Answer:
A. fuel mileage and longevity
Explanation:
For a person purchasing a car, car longevity is one of the main concern. They are also interested in many things such as maximum mileage and service life.
By properly monitoring and assessing few measures one can maintain the efficiency and longevity of the car. One such thing is by monitoring the liquid levels of the car. Certain liquids like the coolant or radiator water level should be well maintain in proper level in order to run the car economically.
Thus by doing this, one can optimize the car's longevity and the fuel mileage.
Hence the correct option is (A).
<em>Soil can be described as the C. Loose covering of weathered rocks and decaying organic matter.</em>
Answer:
n = 2.0686
Explanation:
When an unpolarized ray of light is reflected on a surface, the reflected ray is partially polarized, complete polarization occurs when it is true that between the transmitted and reflected ray one has 90, the relationship is
n = so tea
let's calculate
n = tan 64.2
n = 2.0686
We will apply the concepts related to Newton's second law. At the same time we will convert everything to the system of international units.
The values of the velocities are,
We know that the acceleration is equivalent to the change of the speed in a certain time therefore
Now applying the Newton's second law we have,
Therefore the approximate magnitude is 8516.36N
Answer:
I_v = 2,700 W / m^2
I_m = 610 W / m^2
I_s = 16 W / m^2
Explanation:
Given:
- The Power of EM waves emitted by Sun P_s = 4.0*10^26 W
- Radius of Venus r_v = 1.08 * 10^11 m
- Radius of Mars r_m = 2.28 * 10^11 m
- Radius of Saturn r_s = 1.43 * 10^12 m
Find:
Determine the intensity of electromagnetic waves from the sun just outside the atmospheres of (a) Venus, (b) Mars, and (c) Saturn.
Solution:
- We know that Power is related to intensity and surface area of an object follows:
I = P / 4*pi*r^2
Where, A is the surface area of a sphere models the atmosphere around the planets.
a)
- The intensity at the surface of Venus is calculated as:
I_v = P_s / 4*pi*r^2_v
I_v = 4.0*10^26 / 4*pi*(1.08*10^11)^2
I_v = 2,700 W / m^2
b)
- The intensity at the surface of Mars is calculated as:
I_m = P_s / 4*pi*r^2_m
I_m = 4.0*10^26 / 4*pi*(2.28*10^11)^2
I_m = 610 W / m^2
c)
- The intensity at the surface of Saturn is calculated as:
I_s = P_s / 4*pi*r^2_s
I_s = 4.0*10^26 / 4*pi*(1.43*10^12)^2
I_s = 16 W / m^2
