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3 years ago

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If a golf ball and a ping-pong ball both move with the same kinetic energy, which has the greater speed? 1. the golf ball 2. can

not be determined 3. the ping-pong ball 4. the two balls have the same speed.

1 answer:

schepotkina [342]3 years ago

4 0

Think about it like this, the more mass there is, the faster its going to go. If you took a golf ball and a ping pong ball and you held them each separately, you would notice that the golf ball is heavier. If they move with the same kinetic energy, but the golf ball WEIGHS more, then the golf ball will have the greater speed. If you think about it, the ping pong ball may be taking its time to get to wherever its going.

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3 years ago

Read 2 more answers

Earth orbits the sun at an average circular radius of about 149.60 million kilometers every 365.26 Earth days.

kobusy [5.1K]

The Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x $10^{24}$ Kg

<h3>Relationship between Linear and angular speed</h3>

Linear speed is the product of angular speed and the maximum displacement of the particle. That is,

V = Wr

Where

V = Linear speed

W = Angular speed

r = Radius

Given that the earth orbits the sun at an average circular radius of about 149.60 million kilometers every 365.26 Earth days.

a) To determine the Earth’s average orbital speed, we will make use of the below formula to calculate angular speed

W = 2 $\pi$ /T

W = (2 x 3.143) / (365.26 x 24)

W = 6.283 / 876624

W = 7.2 x $10^{-4}$ Rad/hr

The Earth’s average orbital speed V = Wr

V = 7.2 x $10^{-4}$ x 149.6 x $10^{6}$

V = 107225.5 kilometers per hours.

b) Based on the information given in this question, to calculate the approximate mass of the Sun, we will use Kepler's 3rd law

M = (4 $\pi ^{2}$ $r^{3}$ ) / G $T^{2}$

M = (4 x 9.8696 x 3.35 x $10^{24}$ ) / (6.67 x $10^{-11}$ x 7.68 x $10^{11}$ <em>)</em>

<em>M = 1.32 x </em> $10^{26}$ / 51.226

M = 2.58 x $10^{24}$ Kg

Therefore, the Earth’s average orbital speed expressed in kilometers per hours is 107225.5 Km/hr and the mass of the sun is 2.58 x $10^{24}$ Kg

Learn more about Orbital Speed here: brainly.com/question/22247460

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3 0

1 year ago

I need both parts please (a) Given a material with an attenuation coefficient (a) of 0.6/cm, what is the intensity of a beam (wi

Masteriza [31]

Answer:

<h3>a.</h3>

After it has traveled through 1 cm : $I(1 \ cm) = 0.5488 I_0$

After it has traveled through 2 cm : $I(2 \ cm) = 0.3012 I_0$

<h3>b.</h3>

After it has traveled through 1 cm : $od( 1\ cm) = 0.2606$

After it has traveled through 2 cm : $od( 2\ cm) = 0.5211$

Explanation:

<h2>a.</h2>

For this problem, we can use the Beer-Lambert law. For constant attenuation coefficient $\mu$ the formula is:

$I(x) = I_0 e^{-\mu x}$

where I is the intensity of the beam, $I_0$ is the incident intensity and x is the length of the material traveled.

For our problem, after travelling 1 cm:

$I(1 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 1 cm}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = I_0 e^{- 0.6}$

$I(1 \ cm) = 0.5488 \ I_0$

After travelling 2 cm:

$I(2 \ cm) = I_0 e^{- 0.6 \frac{1}{cm} \ 2 cm}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = I_0 e^{- 1.2}$

$I(2 \ cm) = 0.3012 \ I_0$

<h2>b</h2>

The optical density od is given by:

$od(x) = - log_{10} ( \frac{I(x)}{I_0} )$ .

So, after travelling 1 cm:

$od( 1\ cm) = - log_{10} ( \frac{0.5488 \ I_0}{I_0} )$

$od( 1\ cm) = - log_{10} ( 0.5488 )$

$od( 1\ cm) = - ( - 0.2606)$

$od( 1\ cm) = 0.2606$

After travelling 2 cm:

$od( 2\ cm) = - log_{10} ( \frac{0.3012 \ I_0}{I_0} )$

$od( 2\ cm) = - log_{10} ( 0.3012 )$

$od( 2\ cm) = - ( - 0.5211)$

$od( 2\ cm) = 0.5211$

3 0

3 years ago