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3 years ago

Which forces are capable of affecting particles or objects from large distance

Physics

1 answer:

Ganezh [65]3 years ago

8 0

Answer:

only long-range force that affects all particles is the gravitational force.

Explanation:

In nature there are four fundamental forces: nuclear, weak, gravitational and electrical.

The last two are long-range, that is, the forces are zero for infinite distances, the current gravitational on all the particles and the electric one acts on the charged particles, without the chosen charge it is zero, the forces is also zero.

Consequently the only long-range force that affects all particles is the gravitational force.

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What is the relationship between the temperature of a star and its luminosity?

Sladkaya [172]

The Luminosity of a star is proportional to its Effective Temperature to the 4th power and its Radius squared.

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3 years ago

An object weighs 39.36 newtons. What is its mass if a gravitometer indicates that g = 9.83 m/s2?

CaHeK987 [17]

The force applied to an object is said to be a product of its mass and the acceleration. For this case, acceleration is the reading on the gravitometer. We calculate as follows:

F = mg
39.36 N = m(9.83 m/s^2)
m = 4.00 kg

Hope this answers the question. Have a nice day.

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4 years ago

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Sveta_85 [38]

Answer:

the first one

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3 years ago

What best describes the volcanic process of magma with dissolved water vapor and other gases as it nears the Earth surface?

Assoli18 [71]

Answer:

Pressure on the molten rock lessens and the gases dissolved in rock can bubble and expand rapidly causing violent eruptions.

Explanation:

that's just how it works lol. hope this helps :]

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3 years ago

One day you are driving your friends around town and you drive quickly around a corner without slowing down. You are going at a

Novay_Z [31]

Answer:

1.60 g

Explanation:

From the attached file below:

we can deduce that:

$v = v_x =v_y = 20 \ m/s \\t = 2s$

The distance traveled by 2 s will be:

x = vt

x = 20 m/s × 2 s

x = 40 m

The length is quarter of the circle with radius r,

so; if 2 πr = 4 x

Then radius (r) will be:

$r = \frac{4x}{2 \pi}\\\\r = \frac{4*40}{2 \pi}$

r = 25.5 m

The centripetal acceleration can be expressed as:

$a = \frac{v^2}{r}$

so;

$a = \frac{(20 \ m/s^2)}{25.5 \ m}$

a = 15.7 m/s²

the magnitude of the acceleration experienced by your unfortunate passengers in terms of acceleration due to gravity is then determined by the equation:

$a' = \frac{a}{g} g$

$a' = \frac{15.7 \m/s ^2}{9.81 \ m/s^2}\ g$

$a' = 1.60 \ g$

∴ The magnitude of the acceleration experienced by your unfortunate passengers during the turn = 1.60 g

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4 years ago

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