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sdas [7]
2 years ago
15

A 2 kg stone is tied to a 0.5 m string and swung around a circle at a constant angular velocity of 12 rad/s. the angular momentu

m of the stone about the center of the circle is:
Physics
1 answer:
Illusion [34]2 years ago
3 0
Starting from the angular velocity, we can calculate the tangential velocity of the stone:
v=\omega r= (12 rad/s)(0.5 m)= 6 m/s

Then we can calculate the angular momentum of the stone about the center of the circle, given by
L=mvr
where
m is the stone mass
v its tangential velocity
r is the radius of the circle, that corresponds to the length of the string.

Substituting the data of the problem, we find
L=(2 kg)(6 m/s)(0.5 m)=6 kg m^2 s^{-1}
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An object can be broken up by a planet's gravity once it passes the _______. The Jovian planets are composed primarily of ______
Rina8888 [55]

Answer:1. Roche limit

2.hydrogen

3.atmosphere

4.mercury

5.venus

6.when an object passes the Roche limit, the strength of gravity on the object increases. If the density of the planet is higher, then the object can break up farther away from the planet. If the density is lower, then the Roche limit is located closer to the planet

7.Farther our in the solar system, beyond the frost line, hydrogen was at a low enough temperature that it could condense. This allowed hydrogen to accumulate under gravity, eventually forming the Jovian planets

Explanation:

3 0
2 years ago
Read 2 more answers
A student measures the pH of a solution to be 6.8. Which should the student add if she wants to decrease the pH of the solution?
zloy xaker [14]
The neutral pH is 7. Less than 7 indicates an acid and more than 7 indicates a base (up to 14).
<span> NaCl - it's a salt (we can't measure the pH)
H2O - it can be an acid but also a base  (the pH it is almost neutral,meaning close to 7 )
HF - it is a strong acid
</span><span> KOH  - it is a strong base (pH=14)
</span>
                        ↓

He needs to use HF (Hydrogen fluoride) to decrease the pH.


7 0
2 years ago
Read 2 more answers
A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur
Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

5 0
2 years ago
What is the absolute value of the horizontal force that each athlete exerts against the ground?
alexandr402 [8]
Refer to the diagram shown below.

When an athlete is in motion, he/she exerts a vertical force (the person's weight, W) on the ground. The ground exerts an equal and opposite force, N, the normal reaction on the athlete, so that W = N.

At the same time, the ground exerts a horizontal force, F, o n the athlete so that he/she does not slip.
The magnitude of the horizontal force is
F = μN = μW
where μ = the dynamic coefficient of friction.

Answer:  
The horizontal force is μW,
where
W = the weight of the athlete and,
μ = the dynamic coefficient of friction.

6 0
2 years ago
One car travels 40. meters due east in 5.0 seconds, and a second car travels 64 meters due west in 8.0 seconds. During their per
KengaRu [80]

Answer:

They had the same speed.

Explanation:

It won't be velocity, because velocity is a vector quantity. Speed is scalar.

5 0
2 years ago
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