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3 years ago

In scientific notation, (6.2 x 10^4) x (3.3 x 10^2 ) equals ____________________.

Physics

2 answers:

gregori [183]3 years ago

8 0

For this case we have the following multiplication:

$(6.2 * 10 ^ 4) * (3.3 * 10 ^ 2)$

For associative property, we can rewrite the expression as:

$(6.2 * 3.3) * (10 ^ 4 * 10 ^ 2)$

Then, by properties of exponents we have:

In the multiplication, if we have the same base, the exponents are added.

Therefore, by rewriting the expression we have:

$(6.2 * 3.3) * (10 ^{4 + 2})$

Finally:

$(20.46) * (10 ^ 6)$

Then, we rewrite the result correctly in scientific notation.

To do this, we move the point one space to the left and increase the degree of the exponent one unit:

$2.046 * 10 ^ 7$

Answer:

In scientific notation, $(6.2 * 10 ^ 4) * (3.3 * 10 ^ 2)$ equals $2.046 * 10 ^ 7$

castortr0y [4]3 years ago

5 0

You will get 20460000 as your answer which is broken down into, 2.046 x 10^7 as your number has to be between 1-10.

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. Two identical vehicles traveling at the same speed are made to collide with barriers in an insurance company collision test. T

Serggg [28]

Answer:

F₁ / F₂ = 10

therefore the first out is 10 times greater than the second barrier

Explanation:

For this exercise let's use the relationship between momentum and momentum.

I = F t = Δp

in this case the final velocity is zero

F t = 0 -m v₀

F = m v₀ / t

in order to answer the question we must assume that the two vehicles have the same mass and speed

concrete barrier

F₁ = -p₀ / 0.1

F₁ = - 10 p₀

barrier collapses

F₂ = -p₀ / 1

let's look for the relationship of the forces

F₁ / F₂ = 10

therefore the first out is 10 times greater than the second barrier

5 0

3 years ago

How did the continental Shelf form? Please help, thank you!! :)

shutvik [7]

The movements of the tectonic plates

6 0

3 years ago

If the distance between the Earth and Moon were half what it is now, by what factor would the force of gravity between them be c

hichkok12 [17]

Answer:

Explanation:

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

$m_1$ = Mass of Earth

$m_2$ = Mass of Moon

r = Distance between Earth and Moon

Old gravitational force

$F_o=\dfrac{Gm_1m_2}{r^2}$

New gravitational force

$F_n=\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}$

Dividing the equations

$\dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{(\dfrac{1}{2}r)^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=\dfrac{\dfrac{Gm_1m_2}{\dfrac{1}{4}r^2}}{\dfrac{Gm_1m_2}{r^2}}\\\Rightarrow \dfrac{F_n}{F_o}=4$

The ratio is $\dfrac{F_n}{F_o}=4$

The new force would be 4 times the old force

7 0

3 years ago

Tom has two pendulums with him. Pendulum 1 has a ball of mass 0.2 kg attached to it and has a length of 5 m. Pendulum 2 has a ba

mars1129 [50]

Given Information:

Pendulum 1 mass = m₁ = 0.2 kg

Pendulum 2 mass = m₂ = 0.6 kg

Pendulum 1 length = L₁ = 5 m

Pendulum 2 length = L₂ = 1 m

Required Information:

Affect of mass on the frequency of the pendulum = ?

Answer:

The mass of the ball will not affect the frequency of the pendulum.

Explanation:

The relation between period and frequency of pendulum is given by

f = 1/T

The period of pendulum is given by

T = 2π√(L/g)

Where g is the acceleration due to gravity and L is the length of the string

As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.

Bonus:

Pendulum 1:

T₁ = 2π√(L₁/g)

T₁ = 2π√(5/9.8)

T₁ = 4.49 s

f₁ = 1/T₁

f₁ = 1/4.49

f₁ = 0.22 Hz

Pendulum 2:

T₂ = 2π√(L₂/g)

T₂ = 2π√(1/9.8)

T₂ = 2.0 s

f₂ = 1/T₂

f₂ = 1/2.0

f₂ = 0.5 Hz

So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.

3 0

3 years ago

Read 2 more answers

suggest an experiment to prove that the rate of evaporation of a liquid depends on its surface area vapour already present in su

gulaghasi [49]

That's two different things it depends on:

-- surface area exposed to the air
AND
-- vapor already present in the surrounding air.

Here's what I have in mind for an experiment to show those two dependencies:

-- a closed box with a wall down the middle, separating it into two closed sections;

-- a little round hole in the east outer wall, another one in the west outer wall,
and another one in the wall between the sections;
So that if you wanted to, you could carefully stick a soda straw straight into one side,
through one section, through the wall, through the other section, and out the other wall.

-- a tiny fan that blows air through a tube into the hole in one outer wall.

Experiment A:

-- Pour 1 ounce of water into a narrow dish, with a small surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
-- Pour 1 ounce of water into a wide dish, with a large surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
Show that the 1 ounce of water evaporated faster 
when it had more surface area.
============================================
============================================

Experiment B:

-- Again, pour 1 ounce of water into the wide dish with the large surface area.
-- Again, set the dish in the second half of the box ... the one the air passes
through just before it leaves the box.
-- This time, place another wide dish full of water in the first section of the box,
so that the air has to pass over it before it gets through the wall to the wide dish
in the second section. Now, the air that's evaporating water from the dish in the
second section already has vapor in it before it does the job.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
==========================================
Show that it took longer to evaporate when the air 
blowing over it was already loaded with vapor.
==========================================

6 0

3 years ago