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4 years ago

PLEASE HELP!!! WILL MARK BRAINLIEST!! Name and define the four fundamental forces.

Physics

2 answers:

Naily [24]4 years ago

5 0

Answer:

Gravity - is a natural phenomenon by which all things with mass or energy—including planets, stars, galaxies, and even light—are brought toward one another.

Strong Interaction - is the mechanism responsible for the strong nuclear force

Weak Interaction - is the mechanism of interaction between subatomic particles that is responsible for the radioactive decay of atoms

Electromagnetism - is a branch of physics involving the study of the electromagnetic force, a type of physical interaction that occurs between electrically charged particles.

Hope this helps!

alukav5142 [94]4 years ago

4 0

Answer:

Gravity.The weak force. Electromagnetism.The strong force.

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You have 50L of water, it froze. So find its new volume. Density of water is 1000kg/m^3 or 1kg/L, density of ice is 920kg/m^3 or

ad-work [718]

Answer:

D (density) = Mass / Volume

V (ice) = Mass / Density = 50000 g / .92 kg / L

V = 50 kg / .92 kg / L

1 Liter of water weighs 1 kilogram = 50 L weighs 50000 g

V = 54.3 L the mass does not change upon the change of phase (freezing)

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3 years ago

50 points According to kinetic theory, in a closed system, what happens when two atoms collide?

vfiekz [6]

I believe its the 1st option

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4 years ago

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Why does gravity on Earth have a stronger attraction with you than the sun has with you? Plsssssss help I really need this rn

n200080 [17]

Answer:

Although the Earth has a lesser mass than the Sun, it is far closer to you, allowing for a stronger pull.

Explanation:

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3 years ago

A solid 0.4550 kg ball rolls without slipping down a track toward a vertical loop of radius R = 0.6750 m . What minimum translat

Talja [164]

Answer:

u = 3.35 m/s

Explanation:

given,

mass , m = 0.455 kg

R = 0.675 m

Height of Loop = 1.021 m

the speed required at the top of loop be v

equating the force vertically

$m g =\dfrac{mv^2}{r}$

$9.81 =\dfrac{v^2}{0.675}$

v² = 6.622

v = 2.57 m/s

Let the initial speed of ball be u

using conservation of energy

$\dfrac{1}{2}mu^2 + \dfrac{1}{2}I\omega^2 + m g h = \dfrac{1}{2}I\omega^2 + \dfrac{1}{2}mv^2+ m g (2 R)$

where, $I =\dfrac{2}{5}mr^2$

$\dfrac{1}{2}mu^2 + \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{u}{r})^2 + m g h = \dfrac{1}{2}(\dfrac{2}{5}mr^2)(\dfrac{v}{r})^2 + \dfrac{1}{2}mv^2+ m g (2 R)$

$0.7 u^2 + g H = 0.7 v^2 + g(2R)$

$0.7 u^2 +9.81 \times 1.021= 0.7\times 2.57^2 + 9.81 \times 2\times 0.675)$

0.7 u² = 7.85092

u² = 11.2156

u = 3.35 m/s

the initial speed is 3.35 m/s

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3 years ago

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Irina18 [472]

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3 years ago