Here, we are required to determine the reaction of the power given to a circuit supplied with a constant voltage as the distance of the circuit decreases.
- In a circuit supposed with a <em>constant voltage</em>, as the resistance of the circuit decreases, the power given to the circuit Increases as evident in the formular: P = V²/R
The power given to a circuit is given by the formula;
Power, P = V × I
where, V = voltage supplied
and I = Current through the circuit.
However, from Ohm's law;
V = I × R
In essence, I = V/R
By substituting, I = V/R into the equation P = V × I
Therefore, we have;
P = V²/R.
Ultimately, if the circuit is supplied with a constant voltage, As the resistance of the circuit decreases, the power given to the circuit Increases.
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Answer:
Surface tension
Explanation:
When liquid rises against gravity in a capillary tube, the energy comes from surface tension.
This is because surface tension is the energy that's needed to increase the liquid surface area.
As a result of hydrogen bonding present in Water, it usually has high surface tension which makes it to possess a tough skin that can make it not to break despite high forces applied to it.
The liquid will be in contact with the capillary tube and as such experiences surface tension which in turn makes the capillary tube to experience an upward force that makes the liquid begin to rise up.
The more the liquid keeps rising, the more it gets to the point where the surface tension becomes balanced from the weight of the liquid.
Answer: 10 m/s
We're told the speed is constant, so it's not changing throughout the time period given to us. So throughout the entire interval, the speed is 10 m/s.
Answer:
The net force acting on the object is doubled while the mass of the object is held constant. What will be the new acceleration? An object has an acceleration of 12.0 m/s^2. The net force acting on the object is halved (decreased to one half its original value) while the mass of the object is held constant.
Answer:
v= 13 m/s
Explanation:
Velocity is defined as the derivative of displacement with respect to time
v= ds/dt
Known data
s(t) = 5t + 2t² : distance that the ball has rolled after t seconds
vi= 5 m/s : initial velocity
t= 2 s
Problem develoment
s(t) = 5t + 2t²
v= ds/dt= 5 + 4t : velocity of the ball in function of the time
We replace t =2 s in the equation of velocity
v= 5 + 4(2)
v= 13 m/s : velocity after 2 seconds
