Answer:
C= 0.354 KJ/K
Explanation:
As we know that heat capacity is the product of mass and specific heat.Heat capacity is the measure of heat which any substance can retain a period os time.
Here given that specific heat(c) = 1.77 KJ/kg.k
Mass m= 200 gm = 0.2 kg
So the heat capacity C = m.c
C= 0.2 x 1.77 KJ/kg
C= 0.354 KJ/K
So the heat capacity 0.354 KJ/K.
Answer:
If you push horizontally with a small force, static friction establishes an equal and opposite force that keeps the book at rest. As you push harder, the static friction force increases to match the force. Eventually maximum static friction force is exceeded and the book moves.
Explanation:
The power rating on the motor is the maximum power it's ABLE to deliver. It can run with LESS power output than the rating, but if you try to run it with MORE than it's rated, the motor will overheat and eventually burn out.
" 10 watts " means " 10 Joules of enrgy per second ".
If the motor is operated at its full maximum rated capacity, then
(500 joules) / (10 joules/sec) = <em>50 seconds</em>
Answer:
1210 N/m between the inner and outer edges (repulsive)
252.59 N/m between inner and line charges (attractive)
1210+252.59=1462.59 N/m effective force/metre on the inner edge
Explanation:
f=k*Qq/r^2 was applied, the charges are concentrated at the edges of a ring both inner and the outer were shape edges with uniform charge densities 6.6 micro C/m of distance 4.5-2.7=1.8 cm
Answer:
R = 2481 Ω
L= 1.67 H
Explanation:
(a) We have an inductor L which has an internal resistance of R. The inductor is connected to a battery with an emf of E = 12.0 V. So this circuit is equivalent to a simple RL circuit. It is given that the current is 4.86 mA at 0.725 ms after the connection is completed and is 6.45 mA after a long time. First we need to find the resistance of the inductor. The current flowing in an RL circuit is given by
i = E/R(1 -e^(-R/L)*t) (1)
at t --> ∞ the current is the maximum, that is,
i_max = E/R
solve for R and substitute to get,
R= E/i_max
R = 2481 Ω
(b) To find the inductance we will use i(t = 0.940 ms) = 4.86 mA, solve (1) for L as,
Rt/L = - In (1 - i/i_max
)
Or,
L = - Rt/In (1 - i/i_max
)
substitute with the givens to get,
L = -(2481 Si) (9.40 x 10-4 s)/ In (1 - 4.86/6.45
)
L= 1.67 H
<u><em>note :</em></u>
<u><em>error maybe in calculation but method is correct</em></u>