Yes, you need to understand refraction, because refraction is when an object bends when going into a different medium, meaning, when you put the spear from air to water, it will bend the shape making it harder to spear the fish.
Hopefully that helps!
As simply put as I believe is possible, it's the immediate space-time co-oridinate in which the observer is not in motion. All measurements by the observer are in relation to that 'place'
Answer:
17.7 cm^3
Explanation:
depth, h = 120 m
density of water, d = 1000 kg/m^3
V1 = 1.4 cm^3
P1 = P0 + h x d x g
P2 = P0
where, P0 be the atmospheric pressure
Let V2 be the volume of the bubble at the surface of water.
P0 = 1.01 x 10^5 Pa
P1 = 1.01 x 10^5 + 120 x 1000 x 9.8 = 12.77 x 10^5 Pa
Use
P1 x V1 = P2 x V2
12.77 x 10^5 x 1.4 = 1.01 x 10^5 x V2
V2 = 17.7 cm^3
Thus, the volume of bubble at the surface of water is 17.7 cm^3.
Answer:
Explanation:
Given parameters:
Electrical energy = 18J
Quantity of charge Q = 12C
Unknown:
Voltage = ?
Solution:
The expression of electrical potential energy is given as:
Electrical potential energy = 
c v²
c is the quantity of charge
v is the voltage
18 = x 12 x v²
18 = 6v²
v² = 3
v = 1.7v
