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2 years ago

A concrete block (B-36 x10 °C-') of volume 100 mat 40°C is cooled to

Physics

1 answer:

ruslelena [56]2 years ago

6 0

T1=40°C=313K
T_2=-10°C=263K

Applying Charles law

$\\ \sf\Rrightarrow \dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

$\\ \sf\Rrightarrow \dfrac{100}{313}=\dfrac{V_2}{263}$

$\\ \sf\Rrightarrow V_2=\dfrac{26300}{313}$

$\\ \sf\Rrightarrow V_2=84.02ml$

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** URGENT** The voltage across the primary winding is 350,000 V, and the voltage across the secondary winding is 17,500 V. If th

Vlad [161]

As we know that in transformers we have

$\frac{V_s}{V_p} = \frac{N_s}{N_p}$

here we know that

$V_s = 17,500 Volts$

$V_p = 350,000 Volts$

$N_s = 600 coils$

now from above equation we will have

$\frac{17500}{350000} = \frac{600}{N_p}$

$N_p = 600\times \frac{350000}{17500}$

$N_p = 12000 coils$

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3 years ago

How to find the time with only the distance or height from the ground

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You use acceleration due to gravity
and 1/2 atsqr=d
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3 years ago

A scientific hypothesis must be related to nature and be..

PSYCHO15rus [73]

Answer:

A scientific hypothesis must be tetable so it can become a scientific theory.

Explanation: I think

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3 years ago

Evaluate tan45/tan45

SIZIF [17.4K]

I'm not sure I completely understand the expression you want evaluated.

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3 years ago

Read 2 more answers

A physics student of mass 43.0 kg is standing at the edge of the flat roof of a building, 12.0 m above the sidewalk. An unfriend

Dmitry_Shevchenko [17]

Answer:

The speed of the student just before she lands, v₂ is approximately 8.225 m/s

Explanation:

The given parameters are;

The mass of the physic student, m = 43.0 kg

The height at which the student is standing, h = 12.0 m

The radius of the wheel, r = 0.300 m

The moment of inertia of the wheel, I = 9.60 kg·m²

The initial potential energy of the female student, P.E.₁ = m·g·h₁

Where;

m = 43.0 kg

g = The acceleration due to gravity ≈ 9.81 m/s²

h = 12.0 h

∴ P.E.₁ = 43 kg × 9.81 m/s² × 12.0 m = 5061.96 J

The kinetic rotational energy of the wheel and kinetic energy of the student supporting herself from the rope she grabs and steps off the roof, K₁, is given as follows;

$K_1 = \dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2$

The initial kinetic energy, 1/2·m·v₁² and the initial kinetic rotational energy, 1/2·m·ω₁² are 0

∴ K₁ = 0 + 0 = 0

The final potential energy of the student when lands. P.E.₂ = m·g·h₂ = 0

Where;

h₂ = 0 m

The final kinetic energy, K₂, of the wheel and student is give as follows;

$K_2 = \dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2$

Where;

v₂ = The speed of the student just before she lands

ω₂ = The angular velocity of the wheel just before she lands

By the conservation of energy, we have;

P.E.₁ + K₁ = P.E.₂ + K₂

∴ m·g·h₁ + $\dfrac{1}{2} \cdot m \cdot v_{1}^2+\dfrac{1}{2} \cdot I \cdot \omega_{1}^2$ = m·g·h₂ + $\dfrac{1}{2} \cdot m \cdot v_{2}^2+\dfrac{1}{2} \cdot I \cdot \omega_{2}^2$

Where;

ω₂ = v₂/r

∴ 5061.96 J + 0 = 0 + $\dfrac{1}{2} \times 43.0 \, kg \times v_{2}^2+\dfrac{1}{2} \times 9.60 \, kg\cdot m^2 \cdot \left (\dfrac{v_2}{0.300 \, m} }\right ) ^2$

5,061.96 J = 21.5 kg × v₂² + 53. $\overline 3$ kg × v₂² = 21.5 kg × v₂² + 160/3 kg × v₂²

v₂² = 5,061.96 J/(21.5 kg + 160/3 kg) ≈ 67.643118 m²/s²

v₂ ≈ √(67.643118 m²/s²) ≈ 8.22454363 m/s

The speed of the student just before she lands, v₂ ≈ 8.225 m/s.

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3 years ago