Answer:
t = 444.125 sec
Explanation:
Given data:
V = 24 volt
I = 0.1 ampere
mass of water mw = 51 gm
cr = 4.18 J/gm degree K^-1
mass of resistor = 8 gm
cr = 3.7 J/gm degree K^-1
we know that power is given as
Power P = VI
But P =E/t
so equating both side we have

solving for t


t = 444.125 sec
Answer:
it is double the temperature change of iron
Answer:
a. 11 m/s at 76° with respect to the original direction of the lighter car.
Explanation:
In this exercise, since both cars make a right angle, let's assume that the lighter car only has a horizontal velocity component (vx) and that the heavier one only has a vertical velocity component (vy). The final velocities for both components for the system can be determined as:

Assume that the lighter car has a 1kg mass and that the heavier car has a 4 kg mass.

The magnitude of the final velocity of the wreck can be found as:
![v_{f}^{2}= v_{fx}^{2}+ v_{fy}^{2}\\v_{f}=\sqrt[]{2.6^{2} + 10.4^{2}} \\v_{f}= 10.72](https://tex.z-dn.net/?f=v_%7Bf%7D%5E%7B2%7D%3D%20v_%7Bfx%7D%5E%7B2%7D%2B%20v_%7Bfy%7D%5E%7B2%7D%5C%5Cv_%7Bf%7D%3D%5Csqrt%5B%5D%7B2.6%5E%7B2%7D%20%2B%2010.4%5E%7B2%7D%7D%20%5C%5Cv_%7Bf%7D%3D%2010.72)
The final velocity has an intensity of roughly 11 m/s
As for the angle, it can be determined in respect to the lighter car (x axis) as follows:

Therefore, the wreck has a velocity with an intensity of 11 m/s at 76° with respect to the original direction of the lighter car.
Answer:
42.9 μF
Explanation:
V = 3.50 V, Q = 150 μC
C = Q/V = 150/3.50 μF = 42.9 μF