A useful formula that gives the free-fall distance from rest in 'T' seconds:
D = (1/2 G) x (T²)
G = 9.81 m/s²
1/2 G = 4.905 m/s²
D (5 seconds) = (4.905 m/s²) x (5 sec)²
= (4.905 m/s²) x (25 sec²)
= 122.625 meters .
Since the tower-top is 100m above ground,
the depth of the well, to the top of the water,
accounts for the additional 22.625 meters.
My question is: How do you know exactly when the stone hit the water ?
You probably stood at the top of the well and listened for the
sound of the 'plop'. But it took some time after the stone hit
the water for the sound of the plop to come back up to you.
Well, can't you just subtract that time ? Yes, but you need
to know how much time to subtract. That depends on the
depth of the well ... which is exactly what you're trying to
determine, so you don't know it yet.
Oh well. That's a deep subject.
Answer:
5.49×10¯⁴ m²
Explanation:
From the question given above, the following data were obtained:
Distance (d) = 2.22×10¯⁴ m
Charge (Q) = 5.24×10¯⁹ C
Potential difference (V) = 240 V
Permittivity of free space (ε₀) = 8.85×10¯¹² F/m
Area (A) =?
Thus, the area of the plate can be obtained as follow:
Q = ε₀AV /d
5.24×10¯⁹ = 8.85×10¯¹² × A × 240 / 2.22×10¯⁴
5.24×10¯⁹ = 2.12×10¯⁹ × A / 2.22×10¯⁴
Cross multiply
2.12×10¯⁹ × A = 5.24×10¯⁹ × 2.22×10¯⁴
Divide both side by 2.12×10¯⁹
A = (5.24×10¯⁹ × 2.22×10¯⁴) / 2.12×10¯⁹
A = 5.49×10¯⁴ m²
Thus, the area of the plates is 5.49×10¯⁴ m².
Explanation:
Friction can't affect mass of a substance(considering for a substance whose mass doesn't change with time),rather it is the mass of an object which can affect friction variously. Let's take some example to understand the situation. So,higher the mass of the object,higher will be the frictional force.
Friction occurs because no surface is perfectly smooth. Rougher surfaces have more friction between them. Heavier objects also have more friction because they press together with greater force. Friction produces heat because it causes the molecules on rubbing surfaces to move faster and have more energy.
