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Akimi4 [234]
2 years ago
8

a bicycle pump contains 20cm3 of air at a pressure of 100kpa the air is then pumped in a single stroke through a valve into a ty

re if volume 100cm3 whihc contains air at the same pressure. calculate the pressure of the air in the tyre aftee the stroke assume the tyre volume does not change
Physics
1 answer:
riadik2000 [5.3K]2 years ago
8 0
If we assume also that the temperature of the air does not change, we can use Boyle's Law:
p₁V₁ = p₂V₂

Now, we know: 
p₁ = 100kPa
V₂ = 100cm³ (the volume of the tyre) 
V₁ = 120cm³ (becuse the air is contained inside the tyre AND the pump)

We can solve for p₂:
p₂ = (p₁V₁)/V₂
    = (100×120)/100
    = 120kPa

Therefore your answer is: 120kPa
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2 years ago
A 20kg mass approaches a spring at a speed of 30 m/s. The mass compresses the spring 12cm before coming to a stop. Calculate the
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Answer:

625000 N/ m

Explanation:

m= 20 kg

v= 30 m/s

x= 12 cm

k = ?

Here when the mass when hits at spring its speed is

Vi= 30 m/s

Finally it comes to rest after compressing for 12 cm

i-e Vf = 0 m/s

Distance= S= 12 cm = 0.12 m

using

2aS= Vf2 - Vi2

==> 2a ×0.12 = o- 30 × 30

==> a = 900 ÷ 0.24 = 3750 m/sec2

Now we know;

F = ma

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5 0
2 years ago
A figure skater rotating at 5.00 rad/s with arms extended has a moment of inertia of 2.25 kg·m2. If the arms are pulled in so t
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a) 6.25 rad/s

The law of conservation of angular momentum states that the angular momentum must be conserved.

The angular momentum is given by:

L=I\omega

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I is the moment of inertia

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Since the angular momentum must be conserved, we can write

L_1 = L_2\\I_1 \omega_1 = I_2 \omega_2

where we have

I_1 = 2.25 kg m^2 is the initial moment of inertia

\omega_1 = 5.00 rad/s is the initial angular speed

I_2 = 2.25 kg m^2 is the final moment of inertia

\omega_2 is the final angular speed

Solving for \omega_2, we find

\omega_2 = \frac{I_1 \omega_1}{I_2}=\frac{(2.25 kg m^2)(5.00 rad/s)}{1.80 kg m^2}=6.25 rad/s

b) 28.1 J and 35.2 J

The rotational kinetic energy is given by

K=\frac{1}{2}I\omega^2

where

I is the moment of inertia

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Applying the formula, we have:

- Initial kinetic energy:

K=\frac{1}{2}(2.25 kg m^2)(5.00 rad/s)^2=28.1 J

- Final kinetic energy:

K=\frac{1}{2}(1.80 kg m^2)(6.25 rad/s)^2=35.2 J

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