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Alex_Xolod [135]
3 years ago
10

The gas waste generated by many power plants is

Physics
2 answers:
Andrej [43]3 years ago
8 0

The answer to your question is sulfur oxide. I hope I helped!!


lapo4ka [179]3 years ago
5 0

carbon dioxide

is the answer

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The measure of the average kinetic energy of the particles of a substance is the ______.
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How do you reduce friction
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To reduce friction between two surfaces, you have to
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A person must pull a rope 20 meters with a force of 200 N what is the work input?
tatiyna
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3 years ago
A diffraction pattern forms when light passes through a single slit. The wavelength of the light is 691 nm. Determine the angle
expeople1 [14]

Explanation:

Given that,

Wavelength of the light, \lambda=691\ nm=691\times 10^{-9}\ m

(a) Slit width, a=3.8\times 10^{-4}\ m

The angle that locates the first dark fringe is given by :

sin\theta=\dfrac{\lambda}{a}

sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-4}}

\theta=0.104^{\circ}

(b) Slit width, a=3.8\times 10^{-6}\ m

The angle that locates the first dark fringe is given by :

sin\theta=\dfrac{\lambda}{a}

sin\theta=\dfrac{691\times 10^{-9}}{3.8\times 10^{-6}}

\theta=10.47^{\circ}

Hence, this is the required solution.

7 0
3 years ago
An ore sample weighs 17.50 N in air. When the sample
zysi [14]

Answer with Explanation:

We are given that

Weight of an ore sample=17.5 N

Tension in the cord=11.2 N

We have to find the total volume and the density of the sample.

We know that

Tension, T=W-F_b

F_b=buoyancy force

T=Tension force

W=Weight

By using the formula

11.2=17.5-F_b

F_b=17.5-11.2=6.3 N

F_b=V_{object}\times \rho_{water}\cdot g

Where

V_{object}=Volume of object

\rho_{water}=1000 kgm^{-3}=Density of water

g=9.8 ms^{-2}=Acceleration due to gravity

Substitute the values then we get

6.3=9.8\times 1000\times V_{object}

V_{object}=\frac{6.3}{9.8\times 1000}=6.43\times 10^{-4} m^3

Volume of sample=6.43\times 10^{-4} m^3

Density of sample,\rho_{object}=\frac{Mass}{volume_{object}}

Where mass of ore sample=1.79 kg

Substitute the values then, we get

\rho_{object}=\frac{1.79}{6.43\times 10^{-4}}=2.78\times 10^3 kg/m^3

Density of the sample=2.78\times 10^{3} kgm^{-3}

7 0
3 years ago
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