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3 years ago

How far will a rubber ball fall in 10 seconds?

Physics

1 answer:

Inessa [10]3 years ago

4 0

If it's not moving at all at the beginning of the 10 seconds, then it falls 490 meters straight down in 10 seconds.

(Note: This is true of all objects on Earth . . . rubber balls, feathers, grains of sand, school buses, battle ships . . . everything. As long as air doesn't hold them back. Anything falling from rest falls 490 meters in the first 10 seconds.)

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Suppose you are at the center of a large freely-rotating horizontal turntable in a carnival funhouse. As you crawl toward the ed

I am Lyosha [343]

Answer:

c. remains the same, but the RPMs decrease.

Explanation:

Because there aren't external torques on the system composed by the person and the turntable it follows that total angular momentum (I) is conserved, that means the total angular momentum is a constant:

$\overrightarrow{L}=constant$

The total angular momentum is the sum of the individual angular momenta, in our case we should sum the angular momentum of the turntable and the angular momentum of a point mass respect the center of the turntable (the person)

$\overrightarrow{L_{turnatble}}+\overrightarrow{L_{person}}=constant$ (1)

The angular momentum of the turntable is:

$\overrightarrow{L_{turnatble}}=I\overrightarrow{\omega}$ (2)

with I the moment of inertia and ω the angular velocity.

The angular momentum of the person respects the center of the turntable is:

$\overrightarrow{L_{person}}=\overrightarrow{r}\times m\overrightarrow{v}$ (3)

with r the position of the person respects the center of the turntable, m the mass of the person and v the linear velocity

Using the fact $v=\omega r$ :

$\overrightarrow{L_{person}}=\overrightarrow{r}\times rm\overrightarrow{\omega}$ (3)

By (3) and (2) on (1) and working only the magnitudes (it's all that we need for this problem):

$I\omega+r^{2}m\omega=constant$

$\omega(I+r^{2}m)=constant$

Because the equality should be maintained, if we increase the distance between the person and the center of the turntable (r), the angular velocity should decrease to maintain the same constant value because I and m are constants, so the RPM's (unit of angular velocity) are going to decrease.

4 0

3 years ago

A mass of 0.40 kg is suspended on a spring which then stretches 10 cm. The mass is then removed and a second mass is placed on t

Vanyuwa [196]

Answer:

x' = 1.01 m

Explanation:

given,

mass suspended on the spring, m = 0.40 Kg

stretches to distance, x = 10 cm = 0. 1 m

now,

we know

m g = k x

where k is spring constant

0.4 x 9.8 = k x 0.1

k = 39.2 N/m

now, when second mass is attached to the spring work is equal to 20 J

work done by the spring is equal to

$W = \dfrac{1}{2}kx'^2$

$20= \dfrac{1}{2}\times 39.2\times x'^2$

x'² = 1.0204

x' = 1.01 m

hence, the spring is stretched to 1.01 m from the second mass.

7 0

3 years ago

What are the Si prefix values

Law Incorporation [45]

centic10-2millim10-3microu [footnote 2]10-6nanon10-9

4 0

3 years ago

4. Identify Problems A student is describing a longitudinal wave in his notebook. He

andrew11 [14]

Answer:

"The distance between crests is 3 cm."

Explanation:

If he writes down "The distance between crests is 3 cm."

That means he is describing the wavelength of a wave and not longitudinal wave. He ought to write something about " direction "

Longitudinal waves are waves in which the displacement of the medium is in the same direction as, or parallel to, the direction of propagation of the wave. While

Wavelength is the distance between the two successfully Crest or trough

3 0

3 years ago

An automobile engine delivers 55.0 hp. How much time will it take for the engine to do 6.22 × 105 J of work? One horsepower is e

Gennadij [26K]

Answer:

15.2 s

Explanation:

Convert hp to W:

55.0 hp × 746 W/hp = 41,030 W

Power = energy / time

41030 W = 6.22×10⁵ J / t

t = 15.2 s

8 0

3 years ago