Assuming that the vectors are acting along the same axis, we
could just simply add or subtract the vectors. Since the F1 is greater than F2,
there would be motion, there would be acceleration, and that the direction of
motion is along the F1.
Answer:
The current in wire resistance 2Ω
a). 8696 A
b). fraction power 15.1% a 115kV
Explanation:
Resistance
Ω/Km*40km
R=2Ω
P=1000 MW
a).
Using law ohm
b).
%
The gravitational force <em>F</em> between two masses <em>M</em> and <em>m</em> a distance <em>r</em> apart is
<em>F</em> = <em>G M m</em> / <em>r</em> ²
Decrease the distance by a factor of 7 by replacing <em>r</em> with <em>r</em> / 7, and decrease both masses by a factor of 8 by replacing <em>M</em> and <em>m</em> with <em>M</em> / 8 and <em>m</em> / 8, respectively. Then the new force <em>F*</em> is
<em>F*</em> = <em>G </em>(<em>M</em> / 8) (<em>m</em> / 8) / (<em>r</em> / 7)²
<em>F*</em> = (1/64 × <em>G M m</em>) / (1/49 × <em>r</em> ²)
<em>F*</em> = 49/64 × <em>G M m</em> / <em>r</em> ²
In other words, the new force is scaled down by a factor of 49/64 ≈ 0.7656, so the new force has magnitude approx. 76.56 N.
NOTE: The given question is incomplete.
<u>The complete question is given below.</u>
A student measures the speed of yellow light in water to be 2.00 x 10⁸ m/s. Calculate the speed of light in air.
Solution:
Speed of yellow light in water (v) = 2.00 x 10⁸ m/s
Refractive Index of water with respect to air (μ) = 4/3
Refractive Index = Speed of yellow light in air / Speed of yellow light in water
Or, The speed of yellow light in air = Refractive Index × Speed of yellow light in water
or, = (4/3) × 2.00 x 10⁸ m/s
or, = 2.67 × 10⁸ m/s ≈ 3.0 × 10⁸ m/s
Hence, the required speed of yellow light in the air will be 3.0 × 10⁸ m/s.
1140x9.8x2.4= 26,812.8 significant figures Make it 27,000
