The answer in the blank is that it is difficult to accelerate at decelerate the vehicle when it is on a fast speed because having a fast speed makes it difficult to adjust the meter as well as if you try to decelerate the vehicle, it could burn out the tires and engine as it is in the fast speed, in accelerating it, it could also be complicated because it would only make the car faster enough that you may no longer control of how to stop it.
This question is incomplete
Complete Question
Three equal point charges are held in place as shown in the figure below
If F1 is the force on q due to Q1 and F2 is the force on q due to Q2, how do F1 and F2 compare? Assume that n=2.
A) F1=2F2
B) F1=3F2
C) F1=4F2
D) F1=9F2
Answer:
D) F1=9F2
Explanation:
We are told in the question that there are three equal point charges.
q, Q1, Q2 ,
q = Q1 = Q2
From the diagram we see the distance between the points d
q to Q1 = d
Q1 to Q2 = nd
Assuming n = 2
= 2 × d = 2d
Sum of the two distances = d + 2d = 3d
F1 is the force on q due to Q1 and
F2 is the force on q due to Q2,
Since we have 3 equal point charges and a total sum of distance which is 3d
Hence,
F1 = 9F2
Explanation:
When Michelson-Morley apparatus is turned through
then position of two mirrors will be changed. The resultant path difference will be as follows.

Formula for change in fringe shift is as follows.
n = 

v = 
According to the given data change in fringe is n = 1. The data is Michelson and Morley experiment is as follows.
l = 11 m
c =
m/s
Hence, putting the given values into the above formula as follows.
v = 
= 
= 
Thus, we can conclude that velocity deduced is
.
Answer:
4 (please see the attached file)
Explanation:
While the angular speed (counterclockwise) remained constant, the angular acceleration was just zero.
So, the only force acting on the bug (parallel to the surface) was the centripetal force, producing a centripetal acceleration directed towards the center of the disk.
When the turntable started to spin faster and faster, this caused a change in the angular speed, represented by the appearance of an angular acceleration α.
This acceleration is related with the tangential acceleration, by this expression:
at = α*r
This acceleration, tangent to the disk (aiming in the same direction of the movement, which is counterclockwise, as showed in the pictures) adds vectorially with the centripetal force, giving a resultant like the one showed in the sketch Nº 4.
Answer:
The gravitational force exerted on the object is 75 N (answer D)
Explanation:
Hi there!
The gravitational force is calculated as follows:
F = m · g
Where:
F = force of gravity.
m = mass of the object.
g = acceleration due to gravity (unknown).
For a falling object moving in a straight line, its height at a given time can be calculated using the following equation:
y = y0 + v0 · t + 1/2 · a · t²
Where:
y = position at time t.
y0 = initial position.
v0 = initial velocity.
t = time.
g = acceleration due to gravity.
Let´s place the origin of the frame of reference at the point where the object is released so that y0 = 0. Let´s also consider the downward direction as negative.
Then, after 2 seconds, the height of the object will be -30 m:
y = y0 + v0 · t + 1/2 · g · t²
-30 m = 0 m + 0 m/s · 2 s + 1/2 · g · (2 s)²
-30 m = 1/2 · g · 4 s²
-30 m = 2 s ² · g
-30 m/2 s² = g
g = -15 m/s²
Then, the magnitude of the gravitational force will be:
F = m · g
F = 5 kg · 15 m/s²
F = 75 N
The gravitational force exerted on the object is 75 N (answer D)
Have a nice day!