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3 years ago

Examine figure 26-1 which area is the convection zoneA. AB. BC. CD. D

Physics

2 answers:

zloy xaker [14]3 years ago

5 0

Answer:

B. B

Explanation:

As we know that convective zone is that part of the sun where heat is primarily transferred by convection mode. In this part of sun major proportion is gas so that heat is transferred by gas molecules.

So here we will have

outer most part of the sun is E = CORONA

then it is the surface of sun which is A = PHOTO SPHERE

Then the inner part to the surface of sun is B = CONVECTIVE ZONE

then the next inner part is vacuum where heat is transfer by C = RADIATE ZONE

Then the innermost part of the sun is its D = CORE

so correct answer will be B

kykrilka [37]3 years ago

4 0

I think the answer is B

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KATRIN_1 [288]

Answer:

T = 1.73 kg

Explanation:

Let's use Newton's second law for this balance problem

We draw a coordinate system with the x axis parallel to the plane and the Y axis perpendicular

The only force we have to lay down is the weight (W)

Wx = W sin θ

Wy = W cos θ

Wx = 2.0 ain 60

Wx = 1.73 kg

Wy = 2.0 cos 60

Wy = 1.0 kg

Y Axis

N-Wy = 0

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3 years ago

A pencil rolls horizontally of a 1 meter high desk and lands .25 meters from the base of the desk. How fast was the pencil rolli

vovikov84 [41]

Answer: 0.55 m/s

Explanation:

This situation is related to projectile motion (also called parabolic motion), where the main equations are as follows:

$x=V_{o} cos\theta t$ (1)

$y=y_{o}+Vo sin \theta t + \frac{g}{2}t^{2}$ (2)

Where:

$x=0.25 m$ is the horizontal displacement of the pencil

$V_{o}$ is the pencil's initial velocity

$\theta=0\°$ since we are told the pencil rolls <u>horizontally</u> before falling

$t$ is the time since the pencil falls until it hits the ground

$y_{o}=1 m$ is the initial height of the pencil

$y=0$ is the final height of the pencil (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity, always acting vertically downwards

Begining with (1):

$x=V_{o} cos(0\°) t$ (3)

$x=V_{o}t$ (4)

Finding $t$ from (2):

$0=1 m+ \frac{-9.8m/s^{2}}{2}t^{2}$ (5)

$t=\sqrt{\frac{-2y_{o}}{g}}$ (6)

Substituting (6) in (4):

$x=V_{o}\sqrt{\frac{-2y_{o}}{g}}$ (7)

Isolating $V_{o}$ :

$V_{o}=\frac{x}{\sqrt{\frac{-2y_{o}}{g}}}$ (8)

$V_{o}=\frac{0.25 m}{\sqrt{\frac{-2(1 m)}{-9.8m/s^{2}}}}$ (9)

Finally:

$V_{o}=0.55 m/s$

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3 years ago

Write a speech about bringing back fidget toys because they're banned in school

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3 years ago

5. Describe the shape of the waveform in the secondary coil for a sine, square and triangle wave in the primary coil. How does t

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3 years ago

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lara [203]

Answer:

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Explanation:

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In the second part of this problem we must find the work done, where the work in physics is known as the product of force by distance, it is important to make it clear that force must be applied in the direction of movement.

$W = F*d$