Formulae for Kinetic energy is:
Kinetic Energy= 1/2xmassx(velocity)^2
For comparison we need to have same units,thus we convert 10g into Kg.
10g/1000=0.01Kg
Input the value of bullet in the formulae;
Kinetic Energy= 1/2x0.01kgx(400)^2
K.E=800J
Input value of the ball:
Kinetic Energy=1/2x80kgx(6.5)^2
K.E=1690J
Which means that th Energy of the ball is more than the bullet.
Answer:
a) + Q charge is inducce that compensates for the internal charge
b) There is no excess charge on the external face q_net = 0
c) E=0
Explanation:
Let's analyze the situation when a negative charge is placed inside the cavity, it repels the other negative charges, leaving the necessary positive charges to compensate for the -Q charge. The electrons that migrated to the outer part of the sphere, as it is connected to the ground, can pass to the earth and remain on the planet; therefore on the outside of the sphere the net charge remains zero.
With this analysis we can answer the specific questions
a) + Q charge is inducce that compensates for the internal charge
b) There is no excess charge on the external face q_net = 0
c) If we create a Gaussian surface on the outside of the sphere the net charge on the inside of this sphere is zero, therefore there is no electric field, on the outside
d) If it is very reasonable and this system configuration is called a Faraday Cage
e) We cannot apply this principle to gravity since there are no particles that repel, in all cases the attractive forces.
Ag on the periodic table is silver.
The atomic number for silver is 47, atomic symbol is Ag, and the atomic weight is 107.9.
Answer:
D. 21 ml
Explanation:
Since, the cylinder is marked and graduated in the intervals if 1 ml. Therefore, the values between two consecutive ml, such as between 30 ml and 31 ml can not be determined. Because, we do not have any scale in between the ml. So, the least count of this instrument is 1 ml. This graduated cylinder can give the answers to zero decimal places, accurately. And it can not determine any decimal value due to its graduating or the marking limitation. So, all the options given, contain a decimal value, except for the option D. In option D there is no decimal value, hence it is a correct answer.
D. <u>21 ml</u>
Answer:
The radius of the wetter area expands at a rate of 
milimeters per second when radius is 150 milimeters.
Explanation:
From Geometry we remember that area of a circle is described by this expression:
(Eq. 1)
Where:
- Radius of the circle, measured in milimeters.
- Area of the circle, measured in square milimeters.
Then, the rate of change of the area in time is derived by concept of rate of change, that is:
(Eq. 2)
Where:
- Rate of change of radius in time, measured in milimeters per second.
- Rate of change of area in time, measured in square milimeters per second.
Now the rate of change of radius in time is cleared within equation above:
If we know that 
and 
, then the rate of change of radius in time is:
![\frac{dr}{dt} = \left[\frac{1}{2\pi\cdot (150\,m)} \right] \cdot \left(4\,\frac{mm^{2}}{s} \right)](https://tex.z-dn.net/?f=%5Cfrac%7Bdr%7D%7Bdt%7D%20%3D%20%5Cleft%5B%5Cfrac%7B1%7D%7B2%5Cpi%5Ccdot%20%28150%5C%2Cm%29%7D%20%5Cright%5D%20%5Ccdot%20%5Cleft%284%5C%2C%5Cfrac%7Bmm%5E%7B2%7D%7D%7Bs%7D%20%5Cright%29)
The radius of the wetter area expands at a rate of milimeters per second when radius is 150 milimeters.
