Answer:
a) v = 1.19 m / s
, b) P₁ = 0.922 10⁵ Pa
Explanation:
1) Let's use the fluid continuity equation
Q = A v
The area of a circle is
A = π r2 = π d²/4
v = Q / A = Q 4 / pi d²
v = 0.006 4/π 0.08²
v = 1.19 m / s
2) write Bernoulli's equation, where point 1 is the bladder and point 2 is the urine exit point
P₁ + ½ rho v₁² + rho g y₁ = P₂ + ½ rho v₂² + rho g y₂
The exercise tell us
P₂ = 1.0013 105 Pa
v₁ = 0
y₁ = 1 m
y₂=0
Rho (water) = 1000 kg / m³
P₁ + rho y₁ = P₂ + ½ rho v₂²
P₁ = P₂ + ½ rho v₂² - rho g y₁
P₁ = 1.013 10⁵ + ½ 1000 (1.19)² - 1000 9.8 1
P₁ = 1.013 10⁵ +708.5 - 9800
P₁ = 92208.5Pa
P₁ = 0.922 10⁵ Pa
Answer: B
Explanation:
Limiting the maximum current through the bulb. This will help in preserving or improving the bulb's lifetime and also this won't have an effect on the brightness of the bulb as brightness is affected by the average value. Although brightness is a factor of current, reducing the maximum current won't have any bearing on the average current the bulb is getting.
Answer:
option C
Explanation:
given,
energy dissipated by the system to the surrounding = 12 J
Work done on the system = 28 J
change in internal energy of the system
Δ U = Q - W
system losses energy = - 12 J
work done = -28 J
Δ U = Q - W
Δ U = -12 -(-28)
Δ U = 16 J
hence, the correct answer is option C
Answer:
A) 12P
Explanation:
The power produced by a force is given by the equation

where
W is the work done by the force
T is the time in which the work is done
At the beginning in this problem, we have:
W = work done by the force
T = time taken
So the power produced is

Later, the force does six times more work, so the work done now is

And this work is done in half the time, so the new time is

Substituting into the equation of the power, we find the new power produced:

So, 12 times more power.