Answer:
more speed means that an object has more energy, now if an object's place is something such as a hill, the potential energy will increase meaning an object will have more speed and acceleration. this is because you have the earth's gravity helping you out when the object goes downhill, giving it the higher potential energy
Answer:
Part 1)
Boat A will win the race
Part 2)
Boat A will win the race by 48 km as the 2nd boat will reach the other end while boat A will just touches the finish line
Part 3)
average velocity must be zero
Explanation:
As we know that the distance moved by the boat is given as

now the time taken by the boat to move to and fro is given as



Time taken by Boat B to cover the distance


Part 1)
Boat A will win the race
Part 2)
Boat A will win the race by 48 km as the 2nd boat will reach the other end while boat A will just touches the finish line
Part 3)
Since the displacement of Boat A is zero
so average velocity must be zero
Answer:
a

b

Explanation:
From the question we are told that
The speed of the spaceship is 
Here c is the speed of light with value 
The length is 
The distance of the star for earth is 
The speed is 
Generally the from the length contraction equation we have that
![l = l_o \sqrt{1 -[\frac{v}{c } ]}](https://tex.z-dn.net/?f=l%20%20%3D%20%20l_o%20%20%5Csqrt%7B1%20-%5B%5Cfrac%7Bv%7D%7Bc%20%7D%20%5D%7D)
Now the when at rest the length is 
So



Considering b
Applying above equation
![l =l_o \sqrt{1 - [\frac{v}{c } ]}](https://tex.z-dn.net/?f=l%20%20%3Dl_o%20%5Csqrt%7B1%20-%20%20%5B%5Cfrac%7Bv%7D%7Bc%20%7D%20%5D%7D)
Here 
So



<span>temperature increases and molecular motion increases while shape becomes less defined.
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Answer:
Explanation:
The voltage of a disconnected charged capacitor increases when the plate area is decreased.
When plate area decreases , capacitance C decreases , but charge Q remains constant .
Q = C V where C is capacitance and V is voltage .
when C decreases , V increases for keeping Q constant .
So the statement is true.
The electric field is dependent on the charge density on the plates.
This statement is true .
The voltage of a connected charged capacitor remains the same when the plate area is decreased .
For a connected capacitor , V or voltage is constant which is equal to voltage of charging battery .
So the statement is true .