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3 years ago

Which equation below is correctly balanced?

Physics

2 answers:

maks197457 [2]3 years ago

6 0

Answer:

I am not sure what the answer is but keep reading.

Explanation:

The answer is not D or A for sure

Amiraneli [1.4K]3 years ago

5 0

Answer:

D H2 + O2 ---> H2O

Explanation:

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A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m, and an

maksim [4K]

Answer:

The value is $\alpha = 0.002 Np/m$

Explanation:

From the question we are told that

The first amplitude of the wave is $E_{max}1 = 98.02 \ V/m$

The first depth is $D_1 = 10 \ m$

The second amplitude is $E_{max}2 = 81.87 \ (V/m)$

The second depth is $D_2 = 100 \ m$

Generally from the spatial wave equation we have

$v(x) = Ae^{-\alpha d}cos(\beta x + \phi_o)$

=> $\frac{v(x)}{v(x)} =\frac{ Ae^{-\alpha d}cos(\beta x + \phi_o)}{ Ae^{-\alpha d}cos(\beta x + \phi_o)}$

So considering the ratio of the equation for the two depth

$\frac{A}{A_S} = \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}$

=> $\frac{98.02}{81.87} = \frac{e^{-10 \alpha }}{e^{-100 \alpha }}$

=> $\alpha = \frac{0.18}{90}$

=> $\alpha = 0.002 Np/m$

4 0

4 years ago

A humanoid skeleton is found buried in the ashes of a volcano that erupted between 10,000 and 12,000 years ago. when scientists

Inga [223]

From reliable sources in the internet, the half-live of carbon-14 is given to be 5,730 years. In a span of 10,000 to 12,000 years, there are almost or little more than 2 half-lives. Thus, there should be
A(t) = A(0)(1/2)^t
where t is the number of half-lives, in this case 2. Thus, only about 1/4 of the original amount will be left.

5 0

3 years ago

Read 2 more answers

PLEASE HELP! Daniel is 50.0 meters away from a building. He observes that his line-of-sight to the tip of the building makes an

zavuch27 [327]

Answer:

The height of building should be 98.13 m plus the height of Daniel. Since the 63° was measured from his eye level.

Explanation:

5 0

3 years ago

Why is an element considered a pure substance?

anastassius [24]