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2 years ago

The temperature of a system drops by 30°F during a cooling process. Express this drop in temperature in K, R, and °C.

Physics

1 answer:

babunello [35]2 years ago

8 0

Answer:

272.05 K; 489.69 °R; -1.11 °C

Explanation:

Fahrenheit can be converted to degrees Celsius using the following formula:

°C = 5/9 x (°F - 32) = -1.11 °C

The temperature in Kelvin is calculated from the temperature in degrees Celsius as follows:

K = °C + 273.15 = 272.05 K

The temperature on the Rankine scale is calculated from Kelvin as follows:

°R = 1.8K = 1.8(272.05) = 489.69 °R

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2 years ago

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2 years ago

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A balloon is at a height of 81m and is ascending upwards with a velocity of 12m/s. A body of 2kg weight is dropped from it. If g

kap26 [50]

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2 years ago

A gas has an initial volume of 24.6 L at a pressure of 1.90 atm and a temperature of 335 K. The pressure of the gas increases to

Tatiana [17]

Answer:

the final temperature of the gas is 785.18 K

Explanation:

The computation of the final temperature of the gas is shown below:

Here we apply the gas law

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Given that

P1 = 1.9 atm

V1 = 24.6 L

T1 = 335 K

P2 = 3.5 atm

V2 = 31.3 L

T2 = ?

Now

P1V1 ÷ T1 = P2V2 ÷ T2

(1.9 × 24.6) ÷ 335 = (3.5 × 31.3)/T2

T2 = 785.18 K

hence, the final temperature of the gas is 785.18 K

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1 year ago

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A stationary particle of charge q = 2.1 × 10-8 c is placed in a laser beam (an electromagnetic wave) whose intensity is 2.9 × 10

alisha [4.7K]

(a) The intensity of the electromagnetic wave is related to the amplitude of the electric field by
$I= \frac{1}{2} c \epsilon_0 E^2$
where
I is the intensity
c is the speed of light
$\epsilon_0$ is the electric permittivity
E is the amplitude of the electric field

By substituting the numbers of the problem and re-arranging the equation, we can find E:
$E= \frac{2 I}{c \epsilon_0} = \frac{2 ( 2.9 \cdot 10^3 Wm^{-2})}{(3 \cdot 10^8 m/s)(8.85 \cdot 10^{-12} Fm^{-1})} =2.2 \cdot 10^6 N/C$

Now that we have the intensity of the electric field, we can calculate the electric force on the charge:
$F=qE=(2.1 \cdot 10^{-8} C)(2.2 \cdot 10^6 N/C)=0.046 N$

(b) We can calculate the amplitude of the magnetic field starting from the amplitude of the electric field:
$B= \frac{E}{c}= \frac{2.2 \cdot 10^6 N/C}{3 \cdot 10^8 m/s}=7.3 \cdot 10^{-3} T$

The magnetic force is given by
$F=qvB \sin \theta$
where v is the particle's speed, B the magnetic field intensity and $\theta$ the angle between B and v.
In this case the charge is stationary, so v=0, and so the magnetic force is zero: F=0.

(c) The electric force has not changed compared to point (a), because it does not depend on the speed of the particle, so we have again F=0.046 N.

(d) This time, the particle is moving with speed $v=3.7 \cdot 10^4 m/s$ , in a direction perpendicular to the magnetic field (so, the angle $\theta$ is $90^{\circ}$ ), and so by using the intensity of the magnetic field we found in point (b), we can calculate the magnetic force on the particle:
$F=qvB \sin \theta = (2.1 \cdot 10^{-8}C)(3.7 \cdot 10^4 m/s)(7.3 \cdot 10^{-3} T)(\sin 90^{\circ} )=$
$=5.7 \cdot 10^{-6} N$

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2 years ago