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2 years ago

Astronomers initially had difficulty identifying the emission lines in quasar spectra at optical wavelengths because

Physics

1 answer:

Rus_ich [418]2 years ago

6 0

No one expected violet & ultraviolet spectral lines to be shifted towards the red.

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Juan was wearing a bright red shirt in a very dark room. What color did his shirt appear to the people with him in the room? A)

ikadub [295]

It would appear black.

Hope I helped.

8 0

2 years ago

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A square loop of wire with initial side length 10 cm is placed in a magnetic field of strength 1 T. The field is parallel to the

Fofino [41]

Answer:

2 x 10⁻³ volts

Explanation:

B = magnetic of magnetic field parallel to the axis of loop = 1 T

$\frac{dA}{dt}$ = rate of change of area of the loop = 20 cm²/s = 20 x 10⁻⁴ m²

θ = Angle of the magnetic field with the area vector = 0

E = emf induced in the loop

Induced emf is given as

E = B $\frac{dA}{dt}$

E = (1) (20 x 10⁻⁴ )

E = 2 x 10⁻³ volts

E = 2 mV

7 0

2 years ago

A woman is straining to lift a large crate, without success because it is too heavy. We denote the forces on the crate as follow

Ymorist [56]

Answer:

Explanation:

Given

Force P is acting upward

C is vertical contact Force

W is the weight of the crate

As P is unable to move the Block therefore Normal reaction keeps on acting on block

thus we can say that

P-W+C=0

P=W-C

7 0

2 years ago

Read 2 more answers

The magnitude J(r) of the current density in a certain cylindrical wire is given as a function of radial distance from the cente

kipiarov [429]

Answer:

$I=68.31\times 10^{-6}\ A$

Explanation:

Given that

J(r) = Br

We know that area of small element

dA = 2 π dr

I = J A

dI = J dA

Now by putting the values

dI = B r . 2 π dr

dI= 2π Br² dr

Now by integrating above equation

$\int_{0}^{I}dI= \int_{r_1}^{r_2}2\pi Br^2 dr$

$I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}$

Given that

B= 2.35 x 10⁵ A/m³

r₁ = 2 mm

r₂ = 2+ 0.0115 mm

r₂ = 2.0115 mm

$I={2\pi B}\times \dfrac{r_2^3-r_1^3}{3}$

By putting the values

$I={2\pi \times 2.35 \times 10^5 }\times \dfrac{(2.0115\times 10^{-3})^3-(2\times 10^{-3})^3}{3}\ A$

$I=68.31\times 10^{-6}\ A$

7 0

2 years ago

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Gravity is the force that keeps us on the Earth. It pulls us towards the center of the Earth. If you were to move from the surfa

Nikitich [7]

<h2>Answer: B. Gravitational potential energy </h2>

Explanation:

<em>The gravitational potential energy is the energy that a body or object possesses, due to its position in a gravitational field. </em>

That is why this energy depends on the relative height of an object with respect to some point of reference and associated with the gravitational force.

In the case of the <u>Earth</u>, in which <u>the gravitational field is considered constant</u>, the value of the gravitational potential energy $U_{p}$ will be:

$U_{p}=mgh$

Where $m$ is the mass of the object, $g$ the acceleration due gravity and $h$ the height of the object.

As we can see, the value of $U_{p}$ is directly proportional to the height.

6 0

2 years ago