Answer: hello your question is incomplete below is the missing part
A spherical cavity is hollowed out of the interior of a neutral conducting sphere. At the center of the cavity is a point charge, of positive charge q.
answer:
- q
Explanation:
Since the spherical cavity was carved out of a neutral conducting sphere hence the electric field inside this conductor = zero
given that there is a point charge +q at the center of the spherical cavity hence for the electric field inside the conductor to be = zero the total surface charge qint on the wall of the cavity will be -q
Answer:
The speed of the plank relative to the ice is:
![v_{p}=-0.33\: m/s](https://tex.z-dn.net/?f=v_%7Bp%7D%3D-0.33%5C%3A%20m%2Fs)
Explanation:
Here we can use momentum conservation. Do not forget it is relative to the ice.
(1)
Where:
- m(g) is the mass of the girl
- m(p) is the mass of the plank
- v(g) is the speed of the girl
- v(p) is the speed of the plank
Now, as we have relative velocities, we have:
(2)
v(g/b) is the speed of the girl relative to the plank
Solving the system of equations (1) and (2)
![45v_{g}+168v_{p}=0](https://tex.z-dn.net/?f=45v_%7Bg%7D%2B168v_%7Bp%7D%3D0)
![v_{g}-v_{p}=1.55](https://tex.z-dn.net/?f=v_%7Bg%7D-v_%7Bp%7D%3D1.55)
![v_{p}=-0.33\: m/s](https://tex.z-dn.net/?f=v_%7Bp%7D%3D-0.33%5C%3A%20m%2Fs)
I hope it helps you!
Answer:
a) 1.20227 seconds
b) 0.98674 m
c) 7.3942875 m/s
Explanation:
t = Time taken
u = Initial velocity = 4.4 m/s
v = Final velocity
s = Displacement
a = Acceleration due to gravity = 9.81 m/s²
![v=u+at\\\Rightarrow 0=4.4-9.81\times t\\\Rightarrow \frac{-4.4}{-9.81}=t\\\Rightarrow t=0.44852\ s](https://tex.z-dn.net/?f=v%3Du%2Bat%5C%5C%5CRightarrow%200%3D4.4-9.81%5Ctimes%20t%5C%5C%5CRightarrow%20%5Cfrac%7B-4.4%7D%7B-9.81%7D%3Dt%5C%5C%5CRightarrow%20t%3D0.44852%5C%20s)
![s=ut+\frac{1}{2}at^2\\\Rightarrow s=4.4\times 0.44852+\frac{1}{2}\times -9.81\times 0.44852^2\\\Rightarrow s=0.98674\ m](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2%5C%5C%5CRightarrow%20s%3D4.4%5Ctimes%200.44852%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20-9.81%5Ctimes%200.44852%5E2%5C%5C%5CRightarrow%20s%3D0.98674%5C%20m)
b) Her highest height above the board is 0.98674 m
Total height she would fall is 0.98674+1.8 = 2.78674 m
![s=ut+\frac{1}{2}at^2\\\Rightarrow 2.78674=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.78674\times 2}{9.81}}\\\Rightarrow t=0.75375\ s](https://tex.z-dn.net/?f=s%3Dut%2B%5Cfrac%7B1%7D%7B2%7Dat%5E2%5C%5C%5CRightarrow%202.78674%3D0t%2B%5Cfrac%7B1%7D%7B2%7D%5Ctimes%209.81%5Ctimes%20t%5E2%5C%5C%5CRightarrow%20t%3D%5Csqrt%7B%5Cfrac%7B2.78674%5Ctimes%202%7D%7B9.81%7D%7D%5C%5C%5CRightarrow%20t%3D0.75375%5C%20s)
a) Her feet are in the air for 0.75375+0.44852 = 1.20227 seconds
![v=u+at\\\Rightarrow v=0+9.81\times 0.75375\\\Rightarrow v=7.3942875\ m/s](https://tex.z-dn.net/?f=v%3Du%2Bat%5C%5C%5CRightarrow%20v%3D0%2B9.81%5Ctimes%200.75375%5C%5C%5CRightarrow%20v%3D7.3942875%5C%20m%2Fs)
c) Her velocity when her feet hit the water is 7.3942875 m/s
Explanation:
The changes can be made in airplane longitudinal control to maintain altitude while the airspeed is being decreased is
We can increase the angle of attack this would compensate for the decreasing lift. As the angle of attack directly controls the distribution of pressure on the wings. Moreover, increase in angle of attack will also cause the drag to increase.