Power delivered = (energy delivered) / (time to deliver the energy)
Power delivered = (4,000 J) / (0.5 sec)
Power delivered = 8,000 watts
I'm a little surprised to learn that Electro draws his power from the mains. This is VERY good news for Spiderman ! It means that Spiderman can always avoid tangling with Electro ... all he has to do is stay farther away from Electro than the length of Electro's extension cord.
But OK. Let's assume that Electro draws it all from the mains. Then inevitably, there must be some loss in Electro's conversion process, between the outlet and his fingertips (or wherever he shoots his bolts from).
The efficiency of Electro's internal process is
<em>(power he shoots out) / (power he draws from the mains) </em>.
So, if he delivers energy toward his target at the rate of 8,000 watts, he must draw power from the mains at the rate of
<em>(8,000 watts) / (his internal efficiency) . </em>
Answer:
170 W
Explanation:
Applying
P = VI.................... Equation 1
Where P = Power generated in watt, V = Voltage supplied to the circuit, I = Current running through the circuit.
From the question,
Given: V = 17 V, I = 10 A
Substitute these values into equation 1
P = (17×10)
P = 170 Watt.
Hence the power generated is 170 W.
The right option is A. 170 W
given that
mass of ball = 0.095 kg
initial velocity of ball towards the wall = 40 m/s
final velocity of the ball after it rebound = 30 m/s
now change in momentum is given as



So change in momentum will be 6.65 kg m/s
Answer:

Explanation:
Impulse-Momentum relation:


We solve the equations in order to find the braking force:

Answer:
A vacuum
Explanation:
Sound waves are examples of mechanical waves. Mechanical waves are waves which are transmitted through the vibrations of the particles in a medium.
For example, sound waves in air consist of oscillations of the air particles, which vibrate back and forth (longitudinal wave) along the direction of propagation of the wave itself.
Given this definition of mechanical wave, we see that such a wave cannot propagate if there is no medium, because there are no particles that would oscillate. Therefore, among the choices given, the following one:
a vacuum
represent the only situation in which a sound wave cannot propagate through: in fact, there are no particles in a vacuum, so the oscillations cannot occur. In all other cases, instead, sound waves can propagate.