Answer:
(D) x = 93.8 m
Explanation:
v^2 = v0^2 + 2ax
(20 m/s)^2 = (5 m/s)^2 + 2(2 m/s^2)x
Solving for x,
x = 93.8 m
For the answer to the question above,
we can get the number of fringes by dividing (delta t) by the period of the light (Which is λ/c).
fringe = (delta t) / (λ/c)
We can find (delta t) with the equation:
delta t = [v^2(L1+L2)]/c^3
Derivation of this formula can be found in your physics text book. From here we find (delta t):
600,000^2 x (11+11) / [(3x10^8)^3] = 2.93x10^-13
2.93x10^-13/ (589x10^-9 / 3x10^8) = 149 fringes
This answer is correct but may seem large. That is because of your point of reference with the ether which is usually at rest with respect to the sun, making v = 3km/s.
Hi there!
We can begin by calculating the time taken to reach its highest point (when the vertical velocity = 0).
Remember to break the velocity into its vertical and horizontal components.
Thus:
0 = vi - at
0 = 16sin(33°) - 9.8(t)
9.8t = 16sin(33°)
t = .889 sec
Find the max height by plugging this time into the equation:
Δd = vit + 1/2at²
Δd = (16sin(33°))(.889) + 1/2(-9.8)(.889)²
Solve:
Δd = 7.747 - 3.873 = 3.8744 m
Right, as you mentioned in the comments, you find 
by plugging in the different values of 
.
For 
, we have
Similarly, for 
, you get
Answer:
<h2>25000 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question we have
force = 5000 × 5
We have the final answer as
<h3>25000 N</h3>
Hope this helps you
