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Elena L [17]
3 years ago
14

What is the magnitude of the electric force of attraction between two point charges + 2.8 mc and -1.2 mc if the distance

Physics
1 answer:
Anit [1.1K]3 years ago
8 0

Answer:

a

Explanation:

it just a

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A solid, cylindrical wire conductor has radius R = 30 cm. The wire carries a current of 2.0 A which is uniformly distributed ove
blondinia [14]

Answer:

Explanation:

The point at which magnetic field is to be found lies outside wire so while applying Ampere's law we shall take the whole of current . If B be magnetic field which is circular around conductor.

Applying Ampere's law :-

∫ B dl = μ₀ I      ; I is current passing through ampere's loop

B x 2π x 2.00 = 4 x π x 10⁻⁷ x 2

B = 2 x 10⁻⁷ T.

7 0
3 years ago
In an inelastic collision, which of the following statements is always true?
xz_007 [3.2K]
Well C is definitely one of the correct answers.
8 0
3 years ago
A stuntman with a mass of 80.5 kg swings across a moat from a rope that is 11.5 m. At the bottom of the swing the stuntman's spe
goldenfox [79]

Answer:

  • No
  • 5.49 m/s

Explanation:

The net force required to accelerate the stuntman in a circular arc of radius 11.5 m will be ...

  F = mv²/r . . . . where this m is the mass being accelerated, v is the tangential velocity, and r is the radius.

Here, the net force needs to be ...

  F = (80.5 kg)(8.45 m/s)²/(11.5 m) . . . . . where this m is meters

  ≈ 499.8175 kg·m/s² = 499.8 N

Gravity exerts a force on the stuntman of ...

  F = mg = (80.5 kg)(9.8 m/s²) = 788.9 kg·m/s² = 788.9 N

Then the tension required in the rope/vine is ...

  499.8 N+788.9 N= 1288.7 N

This is more than the capacity of the rope, so we do not expect the stuntman to make it across the moat.

_____

The allowed net force for centripetal acceleration is ...

  1000 N -788.9 N = 211.1 N

Then the allowed velocity is ...

  211.1 = 80.5v²/11.5

  30.16 = v² . . . .  multiply by 11.5/80.5

  5.49 = v . . . . . . take the square root

The maximum speed the stuntman can have is 5.49 m/s.

_____

<em>Comment on crossing the moat</em>

The kinetic energy at the bottom of the swing translates to potential energy at the end of the swing. At the lower speed, the stuntman cannot rise as high, so will traverse a shorter arc. At 8.45 m/s, the moat could be about 16.8 m wide; at 5.49 m/s, it can only be about 11.5 m wide.

5 0
2 years ago
An electron is accelerated from rest through a potential difference. After acceleration the electron has a de Broglie wavelength
Vladimir [108]

Answer:

3x10⁴v

Explanation:

Using

Wavelength= h/ √(2m.Ke)

880nm = 6.6E-34/√ 2.9.1E-31 x me

Ke= 6.6E-34/880nm x 18.2E -31.

5.6E-27/18.2E-31

= 3 x 10⁴ Volts

4 0
3 years ago
A laser of wavelength 720 nm illuminates a double slit where the separation between the slits is 0.22 mm. Fringes are seen on a
kumpel [21]

Answer:

The appropriate solution is "2.78 mm".

Explanation:

Given:

\lambda = 720 \ nm

or,

  = 720\times 10^{-9} \ m

D=0.85 \ m

d = 0.22 \ mm

or,

  =0.22 \times 10^{-3} \ m

As we know,

Fringe width is:

⇒ \beta=\frac{\lambda D}{d}

hence,

Separation between second and third bright fringes will be:

⇒ \theta=\beta=\frac{\lambda D}{d}

       =\frac{720\times 10^{-9}\times 0.85}{0.22\times 10^{-3}}

       =2.78\times 10^{-3} \ m

or,

       =2.78 \ mm

8 0
2 years ago
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