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:
the magnitude of the displacement after 5s is 137.31 m.
Explanation:
Given;
initial velocity of the projectile, u = 60 m/s
angle of projection, θ = 60°
time of motion, t = 5s
the vertical component of the velocity, 
The magnitude of the displacement after 5s is calculated as;
Therefore, the magnitude of the displacement after 5s is 137.31 m.
Answer:
0.117 m
Explanation:
First of all, we can find the wavelength of the wave in the problem, by using the wave equation:
where:
v = 350 m/s is the speed of the wave
f = 500 Hz is the frequency of the wave
is the wavelength
Solving for ,
This means that the distance between two consecutive points of the wave having a difference of phase of
is 0.7 m.
Here we want to find the distance between two points that have a difference of phase of
So, we can set up the following rule of three:
where d' is the distance we are looking for. Solving for d',
Answer: 8.1 x 10^24
Explanation:
I(t) = (0.6 A) e^(-t/6 hr)
I'll leave out units for neatness: I(t) = 0.6e^(-t/6)
If t is in seconds then since 1hr = 3600s: I(t) = 0.6e^(-t/(6 x 3600) ).
For neatness let k = 1/(6x3600) = 4.63x10^-5, then:
I(t) = 0.6e^(-kt)
Providing t is in seconds, total charge Q in coulombs is
Q= ∫ I(t).dt evaluated from t=0 to t=∞.
Q = ∫(0.6e^(-kt)
= (0.6/-k)e^(-kt) evaluated from t=0 to t=∞.
= -(0.6/k)[e^-∞ - e^-0]
= -0.6/k[0 - 1]
= 0.6/k
= 0.6/(4.63x10^-5)
= 12958 C
Since the magnitude of the charge on an electron = 1.6x10⁻¹⁹ C, the number of electrons is 12958/(1.6x10^-19) = 8.1x10^24 to two significant figures.
