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Crank
3 years ago
6

PLEASE HELP WHICH STATEMENTS ARE CORRECT DO NOT GUESS

Physics
2 answers:
Anni [7]3 years ago
8 0
THe first one and the third one!!
Lelechka [254]3 years ago
3 0

Answer:

Explanation:

Use the diagram and drop-down menus to answer each question.

Which circuit is a parallel circuit?____  [B]

Which circuit is a series circuit?____  [A]

In which circuit do the light bulbs all shine at their maximum brightness? _[B]

Those are the answer for edgenuit.....

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If the diameter of a radar dish is doubled, what happens to its resolving power assuming that all other factors remain unchanged
kirill115 [55]

Answer:

      θ’ = θ₀ / 2

we see that the resolution angle is reduced by half

Explanation:

The resolving power of a radar is given by diffraction, for which we will use the Rayleigh criterion for the resolution of two point sources, they are considered resolved if the maximum of diffraction of one coincides with the first minimum of the other.

The first minimum occurs for m = 1, so the diffraction equation of a slit remains

        a sin θ = λ

in general, the diffraction patterns occur at very small angles, so

        sin θ = θ

          θ = λ / a

in the case of radar we have a circular aperture and the equation must be solved in polar coordinates, which introduces a numerical constant.

        θ = 1.22 λ /a

In this exercise we are told that the opening changes

         a’ = 2 a

we substitute

          θ ‘= 1.22  λ / 2a

          θ' = (1.22 λ / a) 1/2

          θ’ = θ₀ / 2

we see that the resolution angle is reduced by half

8 0
3 years ago
Calculate Gravitational Potential Energy for an object on Earth with a mass of 2 kg and a height of 7 m.
victus00 [196]

Answer:

137.2J

Explanation:

Ep= mgh

Given,

  • m=2kg
  • h=7m

and we know, g = 9.81 N/kg

Ep= 2 × 9.81 × 7

Ep= 137.2J

4 0
3 years ago
1. A mass suspended from a spring oscillates vertically with amplitude of 15 cm. At what distance from the equilibrium position
antiseptic1488 [7]

Answer:

The value of the distance is \bf{14.52~cm}.

Explanation:

The velocity of a particle(v) executing SHM is

v = \omega \sqrt{A^{2} - x^{2}}~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~`~(1)

where, \omega is the angular frequency, A is the amplitude of the oscillation and x is the displacement of the particle at any instant of time.

The velocity of the particle will be maximum when the particle will cross its equilibrium position, i.e., x = 0.

The maximum velocity(\bf{v_{m}}) is

v_{m} = \omega A~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(2)

Divide equation (1) by equation(2).

\dfrac{v}{v_{m}} = \dfrac{\sqrt{A^{2} - x^{2}}}{A}~~~~~~~~~~~~~~~~~~~~~~~~~~~(3)

Given, v = 0.25 v_{m} and A = 15~cm. Substitute these values in equation (3).

&& \dfrac{1}{4} = \dfrac{\sqrt{15^{2} - x^{2}}}{15}\\&or,& A = 14.52~cm

6 0
3 years ago
If a charged particle is moving parallel to a magnetic field, it experiences
aleksandr82 [10.1K]

Answer:it experiences no force

Explanation:

a charge moving in a direction parallel to the magnetic field experience no force.since the angle e is 0,force would also be 0

3 0
3 years ago
when a metal ball is heated through 30°c,it volume becomes 1.0018cm^3 if the linear expansivity of the material of the ball is 2
Vlad1618 [11]

Answer:

The original volume of the metal sphere is approximately 1 cm³

Explanation:

The given parameters are;

The temperature change of the metal ball, ΔT = 30°C = 30 K

The new volume of the metal ball, V₂  = 1.0018 cm³

The linear expansivity of the material ball, α = 2.0 × 10⁻⁵ K⁻¹

We have;

d₂ = d₁·(1 + α·ΔT)

Where;

d₁ = The original diameter of the metal ball

d₂ = The new diameter of the ball

From the volume of the ball, V₂, we have;

V₂ = 1.0018 cm³ = (4/3)×π×r₂³

Where;

r₂ = The new radius = d₂/2

∴ V₂ = 1.0018 cm³ = (4/3)×π×(d₂/2)³

∴ d₂ = ∛(2³ × 1.0018 cm³/((4/3) × π)) ≈ 1.241445 cm

d₁ = d₂/(1 + α·ΔT)

∴ d₁ ≈ 1.241445 cm/(1 + 2.0 × 10⁻⁵·K × 30 K) ≈ 1.24070058 cm

The original volume of the metal ball, V₁ = (4/3)×π×(d₁/2)³

∴ V₁ = (4/3)×π×(1.24070058/2)³ ≈ 0.99999902845 cm³ ≈ 1 cm³

The original volume of the metal sphere, V₁ ≈ 1 cm³.

8 0
2 years ago
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