Answer:
g = 1.64m/s²
Explanation:
1.5m in 0.078s
V = 15 / 0.078
= 19.23m/s
Tension = mg
μ = 3.10 × 10⁻⁴
T = V²μ
mg = V²μ
g = V²μ / m
g = ((19.23)²(3.10 × 10⁻⁴)) / (0.070)
g = 1.64m/s²
The answer is underground
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: Mercury, Mars, Venus, Earth, Neptune, Uranus, Saturn, and Jupiter.
Explanation:
That's all of the planets if you need them. Hope this helps!
1. "Weather is the area’s day-to-day conditions and climate is the area’s average conditions" is how <span>an area’s weather differ from the area’s climate. The correct option among all the options given in the question is option "D".
2. </span>All of the following factors contribute to Earth’s climate except longitude. <span>The correct option among all the options given in the question is the second option.</span>