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3 years ago

6. An unknown material has a normal melting/freezing point of 225.0 °C, and the liquid

Physics

1 answer:

kkurt [141]3 years ago

7 0

Answer:

19.4 kJ/kg

Explanation:

I'm assuming you meant that the melting/freezing point is -25.0 °C.

Heat gained by the unknown material = heat lost by the oil and aluminum

mL + mCΔT = -(mCΔT + mCΔT)

First, let's find the amount of heat gained by the unknown material:

(0.100 kg) L + (0.100 kg) (160 J/kg/°C) (20.0°C − (-25.0°C))

(0.100 kg) L + 720 J

The heat lost by the oil:

(0.100 kg) (2430 J/kg/°C) (20.0°C − 27.0°C)

-1701 J

The heat lost by the aluminum:

(0.150 kg) (910 J/kg/°C) (20.0°C − 27.0°C)

-955.5 J

Therefore:

(0.100 kg) L + 720 J = -(-1701 J + (-955.5 J))

(0.100 kg) L + 720 J = 2656.5 J

(0.100 kg) L = 1936.5 J

L = 19365 J/kg

L = 19.4 kJ/kg

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When the continental drift occured, it moved all the land on earth around, therefore changing climates. If fossils of animals are found on a certain continent that doesn't support that species, it proves the drift happened because the climate used to be suitable for them, but now is not.

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3 years ago

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An object of mass m has these three forces acting on it (there is no normal force, "no surface"). F1 = 1 N, F2 = 9 N, and F3 = 5

hammer [34]

Answer:

solved

Explanation:

a) F_net = (F2 - F3)i - F1 j

b) |Fnet| = sqrt( (F2 - F3)^2 + F1^2)

= sqrt( (9- 5)^2 + 1^2)

= 4.123 N

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2 years ago

(a) Write an equation describing a sinusoidal transverse wave traveling on a cord in the positive of a y axis with an angular wa

Vedmedyk [2.9K]

Missing data: the wave number

$k=60 cm^{-1}$

(a) $z = 0.003 sin (6000y-31.4 t)$

For a transverse wave travelling in the positive y-direction and with vibration along the z-direction, the equation of the wave is

$z = A sin (ky-\omega t)$

where

A is the amplitude of the wave

k is the wave number

$\omega$ is the angular frequency

t is the time

In this situation:

A = 3.0 mm = 0.003 m is the amplitude

$k = 60 cm^{-1} = 6000 m^{-1}$ is the wave number

$T = 0.20 s$ is the period, so the angular frequency is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.20}=31.4 rad/s$

So, the wave equation (in meters) is

$z = 0.003 sin (6000y-31.4 t)$

(b) 0.094 m/s

For a transverse wave, the transverse speed is equal to the derivative of the displacement of the wave, so in this case:

$v_t = z' = -A \omega cos (ky-\omega t)$

So the maximum transverse wave occurs when the cosine term is equal to 1, therefore the maximum transverse speed must be

$v_{t}_{max} =\omega A$

where

$\omega = 31.4 rad/s\\A = 0.003 m$

Substituting,

$v_{t}_{max}=(31.4)(0.003)=0.094 m/s$

(c) 5.24 mm/s

The wave speed is given by

$v=f \lambda$

where

f is the frequency of the wave

$\lambda$ is the wavelength

The frequency can be found from the angular frequency:

$f=\frac{\omega}{2\pi}=\frac{31.4}{2\pi}=5 Hz$

While the wavelength can be found from the wave number:

$\lambda = \frac{2\pi}{k}=\frac{2\pi}{6000}=1.05\cdot 10^{-3} m$

Therefore, the wave speed is

$v=(5)(1.05\cdot 10^{-3} )=5.24 \cdot 10^{-3} m/s = 5.24 mm/s$

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3 years ago

What is the time constant of a 9.0-nm-thick membrane surrounding a 0.040-mm-diameter spherical cell? Assume the resistivity of t

yawa3891 [41]

Given Information:

Diameter of spherical cell = 0.040 mm

thickness = L = 9 nm

Resistivity = ρ = 3.6×10⁷ Ω⋅m

Dielectric constant = k = 9.0

Required Information:

time constant = τ = ?

Answer:

time constant = 2.87×10⁻³ seconds

Explanation:

The time constant is given by

τ = RC

Where R is the resistance and C is the capacitance.

We know that resistivity of of any material is given by

ρ = RA/L

R = ρL/A

Where area of spherical cell is given by

A = 4πr²

A = 4π(d/2)²

A = 4π(0.040×10⁻³/2)²

A = 5.026×10⁻⁹ m²

The resistance becomes

R = (3.6×10⁷*9×10⁻⁹)/5.026×10⁻⁹

R = 6.45×10⁷ Ω

The capacitance of the cell membrane is given by

C = kεoA/L

Where k = 9 is the dielectric constant and εo = 8.854×10⁻¹² F/m

C = (9*8.854×10⁻¹²*5.026×10⁻⁹)/9×10⁻⁹

C = 44.5 pF

C = 44.5×10⁻¹² F

Therefore, the time constant is

τ = RC

τ = 6.45×10⁷*44.5×10⁻¹²

τ = 2.87×10⁻³ seconds

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2 years ago

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den301095 [7]

Less than or equal to the magnitude of the vector

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