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2 years ago

Analyze the chemical reaction below

Physics

1 answer:

34kurt2 years ago

6 0

Answer:

Please find the answer in the explanation.

Explanation:

Given that 16 g CH4 + 64 g 02 - 44 g CO2 + 36g H2O

To explain the law of conservation of mass and describe how the equation represents the law of conservation of mass, let me first start from law.

The law state that: mass can neither be created nor destroyed.

The mass of each element at the reaction side must be equal or the Same with the magnitude of mass at the product

The equation represents the law of conservation of mass because the mass of molecules at the right hand side is equal to or balance with the molecules at the left hand side. For example, the number of Oxygen, and othe elements are the at both side.

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The sound level at a distance of 2.30 m from a source is 115 dB. At what distance will the sound level have the following values

Aleksandr [31]

Answer:

distance is 13 m for 100 dB

distance is 409 km for 10 dB

Explanation:

Given data

distance r = 2.30 m

source β = 115 dB

to find out

distance at sound level 100 dB and 10 dB

solution

first we calculate here power and intensity and with this power and intensity we will find distance

we know sound level β = 10 log(I/ $I_{0}$ ) ......................a

put here value (I/ $I_{0}$ ) = 10^−12 W/m² and β = 115

115 = 10 log(I/10^−12)

I = 0.316228 W/m²

and we know power = intensity × 4π r² ...............b

power = 0.316228 × 4π (2.30)²

power = 21.021604 W

we know at 100 dB intensity is 0.01 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 0.01 × 4π r²

so by solving r

r = 12.933855 m = 13 m

distance is 13 m

and

at 10 dB intensity is 1 × 10^–11 W/m²

so by equation b

power = intensity × 4π r²

21.021604 = 1 × 10^–11 × 4π r²

by solving r we get

r = 409004.412465 m = 409 km

5 0

3 years ago

During Cardiovascular Excercise, what do the heart and lungs do?

skad [1K]

They circulate blood and oxygen throughout your body

6 0

3 years ago

On a highway curve with a radius of 46 meters, the maximum force of static friction that can act on a 1,200 kg car going around

Mekhanik [1.2K]

Answer:

$v\approx 16.956\,\frac{m}{s}$

Explanation:

The motion of the vehicule on a highway curve can be modelled by the following equation of equilibrium:

$\Sigma F = f = m\cdot \frac{v^{2}}{R}$

The maximum speed is:

$v = \sqrt{\frac{f\cdot R}{m} }$

$v = \sqrt{\frac{(7500\,N)\cdot (46\,m)}{1200\,kg} }$

$v\approx 16.956\,\frac{m}{s}$

7 0

3 years ago

0.45 kg soccer ball changes its velocity by 20.0 m/s due to a force applied to it in 0.10 seconds. What force was necessary for

Paladinen [302]

Assuming the accleration applied was constant, we have

$v=v_0+at\implies v_0+20.0\,\dfrac{\mathrm m}{\mathrm s}=v_0+a(0.10\,\mathrm s)$

$\implies20.0\,\dfrac{\mathrm m}{\mathrm s}=a(0.10\,\mathrm s)$

$\implies a=200\,\dfrac{\mathrm m}{\mathrm s^2}$

Then the force applied to the ball is given by

$F=ma=(0.45\,\mathrm{kg})\left(200\,\dfrac{\mathrm m}{\mathrm s^2}\right)$

$\implies F=90\,\dfrac{\mathrm{kg}\,\mathrm m}{\mathrm s^2}=90.\,\mathrm N$

8 0

2 years ago

4. What is the acceleration of the car in each section?<br> b<br> с<br> d<br> a

makvit [3.9K]

Answer:

0-4 acceleration comes at 12 m/s where (B) stagnates at 12 m/s and remains for 4 seconds (C) is breaks being activated slowing the car to 6 m/s in 2 seconds and (D) over the course of 4 seconds brings the car to 10 m/s.

Explanation:

3 0

2 years ago

Read 2 more answers