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oksano4ka [1.4K]

3 years ago

7

Discuss two reasons why people find transition between school and university

1 answer:

dangina [55]3 years ago

6 0

Answer:

Is that your answer

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PLEASE HELP ASAP :|

barxatty [35]

Rub a balloon on a woolen fabric to pick up some electrons, to make the balloon negatively charged, and stick them to a wall, which would be positively charged to make them stick.

Opposites attract and when you stick a negatively charged objects to positively charged objects, they tend to stick together. When you pick up electrons, it increases the number of electrons which will make the object negatively charged.

Note: The first part of the answer is a single sentence. The problem says in a complete sentence, so just in case that you need only one sentence you can take the first part. If you can add in more than a sentence, you can put in more from the second paragraph.

5 0

2 years ago

Two charged concentric spherical shells have radii of 11.0 cm and 14.0 cm. The charge on the inner shell is 3.50 ✕ 10−8 C and th

Sergio039 [100]

Answer:

The magnitude of the electric field are $2.38\times10^{4}\ N/C$ and $1.09\times10^{4}\ N/C$

Explanation:

Given that,

Radius of inner shell = 11.0 cm

Radius of outer shell = 14.0 cm

Charge on inner shell $q_{inn}=3.50\times10^{-8}\ C$

Charge on outer shell $q_{out}=1.60\times10^{-8}\ C$

Suppose, at r = 11.5 cm and at r = 20.5 cm

We need to calculate the magnitude of the electric field at r = 11.5 cm

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Where, q = charge

k = constant

r = distance

Put the value into the formula

$E=\dfrac{9\times10^{9}\times3.50\times10^{-8}}{(11.5\times10^{-2})^2}$

$E=2.38\times10^{4}\ N/C$

The total charge enclosed by a radial distance 20.5 cm

The total charge is

$q=q_{inn}+q_{out}$

Put the value into the formula

$q=3.50\times10^{-8}+1.60\times10^{-8}$

$q=5.1\times10^{-8}\ C$

We need to calculate the magnitude of the electric field at r = 20.5 cm

Using formula of electric field

$E=\dfrac{kq}{r^2}$

Put the value into the formula

$E=\dfrac{9\times10^{9}\times5.1\times10^{-8}}{(20.5\times10^{-2})^2}$

$E=1.09\times10^{4}\ N/C$

Hence, The magnitude of the electric field are $2.38\times10^{4}\ N/C$ and $1.09\times10^{4}\ N/C$

7 0

2 years ago

A student look at ocean waves coming into the beach. An oceanwavd with more energy will

pshichka [43]

<span>C. Travel toward the beach faster</span>

5 0

3 years ago

Read 2 more answers

1. A 61-kg woman doing pull-ups lifts her body a distance of 0.32 meters in 1.8 seconds. What is the power provided by her bicep

Allushta [10]

Answer:

P = mgh/t = 61(9.8)(0.32)/1.8 = 106.275555... ≈ 110 W

Explanation:

Power is the rate of doing work.

The work changes her potential energy.

4 0

2 years ago

Pls help asap asap asap aaaaaaa

Goryan [66]

Bro the picture is too dark I can’t see do it again so I can help you

6 0

2 years ago