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3 years ago

How do you find the net force acting on an object?

1 answer:

6 0

It's D. By "net" they mean the overall force the object experiences, so sum all the force vectors, those in a negative direction (eg friction) should be subtracted.

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A particle vibrates in Simple Harmonic Motion with amplitude. What will be its displacement in one time-period? A 2A 4A 0

yan [13]

'Displacement' is the distance and direction between the starting point and
ending point, regardless of the path followed to get there.

A particle that's executing simple harmonic motion is always in the same place
where it was one time period ago, and where it will be later after another time
period has passed.

So its displacement during exactly one time period is exactly zero.

3 0

4 years ago

Can someone help me?

Cloud [144]

What do we know that might help here ?

-- Temperature of a gas is actually the average kinetic energy of its molecules.

-- When something moves faster, its kinetic energy increases.

Knowing just these little factoids, we realize that as a gas gets hotter, the average speed of its molecules increases.

That's exactly what Graph #1 shows.

How about the other graphs ?

-- Graph #3 says that as the temperature goes up, the molecules' speed DEcreases. That can't be right.

-- Graph #4 says that as the temperature goes up, the molecules' speed doesn't change at all. That can't be right.

-- Graph #2 says that after the gas reaches some temperature and you heat it hotter than that, the speed of the molecules starts going DOWN. That can't be right.

4 0

3 years ago

A uniform line charge of density λ lies on the x axis between x = 0 and x = L. Its total charge is 7 nC. The electric field at x

DedPeter [7]

Answer:

The electric field at x = 3L is 166.67 N/C

Solution:

As per the question:

The uniform line charge density on the x-axis for x, 0< x< L is $\lambda$

Total charge, Q = 7 nC = $7\times 10^{- 9} C$

At x = 2L,

Electric field, $\vec{E_{2L}} = 500N/C$

Coulomb constant, K = $8.99\times 10^{9} N.m^{2}/C^{2}$

Now, we know that:

$\vec{E} = K\frac{Q}{x^{2}}$

Also the line charge density:

$\lambda = \frac{Q}{L}$

Thus

Q = $\lambda L$

Now, for small element:

$d\vec{E} = K\frac{dq}{x^{2}}$

$d\vec{E} = K\frac{\lambda }{x^{2}}dx$

Integrating both the sides from x = L to x = 2L

$\int_{0}^{E}d\vec{E_{2L}} = K\lambda \int_{L}^{2L}\frac{1}{x^{2}}dx$

$\vec{E_{2L}} = K\lambda[\frac{- 1}{x}]_{L}^{2L}] = K\frac{Q}{L}[frac{1}{2L}]$

$\vec{E_{2L}} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{2L}] = \frac{63}{L^{2}}$

Similarly,

For the field in between the range 2L< x < 3L:

$\int_{0}^{E}d\vec{E} = K\lambda \int_{2L}^{3L}\frac{1}{x^{2}}dx$

$\vec{E} = K\lambda[\frac{- 1}{x}]_{2L}^{3L}] = K\frac{Q}{L}[frac{1}{6L}]$

$\vec{E} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{6L}] = \frac{63}{6L^{2}}$

Now,

If at x = 2L,

$\vec{E_{2L}} = 500 N/C$

Then at x = 3L:

$\frac{\vec{E_{2L}}}{3} = \frac{500}{3} = 166.67 N/C$

4 0

4 years ago

I NEED THE RIGHT ANSWER ASAP NO LINKS !!!<br> This is a Science question

Tanya [424]

Answer:

The answer is the top choice. (-1)

Explanation:

We are dealing with a negative number here, so we should start at negative five on the number line. From here we see that we are subtracting a negative number from it, which means we are really adding. (Review your rules for negative signs if this doesn't make sense.) So we add 4 to -5, and we get -1. Thus, we get the top choice.

6 0

3 years ago

How much energy is converted when a 110kw appliance is used for 3.0 hrs?

nadya68 [22]

https://quizlet.com/latest

this website might help you can look up your question

8 0

3 years ago