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3 years ago

Everything we know about stars comes from studying their _____.

2 answers:

8 0

Everything we know about stars comes from studying their radiation/light. By measuring the wavelength or frequency of light coming from objects in the universe, we can learn something about their nature. Starts emit a range of wavelengths. The frequency of light is directly related to its color, so it follows that stars have different colors because they have different peak wavelengths of light.

denis-greek [22]3 years ago

5 0

Light / Radiation. The light radiation defines its color. Thermal radiation for its temperature and the overall appearance of the wavelengths the star emits gives off the info from where astronomers and scientist are able to build up knowledge on the age and current state of the stars

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Jack rolled a marble down a ramp and recorded the potential energy and kinetic energy of the marble at different positions on th

mote1985 [20]

Answer: D

Height of marble from ground

Explanation:

From the formula of kinetic energy and potential energy,

K.E = 1/2mv^2

While

P.E = mgh

From all the parameters given from the question. You can see that mass is constant, acceleration due to gravity is also constant.

Independent variable must be a value that can varies.

Since Jack rolled a marble down a ramp and recorded the potential energy and kinetic energy of the marble at different positions on the ramp to see the effects on both energies.

This different position must be the height which will produce an effect on the potential and kinetic energy of the marble.

Independent variable always provides an effect for dependent variable. Which are kinetic energy and potential energy in this case.

Height of marble from ground is the right answer.

6 0

3 years ago

Why is it not possible to achieve a higher resolution in the light microscope?.

Rudiy27

Higher resolution cannot be achieved with a light microscope because wavelength of light is larger that wavelength of electrons.

<h3>What is light microscope?</h3>

Light microscope is defined as the type of microscope that makes use of visible light to magnify objects into images that can be observed.

The wavelength of light is larger than that of the electrons, therefore, higher resolution cannot be achieved with a light microscope.

Learn more about microscope here:

brainly.com/question/15744335

#SPJ1

3 0

2 years ago

In the figure, particle A moves along the line y = 31 m with a constant velocity v with arrow of magnitude 2.8 m/s and parallel

insens350 [35]

Answer:

59.26°

Explanation:

Since a is the acceleration of the particle B, the horizontal component of acceleration is a" = asinθ and the vertical component is a' = acosθ where θ angle between a with arrow and the positive direction of the y axis.

Now, for particle B to collide with particle A, it must move vertically the distance between A and B which is y = 31 m in time, t.

Using y = ut + 1/2a't² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a' = vertical component of particle B's acceleration = acosθ.

So, y = ut + 1/2a't²

y = 0 × t + 1/2(acosθ)t²

y = 0 + 1/2(acosθ)t²

y = 1/2(acosθ)t² (1)

Also, both particles must move the same horizontal distance to collide in time, t.

Let x be the horizontal distance,

x = vt (2)where v = velocity of particle A = 2.8 m/s and t = time for collision

Also, using x = ut + 1/2a"t² where u = initial velocity of particle B = 0 m/s, t = time taken for collision, a" = horizontal component of particle B's acceleration = asinθ.

So, x = ut + 1/2a"t²

x = 0 × t + 1/2(ainsθ)t²

x = 0 + 1/2(asinθ)t²

x = 1/2(asinθ)t² (3)

Equating (2) and (3), we have

vt = 1/2(asinθ)t² (4)

From (1) t = √[2y/(acosθ)]

Substituting t into (4), we have

v√[2y/(acosθ)] = 1/2(asinθ)(√[2y/(acosθ)])²

v√[2y/(acosθ)] = 1/2(asinθ)(2y/(acosθ)

v√[2y/(acosθ)] = ytanθ

√[2y/(acosθ)] = ytanθ/v

squaring both sides, we have

(√[2y/(acosθ)])² = (ytanθ/v)²

2y/acosθ = (ytanθ/v)²

2y/acosθ = y²tan²θ/v²

2/acosθ = ytan²θ/v²

1/cosθ = aytan²θ/2v²

Since 1/cosθ = secθ = √(1 + tan²θ) ⇒ sec²θ = 1 + tan²θ ⇒ tan²θ = sec²θ - 1

secθ = ay(sec²θ - 1)/2v²

2v²secθ = aysec²θ - ay

aysec²θ - 2v²secθ - ay = 0

Let secθ = p

ayp² - 2v²p - ay = 0

Substituting the values of a = 0.35 m/s, y = 31 m and v = 2.8 m/s into the equation, we have

ayp² - 2v²p - ay = 0

0.35 × 31p² - 2 × 2.8²p - 0.35 × 31 = 0

10.85p² - 15.68p - 10.85 = 0

dividing through by 10.85, we have

p² - 1.445p - 1 = 0

Using the quadratic formula to find p,

$p = \frac{-(-1.445) +/- \sqrt{(-1.445)^{2} - 4 X 1 X (-1)}}{2 X 1} \\p = \frac{1.445 +/- \sqrt{2.088 + 4}}{2} \\p = \frac{1.445 +/- \sqrt{6.088}}{2} \\p = \frac{1.445 +/- 2.4675}{2} \\p = \frac{1.445 + 2.4675}{2} or p = \frac{1.445 - 2.4675}{2} \\p = \frac{3.9125}{2} or p = \frac{-1.0225}{2} \\p = 1.95625 or -0.51125$