Answer:
A waffle iron heated by coils
Explanation:
A waffle iron heated by coils - conduction
Food heated in a microwave oven - radiation
Pavement heated by the sun - radiation
A room heated by moving air - convection
Answer:
7.28×10⁻⁵ T
Explanation:
Applying,
F = BILsin∅............. Equation 1
Where F = magnetic force, B = earth's magnetic field, I = current flowing through the wire, L = Length of the wire, ∅ = angle between the field and the wire.
make B the subject of the equation
B = F/ILsin∅.................. Equation 2
From the question,
Given: F = 0.16 N, I = 68 A, L = 34 m, ∅ = 72°
Substitute these values into equation 2
B = 0.16/(68×34×sin72°)
B = 0.16/(68×34×0.95)
B = 0.16/2196.4
B = 7.28×10⁻⁵ T
The scale is a rest scale reads the support for is not enough net force.
Answer:
0.853 m/s
Explanation:
Total energy stored in the spring = Total kinetic energy of the masses.
1/2ke² = 1/2m'v².................... Equation 1
Where k = spring constant of the spring, e = extension, m' = total mass, v = speed of the masses.
make v the subject of the equation,
v = e[√(k/m')].................... Equation 2
Given: e = 39 cm = 0.39 m, m' = 0.4+0.4 = 0.8 kg, k = 1.75 N/cm = 175 N/m.
Substitute into equation 2
v = 0.39[√(1.75/0.8)
v = 0.39[2.1875]
v = 0.853 m/s
Hence the speed of each mass = 0.853 m/s
Answer:
Temperature at the exit = ![267.3 C](https://tex.z-dn.net/?f=267.3%20C)
Explanation:
For the steady energy flow through a control volume, the power output is given as
![W_{out}= -m_{f}(h_{2}-h_{1} + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})](https://tex.z-dn.net/?f=W_%7Bout%7D%3D%20-m_%7Bf%7D%28h_%7B2%7D-h_%7B1%7D%20%2B%20%5Cfrac%7Bv_%7B2%7D%5E%7B2%7D%7D%7B2%7D%20-%20%5Cfrac%7Bv_%7B1%7D%5E%7B2%7D%7D%7B2%7D%29)
Inlet area of the turbine = ![60cm^{2}= 0.006m^{2}](https://tex.z-dn.net/?f=60cm%5E%7B2%7D%3D%200.006m%5E%7B2%7D)
To find the mass flow rate, we can apply the ideal gas laws to estimate the specific volume, from there we can get the mass flow rate.
Assuming Argon behaves as an Ideal gas, we have the specific volume ![v_{1}](https://tex.z-dn.net/?f=v_%7B1%7D)
as
![v_{1}=\frac{RT_{1}}{P_{1}}=\frac{0.2081\times723}{900}=0.1672m^{3}/kg](https://tex.z-dn.net/?f=v_%7B1%7D%3D%5Cfrac%7BRT_%7B1%7D%7D%7BP_%7B1%7D%7D%3D%5Cfrac%7B0.2081%5Ctimes723%7D%7B900%7D%3D0.1672m%5E%7B3%7D%2Fkg)
![m_{f}=\frac{1}{v_{1}}\times A_{1}V_{1} = \frac{1}{0.1672}\times(0.006)(80)=2.871kg/sec](https://tex.z-dn.net/?f=m_%7Bf%7D%3D%5Cfrac%7B1%7D%7Bv_%7B1%7D%7D%5Ctimes%20A_%7B1%7DV_%7B1%7D%20%3D%20%5Cfrac%7B1%7D%7B0.1672%7D%5Ctimes%280.006%29%2880%29%3D2.871kg%2Fsec)
for Ideal gasses, the enthalpy change can be calculated using the formula
![h_{2}-h_{1}=C_{p}(T_{2}-T_{1})](https://tex.z-dn.net/?f=h_%7B2%7D-h_%7B1%7D%3DC_%7Bp%7D%28T_%7B2%7D-T_%7B1%7D%29)
hence we have
![W_{out}= -m_{f}((C_{p}(T_{2}-T_{1}) + \frac{v_{2}^{2}}{2} - \frac{v_{1}^{2}}{2})](https://tex.z-dn.net/?f=W_%7Bout%7D%3D%20-m_%7Bf%7D%28%28C_%7Bp%7D%28T_%7B2%7D-T_%7B1%7D%29%20%2B%20%5Cfrac%7Bv_%7B2%7D%5E%7B2%7D%7D%7B2%7D%20-%20%5Cfrac%7Bv_%7B1%7D%5E%7B2%7D%7D%7B2%7D%29)
![250= -2.871((0.5203(T_{2}-450) + \frac{150^{2}}{2\times 1000} - \frac{80^{2}}{2\times 1000})](https://tex.z-dn.net/?f=250%3D%20-2.871%28%280.5203%28T_%7B2%7D-450%29%20%2B%20%5Cfrac%7B150%5E%7B2%7D%7D%7B2%5Ctimes%201000%7D%20-%20%5Cfrac%7B80%5E%7B2%7D%7D%7B2%5Ctimes%201000%7D%29)
<em>Note: to convert the Kinetic energy term to kilojoules, it was multiplied by 1000</em>
evaluating the above equation, we have ![T_{2}=267.3C](https://tex.z-dn.net/?f=T_%7B2%7D%3D267.3C)
Hence, the temperature at the exit = ![267.3 C](https://tex.z-dn.net/?f=267.3%20C)