centre of square disrance to each corner found by Pythagoras' theorem.
coulombs law used to clculate field of each charge at centre
fields added vectorially for res
Answer:
The change in potential energy of the mass as it goes up the incline is 0.343 joules.
Explanation:
We must remember in this case that change in the potential energy is entirely represented by the change in the gravitational potential energy. From Work-Energy Theorem and definition of work we get that:
![U_{g}= m\cdot g\cdot \Delta y](https://tex.z-dn.net/?f=U_%7Bg%7D%3D%20m%5Ccdot%20g%5Ccdot%20%5CDelta%20y)
Where:
- Gravitational potential energy, measured in Joules.
- Mass, measured in kilograms.
- Gravitational acceleration, measured in meters per square second.
- Change in vertical height, measured in meters.
This work is the energy needed to counteract effects of gravity at given vertical displacement.
If we know that
,
and
, the change in the potential energy of the mass as it goes up the incline is:
![U_{g} = (0.5\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (0.07\,m)](https://tex.z-dn.net/?f=U_%7Bg%7D%20%3D%20%280.5%5C%2Ckg%29%5Ccdot%20%5Cleft%289.807%5C%2C%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%20%5Cright%29%5Ccdot%20%280.07%5C%2Cm%29)
![U_{g} = 0.343\,J](https://tex.z-dn.net/?f=U_%7Bg%7D%20%3D%200.343%5C%2CJ)
The change in potential energy of the mass as it goes up the incline is 0.343 joules.
It could be warmer solid, or liquid, or gas, depending on the pressure and the amount of heat added.
Answer:
50.5 m
Explanation:
We are given that
![\theta=33.5^{\circ}](https://tex.z-dn.net/?f=%5Ctheta%3D33.5%5E%7B%5Ccirc%7D)
Velocity=v=19.5m
Vertical component of initial velocity=![v_y=vsin\theta=19.5sin33.5=10.8m/s](https://tex.z-dn.net/?f=v_y%3Dvsin%5Ctheta%3D19.5sin33.5%3D10.8m%2Fs)
Height=h=-14.5 m
Acceleration due to gravity=![g=-9.8m/s^2](https://tex.z-dn.net/?f=g%3D-9.8m%2Fs%5E2)
![s=v_yt+\frac{1}{2}gt^2](https://tex.z-dn.net/?f=s%3Dv_yt%2B%5Cfrac%7B1%7D%7B2%7Dgt%5E2)
![-14.5=10.8t+\frac{1}{2}(-9.8)t^2](https://tex.z-dn.net/?f=-14.5%3D10.8t%2B%5Cfrac%7B1%7D%7B2%7D%28-9.8%29t%5E2)
![4.9t^2-10.8t-14.5=0](https://tex.z-dn.net/?f=4.9t%5E2-10.8t-14.5%3D0)
By solving we get
![t=-0.9, t=3.1](https://tex.z-dn.net/?f=t%3D-0.9%2C%20t%3D3.1%20)
Time cannot be negative
Therefore, time=3.1 s
Horizontal component of velocity=![v_x=19.5cos33.5=16.3m/s](https://tex.z-dn.net/?f=v_x%3D19.5cos33.5%3D16.3m%2Fs)
Distance=![x=v_xt=16.3\times 3.1=50.5 m](https://tex.z-dn.net/?f=x%3Dv_xt%3D16.3%5Ctimes%203.1%3D50.5%20m)
Hence, the golf ball travel horizontally before it hits the water=50.5 m