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4 years ago

In the exothermic reaction below, how would adding heat affect the equilibrium of the system?A + B ⇄ C + D + energy

Physics

1 answer:

TEA [102]4 years ago

5 0

Answer:

Shift it to the left toward the reactants.

Explanation:

Le Chatelier's principle states that if a chemical system is in equilibrium and one of the factors involved in the equilibrium is altered (e.g temperature,pressure or concentration) ,the equilibrium will shift, so as to neutralize the effect of the change.

If a chemical system is in equilibrium, in which the reaction is exothermic and the temperature is being increased, the equilibrium will shift to left, favoring the backward reaction (reactants) as the constant K decreases.

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A heavy turntable, used for rotating large objects, is a solid cylindrical wheel that can rotate about its central axle with neg

scoundrel [369]

Answer:

a) $I = 113.014\,kg\cdot m^{2}$ , b) $m = 2075.556\,kg$

Explanation:

a) The turntable has the following physical model by using Newton's laws:

$F \cdot R = I \cdot \alpha$

The moment of inertia is:

$I = \frac{F\cdot R}{\alpha}$

$I = \frac{(300\,N)\cdot(0.33\,m)}{0.876\,\frac{rad}{s^{2}} }$

$I = 113.014\,kg\cdot m^{2}$

b) The moment of inertia for a solid cylinder:

$I = \frac{1}{2}\cdot m \cdot R^{2}$

The mass of the turntable is:

$m = \frac{2 \cdot I}{R^{2}}$

$m = \frac{(2)\cdot (113.014\,kg\cdot m^{2})}{(0.33\,m)^{2}}$

$m = 2075.556\,kg$

8 0

3 years ago

A boy lifted a 50 newton rock 1 meter. How much work was done?

dem82 [27]

Answer:

The answer is 50 Nm

Explanation:

<h3>Given;</h3>

Applied Force = 50 Newton
Total Displacement = 1 meter

Work done = ?

Here,

W = F • d

W = 50 • 1

W = 50 Nm

Thus, Work done is 50 Nm

-TheUnknownScientist 72

5 0

3 years ago

Read 2 more answers

Answer A B C D and E

harina [27]

Answer:

what is the question I cannot click the

Explanation:

4 0

3 years ago

The intensity of the sound from a certain source is measured at two points along a line from the source. The points are separate

Artemon [7]

Answer:

The source is at a distance of 4.56 m from the first point.

Solution:

As per the question:

Separation distance between the points, d = 11.0 m

Sound level at the first point, L = 66.40 dB

Sound level at the second point, L'= 55.74 dB

Now,

$L = 10log_{10}\frac{I}{I_{o}}$

$I = I_{o}10^{\frac{L}{10}} = I_{o}10^{0.1L} = 10^{- 12}\times 10^{0.1\times 66.40} = 10^{- 5.36}$

$L' = 10log_{10}\frac{I'}{I_{o}}$

$I' = I_{o}10^{\frac{L'}{10}} = 10^{- 12}\times 10^{0.1\times 55.74} = 10^{- 6.426}$

where

$I_{o} = 10^{- 12} W/m^{2}$

I = Intensity of sound

Now,

$I = \frac{P}{4\pi R^{2}}$

Similarly,

$I' = \frac{P}{4\pi (R + 11.0)^{2}}$

Now,

$\frac{I}{I'} = \frac{(R + 11.0)^{2}}{R^{2}}$

$\frac{10^{- 5.36}}{10^{- 6.426}} = \frac{(R + 11.0)^{2}}{R^{2}}$

$R ^{2} + 22R + 121 = 11.64R^{2}}$

$10.64R ^{2} - 22R - 121 = 0$

Solving the above quadratic eqn, we get:

R = 4.56 m

8 0

3 years ago

In certain ranges of a piano keyboard, more than one string is tuned to the same note to provide extra loudness. For example, th

diamong [38]

Answer:

3.77 Hz

Explanation:

The beat frequency that is heard can be calculated using the following equation:

$f_{b} = f_{1} - f_{2}$

Where:

f₁ = 110 Hz

$f_{2} = \frac{f_{1}}{\sqrt{\frac{T_{1}}{T_{2}}}}$

With:

T₁ = 594 N

T₂ = 554 N

$f_{2} = \frac{f_{1}}{\sqrt{\frac{T_{1}}{T_{2}}}} = \frac{110 Hz}{\sqrt{\frac{594 N}{554 N}}} = 106.23 Hz$

Hence, the beat frequency is:

$f_{b} = f_{1} - f_{2} = 110 Hz - 106.23 Hz = 3.77 Hz$

Therefore, the beat frequency that is heard when the hammer strikes the two strings simultaneously is 3.77 Hz.

I hope it helps you!

3 0

4 years ago