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2 years ago

Which of the following characteristics of stars is affected by a stars temperature?

Physics

1 answer:

vodomira [7]2 years ago

7 0

Answer:

(2) Color

Explanation:

The color--the wavelength of the visible waves emitted by a star-- is tied to the star's temperature. In general, the hotter the surface the shorter the emitted wavelengths (and the stronger will the blue component be represented in the visible band), and the cooler the surface the longer the wavelength (with a shift toward the red end of the visible band).

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I need help real quick. Please help me!!

dimaraw [331]

<h3><u>Answer;</u></h3>

D) Standing wave

<h3><u>Explanation;</u></h3>

Standing wave also called stationary wave is a wave which oscillates in time but whose peak amplitude profile does not move in space.
A standing wave pattern is a vibrational pattern created within a medium when the vibrational frequency of the source causes reflected waves from one end of the medium to interfere with incident waves from the source.
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4 0

2 years ago

What is the magnitude of the kinetic frictional force

Effectus [21]

The magnitude of the kinetic friction force, ƒk, on an object is. Where μk is called the kinetic friction coefficient and |FN| is the magnitude of the normal force of the surface on the sliding object. The kinetic friction coefficient is entirely determined by the materials of the sliding surfaces. hope it helps

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2 years ago

Bob, Jill, Kim, and Steve measure an object's length, density, mass, and brightness, respectively. Which student must derive a u

netineya [11]

The answer is A. Bob (<span>object's length)

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3 0

2 years ago

Bob, of mass m, drops from a tree limb at the same time that Esther, also of mass m, begins her descent down a frictionless slid

galben [10]

Answer:

Explanation:

They have the same kinetic energy

5 0

2 years ago

A rifle bullet with mass 8.00g strikes and embeds itself in a block with mass 0.992kg that rests on a frictionless, horizontal s

mafiozo [28]

Answer:

block velocity v = 0.09186 = 9.18 10⁻² m/s and speed bollet v₀ = 11.5 m / s

Explanation:

We will solve this problem using the concepts of the moment, let's try a system formed by the two bodies, the bullet and the block; In this system all scaffolds during the crash are internal, consequently, the moment is preserved.

Let's write the moment in two moments before the crash and after the crash, let's call the mass of the bullet (m) and the mass of the Block (M)

Before the crash

p₀ = m v₀ + 0

After the crash

$p_{f}$ = (m + M) v

p₀ = $p_{f}$

m v₀ = (m + M) v (1)

Now let's lock after the two bodies are joined, in this case the mechanical energy is conserved, write it in two moments after the crash and when you have the maximum compression of the spring

Initial

Em₀ = K = ½ m v2

Final

E $m_{f}$ = Ke = ½ k x2

Emo = E $m_{f}$

½ m v² = ½ k x²

v² = k/m x²

Let's look for the spring constant (k), with Hook's law

F = -k x

k = -F / x

k = - 0.75 / -0.25

k = 3 N / m

Let's calculate the speed

v = √(k/m) x

v = √ (3/8.00) 0.15

v = 0.09186 = 9.18 10⁻² m/s

This is the spped of the block plus bullet rsystem right after the crash

We substitute calculate in equation (1)

m v₀ = (m + M) v

v₀ = v (m + M) / m

v₀ = 0.09186 (0.008 + 0.992) /0.008

v₀ = 11.5 m / s

6 0

2 years ago