The unit used<span> to measure weight in the metric system is the gram.</span>
The total distance traveled by Aliaa using her skateboard for 20 revolution of the wheels is equal to 3770 millimeters.
<u>Given the following data:</u>
Diameter of skateboard = 60 mm.
Number of revolution = 20 revolutions.
Radius = diameter/2 = 60/2 = 30 mm.
<h3>What is distance?</h3>
Distance can be defined as the amount of ground covered (traveled) by a physical object over a specific period of time and speed, regardless of its direction, starting point or ending point.
For one revolution of the wheels, the distance traveled by Aliaa using her skateboard is given by:
Distance = 2πr
Distance = 2 × 3.142 × 30
Distance = 188.5 mm.
Therefore, the total distance traveled by Aliaa using her skateboard for 20 revolution of the wheels is given by:
Distance = 188.5 × 20
Distance = 3770 millimeters.
Read more on distance per revolutions here: brainly.com/question/10989073
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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
1) An object with more mass has more gravitational pull.
2) The sun has more gravity than the Earth.
3) The gravitational pull of Earth causes the moon to orbit Earth.
4) The unbalanced force of gravity causes comets, meteors, and asteroids to collide with planets.
5) The force of gravity causes Earth to travel in a sideways path around the sun.
6) a. Most mass = Sun
b. Middle mass = Earth
c. Least mass = Moon
Answer:
(a) charge q=5.33 nC
(b) charge density σ=10.62 nC/m²
Explanation:
Given data
radius r=0.20 m
potential V=240 V
coulombs constant k=9×10⁹Nm²/C²
To find
(a) charge q
(b) charge density σ
Solution
For (a) charge q
As
For (b) charge density
As charge density σ is given as:
σ=q/(4πR²)
σ=(5.333×10⁻⁹) / (4π×(0.20)²)
σ=10.62 nC/m²
