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garik1379 [7]
3 years ago
13

[Please help fast! Offering 100 points if it works!}

Physics
2 answers:
Alex17521 [72]3 years ago
8 0

(8.3 min)/(1 AU) = (T)/(1.52 AU)

(8.3 min)x(1.52 AU) = (T x 1 AU)

T = (8.3 min x 1.52 AU) / (1 AU)

T = 12.62 minutes

solmaris [256]3 years ago
3 0

Answer:

d=1.49×1011m

Explanation:

Velocity is defined as the rate of travel, and can be found using the distance formula.

velocity=distancetime

Rearranging this formula we can solve for distance given velocity and time of travel.

d=vt

We are given velocity and time, and so can solve for distance, but if we plug in the values given;

d=(3.00×108m/s)(8.3minutes)

We can see that the units do not match up. Since seconds are the SI unit for time, we will need to convert 8.3 minutes to seconds.

t=(8.3minutes)(60seconds/minute)=(498s)

Now our units work out and we can solve for distance.

= 15.85

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Brut [27]
Hey there!

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Calculate the acceleration if you push with a 20-N horizontal force on a 2-kg block on a horizontal friction-free air table
Afina-wow [57]
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3 years ago
A vertical spring has a spring constant of 2900 N/m. The spring is compressed 80 cm and a 8 kg spider is placed on the spring. T
Serga [27]

Answer:

a)  k_{e} = 928 J , b)U = -62.7 J , c) K = 0 , d) Y = 11.0367 m,  e)  v = 15.23 m / s  

Explanation:

To solve this exercise we will use the concepts of mechanical energy.

a) The elastic potential energy is

      k_{e} = ½ k x²

      k_{e} = ½ 2900 0.80²

      k_{e} = 928 J

b) place the origin at the point of the uncompressed spring, the spider's potential energy

     U = m h and

     U = 8 9.8 (-0.80)

     U = -62.7 J

c) Before releasing the spring the spider is still, so its true speed and therefore the kinetic energy also

      K = ½ m v²

      K = 0

d) write the energy at two points, maximum compression and maximum height

     Em₀ = ke = ½ m x²

     E_{mf} = mg y

     Emo = E_{mf}

     ½ k x² = m g y

     y = ½ k x² / m g

     y = ½ 2900 0.8² / (8 9.8)

     y = 11.8367 m

As zero was placed for the spring without stretching the height from that reference is

     Y = y- 0.80

     Y = 11.8367 -0.80

     Y = 11.0367 m

Bonus

Energy for maximum compression and uncompressed spring

     Emo = ½ k x² = 928 J

     E_{mf}= ½ m v²

     Emo = E_{mf}

     Emo = ½ m v²

      v =√ 2Emo / m

     v = √ (2 928/8)

     v = 15.23 m / s

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