The car is accelerating at 3 m/s² in the positive direction (to the right). By Newton's second law, the net force on the car in this direction is
∑ F = F[a] - F[f] - F[air] = ma
3100 N - 200 N - F[air] = (650 kg) (3 m/s²)
Solve for F[air] :
F[air] = 3100 N - 200 N - (650 kg) (3 m/s²)
F[air] = 3100 N - 200 N - 1950 N
F[air] = 950 N
Magnets are attracted when each of the different sides, most commonly known as "North" and "South", are facing each other. They repel when North and North, or South and South are facing each other.
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
Answer:
60 N.C
Explanation:
The box will move in the direction of the push and pull with a force of 60 N. C.
Answer:
A vector quantity is a quantity that has both magnitude and direction.
Explanation:
