Answer:
The amount of work that must be done to compress the gas 11 times less than its initial pressure is 909.091 J
Explanation:
The given variables are
Work done = 550 J
Volume change = V₂ - V₁ = -0.5V₁
Thus the product of pressure and volume change = work done by gas, thus
P × -0.5V₁ = 500 J
Hence -PV₁ = 1000 J
also P₁/V₁ = P₂/V₂ but V₂ = 0.5V₁ Therefore P₁/V₁ = P₂/0.5V₁ or P₁ = 2P₂
Also to compress the gas by a factor of 11 we have
P (V₂ - V₁) = P×(V₁/11 -V₁) = P(11V₁ - V₁)/11 = P×-10V₁/11 = -PV₁×10/11 = 1000 J ×10/11 = 909.091 J of work
Answer:
4.41 W
Explanation:
P = IV, V = IR
P = V² / R
Given that P = 0.0625 when V = 1.50:
0.0625 = (1.50)² / R
R = 36
So the resistor is 36Ω.
When the voltage is 12.6, the power consumption is:
P = (12.6)² / 36
P = 4.41
So the power consumption is 4.41 W.
Answer:
Explanation:
Calculate the volume of the lead

Now calculate the bouyant force acting on the lead


This force will act in upward direction
Gravitational force on the lead due to its mass will act in downward direction
Hence the difference of this two force

If V is the volume submerged in the water then bouyant force on the bobber is

Equate bouyant force with the tension and gravitational force

Now Total volume of bobble is

=