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3 years ago

11

Sandra is having difficulty with her reading assignment because she does not fully understand the language. Which online tool wo

uld most likely help her understand the text?
a text-to-speech tool
an interactive tool
a glossary
a translation tool

2 answers:

Fynjy0 [20]3 years ago

7 0

Answer:

The answer is a TRANSLATION TOOL or D

Explanation:

devlian [24]3 years ago

6 0

Answer:

it is D

endgnuity2020

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Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 1.38 m and finds that it makes 441

Tasya [4]

The period of a simple pendulum is given by:
$T=2 \pi \sqrt{ \frac{L}{g} }$
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
$g= \frac{4 \pi^2}{T^2}L$ (1)

We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.

We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
$f= \frac{N}{t}= \frac{441}{1090 s}=0.40 Hz$
And its period is the reciprocal of its frequency:
$T= \frac{1}{f}= \frac{1}{0.40 Hz}=2.47 s$

So now we can use eq.(1) to find the gravitational acceleration of the planet:
$g= \frac{4 \pi^2}{T^2}L = \frac{4 \pi^2}{(2.47 s)^2} (1.38 m) =8.92 m/s^2$

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3 years ago

If we have a new moon today, when we will have the next full moon?.

Sav [38]

29.5 days

It takes 27 days, 7 hours, and 43 minutes for our Moon to complete one full orbit around Earth. This is called the sidereal month, and is measured by our Moon's position relative to distant “fixed” stars. However, it takes our Moon about 29.5 days to complete one cycle of phases (from new Moon to new Moon).

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2 years ago

What is the concentration of H^ + at apH = 2 ? Mol / L What is the concentration of H^ + ions at apH = 6 ? Mol/L How many more H

Nonamiya [84]

Answer:

The concentration of hydrogen ion at pH is equal to 2 : $= [H^+]=0.01 mol/L$

The concentration of hydrogen ion at pH is equal to 6 : $[H^+]'=0.000001 mol/L$

There are 0.009999 more moles of $H^+$ ions in a solution at a pH = 2 than in a solution at a pH = 6.

Explanation:

The pH of the solution is the negative logarithm of hydrogen ion concentration in an aqueous solution.

$pH=-\log [H^+]$

The hydrogen ion concentration at pH is equal to 2 = [H^+]

$2=-\log [H^+]\\$

$[H^+]=10^{-2}M= 0.01 M=0.01 mol/L$

The hydrogen ion concentration at pH is equal to 6 = [H^+]

$6=-\log [H^+]\\\\$

$[H^+]=10^{-6}M= 0.000001 M= 0.000001 mol/L$

Concentration of hydrogen ion at pH is equal to 2 = $[H^+]=0.01 mol/L$

Concentration of hydrogen ion at pH is equal to 6 = $[H^+]'=0.000001 mol/L$

The difference between hydrogen ion concentration at pH 2 and pH 6 :

$= [H^+]-[H^+]' = 0.01 mol/L- 0.000001 mol/L = 0.009999 mol/L$

Moles of hydrogen ion in 0.009999 mol/L solution :

$=0.009999 mol/L\times 1 L=0.009999 mol$

There are 0.009999 more moles of $H^+$ ions in a solution at a pH = 2 than in a solution at a pH = 6.

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2 years ago

In science work is defined as

Artist 52 [7]

According to the Jefferson lab, "The scientific definition of work is: using a force to move an object a distance (when both the force and the motion of the object are in the same direction.)"

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2 years ago

Starting from rest, a basketball rolls from the top to the bottom of a hill, reaching a translational speed of 6.1 m/s. Ignore f

tatiyna

Answer:

a) h=3.16 m, b) v_{cm }^ = 6.43 m / s

Explanation:

a) For this exercise we can use the conservation of mechanical energy

Starting point. Highest on the hill

Em₀ = U = mg h

final point. Lowest point

$Em_{f}$ = K

Scientific energy has two parts, one of translation of center of mass (center of the sphere) and one of stationery, the sphere

K = ½ m $v_{cm }^{2}$ + ½ $I_{cm}$ w²

angular and linear speed are related

v = w r

w = v / r

K = ½ m v_{cm }^{2} + ½ I_{cm} v_{cm }^{2} / r²

Em_{f} = ½ v_{cm }^{2} (m + I_{cm} / r2)

as there are no friction losses, mechanical energy is conserved

Em₀ = Em_{f}

mg h = ½ v_{cm }^{2} (m + I_{cm} / r²) (1)

h = ½ v_{cm }^{2} / g (1 + I_{cm} / mr²)

for the moment of inertia of a basketball we can approximate it to a spherical shell

I_{cm} = ⅔ m r²

we substitute

h = ½ v_{cm }^{2} / g (1 + ⅔ mr² / mr²)

h = ½ v_{cm }^{2}/g 5/3

h = 5/6 v_{cm }^{2} / g

let's calculate

h = 5/6 6.1 2 / 9.8

h = 3.16 m

b) this part of the exercise we solve the speed of equation 1

v_{cm }^{2} = 2m gh / (1 + I_{cm} / r²)

in this case the object is a frozen juice container, which we can simulate a solid cylinder with moment of inertia

I_{cm} = ½ m r²

we substitute

v_{cm } = √ [2gh / (1 + ½)]

v_{cm } = √(4/3 gh)

let's calculate

v_{cm } = √ (4/3 9.8 3.16)

v_{cm }^ = 6.43 m / s

4 0

3 years ago