When an object is acted on by an unbalanced force, then that object will accelerate.
Answer:
1) a substance made by mixing other substances together.
2) diverse in character or content.
3) of the same kind; alike.
4) a means of solving a problem or dealing with a difficult situation.
5) a homogeneous noncrystalline substance consisting of large molecules or ultramicroscopic particles of one substance dispersed through a second substance. Colloids include gels, sols, and emulsions; the particles do not settle, and cannot be separated out by ordinary filtering or centrifuging like those in a suspension.
Answer:
c. is more than that of the fluid.
Explanation:
This problem is based on the conservation of energy and the concept of thermal equilibrium
m= mass
s= specific heat
\DeltaT=change in temperature
let s1= specific heat of solid and s2= specific heat of liquid
then
Heat lost by solid= 
Heat gained by fluid=
Now heat gained = heat lost
therefore,
1000 S_2=800 S_1
S_1=1.25 S_2
so the specific heat of solid is more than that of the fluid.
Answer:
D
Explanation:
We know the formula for Work to be:
W = f * d
Where W is work done
f is force
d is the distance
A)
Work = 50
Distance = 50
So, Force is:
Force = 50/50 = 1
B)
Work = 400
Distance = 80
Force = 400/80 = 5
C)
Work = 365
Distance = 73
Force = 365/73 = 5
D)
Work = 144
Distance = 16
Force = 144/16 = 9
Hence, D is the situation in which the force applied is the greatest.
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.
