Answer:
B
Explanation:
<u>for A,</u> transfer means energy change, from one form to another but remains unchanged.
But if we see, for energy lost to surroundings may be changed l.
<u>F</u><u>o</u><u>r</u><u> </u><u>B</u><u> </u>, Dissipation is the process by which energy is done against dissipative forces, eg air resistance, viscosity.
<u>F</u><u>o</u><u>r</u><u> </u><u>C</u><u>,</u><u> </u> Not that every energy is dissipated, like kinetic and potential energy. But some forms of energy are emitted like heat energy.
The electric flux through the cylinder due to this infinte line is 1 355 x 10' N m²/C, The electric flux through the cylinder is independent of the radius of the cylinder. So, the electric flux through the cylinder is the same as in part (a) which is equal to is 1.355 10 Nm C and The electric flux through the cylinder due to this infinte line is 2.711 x 10' N m² C
Given an imaginary cylinder with a radius of r = 0.185 m and a length of l = 0.440 m that has an infinite line of positive charge running along its axis. The charge per unit length on the line is λ = 4.50 μC/m.
Solve for the electric flux through the cylinder due to this infinte line:
As below equation mentions, the electric flux for a uniform electric filed is given by:
Фₓ = EA cosΦ
Φₓ = λLcosΦ)/ε
Φₓ = (3 x 10^(-6) x 0.4 x cos0)/(8.854 x 10^(-12))
Φₓ = 1.355 x 10^(5) Nm²/C
Hence the electric flux through the cylinder due to this infinte line is 1 355 x 10' N m²/C, The electric flux through the cylinder is independent of the radius of the cylinder. So, the electric flux through the cylinder is the same as in part (a) which is equal to is 1.355 10 Nm C and The electric flux through the cylinder due to this infinte line is 2.711 x 10' N m² C
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Answer
given,
mass of the ball = 300 g
= 0.3 Kg
maximum altitude of the ball = 10.1 m
Potential energy of the softball:
PE = m g h
PE = 0.3 x 9.8 x 10.1
PE = 29.69 J
Maximum kinetic energy at which the ball is released from hand to reach height of 10.1 m is 29.69 J.
Kinetic energy at which ball returns to your hand is equal to 29.69 J.
speed of the ball
v² = 197.93
v = 14.069 m/s
Answer:t=0.54 s
Explanation:
Given
Jacob is traveling 5 m/s in North direction
Jacob throw a ball with a in south direction with a velocity of 5 m/s
Ball is thrown in opposite direction of motion of car therefore it seems as if it is dropped from car as its net horizontal velocity is 5-5=0
Time taken by ball to reach ground
t=0.54 s
Motion of ball will be straight line
Answer:
A. 23.9
B.22.9
C. The levels will be equal
D. Obviously that will be to maintain atmospheric pressure
Explanation:
For mercury the pressure in both tubes at R is same so
P_left = P_right
Thus
=>>>Po + rho_t x g x (5 + R - 1.5) = Po + rho_ m x g x R
rho_t x g x (5 + R - 1.5) = rho_m x g xR
rho_t x (3.5 + R) = rho_m x R
3.5 + R = (rho_m/rho_t) x R
3.5 + R = (13560/867) x R
3.5 + R = 15.64 x R
R x (15.64 - 1) = 3.5
R = 3.5/14.64
= 0.239 m
= 23.9 cm this is for Mercury
ii)water
similarly,
3.5 + R = (rho_w/rho_t) x R
3.5 + R = (1000/867) x R
3.5 + R = 1.153 x R
R X (1.153 - 1) = 3.5
R = 3.5/0.153
= 22.9m for water