Power = (voltage) x (current). The motor consumes (12)x(206)=2,472 watts. Some of it is dissipated as heat, but most of it is used to do useful work by turning the engine over to make it start.
Answer:
Kinetic energy = (1/2) (mass) (speed²)
Original KE = (1/2) (1430 kg) (7.5 m/s)² = 40,218.75 joules
Final KE = (1/2) (1430 kg) (11.0 m/s)² = 86,515 joules
Work done during the acceleration = (40218.75 - 86515) = 46,296.25 joules
Power = work/time = 46,296.25 joules / 9.3 sec = 4,978.1 watts .
Explanation:
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The relationship between the masses of the object and the gravitational force between them is a direct relationship
Explanation:
The gravitational force between two objects is given by the equation:
where
is the gravitational constant
m1, m2 are the masses of the two objects
r is the separation between them
We observe that:
- The gravitational force is proportional to the masses of the two objects, m1 and m2, so if the masses increase, the force will increase as well (so, this is a direct relationship)
- The gravitational force is inversely proportional to the square of the separation between the objects, so if the distance is increased, the force will decrease (so, this is an inverse relationship)
Learn more about gravitational force here:
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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:
