Haven't taken physics but I would assume if her friend is standing in front of her that you would add up the speeds and get 30 km/hr.
Answer:
Over such small distances, digital data may be transmitted as direct, two-level electrical signals over simple copper conductors. This results from the electrical distortion of signals traveling through long conductors, and from noise added to the signal as it propagates through a transmission medium.
It holds A. Less water than when it is warmer.
Answer:
Option D: 21.8 degrees
Explanation:
In a parallel RL circuit, the current in the resistor R and that in the inductor L are separated among themselves 90 degrees as illustrated in the attached image. In the image the current in the resistor is represented in orange, that of the inductor in blue, and the total current (vector addition of the previous two) is represented in red, forming a certain angle (theta) with respect to the current in the resistor. The output voltage is the same as the input voltage as measured over the resistor R.
Therefore, the phase angle that separated output voltage and total current can be obtained using the fact that tan(phase angle) = , therefore the angle is the arctangent of 4/10:
degrees.
Answer:
Option B. 5 nC
Explanation:
From the question given above, the following data were obtained:
Capicitance (C) = 100 pF
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:
1 pF = 1×10¯¹² F
Therefore,
100 pF = 100 pF × 1×10¯¹² F / 1 pF
100 pF = 1×10¯¹⁰ F
Next, we shall determine the quantity of charge. This can be obtained as follow:
Capicitance (C) = 1×10¯¹⁰ F
Potential difference (V) = 50 V
Quantity of charge (Q) =?
Q = CV
Q = 1×10¯¹⁰ × 50
Q = 5×10¯⁹ C
Finally, we shall convert 5×10¯⁹ C to nano coulomb (nC). This can be obtained as follow:
1 C = 1×10⁹ nC
Therefore,
5×10¯⁹ C = 5×10¯⁹ C × 1×10⁹ nC / 1 C
5×10¯⁹ C = 5 nC
Thus, the quantity of charge is 5 nC