A compact disc has a radius of 6 centimeters.a. What is its circumference in meters?
b. If the cd rotates 4 times per second, what is the linear speed of a point on the outer edge of the cd? Give
your answer in meters per second.
c. What is the linear speed of a point 3 centimeters from the center of the cd? (Assume the angular speed
has not changed).

Answers

Answer 1

Answer: Circumference is $2\pi{r}=37.7$ (approx.)
Speed is 4 times that. That is, 150.8 cm/s or 1.508 m/s.
If we get a point at distance 3 cm, we take the radius half of that. The circumference also gets half, so does the linear speed. We get 0.754 m/s or 75.398 cm/s.

Answer 2

Answer:

Answer:75.389

Explanation:

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Four wire AB,BC,CD and DA each resistance 4 ohm and a fifth wire BD of resistance 8 ohm are joined to form a rectangle to which BD is a diagonal.The effective resistance between the points A and B is?

Answers

Since it's so late, and it's way past my bedtime, and I'm tired and hungry,
and I still have real work to do, and I don't think you're really a legitimate
asker of Brainly questions anyway, I'm going to do something that's against
my principles and I almost never do. I'm going to violate my own personal
Prime Directive, and give you the answer without any explanation of where
it comes from.

Ready ?

Here goes:

The effective resistance between points 'A' and 'B' is (2 and 2/3) Ω .

The index of refraction of a material medium must be less than 1 true or flase

Answers

That statement is flase. The index of refraction
of every material is more than 1.0 .

When an object falls through air, there is a drag with dimensions ml/t2 that depends on the ________. 1) size of the object 2) mass of the object 3) velocity of the object 4) density of the air

Answers

When an object falls through air, the drag with dimensions $mL/t^2$ depends on the 1) size of the object, 2) velocity of the object, and 4) density of the air.

What is the speed of a 500 kg elevator that has 4000 J of energy ?

Answers

Energy =sxm
S =4000/500
=Answer

Assume that you have a mass of 50.0kg and earth has a mass of 5.97x10kg.The radius of earth is 6.38x10m.What is the force of gravitational attraction between you and the earth

Answers

The formula for force of attraction between any two bodies in the universe
F = GMm / r^2. (Newton's Universal law of Gravitation).

G = Universal gravitational constant, G = 6.67 * 10 ^ -11 Nm^2 / kg^2.
M = Mass of first object. = 5.97 x 10^24 kg.
m = mass of second object = 50 kg.
r = distance apart, between centers = in this case it is the radius of the earth
= 6.38 x 10^6 m.
( Sorry I could not assume with the values you gave).
(Actually the force of attraction between an object on the Earth and the Earth is the weight of the object our answer should be around 500N which is from mg = 50 *10 = 500)

So F = ((6.67 * 10 ^ -11)*(5.97 x 10^24)*(50)) / (6.38 x 10^6)^2.
Punch it all up in your calculator.
F = 489. 135 N.

Remember we said it would be around 500.
So that's our answer.
F = 489. 135 N. Hurray!!

Final answer:

The force of gravitational attraction between you (of mass 50kg) and the Earth (of mass 5.97x10^24 kg) can be calculated using Newton's Universal Law of Gravitation. Substituting the respective values in the formula, the calculated force is approximately 490.5 N.

Explanation:

The force of gravitational attraction between you and the earth can be calculated using Newton's Universal Law of Gravitation. This law states that the force (F) between two masses (m1 and m2) separated by a distance (r) is given by the formula F = G*(m1*m2/r^2), where G is the gravitational constant and is approximately 6.67x10^-11 N*(m^2/kg^2).

In this case, your mass (m1) is 50 kg, Earth's mass (m2) is 5.97x10^24 kg and the radius of Earth (r) is 6.38x10^6 m. Substituting these values into the formula, we can find the force of gravitational attraction between you and the Earth.

F = (6.67x10^-11 N*(m^2/kg^2)) * ((50 kg) * (5.97x10^24 kg)) / (6.38x10^6 m)^2. After calculating the above equation, you find that F ≈ 490.5 N.