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9966 [12]
3 years ago
5

The blackbody curve for a star named Zeta is shown below. What is the peak wavelength for this star?

Physics
2 answers:
VladimirAG [237]3 years ago
8 0

860 nanometers.

Hope this helps.

hram777 [196]3 years ago
3 0

Answer:

The answer is 860 nm!

Just took the test

Explanation:

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We can use renewable sources
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(3.16_Q2) Which weights would you use on a single thread to create a 6.86 N force? Question 2 options: Weight IDs A, B, C, D Wei
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1. E,F

2. D,E,F

3. B,C,E,G

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To calculate the change in kinetic energy, you must know the force as a function of _______. The work done by the force causes t
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(c) position

Explanation:

From the work-energy theorem, the workdone by a force on a body causes a change in kinetic energy of the body.

But, remember that the work done (W) by a force (F) on a body is the product of the force and the distance d, moved by the body caused by the force. i.e

W = F x d

This distance is a measure of the position of the body at a given instance.

Therefore, the work done is given by the force as a function of distance (or position).

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3 years ago
What happens when calcium reacts with chlorine?
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7 0
3 years ago
An engineer is designing a runway. She knows that a plane, starting at rest, needs to reach a speed of 180mph at take-off. If th
kvv77 [185]

Answer:

The plane would need to travel at least 8,\!580\; {\rm ft} (8.58 \times 10^{3}\; {\rm ft}.)

The 10,\!000\; {\rm ft} runway should be sufficient.

Explanation:

Convert unit of the the take-off velocity of this plane to \rm ft\cdot s^{-1}:

\begin{aligned}v &= 180\; {\rm mph} \\ &= 180\; {\rm mi \cdot hrs^{-1}} \times \frac{1\; {\rm hrs}}{3600\; {\rm s}} \times \frac{5280\; {\rm ft}}{1\; {\rm mi}} \\ &= 264\; {\rm ft \cdot s^{-1}}\end{aligned}.

Initial velocity of the plane: u = 0\; {\rm ft \cdot s^{-1}}.

Take-off velocity of the plane v =264\; {\rm ft\cdot s^{-1}}.

Let x denote the distance that the plane travelled along the runway. Since acceleration is constant but unknown, make use of the SUVAT equation x = ((u + v) / 2) \, t.

Notice that this equation does not require the value of acceleration. Rather, this equation make use of the fact that the distance travelled (under constant acceleration) is equal to duration t times average velocity (u + v) / 2.

The distance that the plane need to cover would be:

\begin{aligned}x &= \left(\frac{u + v}{2}\right)\, t \\ &= \frac{0\; {\rm ft \cdot s^{-1}} + 264\; {\rm ft \cdot s^{-1}}}{2} \times 65.0\; {\rm s} \\ &= 8.58\times 10^{3}\; {\rm ft}\end{aligned}.

4 0
3 years ago
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