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Vera_Pavlovna [14]

2 years ago

13

A gas has an initial volume of 168 cm3 at a temperature of 255 K and a pressure of 1.6 atm. The pressure of the gas decreases to

1.3 atm, and the temperature of the gas increases to 285 K.

2 answers:

erastovalidia [21]2 years ago

7 0

Would presume you are asked to find the volume, since there is no second volume.

By General Gas Law:

P₁V₁/T₁ = P₂V₂/T₂

1.6 * 168 /255 = 1.3*V₂/285

V₂ = 1.6 * 168 * 285 / (1.3*255)

V₂ = 231.095

Final volume ≈ 231 cm³

allsm [11]2 years ago

5 0

Answer: The final volume is $231cm^3$

Explanation: Using ideal gas equation:

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

$P_1$ = initial pressure = 1.6 atm

$V_1$ = initial volume = $168cm^3$

$T_1$ = initial temperature = 255 K

$P_2$ = final pressure= 1.3 atm

$V_2$ = final volume = ?

$T_2$ = final temperature = 285 K

$\frac{1.6\times 168}{255}=\frac{1.3\times V_2}{285}$

$V_2=231cm^3$

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The human ear canal is, on average, 2.5cm long and aids in hearing by acting like a resonant cavity that is closed on one end an

Troyanec [42]

Answer:

$3400$ Hz

Explanation:

We know that

1 cm = 0.01 m

$L$ = Length of the human ear canal = 2.5 cm = 0.025 m

$V$ = Speed of sound = 340 ms⁻¹

$f$ = First resonant frequency

The human ear canal behaves as a closed pipe and for a closed pipe, nth resonant frequency is given as

$f = \frac{(2n - 1)V}{4L}$

for first resonant frequency, we have n = 1

Inserting the values

$f = \frac{(2(1) - 1) 340}{4(0.025)}$

$f = \frac{340}{4(0.025)}$

$f = 3400$ Hz

4 0

2 years ago

In a series circuit, a generator (1300 Hz, 12.0 V) is connected to a 14.0- resistor, a 4.40-μF capacitor, and a 6.00-mH inductor

klemol [59]

Answer:

(a) 2.8 V

(b) 5.6 V

(c) 9.8 V

Explanation:

Given:

Frequency of the generator (f) = 1300 Hz)

Terminal voltage (V) =12.0 V

Resistance of resistor (R) = 14.0 Ω

Capacitance of capacitor (C) = 4.40 μF = 4.40 × 10⁻⁶ F

Inductance of the inductor (L) = 6.00 mH = 6.00 × 10⁻³ H

In order to find the voltages across each, we first need to find the reactance and impedance.

Reactance of the inductor is given as:

$X_L=2\pi f L\\\\X_L=2\times 3.14\times 1300\times 6.00\times 10^{-3}\\\\X_L=49\ \Omega$

Reactance of the capacitor is given as:

$X_C=\frac{1}{2\pi fC}\\\\X_C=\frac{1}{2\times 3.14\times 1300\times 4.40\times 10^{-6}}\\\\X_C =28\ \Omega$

Now, impedance is given as:

$Z=\sqrt{X_L^2+X_C^2}\\\\Z=\sqrt{(49)^2+(28)^2}\\\\Z=\sqrt{3185}=56.4\ \Omega$

Current across the circuit is given as:

$I=\frac{V}{Z}\\\\I=\frac{12}{56.4}=0.2\ A$

As resistor, capacitor and inductor are connected in series, the current across each of them is same and equal to total current in the circuit.

(a)

Voltage across the resistor is given as:

$V_R=IR\\\\V_R=0.2\times 14=2.8\ V$

Therefore, the voltage across resistor is 2.8 V.

(b)

Voltage across the capacitor is given as:

$V_C=IX_C\\\\V_C=0.2\times 28=5.6\ V$

Therefore, the voltage across the capacitor is 5.6 V.

(c)

Voltage across the inductor is given as:

$V_L=IX_L\\\\V_L=0.2\times 49=9.8\ V$

Therefore, the voltage across the inductor is 9.8 V.

6 0

2 years ago

Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those a

garik1379 [7]

Answer:

T=2.94*10^-10 N/m.

