a configuration that uses the fewest capacitors possible.
The 2-mu F capacitor has the following characteristics: none of the aforementioned; half the charge of the 1-mu F capacitor; twice the charge of the 1-mu F capacitor; and half the potential difference of the 1-mu F capacitor.
Q = C V, C = Capacitance of the capacitor gives the charge stored by a capacitor with an applied voltage V. V is the applied voltage.
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Answer:
V = 0.39 m/s
Explanation:
Given that,
Mass of hockey puck, m = 0.2 kg
Mass of goalie = 40 kg
Speed of hockey puck, v = 80 m/s
We need to find the speed with which the goalie slide on the slide. Let V be the speed. Using the conservation of momentum as follows :
So, the required speed is 0.39 m/s.
Answer:
The acceleration is 2 m/s2.
Explanation:
We calculate the acceleration (a), with the data of mass (m) and force (F), through the formula:
F = m x a ---> a= F/m
a = 40 N/20 kg <em> 1N= 1 kg x m/s2</em>
a= 40 kgx m/s2/ 20 kg
<em>a= 2 m/s2</em>
