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:
2. You must be able to precisely measure variations in the star's brightness with time.
5. As seen from Earth, the planet's orbit must be seen nearly edge–on (in the plane of our line-of-sight).
6. You must repeatedly obtain spectra of the star that the planet orbits.
Explanation:
The transit method is a very important and effective tool for discovering new exoplanets (the planets orbiting other stars out of the solar system). In this method the stars are observed for a long duration. When the exoplanet will cross in front of theses stars as seen from Earth, the brightness of the star will dip. To observe this dip following conditions must be met:
1. The orbit of the planet should be co-planar with the plane of our line of sight. Then only its transition can be observed.
2. The brightness of the star must be observed precisely as the period of transit can be less than a second as seen from Earth. Also the dip in brightness depends on the size of the planet. If the planet is not that big the intensity dip will be very less.
3. The spectrum of the star needs to be studied and observe during the transit and normally to find out the details about the planets.
4. Also, the orbital period should be less than the period of observation for the transit to occur at least once.
answer
so unit of velocity is m/s
displacement=600m
5minutes should be converted to seconds
5×60=300 seconds
so,
velocity= displacement÷time
= 600m ÷300s
=2m/s or 2ms^-1
Answer:
Time, t = 0.015 seconds.
Explanation:
Given the following data;
Mass, m = 0.2kg
Force, F = 200N
Initial velocity, u = 40m/s
Final velocity, v = 25m/s
To find the time;
Ft = m(v - u)
Time, t = m(v - u)/f
Substituting into the equation, we have;
Time, t = 0.2(25 - 40)/200
Time, t = 0.2(-15)/200
Time, t = 3/200
Time, t = 0.015 seconds.
Note: We ignored the negative sign because time can't be negative.