Answer: 0.48W
useful power output=total power output*efficency
useful power output=2.4W*0.20=0.48W
The correct answer for the question that is being presented above is this one: "Schmidt-Cassegrain focus." A focal arrangement that has a thin lens that the light passes through before traveling down the tube to the objective mirror is a Schmidt-Cassegrain focus.
Here are the following choices:
a. Cassegrain focus
b. Newtonian focus
c. Schmidt-Cassegrain focus
<span>d. Schmidt focus</span>
8. In soft magnetic materials such as iron, what happens when an external magnetic field is removed?
a. The domain alignment persists.
b. The orientation of domains fluctuates.
c. The material becomes a hard magnetic material.
d. The orientation of domains changes, and the material returns to an unmagnetized state.
9. According to Lenz’s law, if the applied magnetic field changes,
a. the induced field attempts to keep the total field strength constant.
b. the induced field attempts to increase the total field strength.
c. the induced field attempts to decrease the total field strength.
d. the induced field attempts to oscillate about an equilibrium value.
10. The direction of the force on a current-carrying wire in an external magnetic field is
a. perpendicular to the current only.
b. perpendicular to the magnetic field only.
c. perpendicular to the current and to the magnetic field.
d. parallel to the current and to the magnetic field
Answer:

Explanation:
So, we are looking for an expression of the amount of water that has been drained from the tub. The expression is in terms of v that represent the number of gallons of water drained since the plug was pulled. Since we are interested in the pounds of water that has been drained from the tub we need to take into account that for every gallon of water drained, 8.345 pounds have left the tub. Therefore, the expression for the weight of water Q that has been drained from the tub in terms of v is simply :

Where v is the amount of gallons that has been drained from the tub.
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Answer:
Explanation:
All the displacement will be converted into vector, considering east as x axis and north as y axis.
5.3 km north
D = 5.3 j
8.3 km at 50 degree north of east
D₁= 8.3 cos 50 i + 8.3 sin 50 j.
= 5.33 i + 6.36 j
Let D₂ be the displacement which when added to D₁ gives the required displacement D
D₁ + D₂ = D
5.33 i + 6.36 j + D₂ = 5.3 j
D₂ = 5.3 j - 5.33i - 6.36j
= - 5.33i - 1.06 j
magnitude of D₂
D₂²= 5.33² + 1.06²
D₂ = 5.43 km
Angle θ
Tanθ = 1.06 / 5.33
= 0.1988
θ =11.25 ° south of due west.