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Svetach [21]
2 years ago
5

Which would require the greater energy; slowing down of the orbital speed of the Earth so it crashes into the sun, or speeding u

p the orbital speed of the Earth so it escapes the sun
Physics
1 answer:
tresset_1 [31]2 years ago
3 0

Answer: Speeding up the orbital speed of earth so it escapes the sun require the greater energy.

Explanation: To find the answer, we need to know more about the Orbital and escape velocities.

<h3>What is Orbital and Escape velocity?</h3>
  • Orbital velocity can be defined as the minimum velocity required to put the satellite in its orbit around the earth.
  • The expression for orbital velocity near to the surface of earth will be,

                   V_o=\sqrt{gR}

  • Escape velocity can be defined as the minimum velocity with which a body must be projected from the surface of earth, so that it escapes from the gravitational field of earth.
  • The expression for orbital velocity will be,

                    V_e=\sqrt{2gR}

  • If we want to get into the sun, we want to slow down almost completely, so that your speed relative to the sun became almost zero.
  • We need about twice the raw speed to go to the sun than to leave the sun.

Thus, we can conclude that, the speeding up the orbital speed of earth so it escapes the sun require the greater energy.

Learn more about orbital and escape velocity here:

brainly.com/question/28045208

#SPJ4

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Charges of 4.0 μC and −6.0 μC are placed at two corners of an equilateral triangle with sides of 0.10 m. What is the magnitude o
jek_recluse [69]

Answer:

4.763 × 10⁶ N/C

Explanation:

Let E₁ be the electric field due to the 4.0 μC charge and E₂ be the electric field due to the -6.0 μC charge. At the third corner, E₁ points in the negative x direction and E₂ acts at an angle of 60 to the negative x - direction.

Resolving E₂ into horizontal and vertical components, we have

E₂cos60 as horizontal component and E₂sin60 as vertical component. E₁ has only horizontal component.

Summing the horizontal components we have

E₃ = -E₁ + (-E₂cos60) = -kq₁/r²- kq₂cos60/r²

= -k/r²(q₁ + q₂cos60)

= -k/r²(4 μC + (-6.0 μC)(1/2))

= -k/r²(4 μC - 3.0 μC)

= -k/r²(1 μC)

= -9 × 10⁹ Nm²/C²(1.0 × 10⁻⁶)/(0.10 m)²

=  -9 × 10⁵ N/C

Summing the vertical components, we have

E₄ = 0 + (-E₂sin60)

= -E₂sin60

= -kq₂sin60/r²

= -k(-6.0 μC)(0.8660)/(0.10 m)²

= -9 × 10⁹ Nm²/C²(-6.0 × 10⁻⁶)(0.8660)/(0.10 m)²

= 46.77 × 10⁵ N/C

The magnitude of the resultant electric field, E is thus

E = √(E₃² + E₄²) = √[(-9 × 10⁵ N/C)² + (46.77 10⁵ N/C)²) = (√226843.29) × 10⁴

= 476.28  × 10⁴ N/C

= 4.7628 × 10⁶ N/C

≅ 4.763 × 10⁶ N/C

8 0
3 years ago
What is the kinetic energy k of an electron with momentum 1.05×10−24 kilogram meters per second?
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v = momentum ÷ m
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kinetic energy = (mv∧2)÷2
                       = (9.1×10∧-31 × 1,153,846.154∧2) ÷2
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                      = 6.05769×10∧-19 J
4 0
4 years ago
Read 2 more answers
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