Answer:
1. They both uses same energy
2. The 6 kg ball requires more power than 3kg ball
Explanation:
Sample 1
m = 3kg
g= 10m/s^2
h = 2m
t = 2secs
W = mgh = 3 x 10 x 2 = 60J
P= w/t = 60/2 = 30watts
Sample 2
m = 6kg
g= 10m/s^2
h = 1m
t = 1sec
W = mgh = 6 x 10 x 1 = 60J
P= w/t = 60/1 = 60watts
They both uses same energy but different power. The 6 kg ball requires more power than 3kg ball
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
Answer:
The system prevents a consumer from accruing debt via electricity use, as it only allows the customer to use electricity which has been paid for upfront.
this gives the advantage of not allowing the consumer to rack up debt
the disadvantage for the consumer comes when they cannot afford to prepay in a time of financial difficulty - as the system now renders them as having no electricity as well as no money
I believe the balloon would shrink from lack of oxygen.<span />
Answer:
P = n P₀ 4.9 10¹⁴ Pa
Explanation:
The radiation pressure for full absorption is
P = S / c
Where S the pointing vector, which is equal to the intensity of the beam that is defined as the energy per unit area per unit time
The energy of the protons can be calculated
Em = K = ½ m v²
Area
A = π r²
Intensity is
I = n ½ m v² / π r²
I = ½ n m /π v² / r²
We replace
S = U / t A
S = ½ n m /π v² / r² Δt
The pressure is
P = 1/c (½ n m /π (v / r Δt)²2
Δt = 45 10⁻⁹ s
P = n [½ m /πc (v/r)²] 4.9 10¹⁴
The amount in square brackets is the pressure that a proton creates, which is why it is useful
P = n P₀ 4.9 10¹⁴ Pa
Where Po is the pressure created by a proton
