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Bad White [126]
3 years ago
8

Which particles in an atom could demonstrate that opposite charges attract?

Physics
2 answers:
Oxana [17]3 years ago
4 0

Answer:

Choice D. a proton and an electron.

Explanation:

Each proton carries a positive charge of +1 unit.

Each electron carries a negative charge of -1 unit.

As their name suggests, neutrons are neutral. They do not carry any electrostatic charge. Because of that, neutrons can't attract or repel neutrons or protons with electrostatic forces.

On the other hand, electrons and protons carry opposite charges. Therefore, an electron and a proton would attract each other electrostatically.

However, since protons are all positive, two protons would repel each other electrostatically.

kompoz [17]3 years ago
3 0

Answer:

The correct answer is D

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A nonconducting sphere of diameter 10.0 cm carries charge distributed uniformly inside with charge density of +5.50 µC/m3 . A pr
VLD [36.1K]

Answer:

t = 2.58*10^-6 s

Explanation:

For a nonconducting sphere you have that the value of the electric field, depends of the region:

rR:\\\\E=k\frac{Q}{r^2}

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

R: radius of the sphere = 10.0/2 = 5.0cm=0.005m

In this case you can assume that the proton is in the region for r > R. Furthermore you use the secon Newton law in order to find the acceleration of the proton produced by the force:

F=m_pa\\\\qE=m_pa\\\\k\frac{qQ}{r^2}=m_pa\\\\a=k\frac{qQ}{m_pr^2}

Due to the proton is just outside the surface you can use r=R and calculate the acceleration. Also, you take into account the charge density of the sphere in order to compute the total charge:

Q=\rho V=(5.5*10^{-6}C/m^3)(\frac{4}{3}\pi(0.05m)^3)=2.87*10^{-9}C\\\\a=(8.98*10^9Nm^2/C^2)\frac{(1.6*10^{-19}C)(2.87*10^{-9}C)}{(1.67*10^{-27}kg)(0.05m)^2}=9.87*10^{11}\frac{m}{s^2}

with this values of a you can use the following formula:

a=\frac{v-v_o}{t}\\\\t=\frac{v-v_o}{a}=\frac{2550*10^3m/s-0m/s}{9.87*10^{11}m/s^2}=2.58*10^{-6}s

hence, the time that the proton takes to reach a speed of 2550km is 2.58*10^-6 s

3 0
3 years ago
Your answer should be precise to 0.1 m/s. Use a gravitational acceleration of 10 m/s/s. At it lowest point, a pendulum is moving
saw5 [17]

Explanation:

It is given that,

Speed, v₁ = 7.7 m/s

We need to find the velocity after it has risen 1 meter above the lowest point. Let it is given by v₂. Using the conservation of energy as :

\dfrac{1}{2}mv_1^2=\dfrac{1}{2}mv_2^2+mgh

v_2^2=v_1^2-2gh

v_2^2=(7.7)^2-2\times 10\times 1

v_2=6.26\ m/s

So, the velocity after it has risen 1 meter above the lowest point is 6.26 m/s. Hence, this is the required solution.

4 0
3 years ago
Read 2 more answers
I need help in this small question:
Natali5045456 [20]

Answer:

1) The energy released during nuclear fission or fusion, especially when used to generate electricity is called nuclear energy.

2) It is not renewable because it is an element that has no way whatsoever to regenerate or replicate itself, nor gets created by any natural terrestrial means, neither makes itself available by arriving from outer space (like sunlight). There is a limited amount of it available on the Earth, and every bit you use is a bit you’ll never have available again (as Uranium atoms get destroyed by the fission process).

Hope this helps You

Please mark as brainliest

Thank You

5 0
3 years ago
A golfer hits a golf ball upwards at an angle. We can ignore air resistance on the ball.
Natalka [10]

Answer:

Check the diagram from the photo

Explanation:

5 0
3 years ago
The space shuttle usually orbited Earth at altitudes of around 300.0 km. 1) Determine the time for one orbit of the shuttle abou
777dan777 [17]

Answer:

Explanation:

distance of shuttle from centre of the earth = radius of the orbit

= 6300 + 300 = 6600 km

= 6600 x 10³

Formula of time period of the satellite

T = 2π R /v₀ , v₀ is orbital velocity

v₀ = √gR , ( if height  is small with respect to radius )

T = 2π R /√gR

= 2π√ R /√g

= 2 x 3.14 x √ 6600 x 10³ / √9.8

= 2 x 3.14 x 256.9 x 10 / 3.13

= 5154.41 s

= 5154.41 / 60 minutes

= 85.91 m

85.9 minutes.

2 ) No of sunrise per day = no of rotation per day

= 24 x 60 / 85.9

= 16.76

or 17 sunrises.

3 0
3 years ago
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