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3 years ago

From inside to out, describe the components of an atom. What makes one element's atoms differ from another element's atoms?

1 answer:

4 0

From the outside is it’s shell, and on the inside is the nucleus (on the middle) which is made or protons (positive charge) and neutrons (neutral charge which is negate and positive) and Lastly there are electrons (negative charge) that orbit the nucleus like planets orbiting the Sun. The shell gets made as the electrons orbit the nucleus.
How do atoms differ from eachother?- their atoms might have different elections, neutrons, and/ or protons inside of it. Plus the location of the elections might be different too.

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The 26-kg sphere c is released from rest when θ = 0∘ and the tension in the spring is f = 100 n

aleksandr82 [10.1K]

You are given the mass of a sphere that is 26 kg sphere and it is released from rest when θ = 0°. You are also given the force of the spring that is F = 100 N. You are asked to find the tension of the spring. Imagine that the sphere is connected to a spring. The spring exerts a tension and the spring exerts gravitational pull. This will follow the second law of newton.

T - F = ma
T = ma + F
T = 26kg (9.81m/s²) + 100 N
T = 355.06 N

5 0

4 years ago

A −3.0 nC charge is on the x-axis at x=−9 cm and a +4.0 nC charge is on the x-axis at x=16 cm. At what point or points on the y-

alexdok [17]

Answer:

y = 10.2 m

Explanation:

It is given that,

Charge, $q_1=-3\ nC$

It is placed at a distance of 9 cm at x axis

Charge, $q_2=+4\ nC$

It is placed at a distance of 16 cm at x axis

We need to find the point on the y-axis where the electric potential zero. The net potential on y-axis is equal to 0. So,

$\dfrac{kq_1}{r_1}+\dfrac{kq_2}{r_2}=0$

Here,

$r_1=\sqrt{y^2+9^2} \\\\r_2=\sqrt{y^2+15^2}$

So,

$\dfrac{kq_1}{r_1}=-\dfrac{kq_2}{r_2}\\\\\dfrac{q_1}{r_1}=-\dfrac{q_2}{r_2}\\\\\dfrac{-3\ nC}{\sqrt{y^2+81} }=-\dfrac{4\ nC}{\sqrt{y^2+225} }\\\\3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}$

Squaring both sides,

$3\times \sqrt{y^2+225}=4\times \sqrt{y^2+81}\\\\9(y^2+225)=16\times (y^2+81)\\\\9y^2+2025=16y^2-+1296\\\\2025-1296=7y^2\\\\7y^2=729\\\\y=10.2\ m$

So, at a distance of 10.2 m on the y axis the electric potential equals 0.

8 0

3 years ago

Can someone answer this please I’ll give brainliest

Alinara [238K]

Answer:

The last one is false

Explanation:

Energy can be neither created or destroyed. It can only move from one type of energy to another.

6 0

3 years ago

How can a organization encourage total person development among their employees.

Sedbober [7]

Answer:

Paying for employees seminars and workshops related to their careers

Explanation:

To motivate personal development among employees, several things can be done. Among them, giving employees chance to present their own solutions to problems, exposing the employees to several global challenges and how to handle them, paying for employees seminars and workshops related to their own careers for professional development among other things.

8 0

3 years ago

Choose all facts that increase the orbital velocity of a vessel around planet B. Bigger mass of planet B smaller mass of planet

telo118 [61]

Answer:

- Bigger mass of planet B

- orbiting closer to planet B

Explanation:

The orbital velocity of the vessel around the planet can be found by equalizing the force of gravity between the vessel and the planet and the centripetal force:

$G\frac{mM}{r^2}=m\frac{v^2}{r}$

where

G is the gravitational constant

m is the mass of the vessel

M is the mass of the planet

r is the distance between the vessel and the centre of the planet

v is the orbital velocity of the vessel

Re-arranging the formula, we find an expression for v:

$v=\sqrt{\frac{GM}{r}}$

We see that:

- the bigger the mass of the planet, M, the bigger the velocity

- the bigger the distance between the vessel and the planet, r, the smaller the velocity

So, the correct choices that increase the orbital velocity are:

- Bigger mass of planet B

- orbiting closer to planet B

6 0

3 years ago