Answer:
It is calculated by dividing Resistance, R, by Inductive reactance, XL.
Explanation:
Q is called the Q factor of a resonance circuit. In a parallel resonance circuit, it is calculated by finding the ratio of the power stored in the circuit to the power distributed in the circuit. It is a way of measuring the quality of a circuit or how effective the circuit is.
Q factor is the inverse in the resonance series circuit.
Q factor of a resonance parallel circuit,
<h3>
Q = R/XL</h3>
R = Resistance
XL = Inductive reactance
In this problem,
Applied force(F) = 10 N
The object’s mass (m) is 5 kg.
Having said that,
An object’s force is equal to the product of its mass and the acceleration it experiences as a result of the applied force.
i.e., Mass + Acceleration = Force (a)
F= m×a
Therefore,
A= F÷m
A= (10÷5) m/sec²
A= 2 m/sec²
Consequently, the object’s acceleration,
A=2 m/sec²
Concept of force and acceleration:
This states that the rate of velocity change of an object is directly proportional to the applied force and moves in the direction of the applied force.
It can be expressed mathematically as force (N) = mass (kg) x acceleration (m/s2). Therefore, an object with constant mass will accelerate in direct proportion to the applied force.
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It transfers the electrical current from the point of contact on the lightning rod into the ground.
To begin with, we can use the formula that links frequency, wavelength and velocity.
Because you already have the wavelength and the frequency, you just need to solve for velocity. You can do this by multiplying each side of the equation by frequency.
Therefore, 400 x 2.5 = 1000m/s.
Hope this helps :)
Answer:
8000J
Explanation:
The kinetic energy of the car lost during breaking are converted to thermal energy and are gained by the brakes.
Kinetic energy loss by car = thermal energy gained by brakes.
∆K.E = ∆T.E ....1
The Kinetic energy loss by car can be expressed as;
∆K.E = K.E1 - K.E2
Initial K.E = K.E1 = 10000J
Final K.E = K.E2 = 2000J
∆K.E= 10000J - 2000J = 8000J
From equation 1,
∆K.E = ∆T.E
∆T.E = 8,000J
thermal energy gain by brakes = 8,000J
