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2 years ago

Is gas matter? How do you know? Hint: Remember there are 2 things we need to be considered matter.

Physics

1 answer:

astra-53 [7]2 years ago

7 0

Answer:

Gas is a state of matter that has no fixed shape and no fixed volume.

In addition to solids and liquids, gases are also a physical state in which matter can occur. All gases have weight. Unlike solids and liquids, gases will occupy the entire container that encloses them.

matter is "anything that has mass and volume (occupies space)

Gases have mass. The space between gas particles is empty. Gases can be formed as products in chemical reactions. Gas particles can form bonds between them under certain conditions

Gases have volume which isn't fixed (no fixed volume) and no fixed shape. Gases expand to fill the space available. They can also be compressed into a very small space.

Explanation:

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An object is 10 cm from thé mirror, its height is 1 cm and thé focal length is 5 cm. What is thé distance from thé mirror? S1= _

Viefleur [7K]

Note: I assume the mirror is concave, so that its focal length is positive (it is not specified in the text)

1a) We can use the mirror equation to find the distance of the image from the mirror:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
where
$f=5 cm$ is the focal length
$p=10 cm$ is the distance of the object from the mirror
$q$ is the distance of the image from the mirror.

Rearranging the equation, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{5}- \frac{1}{10}= \frac{1}{10 cm}$
so, the distance of the image from the mirror is $q=10 cm$ .

1b) The image height is given by the magnification equation:
$\frac{h_i}{h_o}=- \frac{p}{q}$
where $h_i$ is the heigth of the image and $h_o=1 cm$ is the height of the object. By rearranging the equation and using p and q, we find
$h_i=-h_o \frac{p}{q}=-(1 cm) \frac{10 cm}{10 cm}=-1 cm$
and the negative sign means the image is inverted.

2) As before, we can find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

3) As before, we find the distance of the image from the mirror by using the mirror equation:
$\frac{1}{f}= \frac{1}{p}+ \frac{1}{q}$
Rearranging it, we find
$\frac{1}{q}= \frac{1}{f}- \frac{1}{p}= \frac{1}{2}- \frac{1}{10}= \frac{4}{10 cm}$
so, the distance of the image from the mirror is
$q= \frac{10}{4}cm= 2.5 cm$

And now we can use the magnification equation to find the image height:
$\frac{h_i}{h_o}=- \frac{p}{q}$
Rearranging it, we find
$h_i=-h_o \frac{p}{q}=-(3cm) \frac{10 cm}{2.5 cm}=-12 cm$
and the negative sign means the image is inverted.

5 0

3 years ago

The smallest angle of inclined desk at which a book begins to slide off the desk is 15∘ relative to the horizontal.

Ray Of Light [21]

The question appears to be incomplete.
I assume that we are to find the coefficient of static friction, μ, between the desk and the book.

Refer to the diagram shown below.
m = the mass of the book
mg = the weight of the book (g = acceleration due to gravity)

N = the normal reaction, which is equal to
N = mg cos(12°)

R = the frictional force that opposes the sliding down of the book. It is
R = μN = μmg cos(12°)

F = the component of the weight acting down the incline. It is
F = mg sin(12°)

Because the book is in static equilibrium (by not sliding down the plane), therefore
F = R
mg sin(12°) = μmg cos(12°)
$\frac{sin(12^{o})}{cos(12^{o})} = \frac{\mu mg}{mg} = \mu \\\\
tan(12^{o}) = \mu$

Therefore, the static coefficient of friction is
μ = tan(12) = 0.213

Answer: μ = 0.21 (nearest tenth)

6 0

3 years ago

Read 2 more answers

If the total momentum of a system of an old woman pushing a shopping cart is 450 kg×m/s, both with a velocity of 3 m/s, and the

Liono4ka [1.6K]

Answer:

100kg

Explanation:

Momentum is defined as the product of mass times velocity:

$p=mv$

The total momentum M must be equal to the sum of momentums of the system components, in this case the momentum of the lady plus the momentum of the cart:

$M=m_{woman}v_{woman}+m_{cart}v_{cart}$

since both the woman and the shopping cart have the same velocity:

$v_{woman}=v_{cart}=v$

the equation becomes:

$M=m_{woman}v+m_{cart}v\\M=v(m_{woman}+m_{cart})$

and since we need the mass of the cart we solve for $m{cart}$ :

$m_{woman}+m_{cart}=\frac{M}{v} \\\\m_{cart}=\frac{M}{v} -m_{woman}$

the values given by the problem are the following:

total momentum: $M=450kgm/s$

mass of the woman: $m_{woman}=50kg$

velocity: $v=3m/s$

we substitute all of them in the equation to find the mass of the shopping cart:

$m_{cart}=\frac{450kgm/s}{3m/s} -50kg\\\\m_{cart}=150kg-50kg\\\\m_{cart}=100kg$

the mass of the cart is 100kg.

7 0

3 years ago

I saw glaciers and a desert of snow. I saw a whale jump up from the ocean. My sister says she saw a "white bear," but I didn't b

tensa zangetsu [6.8K]

Polar is the place described

4 0

3 years ago

Read 2 more answers

What happens to the coefficient of friction when the weight is increased? Why is this?

Crazy boy [7]

Answer:

Usually the coefficient of friction remains unchanged

Explanation:

The coefficient of friction should in the majority of cases, remain constant no matter what your normal force is. When you apply a greater normal force, the frictional force increases, and your coefficient of friction stays the same. Here's another way to think about it: because the force of friction is equal to the normal force times the coefficient of friction, friction is increased when normal force is increased.

Plus, the coefficient of friction is a property of the materials being "rubbed", and this property usually does not depend on the normal force.

6 0

4 years ago

Read 2 more answers