Answer:
The water is flowing at the rate of 28.04 m/s.
Explanation:
Given;
Height of sea water, z₁ = 10.5 m
gauge pressure,
= 2.95 atm
Atmospheric pressure,
= 101325 Pa
To determine the speed of the water, apply Bernoulli's equation;

where;
P₁ = 
P₂ = 
v₁ = 0
z₂ = 0
Substitute in these values and the Bernoulli's equation will reduce to;

where;
is the density of seawater = 1030 kg/m³

Therefore, the water is flowing at the rate of 28.04 m/s.
Answer:
Q = 47.06 degrees
Explanation:
Given:
- The transmitted intensity I = 0.464 I_o
- Incident Intensity I = I_o
Find:
What angle should the principle axis make with respect to the incident polarization
Solution:
- The relation of transmitted Intensity I to to the incident intensity I_o on a plane paper with its principle axis is given by:
I = I_o * cos^2 (Q)
- Where Q is the angle between the Incident polarized Light and its angle with the principle axis. Hence, Using the relation given above:
Q = cos ^-1 (sqrt (I / I_o))
- Plug the values in:
Q = cos^-1 ( sqrt (0.464))
Q = cos^-1 (0.6811754546)
Q = 47.06 degrees
Answer:
a
Explanation:
as the copper wire is very dangerous so now if these two thing happens then it would easily help the current flows through it so it might be a little bit easy for the current to flow through it
Answer:
<h2>The pin's final velocity is 5m/s</h2>
Explanation:
Step one:
given data
mass of ball m1=5kg
initial velocity of ball u1=10m/s
mass of pin m2=2kg
initial velocity of pin u2= 0m/s
final velocity of ball v2=8m/s
final velocity of pin v2=?
Step two:
The expression for elastic collision is given as
m1u1+m2u2=m1v1+m2v2
substituting we have
5*10+2*0=5*8+2*v2
50+0=40+2v2
50-40=2v2
10=2v2
divide both sides by 2
v2=10/2
v2=5m/s
The pin's final velocity is 5m/s