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
8500 Hz and Longitudinal
Speed = frequency x wavelength
Speed of sound at 20 degrees Celsius is approximately 340 m/s
The answer is D. Either absorbed or reflected. The reason is because if no light is being shown on the other side, the substance is not letting any light pass through. Since we do not know anything else about the substance, we do not know which one of the two it is doing. The scientist would not see any light on the detector if 100% of the light is reflected and the same thing would happen if 100% of the light was absorbed.
Ans. 5 X 10^6
Explanation:
Here,^ represents 6 times 10 and 5 is multiplied. That is 5000 000.
Hope this helps
M1*V1 + M2*V2 = M1*V + M2*V.
1400*25 + 1800*20[180+40]=1400*V+1800*V.
Divide both sides by 100:
14*25 + 18*20[220o] = 14V + 18V.
350 + 360[220o] = 32V.
350 - 276-231i = 32V.
74 - 231i = 32V.
242.6[-72.2o] = 32V.
V = 7.6m/s[-72.2o]=7.6m/s[72o] S. of E.
