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3 years ago

15

An object is dropped at a height of 6 m from the ground. How fast is it moving just before it hits the ground?

1 answer:

maw [93]3 years ago

3 0

Explanation:

Given:

Δy = 6 m

v₀ = 0 m/s

a = 9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (9.8 m/s²) (6 m)

v ≈ 10.8 m/s

Round as needed.

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in which place does subduction occur? a. at transform boundaries b. where two plates move apart c. at most normal faults d. wher

Evgesh-ka [11]

Answer;

D. where two plates collide

Explanation;

-Subduction zones are plate tectonic boundaries where two plates converge, and one plate is thrust beneath the other. This process results in geohazards, such as earthquakes and volcanoes.

-Subduction zone volcanism occurs where two plates are converging on one another. One plate containing oceanic lithosphere descends beneath the adjacent plate, thus consuming the oceanic lithosphere into the earth's mantle. This on-going process is called subduction.

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3 years ago

Read 2 more answers

Calculate the orbital speed (in m/s) of a satellite that circles the Earth with a time period of 12.00 hours. The mass of the Ea

Hoochie [10]

Answer:

$v = 3869 m/s$

Explanation:

As we know that the orbital speed of the satellite is given as

$v = \sqrt{\frac{GM}{r}}$

also we know that

time period of the revolution is given as

$T = \frac{2\pi r}{v}$

now from above equation we know that

$T = \frac{2\pi (\frac{GM}{v^2})}{v}$

$T = \frac{2\pi GM}{v^3}$

so we will have

$v = (\frac{2\pi GM}{T})^{1/3}$

now plug in all data in this equation

$v = (\frac{2\pi (6.67 \times 10^{-11})(5.97 \times 10^{24})}{12 \times 3600})^{1/3}$

$v = 3869 m/s$

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3 years ago

An asteroid orbiting the Sun has a mass of 4.00×1016 kg. At a particular instant, it experiences a gravitational force of 3.14×1

Ksivusya [100]

<h2>The asteroid is 4.11 x 10¹¹ m far from Sun</h2>

Explanation:

We have gravitational force

$F=\frac{GMm}{r^2}$

Where G = 6.67 x 10⁻¹¹ N m²/kg²

M = Mass of body 1

M = Mass of body 2

r = Distance between them

Here we have

M = Mass of Sun = 1.99×10³⁰ kg

m = Mass of asteroid = 4.00×10¹⁶ kg

F = 3.14×10¹³ N

Substituting

$F=\frac{GMm}{r^2}\\\\3.14\times 10^{13}=\frac{6.67\times 10^{-11}\times 1.99\times 10^{30}\times 4\times 10^{16}}{r^2}\\\\r=4.11\times 10^{11}m$

The asteroid is 4.11 x 10¹¹ m far from Sun

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3 years ago

Tell whether each element has a high or low electronegativity : bromine, chlorine, potassium, lithium

finlep [7]

Answer:

Bromine-2.8 high

Chlorine-3.0 high

Potassium-0.8 low

Lithium-1.0 low

Explanation:

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3 years ago

On what criteria is the Fujita scale for Tornadoes based?

svp [43]

B. Tornado destruction
It is based on the amount of damage

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3 years ago