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dolphi86 [110]
2 years ago
9

earth orbits the sun once every 365.25 days. Find the average angular. speed of Earth about the sun.Answer in units of rad/s.

Physics
2 answers:
anzhelika [568]2 years ago
7 0

Answer:

1.99\cdot 10^{-7} rad/s

Explanation:

The average angular speed of the Earth about the Sun is given by:

\omega = \frac{2 \pi}{T}

where

2 \pi rad is the total angle corresponding to one revolution of the Earth around the Sun

T is the orbital period of the Earth

The orbital period of the Earth is 365.25 d. We must convert it into seconds first:

T=365.25 d \cdot (24 h/d) \cdot (60 min/h) \cdot (60 s/min)=3.16\cdot 10^7 s

And by substituting into the equation above, we find the average angular speed:

\omega=\frac{2 \pi rad}{3.16\cdot 10^7 m/s}=1.99\cdot 10^{-7} rad/s

kirill [66]2 years ago
4 0

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John is carrying a shovelful of snow. The center of mass of the 3.00 kg of snow he is holding is 15.0 cm from the end of the sho
Andru [333]

Answer:

James is correct here as the force of hand pushing upwards is always more than the force of hand pushing down

Explanation:

Here we know that one hand is pushing up at some distance midway while other hand is balancing the weight by applying a force downwards

so here we can say

Upwards force = downwards Force + weight of snow

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then for that force we can say that net torque must be balanced

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so here we have

F_{down} = \frac{L_2}{L_1} (W_{snow})

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. Một con lắc đơn có chiều dài dây treo 1m dao động điều hoà treo trong một xe chạy trên mặt phẳng nghiêng
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What is the potential difference across a parallel-plate capacitor whose plates are separated by a distance of 4.0 mm where each
suter [353]

The potential difference across the parallel plate capacitor is 2.26 millivolts

<h3>Capacitance of a parallel plate capacitor</h3>

The capacitance of the parallel plate capacitor is given by C = ε₀A/d where

  • ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m,
  • A = area of plates and
  • d = distance between plates = 4.0 mm = 4.0 × 10⁻³ m.

<h3>Charge on plates</h3>

Also, the surface charge on the capacitor Q = σA where

  • σ = charge density = 5.0 pC/m² = 5.0 × 10⁻¹² C/m² and
  • a = area of plates.

<h3>The potential difference across the parallel plate capacitor</h3>

The potential difference across the parallel plate capacitor is V = Q/C

= σA ÷ ε₀A/d

= σd/ε₀

Substituting the values of the variables into the equation, we have

V = σd/ε₀

V = 5.0 × 10⁻¹² C/m² × 4.0 × 10⁻³ m/8.854 × 10⁻¹² F/m

V = 20.0 C/m × 10⁻³/8.854 F/m

V = 2.26 × 10⁻³ Volts

V = 2.26 millivolts

So, the potential difference across the parallel plate capacitor is 2.26 millivolts

Learn more about potential difference across parallel plate capacitor here:

brainly.com/question/12993474

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