Answer:
a.) -147V
b.) -120V
c.) 51V
Explanation:
a.) Equation for potential difference is the integral of the electrical field from a to b for the voltage V_ba = V(b)-V(a).
b.) The problem becomes easier to solve if you draw out the circuit. Since potential at Q is 0, then Q is at ground. So voltage across V_MQ is the same as potential at V_M.
c.) Same process as part b. Draw out the circuit and you'll see that the potential a point V_N is the same as the voltage across V_NP added with the 2V from the other box.
Honestly, these things take practice to get used to. It's really hard to explain this.
Answer:
The spring is compressed by 0.275 meters.
Explanation:
For equilibrium of the gas and the piston the pressure exerted by the gas on the piston should be equal to the sum of weight of the piston and the force the spring exerts on the piston
Mathematically we can write
![Force_{pressure}=Force_{spring}+Weight_{piston}](https://tex.z-dn.net/?f=Force_%7Bpressure%7D%3DForce_%7Bspring%7D%2BWeight_%7Bpiston%7D)
we know that
![Force_{pressure}=Pressure\times Area=300\times 10^{3}\times \frac{\pi \times 0.1^2}{4}=750\pi Newtons](https://tex.z-dn.net/?f=Force_%7Bpressure%7D%3DPressure%5Ctimes%20Area%3D300%5Ctimes%2010%5E%7B3%7D%5Ctimes%20%5Cfrac%7B%5Cpi%20%5Ctimes%200.1%5E2%7D%7B4%7D%3D750%5Cpi%20Newtons)
![Weight_{piston}=mass\times g=100\times 9.81=981Newtons](https://tex.z-dn.net/?f=Weight_%7Bpiston%7D%3Dmass%5Ctimes%20g%3D100%5Ctimes%209.81%3D981Newtons)
Now the force exerted by an spring compressed by a distance 'x' is given by ![Force_{spring}=k\cdot x=5\times 10^{3}\times x](https://tex.z-dn.net/?f=Force_%7Bspring%7D%3Dk%5Ccdot%20x%3D5%5Ctimes%2010%5E%7B3%7D%5Ctimes%20x)
Using the above quatities in the above relation we get
![5\times 10^{3}\times x+981=750\pi \\\\\therefore x=\frac{750\pi -981}{5\times 10^{3}}=0.275meters](https://tex.z-dn.net/?f=5%5Ctimes%2010%5E%7B3%7D%5Ctimes%20x%2B981%3D750%5Cpi%20%5C%5C%5C%5C%5Ctherefore%20x%3D%5Cfrac%7B750%5Cpi%20-981%7D%7B5%5Ctimes%2010%5E%7B3%7D%7D%3D0.275meters)
Answer:
electrical
computer
mechanical
and manufacturing .... I think
Answer:
If the turbulent velocity profile in a pipe of diameter 0.6 m may be approximated by u/U=(y/R)^(1/7), where u is in m/s and y is in m and 0.15 m from the pipe.
Explanation:
hope it helps