The popular GPS devices that people use to find directions while driving use "Global Navigation Satellite System (GNSS)".
<u>Explanation:</u>
The umbrella term for all global satellite tracking systems is GNSS i.e Global Satellite Navigation System. This involves satellite constellations circulating over the surface of the earth and continuous signal transmission that allow users to evaluate their location.
A satellite array of 18–30 medium Earth Orbit (MEO) satellites distributed across several orbital planes typically achieves greater coverage for each network. The specific systems differ, but use > 50 ° orbital inclinations and approximately twelve hours orbital cycles.
The answer is the red sidelight on a powerboat should be visible from the front and from the left (port side).
What are Sidelights?
- There is various combinations of lights that must be used on a boat when it is dark, and these are:
- Sidelights: These lights are called combination lights and are red and green. The red sidelight must be visible from the port side and the green light indicates the right side (the starboard).
- Stern light: The stern light is seen at the back end of the vessel.
- Masthead Light: The masthead light is a white light that shines forwards and on all sides of the vessel. All powered vessels must use this light.
- All-Round white light: This light is the major light that is used to join the masthead light and the stern light. This single light is visible to all vessels from all directions.
- Thus, as a rule for a boat rider, he should show the vision of red light and it should be visible from the front and from the left (port side).
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Answer: 1.2m/s^2
Explanation: the force exerted on the car is 900N upwards
The mass of the car is 750kg
According to Newton's third law acceleration is proportional to force
F = ma
900 = 750a
a = 900/750
a = 1.2m/s^2
Answer:
Kf= 36 J
W(net) = 32 J
Explanation:
Given that
m = 2 kg
F= 4 N
t= 2 s
Initial velocity ,u= 2 m/s
We know that rate of change of linear momentum is called force.
F= dP/dt
F.t = ΔP
ΔP = Pf - Pi
ΔP = m v - m u
v= Final velocity
By putting the values
4 x 2 = 2 ( v - 2)
8 = 2 ( v - 2)
4 = v - 2
v= 6 m/s
The final kinetic energy Kf
Kf= 1/2 m v²
Kf= 0.5 x 2 x 6²
Kf= 36 J
Initial kinetic energy Ki
Ki = 1/2 m u²
Ki= 0.5 x 2 x 2²
Ki = 4 J
We know that net work is equal to the change in kinetic energy
W(net) = Kf - Ki
W(net) = 36 - 4
W(net) = 32 J
Answer:
![\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%7D%5B%5Cfrac%7B1%7D%7By%7D%28%5C%5Ey%29%20-%20%5Cfrac%7B1%7D%7Bx%7D%28%5C%5Ex%29%5D)
Explanation:
The electric field created by an infinitely long wire can be found by Gauss' Law.
For the electric field at point (x,y), the superposition of electric fields created by both lines should be calculated. The distance 'r' for the first wire is equal to 'y', and equal to 'x' for the second wire.
![\vec{E} = \vec{E}_1 + \vec{E}_2 = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) + \frac{-\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0 y}(\^y) - \frac{\lambda}{2\pi\epsilon_0 x}(\^x)\\\vec{E} = \frac{\lambda}{2\pi\epsilon_0}[\frac{1}{y}(\^y) - \frac{1}{x}(\^x)]](https://tex.z-dn.net/?f=%5Cvec%7BE%7D%20%3D%20%5Cvec%7BE%7D_1%20%2B%20%5Cvec%7BE%7D_2%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%20y%7D%28%5C%5Ey%29%20%2B%20%5Cfrac%7B-%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%20x%7D%28%5C%5Ex%29%5C%5C%5Cvec%7BE%7D%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%20y%7D%28%5C%5Ey%29%20-%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%20x%7D%28%5C%5Ex%29%5C%5C%5Cvec%7BE%7D%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%5Cpi%5Cepsilon_0%7D%5B%5Cfrac%7B1%7D%7By%7D%28%5C%5Ey%29%20-%20%5Cfrac%7B1%7D%7Bx%7D%28%5C%5Ex%29%5D)
