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2 years ago

They occupy the space surrounding the nucleus of the atom, they have a -1 charge

Physics

1 answer:

kipiarov [429]2 years ago

4 0

Answer:

Electrons.

Explanation:

Look it up.

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Name two energy transformations that occur as Adeline pedals her bicycle up a steep hill and then coasts down the other side.

Juliette [100K]

Potential energy is first transformed into kinetic energy as she pedals, then gravitational as she coasts down the hill.

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3 years ago

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What does altered mean

Elina [12.6K]

Altered means to change something.

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3 years ago

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At what position in its elliptical orbit is the speed of a planet a maximum? when it is closest to the sun when it is farthest f

Phantasy [73]

Answer:

1.when it is closest to the sun

2.when it is midway between its farthest

Explanation:

According to the law of Kepler's

T ² ∝ r³

T=Time period

r=semi major axis

We also know that time period T given as

$T=\dfrac{2\pi r}{v}$

v=Speed

$v=\dfrac{2\pi r}{T}$

$v\alpha \dfrac{r}{T}$

$v^2\alpha \dfrac{r^2}{T^2}$

$v^2\alpha \dfrac{r^2}{r^3}$

$v^2\alpha \dfrac{1}{r}$

$v\alpha \dfrac{1}{\sqrt{r}}$

So we can say that ,when r is more then the speed will be minimum and when r is low then speed will be maximum.

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2 years ago

Help me please. <br> A<br> B<br> C<br> D

Y_Kistochka [10]

I will go to school tomorrow .....is this present tense or past tense or future tense

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1 year ago

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A 225-g object is attached to a spring that has a force constant of 74.5 N/m. The object is pulled 6.25 cm to the right of equil

snow_lady [41]

Answer:

1.137278672 m/s

+5.9 cm or -5.9 cm

Explanation:

A = Amplitude = 6.25 cm

m = Mass of object = 225 g

k = Spring constant = 74.5 N/m

Maximum speed is given by

$v_m=A\omega\\\Rightarrow v_m=A\sqrt{\dfrac{k}{m}}\\\Rightarrow v_m=6.25\times 10^{-2}\times \sqrt{\dfrac{74.5}{0.225}}\\\Rightarrow v_m=1.137278672\ m/s$

The maximum speed of the object is 1.137278672 m/s

Velocity is at any instant is given by

$\dfrac{v_m}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A\omega}{3}=\omega\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A}{3}=\sqrt{A^2-x^2}\\\Rightarrow \dfrac{A^2}{9}=A^2-x^2\\\Rightarrow A^2-\dfrac{A^2}{9}=x^2\\\Rightarrow x^2=\dfrac{8}{9}A^2\\\Rightarrow x=A\sqrt{\dfrac{8}{9}}\\\Rightarrow x=6.25\times 10^{-2}\sqrt{\dfrac{8}{9}}\\\Rightarrow x=\pm 0.0589255650989\ m$

The locations are +5.9 cm or -5.9 cm

4 0

2 years ago