Answer:
v = 120 m/s
Explanation:
We are given;
earth's radius; r = 6.37 × 10^(6) m
Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s
Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.
The angle will be;
θ = ¾ × 90
θ = 67.5
¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.
The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:
v = r(cos θ) × ω
v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)
v = 117.22 m/s
Approximation to 2 sig. figures gives;
v = 120 m/s
Answer:
1000 Hz
Explanation:
<em>The frequency would be 1000 Hz.</em>
The frequency, wavelength, and speed of a wave are related by the equation:
<em>v = fλ ..................(1)</em>
where v = speed of the wave, f = frequency of the wave, and λ = wavelength of the wave.
Making f the subject of the formula:
<em>f = v/λ.........................(2)</em>
Also, speed (v) = distance/time.
From the question, distance = 900 m, time = 3.0 s
Hence, v = 900/3.0 = 300 m/s
Substitute v = 300 and λ = 0.3 into equation (2):
f = 300/0.3 = 1000 Hz
Energy decreases as it moves uptrophic levels because energy is lost as metabolic heat when the organisms from one trophic level are consumed by organisms from the next level.Trophic level transfer efficiency (TLTE) measures the amount of energy that is transferred between trophic levels.
Answer:
v = 10 V and E = 2 10³ N/C
Explanation:
The electrical potentials and the electric field at one point are related by the expression
ΔV = - ∫ E. dS
Where the bold indicates vector quantities, E is the electric field and S is the line of displacement of the load, in general displacement is perpendicular to the equipotential lines, which reduces the product scales to the ordinary product.
If the potential difference is the most usual that is V = 10 V, the electric field is
s = 0.5 cm = 0.5 10⁻² m
E = ΔV / S
E = 10/0.5 10⁻²
E = 2 10³ N / C
