If his speed (<em>decreases by 2.4 meters per second every second</em>)**, then
his speed will be zero in (13.2 / 2.4) = 5.5 seconds.
** That's the meaning of "acceleration of -2.4 m/s² " .
Hi there!
Assuming the track is frictionless:
Cancel out the masses and rearrange to solve for velocity:
Plug in the given height and let g = 9.8 m/s²:
1. As of right now, our solar system only includes asteroids and they are found in Kuiper Belt next to Saturn.
On May 6, 2020, scientists have discovered a mysterious blanket of darkness approximately 1,000 light years away from our solar system. However, in this case, black holes would not be a correct answer.
Even though constellations may seem close, they are actually far away since the size of the stars are immense and can be visible from light years away. Constellations are not in our solar system.
Galaxies are groups of solar systems so therefore galaxies is not an answer.
2. Ptolemy was the man who proposed the early Earth-centered model of the universe and his model was called the Ptolemaic model also known as the geocentric model.
3. The planet that is most similar to Earth in mass and diameter is Venus.
4. The inner planets in our solar system are formed from dense elements and that is why they are able to withstand the intense of the sun. Meanwhile the outer planets were formed from gaseous materials but Pluto is so cold that its elements are hardened and dense.
5. The answer would be D since they orbit in the same direction and on the same plane.
6. The speed of rotation determines the length and duration of day on a planet.
7. The answer would be D since they have to have their own gravity, have a path clear of smaller bodies , and orbit a star,
Answer:
a) the torque required is 10.53 N-m
b) The magnitude force applied tangentially is 12.33 N
Explanation:
Given the data in the question;
mass m = 1.765 kg
radius r = 0.854 m
first we calculate the moment of inertia;
= mr²
we substitute
= × 1.765 × (0.854)²
= 0.514897 kg.m²
a)
Find the torque required to bring the sphere from rest to an angular velocity of 317 rad/s, clockwise, in 15.5 s
ω = 0
ω = 317 rad/s
t = 15.5 s
we know that; ω = ω + ∝t
so we substitute
317 = 0 + ∝(15.5)
∝ = 317 / 15.5
∝ = 20.4514 rad/s²
so
ζ = × ∝
we substitute
ζ = 0.514897 × 20.4514
ζ = 10.53 N-m
Therefore, the torque required is 10.53 N-m
b)
What magnitude force applied tangentially at the equator would provide the needed torque.
ζ = F × r
we substitute
10.53 = F × 0.854
F = 10.53 / 0.854
F = 12.33 N
Therefore, magnitude force applied tangentially is 12.33 N