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ludmilkaskok [199]

3 years ago

13

Can some help me plz

1 answer:

kakasveta [241]3 years ago

5 0

Answer:

D. I'm guessing

The gold-foil experiment showed that the atom consists of a small, massive, positively charged nucleus with the negatively charged electrons being at a great distance from the centre.

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Question 12 (1 point) Question 12 Unsaved

Bezzdna [24]

Slope is your answer

7 0

3 years ago

A parallel circuit has a 125 Volt battery connected with 3 resistors. R1= 20 Ω, R2= 100 Ω, and R3= 50 Ω. Find the total current

Musya8 [376]

Answer:

10A

Explanation:

to calculate r

1/R=1/20+1/100+1/50=2/25

r=25/2=12.5

I=125/12.5=10A

8 0

3 years ago

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please hurry In which state of matter has the LEAST kinetic energy? A) gas B) liquid C) plasma D) solid

anastassius [24]

Molecules in the solid phase have the least amount of energy, while gas particles have the greatest amount of energy.

7 0

3 years ago

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A mass is oscillating with amplitude A at the end of a spring.

Dmitry_Shevchenko [17]

A) $x=\pm \frac{A}{2\sqrt{2}}$

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

$E=\frac{1}{2}kA^2$ (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

$E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2$ (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

$U=\frac{1}{3}K$

Using (2) we can rewrite this as

$U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}$

And using (1), we find

$U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2$

Substituting $U=\frac{1}{2}kx^2$ into the last equation, we find the value of x:

$\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}$

B) $x=\pm \frac{3}{\sqrt{10}}A$

In this case, the kinetic energy is 1/10 of the total energy:

$K=\frac{1}{10}E$

Since we have

$K=E-U$

we can write

$E-U=\frac{1}{10}E\\U=\frac{9}{10}E$

And so we find:

$\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A$

3 0

3 years ago

a skier starts at rest at the top of a hill with 350 J of gravitational potential energy. Assuming energy is conserved, what is

spayn [35]

Answer:

350Joules

Explanation:

According to law of Conservation of energy, the amount of energy at the used up at the start is equal to that at the end.

The initial energy used up is gravitational potential energy

Final energy at the lowest point is kinetic energy.

If the energy is conserved then it means energy is not used up during the process hence;

Initial Potential energy = Final kinetic energy

If the gravitational potential energy is 350Joules then her final kinetic energy at the lowest point will also be 350Joules

3 0

3 years ago