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3 years ago

PLEASE HELP What is it called when the moon passes through the penumbra of Earth’s shadow?(1 point)

Physics

1 answer:

bogdanovich [222]3 years ago

8 0

Partial. a penumbral lunar eclipse is when the moon passes through the earths penumbra

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What is the conversion factor between km/h and mi/h

Aloiza [94]

1 mi = 1.609 km
so X mi/hr = 1.609 * X km/hr hope this helps!!

5 0

3 years ago

In a physics laboratory experiment, a coil with 200 turns enclosing an area of 13.1 cm2 is rotated during the time interval 3.10

sergij07 [2.7K]

Answer:

A) $\Phi=83.84\times 10^{-9}$

B) $\Phi=0 Wb$

C) $emf=5.4090\times 10^{-4}V$

Explanation:

Given that:

no. of turns i the coil, $n=200$

area of the coil, $a=13.1 \times 10^{-4}\,m^2$

time interval of rotation, $t=3.1\times 10^{-2}\,s$

intensity of magnetic field, $B=6.4\times 10^{-5}\,T$

(A)

Initially the coil area is perpendicular to the magnetic field.

So, magnetic flux is given as:

$\Phi=B.a\,cos \theta$ ..................................(1)

$\theta$ is the angle between the area vector and the magnetic field lines. Area vector is always perpendicular to the area given. In this case area vector is parallel to the magnetic field.

$\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 0^{\circ}$

$\Phi=83.84\times 10^{-9} Wb$

(B)

In this case the plane area is parallel to the magnetic field i.e. the area vector is perpendicular to the magnetic field.

∴ $\theta=90^{\circ}$

From eq. (1)

$\Phi=6.4\times 10^{-5}\times 13.1 \times 10^{-4}\, cos 90^{\circ}$

$\Phi=0 Wb$

(C)

According to the Faraday's Law we have:

$emf=n\frac{B.a}{t}$

$emf=\frac{200\times 6.4\times 10^{-5}\times 13.1 \times 10^{-4}}{3.1\times 10^{-2}}$

$emf=5.4090\times 10^{-4}V$

7 0

3 years ago

What acceleration will you give to a 24.5 kg

Ludmilka [50]

Heya!!

For calculate aceleration, lets applicate second law of Newton:

$\boxed{F=ma}$

Δ Being Δ

F = Force = 78,3 N

m = Mass = 24,5 kg

a = Aceleration = ?

⇒ Let's replace according the formula and clear "a":

$\boxed{a=78,3\ N / 24,5\ kg}$

⇒ Resolving

$\boxed{a=3.19\ m/s^{2}}$

Result:

The aceleration is 3,19 meters per second squared (m/s²)

Good Luck!!

4 0

3 years ago

Given a 45 45 90 prism with index of 1.5, immersed in air. The hypotenuse acts as the reflecting face by TIR. A ray of light ent

Genrish500 [490]

Answer:

83.6°

Explanation:

For the ray to be totally internally reflected, at the boundary, the angle of refraction is 90. Using the law of refraction where

n₁sinθ₁ = n₂sinθ₂ where n₁ = refractive index of prism = 1.5, θ₁ = critical angle in prism, n₂ = refractive index of air = 1 and θ₂ = refractive angle = 90°.

So, substituting these values into the equation,

n₁sinθ₁ = n₂sinθ₂

1.5 × sinθ₁ = 1 × sin90

1.5 × sinθ₁ = 1

sinθ₁ = 1/1.5

sinθ₁ = 0.6667

θ₁ = sin*(0.6667)

θ₁ = 41.8°

So, for total internal reflection, an incidence angle of 41.8° is required. So, a full convergence angle of 2 × 41.8° = 83.6° is required for the whole bundle of rays.

5 0

3 years ago

A circuit has a voltage of 10 V and a current of 5 A. What must the resistance be?

azamat

Answer:

R=V/I

R= 2

Explanation:

R = 10V/5A

R = 2ohms

3 0

3 years ago