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2 years ago

Tabitha knows the redshift value of a particular celestial object. What can she use this information most directly to find out?

Physics

1 answer:

AlekseyPX2 years ago

3 0

She can use the information of redshift of a celestial body is to find the age and distance of the object.

<h3>What is red shift?</h3>

'Red shift' is a critical idea for astronomers. The frequency of the light is increased, so the light is observed as moving towards the red side of the range.

Tabitha knows the redshift value of a particular celestial object. She uses this information to observe the consequence of the redshift.

Thus, the red shift value is used to determine the age and distance of the object.

Learn more about red shift.

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Why isn’t a bird sitting on a high-voltage power line electrocuted? Contrast this with the situation in which a large bird hits

Grace [21]

Answer:

The reason the bird is not electrocuted is due to some facts about circuit:

1. Completeness of circuit- This circuit needs to be

complete in order for current to flow. The bird standing on only one wire has not completed the circuit.

2. A potential difference: Another factor deciding

the direction of flow of current is (electric)

potential. Current always flows from a higher

potential to a lower potential. In other words it

can be said that electrons flow from lower

potential to higher one. (the direction of electric

current is opposite to that of the electrons). So

we need the potential difference for current to

flow. The bird standing on only one wire has no potential difference.

3. Path of least Resistance- Factor that decides

the path a current will flow in case of parallel

paths is the (electric) resistance offered by the

path. Current will always flow in the path that

offers least resistance. The leg of a bird has high resistance.

Explanation:

It has no potential difference as both the legs of bird are touching the same wire at same constant potential. ... If the bird would touch the ground while sitting on the wire or flap its wings and touch another electric wire with a different voltage, then it would get shocked and likely die by electrocution.

while in the other hand, the Bird that touches two wires with it wings at the same time will get electrocuted because it has completed a circuit and the its feathers created a potential difference .

8 0

3 years ago

Read 2 more answers

If youre walking from point a to b, the magnitude of your displacement will always be equal or less than or greater than your di

xenn [34]

The magnitude of your displacement can be equal to the distance you covered, or it can be less than the distance you covered. But it can never be greater than the distance you covered.

This is because displacement is a straight line, whereas distance can be a straight line, a squiggly line, a zig-zag line, a line with loops in it, a line with a bunch of back-and-forths in it, or any other kind of line.

The straight line is always the shortest path between two points.

3 0

2 years ago

At a distance D from a very long (essentially infinite) uniform line of charge, the electric field strength is 1000 N/C. At what

Alla [95]

Answer:

The correct option is (a).

Explanation:

We know that, the E is inversely proportional to the distance as follows :

$E=\dfrac{k}{d^2}$

We can write it as follows :

$\dfrac{E_1}{E_2}=(\dfrac{d_2}{d_1})^2$

Put all the values,

$\dfrac{1000}{2000}=(\dfrac{d_2}{d})^2\\\\\sqrt{\dfrac{1000}{2000}}=(\dfrac{d_2}{D})\\\\0.7071=\dfrac{d_2}{d}\\\\d_1=0.7071D\\\\d_1=\dfrac{D}{\sqrt2}$

So, the correct option is (a).

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2 years ago

A 4.87-kg ball of clay is thrown downward from a height of 3.21 m with a speed of 5.21 m/s onto a spring with k = 1570 N/m. The

Yuki888 [10]

Answer:

Approximately $0.560\; {\rm m}$ , assuming that:

the height of $3.21\; {\rm m}$ refers to the distance between the clay and the top of the uncompressed spring.
air resistance on the clay sphere is negligible,
the gravitational field strength is $g = 9.81\; {\rm m\cdot s^{-2}}$ , and
the clay sphere did not deform.

Explanation:

Notations:

Let $k$ denote the spring constant of the spring.
Let $m$ denote the mass of the clay sphere.
Let $v$ denote the initial speed of the spring.
Let $g$ denote the gravitational field strength.
Let $h$ denote the initial vertical distance between the clay and the top of the uncompressed spring.

Let $x$ denote the maximum compression of the spring- the only unknown quantity in this question.

After being compressed by a displacement of $x$ , the elastic potential energy $\text{PE}_{\text{spring}}$ in this spring would be:

$\displaystyle \text{PE}_{\text{spring}} = \frac{1}{2}\, k\, x^{2}$ .

The initial kinetic energy $\text{KE}$ of the clay sphere was:

$\displaystyle \text{KE} = \frac{1}{2}\, m \, v^{2}$ .

When the spring is at the maximum compression:

The clay sphere would be right on top of the spring.
The top of the spring would be below the original position (when the spring was uncompressed) by $x$ .
The initial position of the clay sphere, however, is above the original position of the top of the spring by $h = 3.21\; {\rm m}$ .

Thus, the initial position of the clay sphere ( $h = 3.21\; {\rm m}$ above the top of the uncompressed spring) would be above the max-compression position of the clay sphere by $(h + x)$ .

The gravitational potential energy involved would be:

$\text{GPE} = m\, g\, (h + x)$ .

No mechanical energy would be lost under the assumptions listed above. Thus:

$\text{PE}_\text{spring} = \text{KE} + \text{GPE}$ .

$\displaystyle \frac{1}{2}\, k\, x^{2} = \frac{1}{2}\, m\, v^{2} + m\, g\, (h + x)$ .

Rearrange this equation to obtain a quadratic equation about the only unknown, $x$ :

$\displaystyle \frac{1}{2}\, k\, x^{2} - m\, g\, x - \left[\left(\frac{1}{2}\, m\, v^{2}\right)+ (m\, g\, h)\right] = 0$ .

Substitute in $k = 1570\; {\rm N \cdot m^{-1}}$ , $m = 4.87\; {\rm kg}$ , $v = 5.21\; {\rm m\cdot s^{-1}}$ , $g = 9.81\; {\rm m \cdot s^{-2}}$ , and $h = 3.21\; {\rm m}$ . Let the unit of $x$ be meters.

$785\, x^{2} - 47.775\, x - 219.453 \approx 0$ (Rounded. The unit of both sides of this equation is joules.)

Solve using the quadratic formula given that $x \ge 0$ :

$\begin{aligned}x &\approx \frac{-(-47.775) + \sqrt{(-47.775)^{2} - 4 \times 785 \times (-219.453)}}{2 \times 785} \\ &\approx 0.560\; {\rm m}\end{aligned}$ .