1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
SIZIF [17.4K]
3 years ago
15

4.5 billion km, the average separation between the sun and Neptune (report answer in hours). How long does it take light to trav

el each of the following distances?
Physics
1 answer:
Liula [17]3 years ago
4 0

Answer:

t = 4.17 hours

Explanation:

given,

The distance between Sun and Neptune, d = 4.5 billion Km

                                                                         = 4.5 x 10⁹ Km

                                                                          = 4.5 x 10¹¹ m

The velocity of light, c = 3 x 10⁸ m/s

The velocity is always equal to displacement by the time.

                                           <em>V = d / t    m/s</em>

∴                                           t = d / V

                                               = 4.5 x 10¹¹ m / 3 x 10⁸ m/s

                                               = 15,000 s

                                               = 4.17 h

Hence, the time taken by the light rays to reach the Neptune is, t = 4.17 h

You might be interested in
A 4-kg object is moving with a speed of 5 m/s at a height of 2 m. The kinetic
tatyana61 [14]

Hello!

\large\boxed{KE = 50J}

Use the formula for kinetic energy:

KE = \frac{1}{2}mv^{2}

Plug in the given mass and velocity:

KE = \frac{1}{2} (4)5^{2}

Simplify:

KE = \frac{1}{2} (100)\\\\KE = 50 J

7 0
3 years ago
Light can travel from the sun to ________ in less than ten minutes.
Solnce55 [7]
To the Earth in less than ten minutes. 
5 0
3 years ago
. Consider the equation =0+0+02/2+03/6+04/24+5/120, where s is a length and t is a time. What are the dimensions and SI units of
Olegator [25]

Answer:

See Explanation

Explanation:

Given

s=s_0+v_0t+\frac{a_0t^2}{2}+ \frac{j_0t^3}{6}+\frac{S_0t^4}{24}+\frac{ct^5}{120}

Solving (a): Units and dimension of s_0

From the question, we understand that:

s \to L --- length

t \to T --- time

Remove the other terms of the equation, we have:

s=s_0

Rewrite as:

s_0=s

This implies that s_0 has the same unit and dimension as s

Hence:

s_0 \to L --- dimension

s_o \to Length (meters, kilometers, etc.)

Solving (b): Units and dimension of v_0

Remove the other terms of the equation, we have:

s=v_0t

Rewrite as:

v_0t = s

Make v_0 the subject

v_0 = \frac{s}{t}

Replace s and t with their units

v_0 = \frac{L}{T}

v_0 = LT^{-1}

Hence:

v_0 \to LT^{-1} --- dimension

v_0 \to m/s --- unit

Solving (c): Units and dimension of a_0

Remove the other terms of the equation, we have:

s=\frac{a_0t^2}{2}

Rewrite as:

\frac{a_0t^2}{2} = s_0

Make a_0 the subject

a_0 = \frac{2s_0}{t^2}

Replace s and t with their units [ignore all constants]

a_0 = \frac{L}{T^2}\\

a_0 = LT^{-2

Hence:

a_0 = LT^{-2 --- dimension

a_0 \to m/s^2 --- acceleration

Solving (d): Units and dimension of j_0

Remove the other terms of the equation, we have:

s=\frac{j_0t^3}{6}

Rewrite as:

\frac{j_0t^3}{6} = s

Make j_0 the subject

j_0 = \frac{6s}{t^3}

Replace s and t with their units [Ignore all constants]

j_0 = \frac{L}{T^3}

j_0 = LT^{-3}

Hence:

j_0 = LT^{-3} --- dimension

j_0 \to m/s^3 --- unit

Solving (e): Units and dimension of s_0

Remove the other terms of the equation, we have:

s=\frac{S_0t^4}{24}

Rewrite as:

\frac{S_0t^4}{24} = s

Make S_0 the subject

S_0 = \frac{24s}{t^4}

Replace s and t with their units [ignore all constants]

S_0 = \frac{L}{T^4}

S_0 = LT^{-4

Hence:

S_0 = LT^{-4 --- dimension

S_0 \to m/s^4 --- unit

Solving (e): Units and dimension of c

Ignore other terms of the equation, we have:

s=\frac{ct^5}{120}

Rewrite as:

\frac{ct^5}{120} = s

Make c the subject

c = \frac{120s}{t^5}

Replace s and t with their units [Ignore all constants]

c = \frac{L}{T^5}

c = LT^{-5}

Hence:

c \to LT^{-5} --- dimension

c \to m/s^5 --- units

4 0
3 years ago
Convert the following statement to the language used by physicists, "I am cold, please turn on the heat."​
Vikki [24]

Answer:

Explanation:fog

6 0
3 years ago
A 10kg oil is heated by a source of heat that supplies 800J of heat every minute. Calculate The heat added to oil after 30 minut
goblinko [34]

Answer:

4,200 joules per kilogram per degree Celsius

Explanation:

The specific heat capacity of a material is the energy required to raise one kilogram (kg) of the material by one degree Celsius (°C). The specific heat capacity of water is 4,200 joules per kilogram per degree Celsius (J/kg°C). This means that it takes 4,200 J to raise the temperature of 1 kg of water by 1°C.

4 0
2 years ago
Other questions:
  • Why do you think the government of ancient Rome was unfair?
    5·1 answer
  • Newton's first law states that an object traveling with a constant velocity will remain traveling at a constant velocity unless
    13·1 answer
  • Can someone right me a flooding psa ?
    14·2 answers
  • How much kinetic energy does a 4.50kg block with a speed of 5.00m/s have?
    14·1 answer
  • 1. Rocky lifted a 300 newton rock 2 meters off the ground and carried it 2 -
    8·1 answer
  • The photo shows a system that consists of a person on a diving board. What change could you make to the system that would decrea
    13·2 answers
  • Blank is the the decrease in a materail's volume when it's blank decrease ​
    6·1 answer
  • What is the difference between biomass and fossil fuels main points?
    14·1 answer
  • What determines the type of igneous rock that forms from magma?
    10·2 answers
  • What is 7 times 7 times 7 plus 3
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!