Given the following information, in which medium is the velocity of light the highest?

In order to be neutral, an atom has to contain equal numbers of _______

Vedmedyk [2.9K]

In order to be electrically neutral, an atom has to contain
equal numbers of protons (with their positive charges) and
electrons (with their negative charges).

8 0

4 years ago

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a block is pushed up a frictionless 40 incline. if the initial velocity after the push is 5.00 m/s. how far along the incline do

Lana71 [14]

Answer:

d=1.982m, t=1.019s

Explanation:

There are different approaches we can take to solve this problem. You could either solve this by using conservation of energy or by taking a kinematic approach. I'll solve this by using kinematics. So, the very first thing we need to do in order to solve this is do a drawing of the situation so we can analyze it better. (See attached picture).

So, since we are talking about an inclined plane, we can see that the force of gravity is being split into an x and y components if we incline the axis of coordinates. Taking this into account we can see that:

$\sum F_{x}=ma_{x}$

Since there is no friction in our system, then the only force acting upon the box is the force of gravity, or weight. Since we are taking the upwards direction as the positive direction of movement, we can say that the force of gravity is excerting a negative influence on our box, so this acceleration will be negative, so our sum of forces will now look like this:

$-mg sin(40^{o})=ma$

we can cancel the masses out so we can see that:

a=-g sin(40°)

$a=-9.81m/s^{2} sin(40^{o})$

We have now enough information to solve our problem.

we can take the following equation to find the distance the block travels up the incline:

$x=\frac{V_{f}^{2}-V_{0}^{2}}{2a}$

we know the final velocity must be zero, so we can use the provided data to solve our formula:

$x=\frac{(0)^{2}-(5m/s)^{2}}{2(-9.81m/s^{2})sin 40^{o}}$

which yields:

x=1.982m

In order to find the time it takes for the block to return to its original position we can use the following formula:

$x=V_{0}t+\frac{1}{2}at^{2}$

since x=0 is the starting point we can use that to solve our equation:

$0=5t+\frac{1}{2}(-9.81sin 40^{o})t^{2}$

which simplifies to:

$0=5t-4.905t^{2}$

which can now be solved for t

t(5-4.905t)=0

t=0 and 5-4.905t=0

t=0 and $t=\frac{5}{4.905}=1.019$

so the time it takes the block to return to its original position is

t= 1.019s

6 0

3 years ago

Un libro del peso di 12 N è in equilibrio su un tavolo. Sapendo che il coefficiente di attrito statico vale 0,5, la forza di att

Tanzania [10]

Answer:

60

Explanation:

Translation -

A book weighing 12 N is balanced on a table. Knowing that the static friction coefficient is 0.5, how much is the friction force worth?

Friction force is

f = u * n

f = 0.5 * 12N

f = 60

4 0

4 years ago

Which type of wire would have the least resistance?

telo118 [61]

The answer is E. Short and thick while cold.

4 0

4 years ago

Someone please help. Describe how electric potential energy, kinetic energy, and work change when two charges of opposite sign a

erik [133]

Answer:

The answer is based on the conservation of energy law; something you should really understand by now.

For convenience we can hold one of the two charges still; it becomes the frame of reference. And everything we say is in reference to the designated static charge, call it Q.

So the moving charge, call it q, has total energy TE = PE. It's all potential energy as we start with q not moving.

It has potential energy because in order to separate q from Q, we had to do work, add energy, on q. And from the COE law, that work added is converted into PE.

It's a bit like lifting something off the ground. That's work and it becomes GPE. So there's some work, in separating the two charges in the first place.

But there's more.

Now we let q go. As opposites attract, q is pulled to Q. And that force from Q is working on q, force over distance. Which means the potential energy q started with is being converted into kinetic energy. q is accelerating and picking up speed.

And there's more work, done by the EMF on charge q. That converts the PE into KE and the q charge smashes into Q with some kinetic energy.

5 0

3 years ago