the amount of charge stored per volt
Answer: 30.34m/s
Explanation:
The sum of forces in the y direction 0 = N cos 28 - μN sin28 - mg
Sum of forces in the x direction
mv²/r = N sin 28 + μN cos 28
mv²/r = N(sin 28 + μcos 28)
Thus,
mv²/r = mg [(sin 28 + μ cos 28)/(cos 28 - μ sin 28)]
v²/r = g [(sin 28 + μ cos 28)/(cos 28 - μ sin 28)]
v²/36 = 9.8 [(0.4695 + 0.87*0.8829) - (0.8829 - 0.87*0.4695)]
v²/36 = 9.8 [(0.4695 + 0.7681) / (0.8829 - 0.4085)]
v²/36 = 9.8 (1.2376/0.4744)
v²/36 = 9.8 * 2.6088
v²/36 = 25.57
v² = 920.52
v = 30.34m/s
Answer:
E = 13.2 kWh
, Cost = $ 10.8
Explanation:
We can look for the consumed energy from the expression of the power
P = W / t
The work is equal to the variation of the kinetic energy, for which
P = E / t
E = P t
look for the energy consumed in one day and multiply by the days of the month in the month
E = 110 4 30
E = 13200 W h
E = 13.2 kWh
the cost of this energy is
Cost = 0.9 12
Cost = $ 10.8
Answer:
240 ohms
Explanation:
From Ohms law we deduce that V=IR and making R the subject of the formula then R=V/I where R is resistance, I is current and V is coltage across. Substituting 120 V for V and 0.5 A for A then
R=120/0.5=240 Ohms
Alternatively, resistance is equal to voltage squared divided by watts hence 
The friction is 2.5N. The Net force is 10 N - 2.5 N .= 7.5 N.
acceleration = 7.5 / 5 = 1.5 m/s^2
