Answer:
2 a) it is less dense than the water
2 b) it is more dense than the water
3 a ping pong ball is hollow and less dense than the water so it quickly bounces up to the surface of the water
To solve this problem we will use the work theorem, for which we have that the Force applied on the object multiplied by the distance traveled by it, is equivalent to the total work. From the measurements obtained we have that the width and the top are 14ft and 7ft respectively. In turn, the bottom of the tank is 15ft. Although the weight of the liquid is not given we will assume this value of (Whose variable will remain modifiable until the end of the equations subsequently presented to facilitate the change of this, in case be different). Now the general expression for the integral of work would be given as
Basically under this expression we are making it difficult for the weight of the liquid multiplied by the area (Top and widht) under the integral of the liquid path to be equivalent to the total work done, then replacing
Therefore the total work in the system is
Answer:
Explanation:
To solve this problem we can use the Gauss' Theorem
Hence, we have:
where QN is the total net charge inside the Gaussian surface, r is the point where we are going to compute E and ε0 is the dielectric permitivity. For each value of r we have to take into account what is the net charge inside the Gaussian surface.
a) r=4.80m (r>R2)
QN=+2.50 μC+2.70 μC = 5.2 μC
b) r=0.70m (R1<r<R2)
QN=+2.50 μC
c) r=0.210 (r<R1)
Inside the spherical shell of radius R1 the net charge is zero. Hence
E=0N/C
- For the calculation of the potential we have
Thus, we compute the potential by using the net charge of the Gaussian surface
d) r=0.210 (r<R1)
Inside the spherical shell the net charge is zero, thus
E=0N/C
e) r=1.40m (R1<r<=R2)
In this case we take the net charge from the first spherical shell
QN=+2.50 μC
f) r=0.70m
QN=+2.50 μC
V=3.164*10^{4}Nm/C
g) r=0.52
QN=0
V=0
h) r=0.2
QN=0
V=0
HOPE THIS HELPS!!