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Hitman42 [59]
3 years ago
15

A syringe containing 12.0 mL of dry air at 25 C is placed in a sterilizer and heated to 100.0 C. The syringe is sealed, but th

e plunger can move and the volume can change. What is the volume of the air in the syringe at 100.0 C, assuming no change in pressure?
Physics
1 answer:
monitta3 years ago
5 0

Answer:

15.01 Liters

Explanation:

T₁ = Initial temperature = 25°C = 298.15 K

T₂ = Final temperature = 100°C = 373.15 K

V₁ = Initial volume = 12 mL

Here, pressure is constant so we apply Charles Law

\frac{V_1}{T_1}=\frac{V_2}{T_2}\\\Rightarrow {V_2}=\frac{V_1}{T_1}\times T_2\\\Rightarrow {V_2}=\frac{12}{298.15}\times 373.15\\\Rightarrow {V_2}=15.01 L

∴ Final volume at 100°C is 15.01 Liters.

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15.4 kg.

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From the law of conservation of momentum,

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mu+m'u' = V(m+m').................... Equation 1

Where m = mass of the first sphere, m' = mass of the second sphere, u = initial velocity of the first sphere, u' = initial velocity of the second sphere, V = common velocity of both sphere.

Given: m = 7.7 kg, u' = 0 m/s (at rest)

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7.7(x)+m'(0) = 1/3x(7.7+m')

7.7x = 1/3x(7.7+m')

7.7 = 1/3(7.7+m')

23.1 = 7.7+m'

m' = 23.1-7.7

m' = 15.4 kg.

Hence the mass of the second sphere = 15.4 kg

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The flywheel of a steam engine runs with a constant angular velocity of 140 rev/min. When steam is shut off, the friction of the
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Answer:

A) α = -1.228 rev/min²

B) 7980 revolutions

C) α_t = -8.57 x 10^(-4) m/s²

D) α = 21.5 m/s²

Explanation:

A) Using first equation of motion, we have;

ω = ω_o + αt

Where,

ω_o is initial angular velocity

α is angular acceleration

t is time the flywheel take to slow down to rest.

We are given, ω_o = 140 rev/min ; t = 1.9 hours = 1.9 x 60 seconds = 114 s ; ω = 0 rev/min

Thus,

0 = 140 + 114α

α = -140/114

α = -1.228 rev/min²

B) the number of revolutions would be given by the equation of motion;

S = (ω_o)t + (1/2)αt²

S = 140(114) - (1/2)(1.228)(114)²

S ≈ 7980 revolutions

C) we want to find tangential component of the velocity with r = 40cm = 0.4m

We will need to convert the angular acceleration to rad/s²

Thus,

α = -1.228 x (2π/60²) = - 0.0021433 rad/s²

Now, formula for tangential acceleration is;

α_t = α x r

α_t = - 0.0021433 x 0.4

α_t = -8.57 x 10^(-4) m/s²

D) we are told that the angular velocity is now 70 rev/min.

Let's convert it to rad/s;

ω = 70 x (2π/60) = 7.33 rad/s

So, radial angular acceleration is;

α_r = ω²r = 7.33² x 0.4

α_r = 21.49 m/s²

Thus, magnitude of total linear acceleration is;

α = √((α_t)² + (α_r)²)

α = √((-8.57 x 10^(-4))² + (21.49)²)

α = √461.82

α = 21.5 m/s²

4 0
2 years ago
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