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2 years ago

HELP WILL MEDAL !!!!!!!!!!!!!!!!!!!

2 answers:

8 0

The speed of light in a vaccum is 182,282 miles per second. Light travels through space over a period of time to reach the earth. When a new star becomes visible it means that a star's light has finnaly reached us. It takes the sun's light an average of 8 minutes and 20 seconds to reach earth. Hope this helps.

Katarina [22]2 years ago

7 0

I think this is the best way to explain it
light from the sun takes 8 minutes and 20 seconds to travel to Earth. So it would be the same if the Earth's light to travel to the sun. The sun is described to be a star.

You might be interested in

An object is shot upwards, from the ground, with an initial velocity of 120

Alex777 [14]

Answer: 30 metres

Explanation:

Initial velocity of object = 120m/s

Time taken = 4.0s

Distance covered by object = ?

Recall that distance = (Change in velocity / Time taken)

Distance = (120m/s)/4.0s

= (120m/s) / 4.0s

= 30m

Thus, the object will be 30 metres high

6 0

2 years ago

A hydrogen bond forms by the electrostatic interaction of opposite charges in two molecules. If

Darya [45]

The force between the molecules involved in the bond is 6. 426 *10^-11 Newton

<h3>How to determine the force</h3>

Using the formula:

F = K[q1 x q2]/D^2

where K is coulombs constant =9 *10 ^9 Nm^2/C^2.

q1 and q2 = charges = 1.60x10 -20C

d = distance between the charges = 2x10 -10 m

Substitute the values into the formula

F = $9 * 10^9\frac{ 1.60*10^ -20 * 1.60 *10^ -20}{2x10^ -10^{2} }$

F = $9 *10^9\frac{2. 856* 10^-40}{4* 10^-20}$

F = $9* 10^9 * 7. 14* 10^-21$

F = $6. 426 * 10^-11$ Newton

Thus, the force between the molecules involved in the bond is 6. 426 *10^-11 Newton

Learn more about electrostatic force here:

brainly.com/question/8424563

#SPJ1

5 0

1 year ago

A 0.473 kg ice puck, moving east with a speed of 2.76 m/s, has a head-on collision with a 0.819 kg puck initially at rest. Assum

Gekata [30.6K]

Answer:

The final speed of puck 1 is 0.739 m/s towards west and puck 2 is 2.02 m/s towards east .

Explanation:

Let us consider east as positive direction and west as negative direction .

Given

mass of puck 1 , $m_1= 0.473 kg$

mass of puck 2 , $m_2= 0.819 kg$

initial speed of puck 1 , $u_1=2.76\frac{m}{s}$

initial speed of puck 2 , $u_2=0.00\frac{m}{s}$

Final speed of puck 1 and puck 2 be $v_1\, and\, v_2$ respectively

Apply conservation of linear momentum

$m_1u_1+m_2u_2=m_1v_1+m_2v_2$

=> $0.473\times 2.76+0.0=0.473\times v_1+0.819\times v_2$

=> $1.594=0.5775\times v_1+ v_2$ -----(A)

Since collision is perfectly elastic , coefficient restitution e=1

$u_2-u_1=v_1-v_2$

=> $0-2.76=v_1-v_2$ ------(B)

From equation (A) and (B)

$v_1=-0.739\frac{m}{s}$

and $v_2=2.02\frac{m}{s}$

Thus the final speed of puck 1 is 0.739 m/s towards west and puck 2 is 2.02 m/s towards east .

3 0

3 years ago

2. A jack exerts a vertical force of 4.5 X 103

skad [1K]

Correct Question:-

A jack exerts a vertical force of 4.5 × 10³

newtons to raise a car 0.25 meter. How much

work is done by the jack?

$\\ \\$

Given :-

$\star \sf \small force = 4.5 \times {10}^{3} \: newton$

$\star \sf \small distance = 0.25 \: meter$

$\\ \\$

To find:-

$\sf \star \: work = \: ?$

$\\ \\$

Solution:-

we know :-

$\bf \dag \boxed{ \rm work = force \times distance}$

$\\ \\$

So:-

$\dashrightarrow \sf work = force \times distance$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times 0.5 \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{0 \cancel.5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \cancel \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{4\cancel.5}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{3 - 1} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{1} \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times \cancel{10}}{ \cancel2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times 5}{ 1} \\$

$\\ \\$

$\dashrightarrow \sf work =225 \times 10$

$\\ \\$

$\dashrightarrow \bf work =\red{2250\: joule}$

5 0

2 years ago

You push a shopping cart for fun with a force of 60N. If the shopping cart has a mass of 12kg, what is the

posledela

Answer:

$\huge\boxed{\sf a = 5 \ ms^{-2}}$

Explanation:

<u>Given:</u>

Force = f = 60 N

Mass = m = 12 kg

<u>Required:</u>

Acceleration = a = ?

<u>Formula:</u>

F = ma

<u>Solution:</u>

Rearranging formula

a = F / m

a = 60 / 12

a = 5 ms⁻²

$\rule[225]{225}{2}$

Hope this helped!

<h3>~AH1807</h3><h3>Peace!</h3>

4 0

2 years ago