r = radius of the circle of the ride = 3.00 meters
v = linear speed of the person during the ride = 17.0 m/s
m = mass of the person in angular motion in the ride
L = angular momentum of the person in the ride = 3570 kg m²/s
Angular momentum is given as
L = m v r
inserting the values
3570 kg m²/s = m (17 m/s) (3.00 m)
m = 3570 kg m²/s/(51 m²/s)
m = 7 kg
hence the mass comes out to be 7 kg
Answer:
D. "The net force is zero, so the acceleration is zero"
Explanation:
edge 2020
<span>The chemical formula for the unknown gas is Ne.
Since we're looking for the rate at which a gas escapes through a small hole, we're dealing with effusion. For effusion, the rate is proportional to the velocity of the gas particles.
Kinetic energy
E = 0.5 mv^2
Since the kinetic energy of individual gas particles is the same if their temperatures are the same, we can create the following equality:
0.5 m1(v1)^2 = 0.5 m2(v2)^2
Double each side to make it simplier.
m1(v1)^2 = m2(v2)^2
Divide both sides by m1 and by (v2)^2, giving
(v1)^2/(v2^2) = m2/m1
And take the square root, giving
(v1)/(v2) = sqrt(m2/m1)
Now let's use the value 1 and the atomic weight of Kr for v1 and m1
1/(v2) = sqrt(m2/83.798)
And for v2, we'll use the value 2.04
1/2.04 = sqrt(m2/83.798)
Now solve for m2.
1/2.04 = sqrt(m2/83.798)
1/4.1616 = m2/83.798
83.798/4.1616 = m2
20.13600538 = m2
So the atomic weight of the unknown gas should be close to 20.136. Looking at a periodic table, I find that neon has an atomic weight of 20.18 which is quite close. Additionally, since neon is a noble gas, its gas particles consist of individual atoms. So the unknown gas is neon.</span>
We know that tangential acceleration is related with radius and angular acceleration according the following equation:
at = r * aa
where at is tangential acceleration (in m/s2), r is radius (in m) aa is angular acceleration (in rad/s2)
So the radius is r = d/2 = 1.2/2 = 0.6 m
Then at = 0.6 * 5 = 3 m/s2
Tangential acceleration of a point on the flywheel rim is 3 m/s2
