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4 years ago

The buildup of charges on an object is called

Physics

2 answers:

In-s [12.5K]4 years ago

7 0

what whitey said he is right because he is smart

Nastasia [14]4 years ago

4 0

Static electricity.

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A playground merry-go-round of radius R = 1.80 m has a moment of inertia I = 270 kg · m2 and is rotating at 8.0 rev/min about a

Arte-miy333 [17]

Answer:

$\dot n = 6.042\,rpm$

Explanation:

The final angle speed of the merry-go-round is determined with the help of the Principle of Angular Momentum Conservation:

$(270\,kg\cdot m^{2})\cdot \left(8\,rpm\right) = [270\,kg\cdot m^{2}+(27\,kg)\cdot (1.80\,m)^{2}]\cdot \dot n$

$\dot n = 6.042\,rpm$

3 0

3 years ago

The best method for separating an oil and water mixture would be _____.

Mice21 [21]

Answer:

The answer is density

Explanation:

8 0

3 years ago

Read 2 more answers

A golf ball starts with a speed of 2 m/s and slows at a constant rate of 0.5 m/s2, what is its velocity after 2 s?

jeka57 [31]

Answer:

3m/s

Explanation:

Given parameters:

Initial speed = 2m/s

Acceleration = 0.5m/s²

Time = 2s

Unknown:

Final speed = ?

Solution:

To solve this problem, we apply the right motion equation;

V = U + at

V is the final speed

U is the initial speed

a is the acceleration

t is the time

V = 2 + (2 x 0.5)

V = 2 + 1

V = 3m/s

8 0

3 years ago

A girl and a boy are riding on a merry-go-round that is turning at a constant rate. The girl is near the outer edge, and the boy

skad [1K]

(a) Both the girl and the boy have the same nonzero angular displacement.

Explanation:

The angular displacement of an object moving in uniform circular motion, as the boy and the girl on the merry-go-round, is given by

$\theta= \omega t$

where

$\omega$ is the angular speed

t is the time interval

For a uniform object in uniform circular motion, all the points of the object have same angular speed. This means that the value of $\omega$ is the same for the boy and the girl.

Therefore, if we consider the same time interval t, the boy and the girl will also have same nonzero angular displacement.

(b) The girl has greater linear speed.

Explanation:

The linear (tangential) speed of a point along the merry-go-round is given by

$v=\omega r$

where

$\omega$ is the angular speed

r is the distance of the point from the centre of the merry-go-round

In this problem, the girl is near the outer edge, while the boy is closer to the centre: since the value of $\omega$ is the same for both, this means that the value of r is larger for the girl, so the girl will also have a greater linear speed.

3 0

3 years ago

Keisha drew a diagram to compare two of the fundamental forces. A venn diagram of 2 intersecting circles with the left circle la

slamgirl [31]

Answer: B)

X: Attractive and repulsive

Y: Very small range

Z: Always attractive

I just took the test :)

8 0

3 years ago