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kirill [66]
3 years ago
9

Activity 4: Caste System - The Hierarchy

Physics
1 answer:
Ann [662]3 years ago
5 0
111111111111111111111111111111111
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explain the surprising realization scientists are making about bacteria and what Danino hopes to eventually be able to do using
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your answer for $1 at www.gotit-pro.com

Explanation:

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What physical quantity is a measure of the amount of inertia an object has?
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<span>mass and only mass ................</span>
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Which of the following is a type of vaporization?
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"Vaporization" means "becoming vapor".

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b).  No.  Becoming solid.

c).  Yes.

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4 years ago
You drop a pencil from your desk, which is 1 meter above the floor. How long does it take for the pencil to hit the floor? How f
vova2212 [387]

Answer:

1. 0.45 s.

2. 4.41 m/s

Explanation:

From the question given above, the following data were obtained:

Height (h) = 1 m

Time (t) =?

Velocity (v) =?

1. Determination of the time taken for the pencil to hit the floor.

Height (h) = 1 m

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) =?

h = ½gt²

1 = ½ × 9.8 × t²

1 = 4.9 × t²

Divide both side by 4.8

t² = 1/4.9

Take the square root of both side

t = √(1/4.9)

t = 0.45 s.

Thus, it will take 0.45 s for the pencil to hit the floor.

2. Determination of the velocity with which the pencil hit the floor.

Initial velocity (u) = 0 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Time (t) = 0.45 s.

Final velocity (v) =?

v = u + gt

v = 0 + (9.8 × 0.45)

v = 0 + 4.41

v = 4.41 m/s

Thus, the pencil hit the floor with a velocity of 4.41 m/s

6 0
4 years ago
Suppose a spacecraft orbits the moon in a very low, circular orbit, just a few hundred meters above the lunar surface. The moon
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Spacecraft will be moving in 1700 m/s.

Option C

<h3><u>Explanation: </u></h3>

The diameter of the moon is 3500 km and the free fall acceleration at the surface is given as  1.6\ \mathrm{m} / \mathrm{s}^2

The radius will be half of the diameter of the moon that can be written as:  

r_{\text {moon }}=1.75 \times 10^{6}

By the application of the equation for orbit speed, we get  

\begin{aligned}&v_{\text {orbit}}=\sqrt{r \times q}\\&v_{\text {orbit}}=\sqrt{\left(1.75 \times 10^{6}\right) \times\left(1.6\ \mathrm{m} / \mathrm{s}^{2}\right)}\\&v_{\text {orbit}}=1700\ \mathrm{m} / \mathrm{s}\end{aligned}

5 0
3 years ago
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