<u>Answer:</u>
Option A is the correct answer.
<u>Explanation:</u>
Let the east point towards positive X-axis and north point towards positive Y-axis.
First walking 1.2 km north, displacement = 1.2 j km
Secondly 1.6 km east, displacement = 1.6 i km
Total displacement = (1.6 i + 1.2 j) km
Magnitude = 
Angle of resultant with positive X - axis = 
= 36.87⁰ east of north.
Given Information:
Magnetic field = B = 1×10⁻³ T
Frequency = f = 72.5 Hz
Diameter of cell = d = 7.60 µm = 7.60×10⁻⁶ m
Required Information:
Maximum Emf = ?
Answer:
Maximum Emf = 20.66×10⁻¹² volts
Explanation:
The maximum emf generated around the perimeter of a cell in a field is given by
Emf = BAωcos(ωt)
Where A is the area, B is the magnetic field and ω is frequency in rad/sec
For maximum emf cos(ωt) = 1
Emf = BAω
Area is given by
A = πr²
A = π(d/2)²
A = π(7.60×10⁻⁶/2)²
A = 45.36×10⁻¹² m²
We know that,
ω = 2πf
ω = 2π(72.5)
ω = 455.53 rad/sec
Finally, the emf is,
Emf = BAω
Emf = 1×10⁻³*45.36×10⁻¹²*455.53
Emf = 20.66×10⁻¹² volts
Therefore, the maximum emf generated around the perimeter of the cell is 20.66×10⁻¹² volts
The magnitude of the resultant is
√ (22² + 2.2²) = √ (484 + 4.84) = √488.84 = 22.11 m/s .
The direction of the resultant is
tan⁻¹(22N / 2.2E) = tan⁻¹(10) = 5.71° east of north .
Answer:
speed of light simulating traveling at the speed of light. Speed of light, speed at which light waves propagate through different materials. In particular, the value for the speed of light in a vacuum is now defined as exactly 299,792,458 metres per second
The speed of sound is the distance travelled per unit of time by a sound wave as it propagates through an elastic medium
Answer: 
Explanation:
According to the <u>Third Kepler’s Law</u> of Planetary motion:
(1)
Where;:
is the period of the satellite
is the Gravitational Constant and its value is 
is the mass of the Earth
is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).
On the other hand, the orbital velocity 
is given by:
(2)
Now, from (1) we can find
, in order to substitute this value in (2):
![a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BT%5E%7B2%7DGM%7D%7B4%5Cpi%7D%5E%7B2%7D%7D)
(3)
![a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%281.26%2810%29%5E%7B4%7Ds%29%5E%7B2%7D%286.674%2810%29%5E%7B-11%7D%5Cfrac%7Bm%5E%7B3%7D%7D%7Bkgs%5E%7B2%7D%7D%29%285.98%2810%29%5E%7B24%7Dkg%29%7D%7B4%5Cpi%7D%5E%7B2%7D%7D)
(4)
(5)
Substituting (5) in (2):
(6)
(7) This is the speed at which the satellite travels
