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Delicious77 [7]
3 years ago
8

A bicycle pump contains 200 cm3 of air and is connected to a bicycle tyre. The volume of the tyre is 800 cm3. The pressure of th

e air in the tyre (it is ‘flat’) is 1.0 atmosphere, the same as the air in the pump. (= volume1). What is the total volume of the air initially?
Physics
1 answer:
Zolol [24]3 years ago
7 0

Answer:

The total initial volume of air in the tire is 800 cm³

Explanation:

Given;

volume of air in the bicycle pump, V₁ = 200 cm³

volume of the tire, V₂ = 800 cm³

The total initial volume of air in the tire is 800 cm³ and when the pump is pushed in, the total volume of air in the tire becomes 200 cm³. This occurs because the new air pumped into the tire through the bicycle pump compresses the air that was previously in the tire and raises the pressure of the tire that was initially flat.

From Boyle's law, a decrease in volume causes an increase in pressure.

You might be interested in
65mi/hr South is an example of​
Shtirlitz [24]

Answer:

It's an example of velocity.

8 0
3 years ago
2. A car traveling 300 miles in 5 hours is an example of
Tom [10]
The answer would be a speed
5 0
3 years ago
A 2.2-kg block slides on a horizontal surface with a speed of v=0.80m/s and an
mina [271]

Answer:

μ = 0.33

Equal to 3.2 m/s²

Explanation:

Draw a free body diagram of the block.  There are three forces:

Normal force N pushing up.

Weight force mg pulling down.

Friction force Nμ pushing opposite the direction of motion.

Sum of forces in the y direction.

∑F = ma

N − mg = 0

N = mg

Sum of forces in the x direction.

∑F = ma

Nμ = ma

Substitute.

mgμ = ma

μ = a/g

μ = (3.2 m/s²) / (9.8 m/s²)

μ = 0.33

As found earlier, the acceleration is a = gμ.  Since g and μ are constant, a is also constant, so it does not change with velocity.

5 0
3 years ago
A merry-go-round of radius R, shown in the figure, is rotating at constant angular speed. The friction in its bearings is so sma
mel-nik [20]

The angular speed of the merry-go-round reduced more as the sandbag is

placed further from the axis than increasing the mass of the sandbag.

The rank from largest to smallest angular speed is presented as follows;

[m = 10 kg, r = 0.25·R]

              {} ⇩

[m = 20 kg, r = 0.25·R]

              {} ⇩

[m = 10 kg, r = 0.5·R]

              {} ⇩

[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]

              {} ⇩

[m = 10 kg, r = 1.0·R]

Reasons:

The given combination in the question as obtained from a similar question online are;

<em>1: m = 20 kg, r = 0.25·R</em>

<em>2: m = 10 kg, r = 1.0·R</em>

<em>3: m = 10 kg, r = 0.25·R</em>

<em>4: m = 15 kg, r = 0.75·R</em>

<em>5: m = 10 kg, r = 0.5·R</em>

<em>6: m = 40 kg, r = 0.25·R</em>

According to the principle of conservation of angular momentum, we have;

I_i \cdot \omega _i = I_f \cdot \omega _f

The moment of inertia of the merry-go-round, I_m = 0.5·M·R²

Moment of inertia of the sandbag = m·r²

Therefore;

0.5·M·R²·\omega _i = (0.5·M·R² + m·r²)·\omega _f

Given that 0.5·M·R²·\omega _i is constant, as the value of  m·r² increases, the value of \omega _f decreases.

The values of m·r² for each combination are;

Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²

Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²

Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²

Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²

Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²

Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²

Therefore, the rank from largest to smallest angular speed is as follows;

Combination 3 > Combination 1 > Combination 5 = Combination 6 >

Combination 2

Which gives;

[<u>m = 10 kg, r = 0.25·R</u>] > [<u>m = 20 kg, r = 0.25·R</u>] > [<u>m = 10 kg, r = 0.5·R</u>] > [<u>m = </u>

<u>10 kg, r = 0.5·R</u>] = [<u>m = 40 kg, r = 0.25·R</u>] > [<u>m = 10 kg, r = 1.0·R</u>].

Learn more here:

brainly.com/question/15188750

6 0
2 years ago
what does it mean if the coefficient of friction is greater? for example,a coefficient of friction of 0.1 compared to 0.7
Sergeu [11.5K]
Well, 0.1 is actually less than 0.7, but I understand what you're asking.

The coefficient of friction describes the relationship between two surfaces
that are sliding by each other.  The higher the coefficient of friction is, the
'rougher' the meeting is, and the harder it is for one to slide over the other. 
A skate blade against ice has a very low coefficient of friction.  Sandpaper
against blue jeans has a high coefficient of friction.

A higher coefficient of friction means that when one thing is sliding over
the other one, friction robs more energy from the motion.  It's harder to
push one thing over the other one, and when you let go, the moving one
slows down and stops sooner.

Air resistance is actually an example of friction.  It prevents falling things
from falling as fast as they would if there were no air. The coefficient of
friction when something moves through air is pretty low.  If the same
object were trying to move through molasses or honey, the coefficient
of friction would be greater.

Friction robs energy, and turns it into heat.  So, especially in machinery with
moving parts, we want to make the coefficient of friction between the moving parts
as small as possible.  That's what the OIL in a car's engine is for.
3 0
4 years ago
Read 2 more answers
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