Complete Question
Q. Two go-carts, A and B, race each other around a 1.0km track. Go-cart A travels at a constant speed of 20m/s. Go-cart B accelerates uniformly from rest at a rate of 0.333m/s^2. Which go-cart wins the race and by how much time?
Answer:
Go-cart A is faster
Explanation:
From the question we are told that
The length of the track is 
The speed of A is 
The uniform acceleration of B is 
Generally the time taken by go-cart A is mathematically represented as
=> 
=> 
Generally from kinematic equation we can evaluate the time taken by go-cart B as

given that go-cart B starts from rest u = 0 m/s
So

=>
=>
Comparing
we see that
is smaller so go-cart A is faster
Answer:
<em>1,378.9ms²</em>
Explanation:
Given the following
Distance S = 70.6m
Time t = 0.32secs
Initial velocity = 0m/s
Required
Acceleration
Using the equation of motion
S = ut+1/2at²
Substitute
70.6 = 0+1/2a(0.32)²
70.6 = 0.0512a
a = 70.6/0.0512
a = 1,378.9
<em>Hence the acceleration is 1,378.9ms²</em>
I believe your answer is TRUE!
Hope this helps!:)
Solution
Let a cell of emf E be connected across the entire length L of a potentiometer wire . Now , if the balance point is obtained at a length l during measurement of an unknown voltage
.
The balance point is not on the potentiometer wire - this statement means that
. In that case ,
l > L
V > E
Explanation:
It is given that,
An electron is released from rest in a weak electric field of, 
Vertical distance covered, 
We need to find the speed of the electron. Let its speed is v. Using third equation of motion as :

.............(1)
Electric force is
and force of gravity is
. As both forces are acting in downward direction. So, total force is:



Acceleration of the electron, 


Put the value of a in equation (1) as :


v = 0.010 m/s
So, the speed of the electron is 0.010 m/s. Hence, this is the required solution.