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3 years ago

What is true about the relationship between the kinetic energy of molecules in an object and the objects temperature?

1 answer:

6 0

As the temperature increases the kinetic energy of the molecules increases, if u add more heat you get more kinetic energy.

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Which body system remove carbon dioxide and waste

antiseptic1488 [7]

The excretory system

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3 years ago

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Grace rides her bike at a constant speed of 6 miles per hour. How far can she travel in 2 1/2 hours?

aliina [53]

Answer:

15 miles

Explanation:

6 miles per hour

2 1/2 hours

6 x 2 = 12

6 x 1/2 = 3

12 + 3 = 15

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3 years ago

A blue line with 5 orange tick marks then one red tick mark then 4 orange tick marks. The number zero is above the red tick mark

Natali5045456 [20]

The total displacement of the toy car at the given positions is 0.

The given parameters;

First displacement of the car, = 5 cm left
Second displacement of the car, = 8 cm right
Third displacement of the car, = 3 cm to the left

The total displacement of the car is calculated as follows;

Let the left direction be "negative direction"
Let the right direction be "positive direction"

$\Delta x = - \ 5\ cm \ + \ 8 \ cm \ - \ 3 \ cm \\\\\Delta x = 0$

Thus, the total displacement of the toy car at the given positions is 0.

Learn more about displacement here: brainly.com/question/18158577

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2 years ago

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Yellow-green light has a wavelength of 560 nm. What is its frequency?

Natasha2012 [34]

5.4 x 1014Hz
wavelength x frequency = the speed of light

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3 years ago

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Suppose that a balloon is being filled with air at a rate of 10 cm3/s. (Assume that theballoon is a perfect sphere.) At what rat

Basile [38]

Answer:

Therefore the surface area of the balloon is increased at 4 cm³/s.

Explanation:

The balloon is being filled with air at a rate of 10 cm³/s

It means the volume of the balloon is increased at a rate 10 cm³/s.

i.e $\frac{dv}{dt} =10 cm^3/s$

Consider r be the radius of the balloon.

The volume of of a sphere is

$v=\frac{4}{3} \pi r^3$

Differentiate with respect to t

$\frac{dv}{dt} =\frac{4}{3} \pi \times 3r^2\frac{dr}{dt}$

$\Rightarrow 10 =4\pi r^2\frac{dr}{dt}$

$\Rightarrow \frac{dr}{dt}=\frac{10}{4\pi r^2}$

The surface of area of the balloon is(S) = $4\pi r^2$

$S=4\pi r^2$

Differentiate with respect to t

$\frac{dS}{dt} =4\pi\times2r\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\frac{dr}{dt}$

Putting the value of $\frac{dr}{dt}$

$\Rightarrow \frac{dS}{dt} =8\pi r\times\frac{10}{4\pi r^2}$

$\Rightarrow \frac{dS}{dt} =\frac{20}{ r}$

Given that r = 5 cm

$[\frac{dS}{dt}]_{r=5} =\frac{20}{ 5}$ =4 cm³/s

Therefore the surface area of the balloon is increased at 4 cm³/s.

5 0

3 years ago