Explanation:

Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those at which desirable prey might struggle. Orb spider web silk has a typical diameter of 20μm, and spider silk has a density of 1300 kg/m³.

To have a fundamental frequency at 150Hz , to what tension must a spider adjust a 14cm -long strand of silk?

l=length of the spider silk, 14cm

velocity of wave = √(T/μ)

where T = tension and

μ = mass per unit length)

λ/2=l

for fundamental frequency λ/2 =14cm

(λ= wavelength of standing wave; as there will be no node

except the endpoints of silk strand)

λ = 28 cm = 0.28 m

and since frequency * wavelength = speed of wave. we have,

150 * 0.28 = √(T/μ) ..................(#)

now μ = mass/length = [volume * density]/length = [(length*area) * density] / length = area * density

= [π * (10 * 10^(-6))²] * 1300 = 13π * 10^(-8).

now putting this in equation (#) we get

150 * 0.28 = √(T/[13π * 10^(-8)]).

thus T = [13π * 10^(-8)] * (42)² =

2.94*10^-10 N/m.

6 0

2 years ago

Each metal is illuminated with 400 nm (3.10 eV) light. Rank the metals on the basis of the maximum kinetic energy of the emitted

34kurt

Answer:

K.E(K) > K.E(Cs) > 0 (others)

Explanation:

Given the Work functions of the metal as

Aluminium (Wo)=4eV

Platinum(Wo) =6.4eV

Cesium (Wo) =2.1eV

Beryllium (Wo) = 5.0eV

Magnesium (Wo) = 3.7eV

Potassium (Wo) = 2.3eV

Using the formula:

K.E = hf - Wo........(1)

Wo = hfo..............(2)

From these the fo can be calculated for all the metals

Where K.E =Kinetic Energy

hf = energy of illumination = 3.10eV

h is Planck constant and has the value 6.6 × 10^-34JS^-1

The frequency f of the illumination is given by

f = 3.10 × 1.6 × 10^-19/6.6 × 10^-34

f = 7.51 × 10¹⁴ Hz..........(*)

Now an electron is only ejected if the threshold frequency of the metal is reached.

The work function has a threshold frequency (fo) for all the metals and this minimum frequency required to required to remove an electron from the surface of a metal.

We need to compare f with fo

If fo >= f there is emission, otherwise there is no emission

So using (2) we calculate for all fo and compare with f

K.E(Al) = 3.10 - 4.0 - 3.10 = -0.9eV, fo = 9.70 × 10¹⁴ Hz (no emission)

K.E(Pt) = 3.10 - 6.40 = -3.30eV, fo = 1.55 × 10^15 Hz, ( no emission)

K.E(Cs) = 3.10 - 2.10 = -1.0eV, fo = 5.09×10¹⁴ Hz, (emission)

K.E(Be) =3.10-5.0 = -1.90eV, fo = 12.12 ×10^15 Hz.,(no emission)

K.E(Mg) = 3.10-3.70 = -0.6eV, fo = 8.97 × 10¹⁴Hz, (no emission)

K.E(K) = 3.10 - 2.30= 0.9eV, fo = 5.58 × 10¹⁴ Hz, (emission)

So the metals whose electron gain Kinetic energy are:

Cesium

Potassium

Others have zero kinetic energy since no electron is emitted.

Hence the rank is:

K.E(K) > K.E(Cs) > 0 (others)

6 0

2 years ago

A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s

Leona [35]

Answer:

0.6

Explanation:

The volume of a sphere = $\frac{4}{3} \pi (\frac{D}{2})^3$

Therefore $\pi * r^2 * (\frac{D}{2} ) = \frac{4}{3} \pi (\frac{D}{2})^3$

r of the disc = $1.15(\frac{ D}{2} )$

Using conservation of angular momentum;

The $M_i$ of the sphere = $\frac{2}{5} m \frac{D}{2}^2$

$M_i$ of the disc = $m*\frac{ \frac{1.15*D}{2}^2 }{2}$

$\frac{wd}{ws} = \frac{\frac{2}{5}m * \frac{D}{2}^2}{ m * \frac{(\frac{`.`5*D}{2})^2 }{2} }$

= 0.6

5 0

2 years ago