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Maru [420]
3 years ago
5

The radio signal sent from the Earth to the Moon was received again on Earth 2.6 seconds after it was sent. The signal ran at 30

0,000 km /s. Using this data, calculate the distance from Earth to the Moon.​
Physics
1 answer:
Pavlova-9 [17]3 years ago
3 0

Answer:

390000 km

Explanation:

2.6 / 2 = 1.3 (second)

300000 * 1.3 = 390000 km

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Give two examples of gravity in the solar system
Nikolay [14]
The planets revolving around the sun, The moon revolving around the Earth.

6 0
3 years ago
How do you find the speed of an object given its mass and kinetic energy (what is the formula)?
madam [21]
   v  =   √ { 2*(KE) ] / m } ; 

Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"] ; 
        
and solve for "v".

______________________________________________________
Explanation:
_____________________________________________________
The formula is:  KE = (½) * (m) * (v²) ;
_____________________________________
  
"Kinetic energy" = (½) * (mass) * (velocity , "squared")
________________________________________________
Note:  Velocity is similar to speed, in that velocity means "speed and direction";  however, if you "square" a negative number, you will get a "positive"; since:  a "negative" multiplied by a "negative" equals a "positive".
____________________________________________
So, we have the formula:
___________________________________
KE = (½) * (m) * (v²) ;  to solve for "(v)" ; velocity, which is very similar to                                          the "speed"; 
___________________________________________________
we arrange the formula ;
__________________________________________________
(KE) = (½) * (m) * (v²) ;  ↔  (½)*(m)* (v²) = (KE) ; 
___________________________________________________

→ We have:  (½)*(m)* (v²) = (KE)  ; we isolate, "m" (mass) on one side of the equation:
______________________________________________________
   
→ We divide each side of the equation by: "[(½)* (m)]" ; 
___________________________________________________
    
           →   [ (½)*(m)*(v²) ] /  [(½)* (m)]  = (KE) / [(½)* (m)]<span> ;
</span>______________________________________________________
 to get: 
______________________________________________________
                           →   v²     =   (KE) / [(½)* (m)]
                     
                           →   v²     = 2 KE / m
_______________________________________________________
Take the "square root" of each side of the equation ;
_______________________________________________________
                          →  √ (v²)  =  √ { 2*(KE) ] / m }
________________________________________________________

                          →     v  =   √ { 2*(KE) ] / m } ; 

Now, plug in the known values for "KE" ["kinetic energy"] and "m" ["mass"]; 
       
and solve for "v".

______________________________________________________
8 0
3 years ago
Firecrackers A and B are 600 m apart. You are standing exactly halfway between them. Your lab partner is 300 m on the other side
pishuonlain [190]

Answer:

See the explanation

Explanation:

Given:

Distance of Firecrackers A and B = 600 m

Event 1 = firecracker 1 explodes

Event 2 = firecracker 2 explodes

Distance of lab partner from cracker A = 300 m

You observe the explosions at the same time

to find:

does event 1 occur before, after, or at the same time as event 2?

Solution:

Since the lab partner is at 300 m distance from the firecracker A and Firecrackers A and B are 600 m apart

So the distance of fire cracker B from the lab partner is:

600 m  + 300 m = 900 m

It takes longer for the light from the more distant firecracker to reach so

Let T1 represents the time taken for light from firecracker A to reach lab partner

T1 = 300/c

It is 300 because lab partner is 300 m on other side of firecracker A

Let T2 represents the time taken for light from firecracker B to reach lab partner

T2 = 900/c

It is 900 because lab partner is 900 m on other side of firecracker B

T2 = T1

900 = 300

900 = 3(300)

T2 = 3(T1)

Hence lab partner observes the explosion of the firecracker A before the explosion of firecracker B.

Since event 1 = firecracker 1 explodes and event 2 = firecracker 2 explodes

So this concludes that lab partner sees event 1 occur first and lab partner is smart enough to correct for the travel time of light and conclude that the events occur at the same time.

8 0
3 years ago
What is the atmospheric pressure and temperature at sea level in a standard<br> atmosphere?
Lady bird [3.3K]

Answer:

The tropospheric tabulation continues to 11,000 meters (36,089 ft), where the temperature has fallen to −56.5 °C (−69.7 °F), the pressure to 22,632 pascals (3.2825 psi), and the density to 0.3639 kilograms per cubic meter (0.02272 lb/cu ft). Between 11 km and 20 km, the temperature remains constant

Explanation:

Hope this helped, Have a wonderful day!!

4 0
3 years ago
An Airbus A350 is initially moving down the runway at 6.0 m/s preparing for takeoff. The pilot pulls on the throttle so that the
alexdok [17]

Answer:

t=67.7s

Explanation:

From this question we know that:

Vo = 6m/s

a = 1.8 m/s2

D = 1500m

And we also know that:

X=V_{o}*t + \frac{a*t^{2}}{2}   Replacing the known values:

1500=6t+0.9*t^{2}    Solving for t we get 2 possible answers:

t1 = -44.3s   and t2 = 67.7s    Since negative time represents an instant before the beginning of the movement, t1 is discarded. So, the final answer is:

t = 67.7s

8 0
3 years ago
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