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lisov135 [29]
3 years ago
6

During the class prize-giving ceremony, Anand clapped his hands hard while Kumar clapped his hands softly. Everybody could hear

Anand's clapping while only a few could hear Kumar's clapping. This was because the sound produced by Anand was of____________.
A - higher pitch
B - lower frequency
C - higher volume
D- lower pitch
Physics
1 answer:
algol133 years ago
3 0

Answer:

C - higher volume

Explanation:

The pitch or frequency of sound that an object can produce depends upon its size and configuration . The shape of hand of all are same so the frequency of sound produced by hands of all will be almost same . Hence frequency of sound produced by the hands of Anand and Kumar would have been almost the same .

But the intensity of sound produced by them would have been different . Intensity represents energy a sound carries . Hard hitting clap will produce sound of higher intensity . Intensity of sound is also called high volume sound . So Kumar's clap will carry greater energy and hence greater volume of sound .

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Answer:41.991ml

Explanation:

Equations: 2 H2O → 4H+ + 4e + O2 OXIDATION

2 H+ + 2e → H2 REDUCTION 

Electrolysis is the chemical decomposition of compounds when electricity is made to pass through a molten compound or solution.

from the oxidation reaction:

1moles of oxygen requires 4moles of electrons to be discharged at the product

F=96500C/mol

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=60*60*0.201A

Q=723.6C

Mole=Q/(F*mole ratio of electron)

Mole= 723.6/(4*96500)

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5 0
2 years ago
In the following reaction, which substance is the precipitate?(NH4)2SO4(aq) + Ba(NO3)2(aq) BaSO4(s) + 2NH4NO3(aq
Delvig [45]
The precipitate is the barium sulfate, or 3) BaSO4. This is because it is a solid (as seen by the (s)), unlike the other aqueous product. The fact that it is a solid means that it is insoluble in water and therefore a precipitate.

Hope this helps!
7 0
3 years ago
Read 2 more answers
Water is used to cool down automobile engines when they get hot. Why is water used as a coolant? A. Water is a good conductor. B
Mrac [35]

Water is usually used to cool down automobile engines when they get hot, yes. Therefore, that means water has a high heat capacity.

That makes the answer letter D you provided above.

D) Water has a high heat capacity.

Another example would be trying to put out a fire with a bucket of water. Usually, you can put out the fire debating on the size!

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3 years ago
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6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri
mylen [45]

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a =  - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

4 0
3 years ago
How does the sun’s gravitational attraction impact Earth’s motion?
Tasya [4]
The Sun's gravitational pull keeps our planet orbiting the Sun <span>in a nice nearly-circular orbit.</span>
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2 years ago
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