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katovenus [111]
4 years ago
8

What evidence supports the fact that this image is of a trial? Check all of the boxes that apply. The women are pointing. The wo

men are wearing white hats. It looks like there are guards and judges, as well as a jury.
Physics
2 answers:
tino4ka555 [31]4 years ago
5 0

Answer:The women are pointing and it looks like there are guards and judges ,as well as a jury

Explanation:

I just put these answers on edgenuity and it told me I was right

vesna_86 [32]4 years ago
3 0

Answer:

the last one just took it on edgenuity :)

Explanation:

i got it right!

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Please Help!!! It's for a quiz!!
Sphinxa [80]

Answer:

38.8 m/s

Explanation:

Force F(x) = 6 - 2x + 6x²

work

W=\displaystyle\int_{0}^{13.9}F(x)dx=\displaystyle\int_{0}^{13.9}(6-2x+6x^2)dx

=6x-x^2+2x^3|_{0}^{13.9}\\=5261 J

W = mv²/2=7v²/2 = 3.5v² = 5261

v = 38.8 m/s

3 0
2 years ago
How can we avoid water pollution​
Alexandra [31]

Explanation:

1.Pick up litter and throw it away in a garbage can.

2.Blow or sweep fertilizer back onto the grass if it gets onto paved areas. ...

3.Mulch or compost grass or yard waste. ...

4.Wash your car or outdoor equipment where it can flow to a gravel or grassed area instead of a street.

5.Don't pour your motor oil down the storm drain.

8 0
3 years ago
Read 2 more answers
If a sinusoidal electromagnetic wave with intensity 18 W/m2 has an electric field of amplitude E, then a 36 W/m2 wave of the sam
Neporo4naja [7]

Answer:

The  correct option is D

Explanation:

From the question we are told that

  The intensity of the first  electromagnetic wave is  I =  18 \  W/m^2

  The amplitude of the electric field is  E_{max}_1 =A

   The intensity of the second electromagnetic wave is  I =  36 \  W/m^2

Generally the an electromagnetic wave intensity is mathematically represented as

       I  =  \frac{1}{2} *  \epsilon_o  * c  * E_{max}^2

Looking at this equation we see that

     I \ \ \alpha  \ \ E^2_{max}

=>  \frac{I_1}{I_2}  =  [ \frac{ E_{max}_1}{ E_{max}_2} ] ^2

=>   E_{max}_2 = \sqrt{\frac{x}{y} }  *  E_{max}_1

=>  E_{max}_2 = \sqrt{\frac{36}{18} }  * E        

=>  E_{max}_2 = \sqrt{2 }  E        

5 0
3 years ago
A. Draw the electric field lines around a negative charge.
Alborosie
<h2>a. Answer:</h2>

We use Electric field lines for visualizing electric  fields, so this helps us to see the problem more real. So an electric field line is an imaginary  line or curve drawn through a region of space such that the tangent at any point comes from the direction of the electric-field vector at that point. The electric field lines around a negative charge is shown in the First figure below.

<h2>b. Answer:</h2>

Electric forces can be found by using the Coulomb Law's that states <em>that The magnitude of the electric force between two point charges is directly proportional  to the product of the charges and inversely proportional to the square  of the distance between them. </em>This can be expressed as follows:

F=k\frac{\left | q_{1}q_{2} \right |}{r^2} \\ \\ Where: \\ \\ k=9\times 10^9Nm^2/c^2 \\ \\ q_{1}=0.00150 C \ and \ q_{2}=0.00240 C \\ \\ r=0.900 m

Then:

F=9\times 10^9\frac{\left | 0.00150 \times 0.00240 \right |}{(0.900)^2} \\ \\ \therefore \boxed{F=40000N}

This force is repulsive because the two charges are positive and recall that two positive charges or two negative charges repel each other while a positive charge  and a negative charge attract each other.

<h2>C. Answer:</h2>

From the statement, we have two charged objects. Let's say that this charges are:

q_{1} \ and \ q_{2}

If the amount of charge on one of the objects is tripled, let's say this is the charge q_{2}, then the new charge is:

q_{N}=3q_{2}

In the formula of Coulomb:

F=k\frac{\left | q_{1}q_{N} \right |}{r^2} \\ \\ \therefore F=k\frac{\left | q_{1}(3q_{2}) \right |}{r^2} \\ \\ \therefore \boxed{F=3k\frac{\left | q_{1}q_{2} \right |}{r^2}}

<em>The conclusion is that if the amount of charge on one of the objects is tripled, the electric force between two charged objects is also tripled</em>

<h2>d. Answer:</h2>

Let's use the Coulomb's Law again to solve this problem. We want to know how the electric force between two charged objects changes if the charges are moved closer together:

F=k\frac{\left | q_{1}q_{N} \right |}{r^2}

<em>By saying that the charges are moved closer together, we want to express that r becomes smaller. Since r is in the denominator, this implies that the electric force between these two charged objects becomes greater.</em>

<h2>e. Answer:</h2>

From the figure, we can see a metal sphere on a stand. There we have both positive and negative charges. We can say that the positive charge of this sphere is +10q and the negative and the negative charge is -10q. Since the electric charge is conserved, then the algebraic sum of all the electric charges in any closed system is constant. In conclusion, <em>the sphere has no net charge.</em>

<h2>f. Answer:</h2>

Here we want to know how the negative charges in the same sphere are redistributed when a positively charged rod is brought near it. Therefore, positive charge on rod  repels positive charges on the sphere, creating  zones of negative and  positive charge as indicated in the second Figure.

7 0
3 years ago
Read 2 more answers
a student conducts an experiment to determine how the additional of salt to water affects the density of the water. the student
Lera25 [3.4K]
The water would be because however much salt you add the water rises
3 0
4 years ago
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