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Black_prince [1.1K]
2 years ago
6

Which of the following statements describes a perfectly inelastic collision?

Physics
1 answer:
Lyrx [107]2 years ago
7 0

Answer: C

Explanation:

A collision in which the objects stick together is sometimes called “perfectly inelastic“

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A comet of mass 1.20 × 10¹⁰kg moves in an elliptical orbit around the Sun. Its distance from the Sun ranges between 0.500 AU and
irina [24]

The eccentricity of its orbit is $$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

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The length of the semi-major axis is calculated as follows:

where, $G=6.67 \times 10^{-1} \mathrm{~m}^3 / \mathrm{kgs}$

$M=1.99 \times 10^{30} \mathrm{~kg}=$mass of sur

$m=1.20 \times 10^{10} \mathrm{~kg}$ - a mass of the comet

$$\begin{aligned}\therefore \quad \text { At aphelion, } r &=50 \times U \\&=50 \times 1.496 \times 10^{11} \mathrm{~m} . \\U=-\frac{6.67 \times 10^{-11} \times \mathrm{m} 1.99 \times 10^{30} \times 1.20 \times 10^{10}}{50 \times 1.496 \times 10^{11}} \\U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

$$U=-2.13 \times 10^{17} \mathrm{~J} .\end{aligned}$$

To learn more about mass, refer to:

brainly.com/question/3187640

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3 years ago
A 30-gram bullet is fired and a 50-gram bullet is dropped simultaneously from the same height. Which will hit the ground first?
Marina86 [1]

Answer: the 30gram will hit the ground first

Explanation:the 30gram bullet will hit the ground first because it is fired

4 0
3 years ago
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What is a non contact force that attracts all objects to the centre of the earth
beks73 [17]

Answer:

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<em>Hope</em><em> </em><em>this</em><em> </em><em>helps</em><em>,</em><em> </em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>x</em>

4 0
3 years ago
What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC?
Nastasia [14]

Answer:

F=1.26*10^{-3}N

Explanation:

Assuming the pith balls as point charges, we can calculate the repulsive force between them, using Coulomb's law:

F=\frac{kq_1q_2}{d^2}

We observe that the magnitude of the electric force is directly proportional to the product of the magnitude of both signed charges(q_1,q_2) and inversely proportional to the square of the distance(d) that separates them.

Replacing the given values, where k is the Coulomb constant:

F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-30*10^{-9}C)(-30*10^{-9}C)}{(8*10^{-2}m)^2}\\F=1.26*10^{-3}N

8 0
3 years ago
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