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Oksana_A [137]
3 years ago
9

What some nonexamples of diffractions

Physics
1 answer:
Julli [10]3 years ago
5 0

Dispersion, refraction, reflection, propagation, violation, distraction, reaction, acceleration, attenuation, clarification, obfuscation, bloviation, attraction, reception, concentration, rarefaction, evaporation, contraction, accumulation, evisceration, ovulation, lactation, fascination, ejaculation, fertilization, fermentation, intoxication, gravitation, nucleation, publication, privatization, quotation, plagiarization, composition, satiation, tintinnabulation, emasculation, matriculation, graduation, and discussion, are a few examples of NOT diffraction.

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What is the momentum of a 12,000 kg car moving at a speed of 40 m/s?
Harman [31]

Momentum = (mass) x (speed)

Momentum = (12,000 kg) x (40 m/s)

Momentum = (12,000 x 40) kg-m/s

<em>Momentum = 480,000 kg-m/s </em>

8 0
4 years ago
Constanza is on a commuter train between Richville and Shoptown. The train takes 35 minutes to cover the distance between the tw
just olya [345]

Answer:

1.) Displacement = 32.08m

2.) Velocity = 55km/h

3.) Acceleration = 94.29 km/h^2

Explanation:

Given that the train takes 35 minutes to cover the distance between the two towns. The train's average speed is 55 kilometers per hour. That is,

Time t = 35 minute

Convert it to hours by dividing it by 60

35/60 = 7/12 hours

Speed = 55 km/h

The displacement of the car is:

Displacement = 55 × 7/12 = 32.08 m

The velocity of the car will be 55 km/h

The acceleration of the car will be:

Acceleration = velocity/time

Substitute velocity and time into the formula

Acceleration = 55 ÷ 7/12

Acceleration = 94.29 km/h^2

5 0
4 years ago
Prove that..<br>please help<br>​
GaryK [48]

\large{ \boxed{ \bf{ \color{red}{Universal \: law \: of \: gravitation}}}}

Every object in the universe attracts every other object with a force which is proportional to the product of their masses and inversely proportional to the square of the distance between them. The forces along the line joining the centre of the two objects.

❍ Let us consider two masses m1 and m2 line at a separation distance d. Let the force of attraction between the two objects be F.

According to universal law of gravitation,

\large{ \longrightarrow{ \rm{F \propto m_1 m_2}}}

Also,

\large{ \longrightarrow{ \rm{ F \propto  \dfrac{1}{ {d}^{2} } }}}

Combining both, We will get

\large{ \longrightarrow{ \rm{F  \propto  \dfrac{ m_1 m_2}{ {d}^{2}}}}}

Or, We can write it as,

\large{ \longrightarrow{ \rm{F  \propto  \:  G \dfrac{ m_1 m_2}{ {d}^{2} }}}}

Where, G is the constant of proportionality and it is called 'Universal Gravitational constant'.

☯️ Hence, derived !!

<u>━━━━━━━━━━━━━━━━━━━━</u>

8 0
3 years ago
sheila uses a 45N force on her bowling ball across a 15M lane. what work did she do on the bowling ball? show your work.​
mihalych1998 [28]

Answer:

Work=force × distance

=45×15

=675

Explanation:

need thanks and make me brainiest if it helps you

7 0
3 years ago
Need help ??? Please
Yanka [14]
Is there information in the previous question which relates to this one?
8 0
3 years ago
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