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Kaylis [27]
4 years ago
5

Why is the distance between the sun and Neptune dim light years

Physics
2 answers:
Bingel [31]4 years ago
4 0
.00004755 light years... Its really not far enough to measure in light years.
djyliett [7]4 years ago
3 0

The average distance from the Sun to Neptune is about 2.795 billion miles.

That's roughly  0.00048  of a light year .

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Suppose you're studying levels of job stress and measures of job efficiency in a nursing home. Which one of the following statem
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B. As job efficiency increases, stress levels decline (THIS IS WRONG)I think its between A. As job efficiency decreases, stress levels increaseD. As stress levels increase, job efficiency increases
6 0
3 years ago
Read 2 more answers
Blue bumper car, traveling east with a force of 100 N, runs into a yellow bumper car traveling west with a force of 300 N. Becau
Pani-rosa [81]

Answer:

the blue bumper car

Explanation:

because the yellow bumper car is coming in with much more force

and according to Newton, actions and reactions are equal and opposite, so the yellow bumper car will five out 300N while the blue car will give out only 100N

then it almost because a fight between 300N vs 100N

and obviously, the 300N will win and the 100N will be pushed back

4 0
3 years ago
In a double slit experiment, if the separation between the two slits is 0.050 mm and the distance from the slits to a screen is
meriva

The spacing between the first-order and second-order bright fringes is 3 cm. Hence, this is the required solution.

<h3>What is double slit experiment?</h3>

The double-slit experiment serves as a proof in current physics that both light and matter may exhibit properties of classically defined waves and particles. It also illustrates the inherently probabilistic nature of quantum mechanical events. Thomas Young carried out the first experiment of this kind employing light in 1802, illustrating how light behaves like a wave. It was formerly believed that light was made up of either waves or particles. About a century later, with the advent of modern physics, it was discovered that light may in fact exhibit behaviour like that of both waves and particles. The identical behaviour of electrons was first shown by Davisson and Germer in 1927, and it was later extended to atoms and molecules.

The separation between the slits, d = 0.05mm = 5×10⁻⁵ m

The distance from the slits to a screen, D = 2.5 m

Let x is the spacing between the first-order and second-order bright fringes when coherent light of wavelength 600 nm illuminates the slits,

λ = 600nm = 6× 10⁻⁷ m  

We know that the bright fringe is given by :

y = nλD/d

So, the spacing between the first-order and second-order bright fringes is :

x = 2λD/d - λD/d

x =  λD/d

x = 6 × 10⁻⁷ × 2.5/5 × 10⁻⁵

x = 0.03 m

or

x = 3 cm

So, the spacing between the first-order and second-order bright fringes is 3 cm. Hence, this is the required solution.

to learn more about double slit experiment go to -

brainly.com/question/28108126

#SPJ4

8 0
1 year ago
A linear accelerator uses alternating electric fields to accelerate electrons to close to the speed of light. A small number of
diamong [38]

Answer:i

9E13 ELECTRONS

Explanation:

First we find the number of charges in two hrs

Which is - 2x 7200=- 14400NC

So no of electrons is now q/e

-144x10^-7/-1.6*10^-19

= 9*10^13

7 0
3 years ago
A particle moves along the x axis. It is initially at the position 0.270 m, moving with velocity 0.140 m/s and acceleration 20.3
mixas84 [53]

Answer:

a) -2.34 m

b) -1.3 m/s

c) 0.056 m

d) 0.320 m/s

Explanation:

part a

Given:

s(0) = 0.27 m

v(0) = 0.14 m/s

a = -0.320 m/s^2

t = 4.50s

Using kinematic equation of motion for constant acceleration:

s (t) = s(0) + v(0)*t + 0.5*a*t^2

s ( 4.5 s ) = 0.27 + 0.14*4.5 + 0.5*-0.32*4.5^2

s(4.5) = -2.34 m

part b

Using kinematic equation of motion for constant acceleration:

v(t) = v(0) + a*t

v(4.5s) = (0.14) + (-0.32)(4.5) = -1.3 m/s

part c

Use equation for simple harmonic motion:

s(t) = A*cos(w*t)

v(t) = -A*w*sin (w*t)

a(t) = -A*(w)^2 * cos (w*t)

0.27 m = A*cos(w*t)   .... Eq 1

0.14 m/s = - A*w*sin (w*t)  .....Eq2

-0.320 = -A*(w)^2 * cos (w*t)   .... Eq3

Solve the three equations above for A, w, and t

Divide 1 and 3:

w^2 = 0.32 / 0.27

w = 1.0887 rad / s

Divide 2 and 1:

w*tan(wt) = 0.14 / 0.27

tan(1.0887*t) = 0.476289

t = 0.4083 s

A = 0.27 / cos (1.0887*0.4083) = 0.3 m

Hence, the SHM is s(t) = 0.3*cos(1.0887*t)

s(4.5) = 0.3*cos(1.0887*4.5) = 0.056 m

part d

v(t) = - 0.32661 * sin (1.0887*t)

v(4.5) = - 0.32661 * sin (1.0887*4.5) = 0.320 m/s

3 0
3 years ago
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