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kolbaska11 [484]
3 years ago
15

In which decade was Sputnik I launched? A. 1930-1939 B. 1950-1959 C. 1960-1969 D. 1980-1989

Physics
1 answer:
Olenka [21]3 years ago
6 0
B) 1950-1959
Sputnik I was launched on October 4, 1957
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Which occurrence would lead you to conclude that lights are connected in a
skelet666 [1.2K]

Answer:B When one bulb burns out, all the others lights stay lit.

Explanation:

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3 years ago
A particle with charge 3.01 µC on the negative x axis and a second particle with charge 6.02 µC on the positive x axis are each
ra1l [238]

Answer:

The third particle should be at 0.0743 m from the origin on the negative x-axis.

Explanation:

Let's assume that the third charge is on the negative x-axis. So we have:

E_{1}+E_{3}-E_{2}=0

We know that the electric field is:

E=k\frac{q}{r^{2}}

Where:

  • k is the Coulomb constant
  • q is the charge
  • r is the distance from the charge to the point

So, we have:

k\frac{q_{1}}{r_{1}^{2}}+k\frac{q_{3}}{r_{3}^{2}}-k\frac{q_{2}}{r_{2}^{2}}=0

Let's solve it for r(3).

\frac{3.01}{0.0429^{2}}+\frac{9.03}{r_{3}^{2}}-\frac{6.02}{0.0429^{2}}=0

r_{3}=0.0743\:  

Therefore, the third particle should be at 0.0743 m from the origin on the negative x-axis.

I hope it helps you!

 

3 0
3 years ago
A slit of width 2.0 μm is used in a single slit experiment with light of wavelength 650 nm. If the intensity at the central maxi
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Answer:

The intensity at 10° from the center is 3.06 × 10⁻⁴I₀

Explanation:

The intensity of light I = I₀(sinα/α)² where α = πasinθ/λ

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θ = angle at intensity point = 10°

λ = wavelength of light = 650 nm = 650 × 10⁻⁹ m

α = πasinθ/λ

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Now, the intensity I is

I = I₀(sinα/α)²

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= I₀(0.0293/1.679)²

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= 0.0003063I₀

= 3.06 × 10⁻⁴I₀

So, the intensity at 10° from the center is 3.06 × 10⁻⁴I₀

5 0
3 years ago
A track is mounted on a large wheel that is free to turn with negligible friction about a vertical axis (Fig. 11-48).A toy train
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Answer:

0.166 rad/s

Explanation:

See attachment for calculations

5 0
3 years ago
Read 2 more answers
Why current remains same in series combination of resistors in all resistors and p.d. remains different?
IgorC [24]
Current at all points of a series circuit must be the same, because there's no place in the circuit where electrons are being manufactured, and no place where they're leaking out and falling on the floor. The nimber of electrons that leaves the loop is the same number that entered it. I'm not sure what is nmeant by "p.d. remains different" .
4 0
3 years ago
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