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SashulF [63]
3 years ago
9

What is most likely the author’s motive for writing this article?

Physics
2 answers:
netineya [11]3 years ago
5 0

Answer:

D

Explanation:

Rufina [12.5K]3 years ago
4 0

Answer

Its D

Explanation:

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A force of 125 N is applied to a 50 kg. Mass. What is the acceleration of the mass?
grandymaker [24]
The acceleration would be 2.5 m/sec^2
A= acceleration
F= force
M= mass
A = f/m
125N/50kg=2.5 m/s^2
3 0
4 years ago
Give an example of how the relationship between force and acceleration might affect YOU.
IrinaK [193]
Take a car collision as an example, the more you speed up as means of acceleration, the more force will be on impact.
5 0
3 years ago
What do all types of electromagnetic radiation have in common?
Andrew [12]
<span>They are emitted by the unstable nuclei of certain atoms.
 That's all I could find out; Sorry I couldn't be more of an help.</span>
5 0
3 years ago
A bowling ball with a mass of 7.0kg strikes a pin that had a mass of 2.0kg the pin flies forward with a velocity of 6.0m/s, and
Natalka [10]

The conservation of momentum P states that the amount of momentum remains constant when there are not external forces.

We don't have external forces, so:

P_0 = P_1\\m_bv_{0b}+m_pv_{0p}=m_bv_{1b}+m_pv_{1p}\\

Where:

  • mb is the mass of the bowling ball
  • mp the mass of the pin
  • v_{0b}\quad and\quad v_{0p} the initial velocities of the bowling ball and the pin.
  • v_{1b}\quad and\quad v_{1p} the final velocities of the bowling ball and the pin.

Solving for v0b:

v_{0b} =\dfrac{m_bv_{1b}+m_pv_{1p}- m_pv_{0p}}{m_{b}}\\\\v_{0b} =\dfrac{(7\;kg)(4\;m/s)+(2\;kg)(6\;m/s)- (2\;kg)(0 \;m/s)}{7\;kg}\\v_{0b}=\dfrac{40}{7}\;m/s\\\\\boxed{v_{0b}\approx5.71\;m/s}

<h2>R/ The original velocity of the ball was 5.71 m/s.</h2>
6 0
3 years ago
The red giant Betelgeuse has a surface temperature of 3000 K and is 600 times the diameter of our sun. (If our sun were that lar
Nat2105 [25]

Answer:

Explanation:

a )

Radius of the sun = .69645 x 10⁹ m .

600 times = 600 x .69645 x 10⁹ m

= 4.1787 x 10¹¹ m .

surface area A = 4π (4.1787 x 10¹¹)²

= 219.317 x 10²²

energy radiated E = σ A Τ⁴

= 5.67 x 10⁻⁸ x 219.317 x 10²² x (3000)⁴

= 100695 x 10²⁶ J

To know the wavelength of photon emitted

\lambda_mT= b

\lambda_m= \frac{b}{T}

= 2.89777 x 10⁻³ / 3000

= 966 nm

= 1275 /966 eV

1.32 x 1.6 x 10⁻¹⁹ J

= 2.112 x 10⁻¹⁹ J

No of photons radiated = 100695 x 10²⁶ / 2.112 x 10⁻¹⁹

= 47677.5 x 10⁴⁵

= .476 x 10⁵⁰ .

b )

energy radiated by our sun per second

E₂ = σ A 5800⁴

energy radiated by Betelgeuse per second

E₁ = σ  x 600²A x  3000⁴

E₁ / E₂  = σ  x 600²A x  3000⁴ / σ A 5800⁴

= 36 X 10⁴ x 3⁴ x 10¹² / 58⁴ x 10⁸

= 25.76 x 10⁸ x 10⁻⁵

= 25760 times .

4 0
4 years ago
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