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3 years ago

ANSWER THIS AND YOU WILL BE THE BRAINLIEST

Physics

1 answer:

VladimirAG [237]3 years ago

6 0

Answer:

Explanation:

it is correct

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Name three bright Saturn ring features, and explain why they are so bright.

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The F Ring, the Cassini Division, and the C Ring are bright ring features. They are bright due to the low concentration of materials within them, which allows sunlight to shine through.

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3 years ago

At present most of the world's energy needs are supplied by what kind of energy

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Fossil fuels . . . coal, oil, natural gas

Among primitive cultures, wood is an important source.

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4 years ago

Gordon throws a baseball into the air. It rises, stops when it reaches its greatest height, and then falls back to the ground. A

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3 years ago

The radioactive element radium (Ra) decays by a process known asalpha decay,in which the nucleus emits a helium nucleus. (Factoi

tamaranim1 [39]

<u>Answer:</u> The energy released in the decay process is $4.6800\times 10^{11}J$

<u>Explanation:</u>

The equation for the alpha decay of Ra-226 follows:

$_{88}^{226}\textrm{Ra}\rightarrow _{2}^{4}\textrm{He}+_{86}^{222}\textrm{Rn}$

To calculate the mass defect, we use the equation:

Mass defect = Sum of mass of product - Sum of mass of reactant

$\Delta m=(m_{Rn}+m_{He})-(m_{Ra})$

We know that:

$m_{Rn}=222.0176u\\m_{Ra}=226.0254u\\m_{He}=4.0026u$

Putting values in above equation, we get:

$\Delta m=(222.0176+4.0026)-(226.0254)=-0.0052g=-5.2\times 10^{-6}kg$

(Conversion factor: 1 kg = 1000 g )

To calculate the energy released, we use Einstein equation, which is:

$E=\Delta mc^2$

$E=(-5.2\times 10^{-6}kg)\times (3\times 10^8m/s)^2$

$E=-4.6800\times 10^{11}J$

Hence, the energy released in the decay process is $4.6800\times 10^{11}J$

8 0

3 years ago

Write the y-equation for a wave traveling in the negative x-direction with wavelength 50 cm, speed 4.0 m/s, and amplitude 5.0 cm

Zanzabum

Answer:

$y = A sin(2\pi(\dfrac{x}{50})+ 8 t)$

Explanation:

given,

wavelength, λ = 50 cm

speed, v = 4 m/s

Amplitude, A = 5 cm

general equation of the wave along x- axis

$y = A sin(2\pi(\dfrac{x}{\lambda})\pm \nu t)$

sign is positive when wave is traveling in negative direction

now,

$\nu = \dfrac{v}{\lambda}$

$\nu = \dfrac{4}{0.5}$

$\nu = 8\ s^{-1}$

inserting all the values

$y = A sin(2\pi(\dfrac{x}{50})+ 8 t)$

Hence, the y-equation of wave is equal to $y = A sin(2\pi(\dfrac{x}{50})+ 8 t)$

7 0

3 years ago