D !.!.!.!.!.!!.!.!.!.!.!.!.!.!.!.!.!.!.!.!.!.!.!!..!!..!
Answer:
ΔP.E = 6.48 x 10⁸ J
Explanation:
First we need to calculate the acceleration due to gravity on the surface of moon:
g = GM/R²
where,
g = acceleration due to gravity on the surface of moon = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
M = Mass of moon = 7.36 x 10²² kg
R = Radius of Moon = 1740 km = 1.74 x 10⁶ m
Therefore,
g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²
g = 2.82 m/s²
now the change in gravitational potential energy of rocket is calculated by:
ΔP.E = mgΔh
where,
ΔP.E = Change in Gravitational Potential Energy = ?
m = mass of rocket = 1090 kg
Δh = altitude = 211 km = 2.11 x 10⁵ m
Therefore,
ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)
<u>ΔP.E = 6.48 x 10⁸ J</u>
Answer:
ε = 6.617 V
Explanation:
We are given;
Number of turns; N = 40 turns
Diameter;D = 18cm = 0.18m
magnetic field; B = 0.65 T
Time;t = 0.1 s
The formula for the induced electric field(E.M.F) is given by;
ε = |-NAB/t|
A is area
ε is induced electric field
While N,B and t remain as earlier described.
Area = π(d²/4) = π(0.18²/4) = 0.02545
Thus;
ε = |-40 × 0.02545 × 0.65/0.1|
ε = 6.617 V
(we ignore the negative sign because we have to take the absolute value)
B. Energy
A power company charges its customers for electricity based upon B. Energy.
<h3>
Explanation:</h3>
Kilo-watt Hours (kWh) is the unit that measures the electricity consumption of customers. Since Power is defined as the rate at which electrical energy is transferred by an electrical circuit per unit time,
If energy is transmitted at a constant rate over a period of time, the total energy in kilowatt hours is the product of power in kilowatts(kW) and time in hours (h)
Explanation:
The time taken by a wave crest to travel a distance equal to the length of wave is known as wave period.
The relation between wave period and frequency is as follows.
T = \frac{1}{f}T=
f
1
where, T = time period
f = frequency
It is given that wave period is 18 seconds. Therefore, calculate the wave period as follows.
T = \frac{1}{f}T=
f
1
or, f = \frac{1}{T}f=
T
1
= \frac{1}{18 sec}
18sec
1
= 0.055 per second (1cycle per second = 1 Hertz)
or, f = 5.5 \times 10^{-2} hertz5.5×10 −2 hertz
<h3>Thus, we can conclude that the frequency of the wave is 5.5 \times 10^{-2} hertz5.5×10 −2 hertz .</h3>
