Answer:
The pressure on the floor is 12 Pascal(Pa).
Explanation:
Given,
Force (F) = 240N
Area (A) = 20 cm^2
Pressure (P) = ?
we know that,
P = F/A
= 240/ 20
= 12 Pa
Gravitational potential energy can be calculated using the formula <span>PE = m × g × h, where g is the gravitational acceleration and is constant hence the energy is dependent directly to mass and the height of the object. Hence more PE is registered when the object is heavier and/or at greater initial height. </span>
Given:
The balanced chemical reaction of the synthesis of phosphorus trichloride:
2P + 3Cl2 ===> 2PCl3
Initial amount of phosphorus = 15 grams
The amount of product produced from 15 grams of phosphorus:
15 grams / 31 g/mol * (2/2) = 66.46 grams PCl3
The amount of chlorine is 44.31 grams, nearest to 45 grams.
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.
Solution :
Let 
kg
m/s
Let 
and 
are the speeds of the disk 
and 
after the collision.
So applying conservation of momentum in the y-direction,
Therefore, the disk 2 have greater velocity and hence more kinetic energy after the collision.
Now applying conservation of momentum in the x-direction,
m/s
So, 
= 4.33 m/s
Therefore, speed of the disk 2 after collision is 4.33 m/s
