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3 years ago

7

Perhaps you have attended a party where balloons floated. A balloon filled with air does not rise above your head, but a balloon

filled with helium gas rises in the air to the end of its string. What does this tell you about the density of helium?

1 answer:

Drupady [299]3 years ago

3 0

Helium has a low dense than air

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The drawing shows two situations in which charges are placed on the x and y axes. They are all located at the same distance of 5

ra1l [238]

Answer:

For situation (a)

net charge E = E₊₂ + E₋₅ + E₋₃

E = K(q/d²)

where K = 8.99e9

d = 5.7cm = 5.7e-2m

Therefore,

E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)

E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) = 7.88e5(+y)

E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) = 4.73e5(+x)

thus

E = E₊₂ + E₋₅ + E₋₃

= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)

= 7.88e6(x) + 7.88e6(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.89e5)^{2} + (7.89e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) net magnitude = 1.115e6 $\frac{N}{C}$ @ 45° above +x axis

for situation (b)

net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆

E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)

E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)

E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)

E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)

thus,

E = E₊₄ + E₊₁ + E₋₁ + E₊₆

= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)

= 7.88e5(x) + 7.88e5(y)

use Pythagorean theorem

I <em>E </em>I = $\sqrt{(7.88e5)^{2} + (7.88e5)^{2}}$ = 1.242e6 $\frac{N}{C}$

∅ = $tan^{-1}$ $(\frac{7.88e5}{7.88e5} )$ = $tan^{-1}(1)$ = 45°

Thus for (a) and (b) the net magnitude = 1.242e6 $\frac{N}{C}$ @ 45° above +x axis

Explanation:

I attached a sample image, i hope that corresponds to your question

5 0

3 years ago

When you whisper you produce a 10-dB sound.

masha68 [24]

All you would do is for a, 10 times 2 is 20 so it would be 20-dB
For b, 10 times 4 is 40 so it would be 40-dB
<span>For c, 10 times 8 is 80 so it would be 80-dB</span>

4 0

4 years ago

Calculate the force needed to accelerate a car of a mass 1000 kg by 3 m/s2

beks73 [17]

Answer:

<h2>3000 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 1000 × 3

We have the final answer as

<h3>3000 N</h3>

Hope this helps you

5 0

3 years ago

Please help me find the equations guys

Alexeev081 [22]

The line at the bottom of the picture ... probably the first line on a list of choices .. is the correct equation.

4 0

3 years ago

A plastic rod is charged up by rubbing a wool cloth, and brought to an initially neutral metallic sphere that is insulated from

andrew-mc [135]

Answer:

Yes option A is right.

Explanation:

As we know that the "Opposite charges attract and like charges repel eachother". So based upon that fact we find out the sphere will be repelled or attract by the rod. As in this case metallic sphere was neutral initially but then we touched the rod with it. Although it was for few seconds but the charge is transferred to the sphere. Now both sphere and the rod have charge. After the seperation we look towards their respond If both have the opposite charge they will attract eachother. But here in this case they repel because they have the same charge, as we have charged the neutral sphere with the rod so we already know that they have the same charges that is why they are repelling eachother.

Insulation from the ground means that blocking the way of charges or free electrons from earth to metallic sphere and vice versa. As there exists free electrons and charges in earth they would flow into the metallic objects. So for more precise and accurate experiments we insulate the metals or prevent the metals from touching the earth surface to avoid the flow of charges through them. I hope it will help you.

5 0

3 years ago