Answer:
It will take 15.55s for the police car to pass the SUV
Explanation:
We first have to establish that both the police car and the SUV will travel the same distance in the same amount of time. The police car is moving at constant velocity and the SUV is experiencing a deceleration. Thus we will use two distance fromulas (for constant and accelerated motions) with the same variable for t and x:
1. 
2. 
Since both cars will travel the same distance x, we can equal both formulas and solve for t:

We simplify the fraction present and rearrange for our formula so that it equals 0:

In the very last step we factored a common factor t. There is two possible solutions to the equation at
and:

What this means is that during the displacement of the police car and SUV, there will be two moments in time where they will be next to each other; at
(when the SUV passed the police car) and
(when the police car catches up to the SUV)
Answer:
Gamma-rays have the smallest wavelengths and the most energy of any other wave in the electromagnetic spectrum. These waves are generated by radioactive atoms and in nuclear explosions.
Explanation:
Gamma-rays can kill living cells, a fact which medicine uses to its advantage, using gamma-rays to kill cancerous cells.
Hope this helps!
Brain-LIst?
I think it's 3. within an outer arm
Answer:
-67,500 kgm/s
Explanation:
1300 * 20 + 1100 * (-85) = -67,500 kgm/s
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.