Answer:
F = - 2 A x - B
Explanation:
The force and potential energy are related by the expression
F = - dU / dx i ^ -dU / dy j ^ - dU / dz k ^
Where i ^, j ^, k ^ are the unit vectors on the x and z axis
The potential they give us is
U (x) = A x² + B x + C
Let's calculate the derivatives
dU / dx = A 2x + B + 0
The other derivatives are zero because the potential does not depend on these variables.
Let's calculate the strength
F = - 2 A x - B
A voltmeter is the instrument used to measure a potential difference between two points in an electric circuit
Correct temperature is 80°F
Answer:
T_f = 38.83°F
Explanation:
We are given;
Volume; V = 8 ft³
Initial Pressure; P_i = 100 lbf/in² = 100 × 12² lbf/ft²
Initial temperature; T_i = 80°F = 539.67 °R
Time for outlet flow; t_o = 90 s
Mass flow rate at outlet; m'_o = 0.03 lb/s
Final pressure; P_f = 30 lbf/in² = 30 × 12² lbf/ft²
Now, from ideal gas equation,
Pv = RT
Where v is initial specific volume
R is ideal gas constant = 53.33 ft.lbf/°R
Thus;
v = RT/P
v_i = 53.33 × 539.67/(100 × 12²)
v_i = 2 ft³/lb
Formula for initial mass is;
m_i = V/v_i
m_i = 8/2
m_i = 4 lb
Now change in mass is given as;
Δm = m'_o × t_o
Δm = 0.03 × 90
Δm = 2.7 lb
Now,
m_f = m_i - Δm
Thus; m_f = 4 - 2.7
m_f = 1.3 lb
Similarly in above;
v_f = V/m_f
v_f = 8/1.3
v_f = 6.154 ft³/lb
Again;
Pv = RT
Thus;
T_f = P_f•v_f/R
T_f = (30 × 12² × 6.154)/53.33
T_f = 498.5°R
Converting to °F gives;
T_f = 38.83°F
P=I^2 *R
600 =5.0^2 *R
R=24
Answer: 24 ohms
I hope it’s correcttttttt...
