Answer:
(a) ε = 1373.8.
(b) The wingtip which is at higher potential.
Explanation:
(a) Finding the potential difference between the airplane wingtips.
Given the parameters
wingspan of the plane is = 18.0m
speed of the plane in north direction is = 70.0m/s
magnetic field of the earth is = 1.20μT
The potential difference is given as:
ε = Blv
where ε = potential difference of wingtips
B = magnetic field of earth
l = wingspan of airplane
v = speed of airplane
ε = 1.2 x 18.0 x 63.6
ε = 1373.8
(b) Which wingtip is at higher potential?
The wingtip which is at higher potential.
55 meters. If she started at 10 meters and ran 45 more, 10+45=55.
The time when the particle is at rest is at 1.63 s or 3.36 s.
The velocity is positive at when the time of motion is at
.
The total distance traveled in the first 10 seconds is 847 m.
<h3>When is a particle at rest?</h3>
- A particle is at rest when the initial velocity of the particle is zero.
The time when the particle is at rest is calculated as follows;
s(t) = 2t³ - 15t² + 33t + 17

The velocity is positive at when the time of motion is as follows;
.
The total distance traveled in the first 10 seconds is calculated as follows;

Learn more about motion of particles here: brainly.com/question/11066673
Incomplete question.The complete question is attached below as screenshot along with figure
Answer:

Force is repulsive
Explanation:
Given data
Current I₁=5.00A
Current I₂=2.00A
Length L=1.20 m
Radius r=0.400m
To find
Force F
Solution
As the force is repulsive because currents are in opposite direction
From repulsive force we know that:

Substitute the given values
Answer:
It corresponds to 1mm-10 mm range.
Explanation:
- Electromagnetic waves (such as the millimeter-wave radiation) travel at the speed of light, which is 3*10⁸ m/s in free space.
- As in any wave, there exists a fixed relationship between speed, frequency and wavelength, as follows:

- Replacing v= c=3*10⁸ m/s, and the extreme values of f (which are givens), in (1) and solving for λ, we can get the free-space wavelengths that correspond to the 30-300 GHz range, as follows:

