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4 years ago

7

Gary is trying to think of an object that could model a sunspot. His teacher tells him that such an object might be found right

on his face. Which of the following most closely models the behavior of a sunspot? A. an eyelash B. a freckle C. a birthmark D. a lip

2 answers:

alexgriva [62]4 years ago

7 0

Ans. (B). a freckle.

Sunspots can be defined as temporary phenomena on the photosphere of sun, which appears as spot darker than its surrounding regions. Freckles on the face can model the behavior of sunspots are they represent spots brown spots on skin, darker than face complexion.

Freckles are resulted due to concentrated melaninized cells, having melanin (a type of skin pigment).

Thus, the correct answer is option (B).

const2013 [10]4 years ago

6 0

the answer is B a freckle I just had this as a Study Island and got it right.

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3 years ago

You dip your finger into a pan of water twice each second, producing waves with crests that are separated by 0.15 m. Determine t

PIT_PIT [208]

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3 years ago

A tube of water is open on one end to the environment while the other end is closed. The height of the water relative to the bas

Andreas93 [3]

Answer:

<em>1.06 atm</em>

Explanation:

On the open end of the tube, the pressure will be the sum of atmospheric pressure and the pressure due to the height of water

The pressure due to a height of water = ρgh

where ρ is the density of water = 1000 kg/m^3

g is the acceleration due to gravity = 9.81 m/s^2

h is the height of the water column

The height of water column on the open end = 100 cm = 1 m

pressure on this end = ρgh = 1000 x 9.81 x 1 = 9810 Pa

Atmospheric pressure = 101325 Pa

The total pressure on the open end = 101325 Pa + 9810 Pa = <em>111135 Pa</em>

The pressure due to the water column on the closed end = ρgh

The height of the water in the closed end = 40 cm = 0.4 m

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The resultant pressure on the water on the top of the closed end of the tube = 111135 Pa - 3924 Pa =<em> 107211 Pa</em>

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3 years ago

Which of the following is NOT a characteristic of an inner planet?. . A.. rocky. . B.. solid surface. . C.. near the sun. . D..

sveticcg [70]

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3 years ago

Read 2 more answers

An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic

malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

$\omega_f = \omega_0 + \alpha t$

Where,

$\omega_f =$ Final Angular Velocity

$\omega_0 =$ Initial Angular velocity

$\alpha =$ Angular acceleration

t = time

The relation between the tangential acceleration is given as,

$a = \alpha r$

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

$\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s$

$\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s$

$t = 11s$

Replacing the previous equation with our values we have,

$\omega_f = \omega_0 + \alpha t$

$9.8436 = 6.5973 + \alpha (11)$

$\alpha = \frac{9.8436- 6.5973}{11}$

$\alpha = 0.295rad/s^2$

The tangential velocity then would be,

$a = \alpha r$

$a = (0.295)(0.2)$

$a = 0.059m/s^2$

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

$\omega_f^2=\omega_0^2+2\alpha\theta$

Replacing with our values and re-arrange to find $\theta,$

$\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}$

$\theta = \frac{9.8436^2-6.5973^2}{2*0.295}$

$\theta = 90.461rad$

That is equal in revolution to

$\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev$

The linear displacement of the system is,

$x = \theta*(2\pi*r)$

$x = 14.397*(2\pi*\frac{0.25}{2})$

$x = 11.3m$

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3 years ago