Answer:
Explanation:
The formula P=VI, just rearrange to find I
So
I=P/V
=75/120
=0.625amps
625mA
Answer:

Explanation:
Given:
Ratio of lift force to drag force is, 
Lift force on a short section is, 
Magnitude of resultant, 
The angle of 'R' with the horizontal is, 
We know that, lift force and drag are at right angles to each other. So, the resultant can be computed using Pythagoras theorem.
For calculating 'R', we first compute drag force 'D'.
As per question:

Now, the magnitude of resultant 'R' is given as:

Plug in the given values and solve for 'R'. This gives,

Therefore, the magnitude of the resultant force 'R' is 64.32 lb.
Now, the angle
is given as the arctan of the ratio of the lift and drag force.
Therefore,

Therefore, the angle made with the horizontal is 84.3°.
Porque la matematica es imposible
Answer:
how is that a question?
Explanation:
yeah i dunno the answer cause thats not a question
Answer:
Option 10. 169.118 J/KgºC
Explanation:
From the question given above, the following data were obtained:
Change in temperature (ΔT) = 20 °C
Heat (Q) absorbed = 1.61 KJ
Mass of metal bar = 476 g
Specific heat capacity (C) of metal bar =?
Next, we shall convert 1.61 KJ to joule (J). This can be obtained as follow:
1 kJ = 1000 J
Therefore,
1.61 KJ = 1.61 KJ × 1000 J / 1 kJ
1.61 KJ = 1610 J
Next, we shall convert 476 g to Kg. This can be obtained as follow:
1000 g = 1 Kg
Therefore,
476 g = 476 g × 1 Kg / 1000 g
476 g = 0.476 Kg
Finally, we shall determine the specific heat capacity of the metal bar. This can be obtained as follow:
Change in temperature (ΔT) = 20 °C
Heat (Q) absorbed = 1610 J
Mass of metal bar = 0.476 Kg
Specific heat capacity (C) of metal bar =?
Q = MCΔT
1610 = 0.476 × C × 20
1610 = 9.52 × C
Divide both side by 9.52
C = 1610 / 9.52
C = 169.118 J/KgºC
Thus, the specific heat capacity of the metal bar is 169.118 J/KgºC