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zmey [24]
3 years ago
8

The attraction between the Sun's gravity and Earth's gravity holds Earth in is ?

Physics
1 answer:
soldi70 [24.7K]3 years ago
5 0
The attraction between the suns gravity and earths gravity holds earth in it's orbit
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Position vs Time
Annette [7]

Answer:

Determining the Slope on a p-t Graph. It was learned earlier in Lesson 3 that the slope of the line on a position versus time graph is equal to the velocity of the object. If the object is moving with a velocity of +4 m/s, then the slope of the line will be +4 m/s

7 0
3 years ago
Were there experiences in his life that led to his interest in science and the study of the atom?
charle [14.2K]

Answer:

The right answer is Democritus

Explanation:

He experienced bonding with his father which strongly resulted his interest in science and to the study of the atoms ;)

Your welcome

5 0
2 years ago
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Is this wave transverse or longitudinal
lys-0071 [83]
It depends on if it is a sound wave or light wave
8 0
3 years ago
Two infinite nonconducting sheets of charge are parallel to each other, with sheet A in the x = -2.15 plane and sheet B in the x
GalinKa [24]

Answer:

a) (-367231.63i ,  367231.63i, 0) N/C

b) (0 , 0  , 367231.63i ) N/C

Explanation:

a)

Case x < -2.15

E =E_{+} + E_{+} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = - 367231.63 i

Case x > 2.15

E =E_{+} + E_{+} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = +367231.63 i

Case -2.15 < x <+2.15

E =E_{+} - E_{+} \\\\E = 0

b)

Case x < -2.15

E =E_{+} - E_{+} \\\\E = 0

Case x > 2.15

E =E_{+} - E_{+} \\\\E = 0

Case -2.15 < x <+2.15

E =E_{+} + E_{-} \\E = 2*\frac{sigma}{2e_{0} } = \frac{sigma}{e_{0} }\\\\E = \frac{3.25*10^(-6)}{8.85*10^(-12) }\\\\E = +367231.63 i

7 0
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

8 0
4 years ago
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