Answer:
1.) h = 164.8 m
2.) U = 49.1 m/s
3.) t = 1.43 seconds
Explanation:
1.) A soccer ball is dropped from the top of a building. It takes 5.8 seconds to fall to the ground. The height of the building is...?
Since the soccer ball is dropped from the building, the initial velocity U will be equal to zero
Using second equation of motion
h = Ut + 1/2gt^2
Substitutes the time into the formula
h = 1/2 × 9.8 × 5.8^2
h = 164.8 m
2. The Falcon 9 launches to a height of 123 meters. What is its vertical initial velocity?
At maximum height final velocity = 0
Using the third law of motion
V^2 = U^2 - 2gH
0 = U^2 - 2 × 9.8 × 123
U^2 = 2410.8
U = 49.1 m/s
3. An apple falls from rest off a 10.m m tree. How long will it take before it hits the ground?
Since the apple fall from rest, the initial velocity U will be equal to zero
Using the second equation of motion,
h = Ut + 1/2gt^2
substitute all the parameters into the formula
10 = 1/2 × 9.8 × t^2
10 = 4.9t^2
t^2 = 10/4.9
t^2 = 2.04
t = 1.43 seconds
Answer:
i think D
hope this helps
let me know if i'm wrong i will change the answer
Explanation:
Answer: The answer is 333.3333 repeating
Explanation:
Divide the mass by the volume.
Answer:
Frequency, 
Explanation:
Given that,
Wavelength of the light,
We need to find the frequency of light. We know that light is an electromagnetic wave. It moves with the speed of light. So,
f is the frequency of light
So, the frequency of light is 
. Hence, this is the required solution.
Answer:
So, the force, F is the agent which provides the basic property of motion or rest to the body.While, the car is more obviously to have a mass, m and that the angle withe road or surface is also given which is normal to the road(i.e angle =90 degree). Then we say that lets say that the car is moving with the constant velocity of 20 m/sec and its kept unchanged by the car. So, we have the mass, m as 1 kg for the car and the value of the gravity we have the g=9.8 m/sec.
Now,
We have F=ma,
and a=v/t,
so we can have another equation for it as,
Now, providing the required data to, it; ∴t =2 sec,
F=(1)×(20/2),
- So, the car would be acting the force,F of about 10 N while the car is present on the lower region of the track.
