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otez555 [7]
3 years ago
9

Be sure to answer all parts. Assume the diameter of a neutral helium atom is 1.40 × 102 pm. Suppose that we could line up helium

atoms side by side in contact with one another. How many atoms would it take to make the distance 2.60 cm from end to end? × 10 He atoms (Enter your answer in scientific notation.)
Physics
1 answer:
ss7ja [257]3 years ago
8 0

Answer:

The number of atoms are 1.86\times10^{8}.

Explanation:

Given that,

Diameter D = 1.40\times10^{2}\ pm

D=1.40\times10^{2}\times10^{-12}\ m

Distance = 2.60 cm

We calculate the number of atoms

Using formula of numbers of atoms

Number\ of\ atoms =\dfrac{2.60\times10^{-2}}{1.40\times10^{2}\times10^{-12}}

Number\of\atoms =1.86\times10^{8}

Hence, The number of atoms are 1.86\times10^{8}.

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A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.
Advocard [28]

The potential across the capacitor at t = 1.0 seconds, 5.0 seconds, 20.0 seconds respectively is mathematically given as

  • t=0.476v
  • t=1.967v
  • V2=4.323v

<h3>What is the potential across the capacitor?</h3>

Question Parameters:

A 1. 0 μf capacitor is being charged by a 9. 0 v battery through a 10 mω resistor.

at

  • t = 1.0 seconds
  • 5.0 seconds
  • 20.0 seconds.

Generally, the equation for the Voltage is mathematically given as

v(t)=Vmax=(i-e^{-t/t})

Therefore

For t=1

V=5(i-e^{-1/10})

t=0.476v

For t=5s

V2=5(i-e^{-5/10})

t=1.967

For t=20s

V2=5(i-e^{-20/10})

V2=4.323v

Therefore, the values of voltages at the various times are

  • t=0.476v
  • t=1.967v
  • V2=4.323v

Read more about  Voltage

brainly.com/question/14883923

Complete Question

A 1.0 μF capacitor is being charged by a 5.0 V battery through a 10 MΩ resistor.

Determine the potential across the capacitor when t = 1.0 seconds, 5.0 seconds, 20.0 seconds.

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2 years ago
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Explanation:

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3 years ago
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During an experiment of momentum, trolley, X, of mass (2.34 ± 0.01) kg is moving away from another trolley, Y, of mass (2.561 ±
Alla [95]

Answer:

P = 1 (14,045 ± 0.03 )  k gm/s

Explanation:

In this exercise we are asked about the uncertainty of the momentum of the two carriages

            Δ (Pₓ / Py) =?

 Let's start by finding the momentum of each vehicle

car X

        Pₓ = m vₓ

        Pₓ = 2.34 2.5

        Pₓ = 5.85 kg m

car Y

        Py = 2,561 3.2

        Py = 8,195 kgm

How do we calculate the absolute uncertainty at the two moments?

          ΔPₓ = m Δv + v Δm

          ΔPₓ = 2.34 0.01 + 2.561 0.01

          ΔPₓ = 0.05 kg m

         ΔP_{y} = m Δv + v Δm

         ΔP_{y} = 2,561 0.01+ 3.2 0.001

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now we have the uncertainty of each moment

          P = Pₓ / P_{y}

          ΔP = ΔPₓ/P_{y} + Pₓ ΔP_{y} / P_{y}²

          ΔP = 8,195 0.05 + 5.85 0.03 / 8,195²

          ΔP = 0.006 + 0.0026

          ΔP = 0.009 kg m

The result is

           P = 14,045 ± 0.039 = (14,045 ± 0.03 )  k gm/s

7 0
3 years ago
20 N effort is used in the 20 cm long spanner to unscrew a nut, then calculate momentum to unscrew the nut. [ Ans=4 Nm]
irga5000 [103]

Answer:

Refer to the attachment.

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