The kinetic energy would be 53,775J:)
m = mass of the car moving in horizontal circle = 1750 kg
v = Constant speed of the car moving in the horizontal circle = 15 m/s
r = radius of the horizontal circular track traced by the car = 45.0 m
F = magnitude of the centripetal force acting on the car
To move in a circle . centripetal force is required which is given as
F = m v²/r
inserting the above values in the formula
F = (1750) (15)²/(45)
F = (1750) (225)/(45)
F = 1750 x 5
F = 8750 N
Answer:
Thermal conduction is the transfer of internal energy by microscopic collisions of particles and movement of electrons within a body. The colliding particles, which include molecules, atoms and electrons, transfer disorganized microscopic kinetic and potential energy, when joined known as internal energy.
Explanation:
sana makatulong
Answer : The significant digit is 6
Explanation :
Multiply 
by 
Now, on multiplying
Now, the significant digit is 6.
Hence, this is the required solution.
Answer:
The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)
Explanation:
The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. 
, the amount of charge stored in this capacitor, will stay the same.
The formula 
relates the electric potential across a capacitor to:
- , the charge stored in the capacitor, and
, the capacitance of this capacitor.
While stays the same, moving the two plates apart could affect the potential 
by changing the capacitance of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:
,
where
is the permittivity of the material between the two plates.
is the area of each of the two plates.
is the distance between the two plates.
Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of . Neither will that change the area of the two plates.
However, as (the distance between the two plates) increases, the value of will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.
On the other hand, the formula can be rewritten as:
.
The value of (charge stored in this capacitor) stays the same. As the value of becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.
