Mechanical and electrical
Answer:
d) -4.0
Explanation:
The magnification of a lens is given by

where
M is the magnification
q is the distance of the image from the lens
p is the distance of the object from the lens
In this problem, we have
p = 50 cm is the distance of the object from the lens
q = 250 cm - 50 cm is the distance of the image from the lens (because the image is 250 cm from the obejct
Also, q is positive since the image is real
So, the magnification is

Kepler's third law is used to determine the relationship between the orbital period of a planet and the radius of the planet.
The distance of the earth from the sun is
.
<h3>
What is Kepler's third law?</h3>
Kepler's Third Law states that the square of the orbital period of a planet is directly proportional to the cube of the radius of their orbits. It means that the period for a planet to orbit the Sun increases rapidly with the radius of its orbit.

Given that Mars’s orbital period T is 687 days, and Mars’s distance from the Sun R is 2.279 × 10^11 m.
By using Kepler's third law, this can be written as,


Substituting the values, we get the value of constant k for mars.


The value of constant k is the same for Earth as well, also we know that the orbital period for Earth is 365 days. So the R is calculated as given below.



Hence we can conclude that the distance of the earth from the sun is
.
To know more about Kepler's third law, follow the link given below.
brainly.com/question/7783290.
We can use the equation vf (the final velocity) =vi (the initial velocity) +at (aceleration times time)
We know the final velocity 100m/s, the initial velocity 0, and the acceleration (gravity) 9.8m/s^2. So, 100=0+9.8t. t=100/9.8
Answer:
deep scattering layer
Explanation:
I hope this is right, I kind of learned this like a year ago