Answer:
F' = (4/9)F
Explanation:
The electrostatic force between two charged objects is given by Coulomb's Law:
F = kq₁q₂/r² -------------------- equation (1)
where,
F = Electrostatic Force
k = Coulomb's Constant
q₁ = magnitude of first charge
q₂ = magnitude of second charge
r = distance between charges
Now, when the charges and distance altered as follows:
q₁' = 2q₁
q₂' = 2q₂
r' = 3r
Then,
F' = kq₁'q₂'/r'²
F' = k(2q₁)(2q₂)/(3r)²
F' = (4/9)kq₁q₂/r²
using equation (1):
<u>F' = (4/9)F</u>
Answer:
The minimum frequency of the coil is 7.1 Hz
Explanation:
Given;
number of turns, N = 200 turns
cross sectional area, A = 300 cm² = 300 x 10⁻⁴ m²
magnitude of magnetic field strength, B = 30 x 10⁻³ T
maximum value of the induced emf, E = 8 V
Maximum induced emf is given as;
E = NBAω
where
ω is angular velocity (ω = 2πf)
E = NBA2πf
where;
f is the minimum frequency, measured in hertz (Hz)
f = E / (NBA2π)
f = 8 / (200 x 30 x 10⁻³ x 300 x 10⁻⁴ x 2 x 3.142)
f = 7.073 Hz
f = 7.1 Hz
Therefore, the minimum frequency of the coil is 7.1 Hz
Answer:
<em>I belive it is 38 degrees</em>
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Explanation:
Plate Tectonics cause most earthquakes.
Answer:
1.08 s
Explanation:
From the question given above, the following data were obtained:
Height (h) reached = 1.45 m
Time of flight (T) =?
Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:
Height (h) = 1.45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
1.45 = ½ × 9.8 × t²
1.45 = 4.9 × t²
Divide both side by 4.9
t² = 1.45/4.9
Take the square root of both side
t = √(1.45/4.9)
t = 0.54 s
Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).
Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:
Time (t) taken to reach the height = 0.54 s
Time of flight (T) =?
T = 2t
T = 2 × 0.54
T = 1.08 s
Therefore, it will take the kangaroo 1.08 s to return to the earth.
