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Svet_ta [14]
3 years ago
10

What is the resulting value for the Hubble constant?Write your answer in scientific notation

Physics
1 answer:
KonstantinChe [14]3 years ago
5 0

\bf{Given : Hubble\;Constant(H_0) = 73(\frac{km}{s \times Mpc})}

\bf{Given : Conversion\;Factor = (\frac{1Mpc}{3.086 \times 10^1^9\;km})}

\bf{\implies Resulting\;Hubble\;Constant = 73(\frac{km}{s \times Mpc})(\frac{1Mpc}{3.086 \times 10^1^9\;km})}

\bf{\implies Resulting\;Hubble\;Constant = 73(\frac{1}{s \times 3.086 \times 10^1^9})}

\bf{\implies Resulting\;Hubble\;Constant = (\frac{23.65}{s \times 10^1^9})}

\bf{\implies Resulting\;Hubble\;Constant = 2.365 \times 10^-^1^8(1/s)

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8 0
3 years ago
Equipotential surface A has a potential of 5650 V, while equipotential surface B has a potential of 7850 V. A particle has a mas
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Answer:

0.247 J = 247 mJ

Explanation:

From the principle of conservation of energy, the workdone by the applied force, W = kinetic energy change + electric potential energy change.

So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁) where m = mass of particle = 5.4 × 10⁻² kg, q = charge of particle = 5.10 × 10⁻⁵ C, v₁ = initial speed of particle = 2.00 m/s, v₂ = final speed of particle = 3.00 m/s, V₁ = potential at surface A = 5650 V, V₂ = potential at surface B = 7850 V.

So, W = ΔK + ΔU =1/2m(v₂² - v₁²) + q(V₂ - V₁)

          = 1/2 × 5.4 × 10⁻²kg × ((3m/s)² - (2 m/s)²) + 5.10 × 10⁻⁵ C(7850 - 5650)

          = 0.135 J + 0.11220 J

          = 0.2472 J

          ≅ 0.247 J = 247 mJ

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