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2 years ago

5

Hat is the processes called when matter, that is a liquid, transforms into a gas?

1 answer:

nirvana33 [79]2 years ago

8 0

Answer:

vaporization ,( Evaporation)

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VMariaS [17]

A. the light bulb goes out once the circuit is open since it causes the flow of electricity to cut off. the light bulb dosent get the energy it needs to light up

Explanation:

B. a simple example of this in our every day life is a light switch. when you switch the light on then the circuit is closed and the energy transfers to the light bulb, when u switch the light off then you cut off the lights source of energy which causes the light to turn off.

5 0

1 year ago

A national college researcher reported that 67%of students who graduated from high school in 2012 enrolled in college. Twenty-ni

dsp73

Answer:

Explanation:

Theorem of Binomial Distribution will apply here.

n = 29 , p = .67 , q = 0.33

mean = np = 29 x .67 = 19.43

Standard Deviation = √npq

= √29 x .67 x .33

= √6.4

= 2.53

=

4 0

2 years ago

When describing the interaction between magnetic poles we could say that

kifflom [539]

The Answer Is A
Hope This Helps !

8 0

2 years ago

Read 2 more answers

Two Mississippi companies, Howard Industries and Kuhlman Electric Company, are responsible for most of the transformers and powe

yanalaym [24]

I think b but im not sure just helping

5 0

3 years ago

Read 2 more answers

An electron is released from rest at a distance of 0.470 m from a large insulating sheet of charge that has uniform surface char

ArbitrLikvidat [17]

Answer:

Part a)

$W = 1.58 \times 10^{-20} J$

Part b)

$v = 1.86 \times 10^5 m/s$

Explanation:

Part a)

Electric field due to large sheet is given as

$E = \frac{\sigma}{2\epsilon_0}$

$\sigma = 4.00 \times 10^{-12} C/m^2$

now the electric field is given as

$E = \frac{4.00 \times 10^{-12}}{2(8.85 \times 10^{-12})}$

$E = 0.225 N/C$

Now acceleration of an electron due to this electric field is given as

$a = \frac{eE}{m}$

$a = \frac{(1.6 \times 10^{-19})(0.225)}{9.1 \times 10^{-31}}$

$a = 3.97 \times 10^{10}$

Now work done on the electron due to this electric field

$W = F.d$

$d = 0.470 - 0.03$

$d = 0.44 m$

So work done is given as

$W = (ma)(0.44)$

$W = (9.11 \times 10^{-31})(3.97 \times 10^{10})(0.44)$

$W = 1.58 \times 10^{-20} J$

Part b)

Now we know that work done by all forces = change in kinetic energy of the electron

so we will have

$W = \frac{1}{2}mv^2 - 0$

$1.58 \times 10^{-20} = \frac{1}{2}(9.1\times 10^{-31})v^2$

$v = 1.86 \times 10^5 m/s$

7 0

3 years ago