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2 years ago

Why do some scientists think that Irr II galaxies have irregular, distorted shapes?

Physics

1 answer:

laila [671]2 years ago

3 0

They believe the distortions happened when two galaxies collided.

Hope This Helps :)

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F PHY112 Chapter 29 Homework Question 2 of 4 > -/1 View Policies Current Attempt in Progress An incident X-ray photon of wave

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Answer: i dont know

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2 years ago

Which statement about the ocean is true? A. No evaporation or precipitation in the water cycle occurs over the ocean. B. Most ev

OLEGan [10]

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Explanation:

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2 years ago

What element has 31 protons and 37 neutrons???

Lelechka [254]

You can tell the neutrons, protons, and electrons of an element just by looking at a periodic table. The number on top of an element is its charge; for fluorine this would be 9, for rubidium<span> this would be 37

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2 years ago

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If the car passes point A with a speed of 20 m/s and begins to increase its speed at a constant rate of at = 0.5 m/s2 , determin

IceJOKER [234]

Answer:

$1.68 \frac{m}{s^2}$

Explanation:

Please find the image for the question as attached file.

Solution -

Given -

First of all we will calculate the velocity at point C,

As per newton's third law of motion-

$V_C^2 = V_A^2 + 2 a_t (S_C - S_A)\\$

Substituting the given values in above equation, we get -

$V_C^2 = 20^2 + 2*0.5*(100-0)\\V_C = 22.361 \frac{m}{s}$

Now we will determine the radius of curvature for the curve shown in the attached image

$Y = 16 - \frac{1}{625} X^2\\$

Differentiating on both the sides, we get -

$\frac{dy}{dx} = -3.2 (10^-3) X\\\frac{d^2y}{d^2x} = -3.2 (10^-3)\\Curve = \frac{[1+(\frac{dy}{dx})^2]^{\frac{3}{2}} }{\frac{d^2y}{d^2x}} \\Curve = 312.5$ meter

Acceleration on curved path

$a = \frac{V_C^2}{Curve} \\a = \frac{22.361^2}{312.5} \\a= 1.60 \frac{m}{s^2}$

Final acceleration

$a_f = \sqrt{0.5^2 + 1.6^2} \\a_f = 1.68\frac{m}{s^2}$

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2 years ago

A skier with a potential energy of 137,200 J waits on top of a ski jump that is 200 m high. What is the mass of the skier?

wolverine [178]

Remember that the equation for gravitational potential energy is PE = mgy, where PE = potential energy, m=mass, g=acceleration of gravity, and y=distance above your reference level.

You know that PE = 137200J, y = 200m, and g = 9.8 $m/s^{2}$ . Plug these values into the equation and solve for m:
$PE = mgy\\
137200\:J = m(9.8 \: m/s^{2})(200\:m)\\
m = 70 kg$

The mass of the skier is 70kg.

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3 years ago

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