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olya-2409 [2.1K]
3 years ago
6

How an elastic material can no longer be elastic but plastic instead?

Physics
1 answer:
Dvinal [7]3 years ago
3 0
<span><span>deformação elástica – é aquela em que removidos os esforços atuando sobre o corpo, ele volta a sua forma original</span><span>deformação plástica – é aquela em que removidos os esforços, não há recuperação da forma original.</span></span>
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What happens to the force between two charges if the magnitude of both charges is doubled and the distance between them is doubl
vlabodo [156]

Answer:the force will remain same

Explanation:

because force is equal to the ratio of magnitude and distance

4 0
2 years ago
A racecar drives at a constant speed down a straight track. The car is in _?_
vagabundo [1.1K]

Answer:

the answer is

the car is in motion

7 0
2 years ago
Mali loves to make herself dizzy. She spins in place 7 times before falling down right where she was standing. Find her distance
Mila [183]
The answer is 0 bc she didint move
Explanation:
8 0
3 years ago
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Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical
lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}} (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}

\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2 (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio \frac{T^{2}}{a^{3}} is constant for all the planets orbiting the same sun, so we can said that:

\frac{T_{A}^{2}}{a_{A}^{3}}=\frac{T_{B}^{2}}{a_{B}^{3}}

\frac{T_{B}^{2}}{T_{A}^{2}}=\frac{a_{B}^{3}}{a_{A}^{3}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{a_{B}^{3}}{a_{A}^{3}}}=\sqrt{\frac{(2a_{A})^{3}}{a_{A}^{3}}}

\frac{T_{B}}{T_{A}}=\sqrt{\frac{2^3}{1}}=2.83

6 0
3 years ago
Read 2 more answers
PLEASE HELP MEE PLEASE I BEG
TEA [102]

Explanation: (I think)

Plug your values into the momentum equation.

So m1= 63kg

m2 = 10 kg

V1 = 12 m/s

And then plug in your values and solve for your unknown (v2)

8 0
2 years ago
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