Ask question

Ask question

All categories

3 years ago

15

How long does light take to travel from the sun to earth? heres the exact question: light travels at 300,000 km.s. The sun is a

total of 148 million kilometres from the earth. How long does it take light to travel from the sun to the earth

1 answer:

Artist 52 [7]3 years ago

4 0

I've heard it takes around 8 min

You might be interested in

What is the acceleration of the the object during the first 4 seconds?

AVprozaik [17]

Answer:

Velocity (m/s) over time (s) graph

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second

Explanation:

What this formula is telling us is that if we know the acceleration of an object, and the ... we can plug in our acceleration of 12.5 m/s2 for a, and 4 seconds for t.

Velocity (m/s) over time (s) graph

Velocity (m/s) over time (s) graph

We could write out our average acceleration as:

a = Δv/ Δta=Δv/Δta, equals, Δ, v, slash, Δ, t

a = (15 m/s - 0 m/s) / 0.2 seconds

a = 15 m/s / 0.2 seconds

a = 75 m/s / second

6 0

3 years ago

7. When initially-unpolarized light passes through three polarizing filters, each oriented at 45 degree angles from the precedin

GREYUIT [131]

Answer:

Explanation:

Let the intensity of unpolarised light be I₀ . After passing through the first polarising filter , the intensity is I₀ / 2 .

After second filter , the intensity will be I₀ / 2 x cos²45 = I₀ / 4

After third filter , the intensity will be I₀ / 4 x cos²45 = I₀ / 8 .

So,

1 / 8 the of initial light passes through the last filter .

7 0

3 years ago

Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.7 cm2 , 2.5 cm2 , 2.2 cm2 , and 3 c

Travka [436]

Answer:

22.1 V

Explanation:

We are given that

$A_1=0.7 cm^2=0.7\times 10^{-4} m^2$

$A_2=2.5 cm^2=2.5\times 10^{-4} m^2$

$A_3=2.2 cm^2=2.2\times 10^{-4} m^2$

$A_4=3 cm^2=3\times 10^{-4} m^2$

Using $1cm^2=10^{-4} m^2$

We know that

$R=\frac{\rho l}{A}$

In series

$R=R_1+R_2+R_3+R_4$

$R=\frac{\rho l}{A_1}+\frac{\rho l}{A_2}+\frac{\rho l}{A_3}+\frac{\rho l}{A_4}$

$R=\frac{\rho l}{\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4}}$

Substitute the values

$R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})$

$R=\rho l(2.62\times 10^4)$

$V=145 V$

$I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}$

Voltage across the 2.5 square cm wire= $IR=I\times \frac{\rho l}{A_2}$

Voltage across the 2.5 square cm wire= $\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V$

Voltage across the 2.5 square cm wire=22.1 V

6 0

3 years ago

How to know which side is positive or negative on resistors.

Semenov [28]

Answer:

"look for a large stripe or a minus sign (or both) on one side of the capacitor."

Explanation:

i hope it helped :)

5 0

2 years ago

A plane travels at a constant speed of 250 m/s as it flies once around a horizontal circle whose radius is 2542m.

hichkok12 [17]

C has to be the correct answer

5 0

3 years ago