V(voltage) = I(current)R(resistance)
substitute in the values
V = 15 * 0.10
V = 1.5 volts
Answer:
If by 1.5 MJ you mean 1.5E6 Joules then
W = P t = power X time
W / t = P power
P = 1.5E6 J / 600 sec = 2500 J / s
P = I V
a) I = 2500 J/s / (240 J/c) = 10.4 C / sec = 10.4 amps
b) Q = I t = 10.4 C / sec * 300 sec = 3120 Coulombs
c) E = P * t = 2500 J / sec * 100 hr * 3600 sec / hr = 9.0E8 Joules
There are two torques t1 and t2 on the beam due to the weights, one torque t3 due to the weight of the beam, and one torque t4 due to the string.
You need to figure out t4 to know the tension in the string.
Since the whole thing is not moving t1 + t2 + t3 = t4.
torque t = r * F * sinФ = distance from axis of rotation * force * sin (∡ between r and F)
t1 =3.2 * 44g
t2 = 7 * 49g
t3 = 3.5 * 24g
t4 = t1 + t2 + t3 = 5570,118
The t4 also is given by:
t4 = r * T * sin Ф
r = 7
Ф = 32°
T: tension in the string
T = t4 / (r * sinФ)
T = t4 / (7 * sin(32°))
T = 1501,6 N
Answer:
619.8 N
Explanation:
The tension in the string provides the centripetal force that keeps the rock in circular motion, so we can write:
where
T is the tension
m is the mass of the rock
v is the speed
r is the radius of the circular path
At the beginning,
T = 50.4 N
v = 21.1 m/s
r = 2.51 m
So we can use the equation to find the mass of the rock:
Later, the radius of the string is decreased to
r' = 1.22 m
While the speed is increased to
v' = 51.6 m/s
Substituting these new data into the equation, we find the tension at which the string breaks:
