What is the strength of the electric field between two parallel conducting plates separated by 1.00 cm and having a potential difference (voltage) between them of 1.50×10^4v ?
Answer:
Explanation:
Velocity of a wave is describe as
velocity =Frequency × Wavelength
Mathematically
v = fλ
Hence, Frequency, F = v / λ
Wavelength λ = v/f
So, if the frequency is kept constant, wavelength of the wave becomes directly proportional to velocity of the wave.
And this implies that, as the speed double, the wavelength is double.
Kepler's 3rd law is given as
P² = kA³
where
P = period, days
A = semimajor axis, AU
k = constant
Given:
P = 687 days
A = 1.52 AU
Therefore
k = P²/A³ = 687²/1.52³ = 1.3439 x 10⁵ days²/AU³
Answer: 1.3439 x 10⁵ (days²/AU³)
A child height can be very tall or short
Answer:
540C.
Explanation:
A capacitor of capacitance C when charged to a voltage of V will have a charge Q given as follows;
Q = CV ----------(i)
From the question, the initial charge on the capacitor is the charge on it before it was connected to the resistor. In other words, the initial charge on the capacitor will have a maximum value which can be calculated using equation (i) above.
Where;
C = 6F
V = 90V
Substitute these values into equation (i) as follows;
Q = 6 x 90
Q = 540 C
Therefore, the initial charge on the capacitor is 540C.
