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3 years ago

Read the description of the centrioles. What is their function?

Physics

2 answers:

KATRIN_1 [288]3 years ago

7 0

Answer:

Typically found in eukaryotic cells, centrioles are cylindrical (tube-like) structures/organelles composed of microtubules. In the cell, centrioles aid in cell division by facilitating the separation of chromosomes. For this reason, they are located near the nucleus.

Brut [27]3 years ago

3 0

Answer:

sorry im just gettig points but the actaul anserw is on top of this oneExplanation:

Bye

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Kruka [31]

Answer:

Immunization, or immunisation, is the process by which an individual's immune system becomes fortified against an infectious agent.

Explanation:

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3 years ago

8. Chemical weathering occurs more rapidly in what kind of climates?

Lady bird [3.3K]

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3 years ago

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The asteroid, Ida, has a small moon, Dactyl, that orbits at a speed of 5.66 m/s in an orbit of radius 90, 000m . What is Ida's m

cricket20 [7]

Answer:

4.3 x 10^16 kg

Explanation:

M = rv^2/G =[90,000 x 5.66^2] / [6.67 x 10^-11]

M = 43,226,446,776,611,694 = 4.3 x 10^16 kg - Ida's mass.

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3 years ago

Which ones are all renewable resources

Aleksandr-060686 [28]

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3 years ago

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Two manned satellites approach one another at a relative velocity of v=0.190 m/s, intending to dock. The first has a mass of m1=

drek231 [11]

Answer:

Their final relative velocity is 0.190 m/s

Explanation:

The relative velocity of the satellites, v = 0.190 m/s

The mass of the first satellite, m₁ = 4.00 × 10³ kg

The mass of the second satellite, m₂ = 7.50 × 10³ kg

Given that the satellites have elastic collision, we have;

$v_2 = \dfrac{2 \cdot m_1}{m_1 + m_2} \cdot u_1 - \dfrac{m_1 - m_2}{m_1 + m_2} \cdot u_2$

$v_2 = \dfrac{ m_1 - m_2}{m_1 + m_2} \cdot u_1 + \dfrac{2 \cdot m_2}{m_1 + m_2} \cdot u_2$

Given that the initial velocities are equal in magnitude, we have;

u₁ = u₂ = v/2

u₁ = u₂ = 0.190 m/s/2 = 0.095 m/s

v₁ and v₂ = The final velocities of the satellites

We get;

$v_1 = \dfrac{2 \times 4.0 \times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 - \dfrac{4.0 \times 10^3- 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095$

$v_2 = \dfrac{ 4.0 \times 10^3 - 7.50\times 10^3}{4.0 \times 10^3 + 7.50 \times 10^3} \times 0.095 + \dfrac{2 \times 7.50\times 10^3}{4.0 \times 10^3+ 7.50\times 10^3} \times 0.095 = 0.095$

The final relative velocity of the satellite, $v_f$ = v₁ + v₂

∴ $v_f$ = 0.095 + 0.095 = 0.190

The final relative velocity of the satellite, $v_f$ = 0.190 m/s

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2 years ago