Answer:
-5 V
Explanation:
The charged particle (which is positively charged) moves from point A to B, and its kinetic energy increases: it means that the particle is following the direction of the field, so its potential energy is decreasing (because it's been converted into potential energy), therefore it is moving from a point at higher potential (A) to a point at lower potential (B). This means that the value
vb−va
is negative.
We can calculate the potential difference between the two points by using the law of conservation of energy:

where:
is the change in kinetic energy of the particle
is the charge of the particle
is the potential difference
Re-arranging the equation, we can find the value of the potential difference:

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
"When we do experiments it's a good idea to do multiple trials, that is, do the same experiment lots of times. When we do multiple trials of the same experiment, we can make sure that our results are consistent and not altered by random events. Multiple trials can be done at one time."
Answer:
(A). The order of the bright fringe is 6.
(B). The width of the bright fringe is 3.33 μm.
Explanation:
Given that,
Fringe width d = 0.5 mm
Wavelength = 589 nm
Distance of screen and slit D = 1.5 m
Distance of bright fringe y = 1 cm
(A) We need to calculate the order of the bright fringe
Using formula of wavelength


Put the value into the formula


(B). We need to calculate the width of the bright fringe
Using formula of width of fringe

Put the value in to the formula



Hence, (A). The order of the bright fringe is 6.
(B). The width of the bright fringe is 3.33 μm.
Answer:
He could jump 2.6 meters high.
Explanation:
Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:

With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.
Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:

This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.