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3 years ago

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A student is given a red and a blue liquid. The two samples of liquids are heated in identical beakers over identical Bunsen bur

ners. The red liquid shows a dramatically faster increase in temperature. Which of these is the most likely explanation for this?
A) The red liquid has a greater mass.
B) The blue liquid has a smaller mass.
C) The red liquid has a greater specific heat.
D) The red liquid has a smaller specific heat.

2 answers:

ANTONII [103]3 years ago

3 0

I think the answer is a

kozerog [31]3 years ago

3 0

The answer is most likely D

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When a man returns to his well-sealed house on a summer day, he finds that the house is at 35°C. He turns on the air conditioner

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Answer:

1353.38 Watt

Explanation:

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T₂ = Final temperature of the house = 20°C

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Cp = Specific heat at constant pressure = 1.0 kJ/kgK

Heat removed

q = mCvΔT

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A slit has a width of W1 = 4.4 × 10-6 m. When light with a wavelength of λ1 = 487 nm passes through this slit, the width of the

Vitek1552 [10]

Answer:

The width of the central bright fringe on the screen is observed to be unchanged is $4.48*10^{-6}m$

Explanation:

To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes. Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

$w sin\theta = m\lambda$

Where,

w = width

$\lambda =$ wavelength

m is an integer, m = 1, 2, 3...

We here know that as $sin\theta$ as w are constant, then

$\frac{w_1}{\lambda_1} = \frac{w_2}{\lambda_2}$

We need to find $w_2$ , then

$w_2 = \frac{w_1}{\lambda_1}\lambda_2$

Replacing with our values:

$w_2 = \frac{4.4*10^{-6}}{487}496$

$w_2 = 4.48*10^{-6}m$

Therefore the width of the central bright fringe on the screen is observed to be unchanged is $4.48*10^{-6}m$

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3 years ago

A commuter train passes a passenger platform at a constant speed of 40.4 m/s. The train horn is sounded at its characteristic fr

mihalych1998 [28]

(a) -83.6 Hz

Due to the Doppler effect, the frequency of the sound of the train horn appears shifted to the observer at rest, according to the formula:

$f' = (\frac{v}{v\pm v_s})f$

where

f' is the apparent frequency

v = 343 m/s is the speed of sound

$v_s$ is the velocity of the source of the sound (positive if the source is moving away from the observer, negative if it is moving towards the observer)

f is the original frequency of the sound

Here we have

f = 350 Hz

When the train is approaching, we have

$v_s = -40.4 m/s$

So the frequency heard by the observer on the platform is

$f' = (\frac{343 m/s}{343 m/s - 40.4 m/s})(350 Hz)=396.7 Hz$

When the train has passed the platform, we have

$v_s = +40.4 m/s$

So the frequency heard by the observer on the platform is

$f' = (\frac{343 m/s}{343 m/s + 40.4 m/s})(350 Hz)=313.1 Hz$

Therefore the overall shift in frequency is

$\Delta f = 313.1 Hz - 396.7 Hz = -83.6 Hz$

And the negative sign means the frequency has decreased.

(b) 0.865 m

The wavelength and the frequency of a wave are related by the equation

$v=\lambda f$

where

v is the speed of the wave

$\lambda$ is the wavelength

f is the frequency

When the train is approaching the platform, we have

v = 343 m/s (speed of sound)

f = f' = 396.7 Hz (apparent frequency)

Therefore the wavelength detected by a person on the platform is

$\lambda' = \frac{v}{f'}=\frac{343 m/s}{396.7 Hz}=0.865m$

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3 years ago