Answer:(a)10 ft/s
(b)4 ft/s
Explanation:
Given
height of light 
height of man

From diagram



differentiate both sides

Tip of shadow is moving at the rate of

(b)rate at which length of his shadow is changing
Length of shadow is 
differentiating w.r.t time


Answer:
fundamental formula for density = Kg/m³
Explanation:
density = mass/volume
Answer: Option A: The number of trees sampled.
The accuracy can be understood as how close is the measured value to the true value. The aim is to monitor the population size of the insect pest in a 50 square kilometer. Random trees are selected, and number of eggs and larvae are counted. So, the measured value would be closer to actual value when the number of trees sampled are increased. More the number of trees sampled, less would be the chances of error and the accuracy of the estimate would increase.
We need to find the volume of a spherical shell with a radius of
6.37 million meters and a thickness of 0.95 mile.
The technically correct way to do this is to find the volume of the
outside of the shell, then find the volume of the inside of the shell,
and subtract the inside volume from the outside volume. That's
the REAL way to do it.
But look. This 'shell' (the 0.95 mile of water) is only about 1530 meters thick,
on a sphere with a radius of 6.37 million meters. The depth of the water is like
0.024 percent of the radius ! There's not a whole lot of difference between the
sphere outside the water and the sphere inside it.
So I want to do this problem the easier way ... Let's say that the volume
of the water is going to be
(the surface area that it covers on the Earth)
times
(the thickness of the coating of water) .
The area of a sphere is 4 pi Radius² .
That's
(4 pi) x (6.37 x 10⁶ m)²
= (4 pi) x (40.58 x 10¹² m²)
We're only interested in 70% of the total surface area.
= (0.7) x (4 pi) x (40.58 x 10¹²) m²
= 3.57 x 10¹⁴ square meters of Earth's surface.
The volume of the water covering that area is
(the area) times (average depth of 0.95 mile) .
We have to change that 0.95 mile to meters.
The question reminds us that 1 mile = 1609 meters .
So the volume of the water is
(the area) times (0.95 x 1609 meters).
But we're not there yet. The question isn't asking for the volume.
It's asking for the mass of the water.
We're ready to get the volume in cubic meters.
We're supposed to know that each cubic meter is 1,000 liters,
and the mass of 1 liter of water is 1 kilogram.
So each cubic meter of volume is 1,000 kilograms of mass.
Now we're ready to dump all the numbers into the machine and
turn the crank. The mass of all this water will be
(the surface area) x (0.95 x 1609 meters) x (1,000 kg/m³)
= (3.57 x 10¹⁴ m²) x (1528.6 m) x (1,000 kg/m³)
= 5.457 x 10²⁰ kilograms .
This is my answer, and I'm stickin to it.
But ... just like all the other problems you get in high school, the
answer doesn't matter. The teacher doesn't need the answer,
and YOU don't need the answer. The reason you got this problem
for an assignment is to give you practice in HOW TO FIND the
answer ... how to plan what you're going to do with the problem,
and then how to carry it out.
I don't know how much effort you put into this problem, but somewhere
along the way, you chickened out and posted it on Brainly. So far, the
result of that decision was: The person who got all the practice was ME.
I got the good stuff, and all YOU got was the answer.
I hope my work is clear enough that you can go through it, and pick up
some of the good stuff for yourself.
To solve this problem it is necessary to apply the concepts related to momentum, momentum and Force. Mathematically the Impulse can be described as

Where,
F= Force
t= time
At the same time the moment can be described as a function of mass and velocity, that is

Where,
m = mass
v = Velocity
From equilibrium the impulse is equal to the momentum, therefore


PART A) Since the body ends at rest, we have the final speed is zero, so the momentum would be



Therefore the magnitude of the person's impulse is 1125Kg.m/s
PART B) From the equation obtained previously we have that the Force would be:



Therefore the magnitude of the average force the airbag exerts on the person is 45000N