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Ugo [173]
3 years ago
10

During takeoff, an airplane goes from 0 to 56 m/s in 9 s. How fast is it going after 4 s?

Physics
1 answer:
meriva3 years ago
8 0
24 miles per second
56/9=6
6+6+6+6=24
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What type of wave is shown below?
Elena-2011 [213]

Answer:

<u>B</u><u>.</u><u> </u><u>Transverse</u><u> </u><u>wave</u><u>.</u>

Explanation:

Because it has troughs and crests.

5 0
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Traveling into space from Earth, which of these would he be able to reach first?
mr Goodwill [35]

You didn't Provide choices to choose from. Please fix question and I'd love to be able to help.

5 0
3 years ago
Read 2 more answers
0.01
katen-ka-za [31]

Answer:

A

Explanation:

60 + 30 = 90

90 divided by 3 = 30

30 divided by 60 = 0.5

so your answer is 0.5 m/s

4 0
3 years ago
A typical human consumes 2500 Kcal of energy during a day. This is the equivalent to 10,450,000 J! Say you decided to run stairs
defon

Answer:

2940.1 joules  would you burn in climbing stairs all day.

Explanation:

Work = W = F\times d

going up stairs  would be against force of gravity

W = mgh

where h is the  height

the question is not complete because we need speed or distance

h =  v \times t

so assuming 1 step per second

h = 86,400 steps  \times 7inchs/step \times 0.0254 m/inch

h = 15362 m

so from this    

W = 800 N \times 15362

   = 12289600 J  

that means YOU  need 12289600 J to walk 1 step per second all day

divide that by 4180 J /Kcal  

Kcal =  \frac{W}{(J/Kcal)}

       = \frac{12289600}{4180}

       = 2940.1 Kcal

if you ran faster you would use more energy  2 steps per second would mean 5880 Kcal.

8 0
3 years ago
Two spherical conductors are separated by a distance much larger than either of their radii. Sphere A has a radius of 11.5 cm an
bonufazy [111]

Explanation:

As the given spheres are connected by a thin wire so, the potential on the spheres are the same.

          \frac{q_{1}}{r_{1}} = \frac{q_{2}}{r_{2}} ......... (1)

Hence, total charge will be as follows.

              q_{1} + q_{2} = Q = -95.5 nC .......... (2)

Using the above two equations, the final equation will be as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

and,    q_{1} = \frac{Qr_{1}}{r_{1} + r_{2}}

Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

                     = \frac{-95.5 \times 74.4 cm}{(11.5 + 74.4) cm}

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Thus, we can conclude that the charge on sphere B after equilibrium has been reached is 82.714 nC.

                       

5 0
3 years ago
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