If a body p with a positive charge is placed in contact with a body q (initially uncharged), then the nature of charge gained by q must be positive, because rubbing an uncharged body with a charged body or placed in contact with a positive charged body, helps gain a charge to the uncharged body.
There are a variety of methods to charge an object. One method is known as induction. In the induction process, a charged object is brought near but not touched to a neutral conducting object.
Let's know, how a element gain positive charge?
A positive charge occurs when the number of protons exceeds the number of electrons. A positive charge may be created by adding protons to an atom or object with a neutral charge. A positive charge also can be created by removing electrons from a neutrally charged object.
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It depends on the steady-state frequency. At zero frequency an inductor behaves like an open circuit. As the frequency increases, the inductor acts more like an open circuit and a capacitator acts more like a short circuit
Answer:
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Explanation:
Answer:
The value of resistance of each resistor, R is 2.25 Ω
Explanation:
Given;
voltage across the three resistor, V = 1.5 V
power dissipated by the resistors, P = 3.00 W
the resistance of each resistor, = R
The effective resistance of the three resistors is given by;
R(effective) = R/3
Apply ohms law to determine the current delivered by the source;
V = IR
I = V/R
I = 3V/R
Also, power is calculated as;
P = IV
P = (3V/R) x V
P = 3V²/R
R = 3V² / P
R = (3 x 1.5²) / 3
R = 2.25 Ω
Therefore, the value of resistance of each resistor, R is 2.25 Ω
Answer:
Approximately
(rounded down,) assuming that
.
The number of repetitions would increase if efficiency increases.
Explanation:
Ensure that all quantities involved are in standard units:
Energy from the cookie (should be in joules,
):
.
Height of the weight (should be in meters,
):
.
Energy required to lift the weight by
without acceleration:
.
At an efficiency of
, the actual amount of energy required to raise this weight to that height would be:
.
Divide
by
to find the number of times this weight could be lifted up within that energy budget:
.
Increasing the efficiency (the denominator) would reduce the amount of energy input required to achieve the same amount of useful work. Thus, the same energy budget would allow this weight to be lifted up for more times.