Moira is drawing the electric field lines around a pair of charges. One charge is positive, and the other charge is negative.

1 answer:

8 0

The best answer is B. If you curve them away, they have a less tendency to be attracted to the pair of charges and would stay around the pair instead of interacting with them.

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A force of 16 N to the west is applied to each object below. Which object will

Montano1993 [528]

41kg object that is moving east at 5 m s

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2 years ago

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Two models of the same compound are shown. In what way is Model A better than Model B?

astra-53 [7]

The correct answer is Model A shows the three-dimensional shape of the molecule, but Model B does not.

Explanation:

Model A and B show the structure of a molecule. In the case of model A, the structure is represented through the use of three-dimensional shapes, while in model B the structure is represented using the letters of each element and showing how each element is connected to others.

In this context, one feature that makes model A better is that this represents the molecule using a 3D model, which is better to understand how the molecule looks like and what is its structure. Moreover, both models are alike because they show the number of atoms of each element, although model A does not show the types of elements.

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3 years ago

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Me ajudem, Por favor!!!!!!

zzz [600]

Answer:

a) $a=4\,\frac{m}{s^2}$

b) $V(t)=4\,t\,+3$

c) $V(1)=7 \,\frac{m}{s} \\$

d) Displacement = 22 m

e) Average speed = 11 m/s

Explanation:

Notice that the acceleration is the derivative of the velocity function, which in this case, being a straight line is constant everywhere, and which can be calculated as:

$slope= \frac{15=3}{3-0} =4\,\frac{m}{s^2}$

Therefore, acceleration is $a=4\,\frac{m}{s^2}$

b) the functional expression for this line of slope 4 that passes through a y-intercept at (0, 3) is given by:

$y=m\,x+b\\V(t)=4\,t\,+3$

c) Since we know the general formula for the velocity, now we can estimate it at any value for 't", for example for the requested t = 1 second:

$V(t)= 4\,t+3\\V(1)=4\,(1)+3\\V(1)=7 \,\frac{m}{s}$

d) The displacement between times t = 1 sec, and t = 3 seconds is given by the area under the velocity curve between these two time values. Since we have a simple trapezoid, we can calculate it directly using geometry and evaluating V(3) (we already know V(1)):

Displacement = $\frac{(7+15)\,2}{2} = 22\,\,m$