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katrin2010 [14]
3 years ago
10

What was the physical principle used by edwin hubble to demonstrate that the "spiral nebulae" are very distant and large?

Physics
1 answer:
alekssr [168]3 years ago
4 0
<span>The physical principle used by Edwin Hubble to demonstrate that the "spiral nebulae" are very distant and large was the inverse-square law of light. The Cepheid variable stars in the spiral nebulae obey a period-luminosity relation; by measuring the period of a Cepheid star, its average luminosity can be determined, and then, by comparing this with its average apparent brightness, its distance can be calculated.</span>
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Can someone explain to me what Murphy's law is and how it works?
Vikki [24]
<span>Murphy's law is an adage or epigram that is typically stated as: Anything that can go wrong, will go wrong.
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3 0
4 years ago
An object is thrown straight up with a velocity, in ft/s, given by v(t)= -32t + 83, where t is in seconds, from a height of 46 f
____ [38]
<h2>a) Initial velocity = 83 ft/s</h2><h2>b) Object's maximum speed = 99.4 ft/s</h2><h2>c) Object's maximum displacement = 153.64 ft</h2><h2>d) Maximum displacement occur at t = 2.59 seconds.</h2><h2>e) The displacement is zero when t = 5.70 seconds</h2><h2>f) Object's maximum height = 153.64 ft</h2>

Explanation:

We have velocity

             v(t)= -32t + 83

Integrating

              s(t) = -16t²+83t+C

At t = 0 displacement is 46 feet

              46 = -16 x 0²+83 x 0+C

                 C = 46 feet

So displacement is

              s(t) = -16t²+83t+46

a) Initial velocity is

                 v(0)= -32 x 0 + 83 = 83 ft/s

       Initial velocity = 83 ft/s

b) Maximum velocity is when the object reaches ground, that is s(t) = 0 ft

Substituting

             0 = -16t²+83t+46

             t = 5.70 seconds

Substituting in velocity equation

           v(t)= -32 x 5.70 + 83 = -99.4 ft/s

           Object's maximum speed = 99.4 ft/s

c) Maximum displacement is when the velocity is zero

   That is

                 -32t + 83 = 0

                       t = 2.59 s

Substituting in displacement equation

                s(2.59) = -16 x 2.59²+83 x 2.59+46 = 153.64 ft

Object's maximum displacement = 153.64 ft

d) Maximum displacement occur at t = 2.59 seconds.

e) Refer part b

   The displacement is zero when t = 5.70 seconds

f) Same as option d

   Object's maximum height = 153.64 ft

5 0
4 years ago
Which area of Earth is most similar to the sun’s convection zone?
Pavlova-9 [17]
I would say the correct answer is the third option. The area of the Earth that is most similar to the Sun's convection zone would be the mantle. The convection zone of the sun is its outermost layer where heat transfer by convection happens which is similar to the Earth's mantle.
7 0
3 years ago
Read 2 more answers
Which definition is correct and uses only quantities rather than units? A. Density is mass per unit volume B. potential differen
kvv77 [185]
I think it is D, let me know if I'm wrong.
4 0
4 years ago
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The second-order bright fringe in a single-slit diffraction pattern is 1.35 mm from the center of the central maximum. The scree
Dafna11 [192]

Answer:

The wavelength is 754.2 nm.

Explanation:

Given that,

Diffraction pattern y= 1.35 mm

Width = 0.838 mm

Distance D= 75 cm

We need to calculate the wavelength

Using formula of diffraction pattern

y=\dfrac{m\lambda D}{d}

\lambda=\dfrac{yd}{mD}

Where, y = diffraction pattern

m = order

d = width

D = distance

Put the value into the formula

\lambda=\dfrac{1.35\times10^{-3}\times0.838\times10^{-3}}{2\times75\times10^{-2}}

\lambda=7.542\times10^{-7}\ m

\lambda=754.2\ nm

Hence, The wavelength is 754.2 nm.

4 0
4 years ago
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