Answer:
(a) The magnetic energy density in the field is 6.366 J/m³
(b) The energy stored in the magnetic field within the solenoid is 5 kJ
Explanation:
magnitude of magnetic field inside solenoid, B = 4 T
inner diameter of solenoid, d = 6.2 cm
inner radius of the solenoid, r = 3.1 cm = 0.031 m
length of solenoid, L = 26 cm = 0.26 m
(a) The magnetic energy density in the field is given by;
(b) The energy stored in the magnetic field within the solenoid
IM sure there is C, D, and E in kuiper belts, but not really sure of silicon and iron
A) The friction force while the box is stationary is (the coefficient of static friction)*(the normal force). In this case, the normal force is equal to the gravitational force, or the weight. To move the box, we need a minimum horizontal force that is equal to the friction force. The weight is (500 kg)*(9.81 m/s^2)= 4905 N. So, (0.45)*(4905 N) = 2207.25 N.
b) The acceleration will be the horizontal force - the kinetic friction force (since they act in opposite directions) divided by the mass. Kinetic friction force = (coefficient of kinetic friction)*(normal force or weight).
F(net) = (2207.25 N)-(0.30)(4905 N) = 735.75 N
a = (735.75 N)/(500kg)= 1.4715 m/s^2
Answer:
T_ww = 43,23°C
Explanation:
To solve this question, we use energy balance and we state that the energy that enters the systems equals the energy that leaves the system plus losses. Mathematically, we will have that:
E_in=E_out+E_loss
The energy associated to a current of fluid can be defined as:
E=m*C_p*T_f
So, applying the energy balance to the system described:
m_CW*C_p*T_CW+m_HW*C_p*T_HW=m_WW*C_p*T_WW+E_loss
Replacing the values given on the statement, we have:
1.0 kg/s*4,18 kJ/(kg°C)*25°C+0.8 kg/s*4,18 kJ/(kg°C)*75°C=1.8 kg/s*4,18 kJ/(kg°C)*T_WW+30 kJ/s
Solving for the temperature Tww, we have:
(1.0 kg/s*4,18 kJ/(kg°C)*25°C+0.8 kg/s*4,18 kJ/(kg°C)*75°C-30 kJ/s)/(1.8 kg/s*4,18 kJ/(kg°C))=T_WW
T_WW=43,23 °C
Have a nice day! :D