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3 years ago

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Using the set-up seen here, Ms. Garcia places a golf ball between the globe and the flashlight. Turning off the lights in the ro

om, she shines a flashlight from behind the golf ball. The students observe the shadow of the golf ball falling on the globe. What is this setup trying to model?
A) a solar eclipse
B) the solar system
C) the idea of gravity
D) the surface of the moon

1 answer:

Julli [10]3 years ago

7 0

This is a solar Eclipse.

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Can someone please describe surface tension to me. Please don't give me the dictionary definition. Thank you

11111nata11111 [884]

Surface tension - My definition -

It's exactly what it says - The tension of a surface with a liquid (such as water), caused by the attraction of the surfaces layer ---- I hope this helps ---- I actually did research it and got some of this from a dictionary, but I changed some of it, too.... Sorry if this doesn't help :)

6 0

3 years ago

An oxygen atom picks up two additional, free-floating electrons. Is the charge of the newly formed oxygen ion positive, negative

MariettaO [177]

Answer: Negative

Explanation:

An oxygen atom has the chemical symbol O. Its atomic number is 8, with an electronic configuration of 1s2, 2s2 2p4.

Now, due to the incompletely filled p-orbital, oxygen picks up two additional, free-floating electrons (2e-) to complete its outermost shell. And since each electrons received carries a negative charge, oxygen atom becomes a negatively charged divalent ion, O2-

See the equation below

O + 2e- ---> O2-

4 0

3 years ago

Four traveling waves are described by the following equations, where all quantities are measured in SI units and y represents th

agasfer [191]

Answer:

$T_1=T_3=\dfrac{2\pi}{21}$

$T_2=T_4=\dfrac{2\pi}{42}$

Explanation:

Wave 1, $y_1=0.12\ cos(3x-21t)$

Wave 2, $y_2=0.15\ sin(6x+42t)$

Wave 3, $y_3=0.13\ cos(6x+21t)$

Wave 4, $y_4=-0.27\ sin(3x-42t)$

The general equation of travelling wave is given by :

$y=A\ cos(kx\pm \omega t)$

The value of $\omega$ will remain the same if we take phase difference into account.

For first wave,

$\omega_1=21$

$\dfrac{2\pi }{T_1}=21$

$T_1=\dfrac{2\pi}{21}$

For second wave,

$\omega_2=42$

$\dfrac{2\pi }{T_2}=42$

$T_2=\dfrac{2\pi}{42}$

For the third wave,

$\omega_3=21$

$\dfrac{2\pi }{T_3}=21$

$T_3=\dfrac{2\pi}{21}$

For the fourth wave,

$\omega_4=42$

$\dfrac{2\pi }{T_4}=42$

$T_4=\dfrac{2\pi}{42}$

It is clear from above calculations that waves 1 and 3 have same time period. Also, wave 2 and 4 have same time period. Hence, this is the required solution.

3 0

3 years ago

Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d

vodka [1.7K]

Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that $F = k \frac{Q_1Q_2}{d^2}$ so if we double the charge on $Q_1$ and double the distance to $(2d)$ we plug these into the equation to find

<span> $F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}$ </span>

So we see the new force is exactly 1/2 of the old force so your answer should be $\frac{1}{2}F$ if I can remember my physics correctly.

8 0

3 years ago

Read 2 more answers

The charge on the square plates of a parallel-plate capacitor is Q. The potential across the plates is maintained with constant

Flauer [41]

Answer:

D. Q/2

Explanation:

See attached file

8 0

3 years ago