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3 years ago

8

Explain how the data collected and the calculations for the first and second resonance points in today's experiment would change

if you collect the data in death valley in july when the temperature is 47 C

1 answer:

grandymaker [24]3 years ago

8 0

Answer:

tssths

Explanation:

hgst

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The take-up reel of a cassette tape has an average radius of 1.5 cm. Find the length of tape (in meters) that passes around the

Rudik [331]

Answer:

1.28 m

Explanation:

Given;

Radius, r = 1.5 cm = 0.015 m

Time, t = 19 s

Average angular speed = 4.5 rad/s

Consider a point when the tape is moving at a constant velocity along the circumference of the circular reel of radius r. The linear velocity v at this point is given by;

v = rω ----(1)

Where

v is the linear velocity of the circular motion

r is the radius of the reel

ω is the the angular velocity.

At a point the tap undergoes a linear motion before passing round the reel of the cassette. The linear velocity v at this point is given by;

v = L/t ----(2)

where;

v is the velocity of the linear motion

L is the length of the tape (distance covered by the tape)

t is the time taken

Equating equation(1) and equation (2)

L/t = rω

L = rωt

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3 years ago

An ideal gas initially at 4.00atm and 350 K is permitted

Nuetrik [128]

Explanation:

It is given that initially pressure of ideal gas is 4.00 atm and its temperature is 350 K. Let us assume that the final pressure is $P_{2}$ and final temperature is $T_{2}$ .

(a) We know that for a monoatomic gas, value of $\gamma$ is \frac{5}{3}[/tex].

And, in case of adiabatic process,

$PV^{\gamma}$ = constant

also, PV = nRT

So, here $T_{1}$ = 350 K, $V_{1} = V$ , and $V_{2} = 1.5 V$

Hence, $\frac{T_{2}}{T_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma -1}$

$\frac{T_{2}}{350 K} = (\frac{V}{1.5V})^{\frac{5}{3} -1}$

$T_{2}$ = 267 K

Also, $P_{1}$ = 4.0 atm, $V_{1} = V$ , and $V_{2} = 1.5 V$

$\frac{P_{2}}{P_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma}$

$\frac{P_{2}}{4.0 atm} = (\frac{V}{1.5V})^{\frac{5}{3}}$

$P_{2}$ = 2.04 atm

Hence, for monoatomic gas final pressure is 2.04 atm and final temperature is 267 K.

(b) For diatomic gas, value of $\gamma$ is \frac{7}{5}[/tex].

As, $PV^{\gamma}$ = constant

also, PV = nRT

$T_{1}$ = 350 K, $V_{1} = V$ , and $V_{2} = 1.5 V$

$\frac{T_{2}}{T_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma -1}$

$\frac{T_{2}}{350 K} = (\frac{V}{1.5V})^{\frac{7}{5} -1}$

$T_{2}$ = 289 K

And, $P_{1}$ = 4.0 atm, $V_{1} = V$ , and $V_{2} = 1.5 V$

$\frac{P_{2}}{P_{1}} = (\frac{V_{1}}{V_{2}})^{\gamma}$

$\frac{P_{2}}{4.0 atm} = (\frac{V}{1.5V})^{\frac{7}{5}}$

$P_{2}$ = 2.27 atm

Hence, for diatomic gas final pressure is 2.27 atm and final temperature is 289 K.

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3 years ago