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2 years ago

The primary gas in a volcano is: carbon dioxide, sulfur dioxide, water vapor, nitrogen.

Physics

2 answers:

vaieri [72.5K]2 years ago

7 0

Answer:

The primary components in volcanic gas are water vapor, carbon dioxide and sulfur (either sulfur dioxide or hydrogen sulfide). But you can also find nitrogen, argon, helium, neon, methane, carbon dioxide and hydrogen. Approximately 60% of total emissions released by volcanoes is water vapor, and carbon dioxide accounts for 10 to 40% of emissions.

Explanation:

Mice21 [21]2 years ago

6 0

Answer:

Water vapor

Explanation:

The magma consists of dissolved gases when these gases produce the force the volcanic eruption takes place. The volcanic gases come out and their volume is increased tremendously. The gases present in the volcano are listed below:

The volcanic gases consist of water vapors, carbon dioxide, and sulfur.

These three are the primary gases but the water is present in a higher amount.

The percentage of water is 60%.

The carbon dioxide present in 10-40%.

Other gases present in volcanos are nitrogen, argon, helium, neon methane, and hydrogen.

Hope this helps, Let me know if I am wrong ( I am pretty sure that am not) but let me know anyway..., OH! and Good LUCK! :D ;P

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A student at a window on the second floor of a dorm sees her physics professor walking on the sidewalk beside the building. she

klasskru [66]

Refer to the diagram shown below.

In order for the balloon to strike the professor's head, th balloon should drop by 18 - 1.7 = 16.3 m in the time at the professor takes to walk 1 m.
The time for the professor to walk 1 m is
t = (1 m)/(0.45 m/s) = 2.2222 s

The initial vertical velocity of the balloon is zero.
The vertical drop of the balloon in 2.2222 s is
h = (1/2)*(9.8 m/s²)*(2.2222 s)² = 24.197 m

Because 24.97 > 16.3, the balloon lands in front of the professor, and does not hit the professor.

The time for the balloon to hit the ground is
(1/2)*(9.8)*t² = 18
t = 1.9166 s

The time difference is 2.2222 - 1.9166 = 0.3056 s
Within this time interval, the professor travels 0.45*0.3056 = 0.175 m
Therefore the balloon falls 0.175 m in front of the professor.

Answer:
The balloon misses the professor, and falls 0.175 m in front of the professor.

8 0

3 years ago

A small rock is thrown straight up with initial speed v0 from the edge of the roof of a building with height H. The rock travels

Crank

Answer:

$v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} }$

Explanation:

The average velocity is total displacement divided by time:

$v_{avg} =\dfrac{D_{tot}}{t}$

And in the case of vertical $v_{avg}$

$v_{avg}=\dfrac{y_{tot}}{t}$

where $y_{tot}$ is the total vertical displacement of the rock.

The vertical displacement of the rock when it is thrown straight up from height $H$ with initial velocity $v_0$ is given by:

$y=H+v_0t-\dfrac{1}{2} gt^2$

The time it takes for the rock to reach maximum height is when $y'(t)=0$ , and it is

$t=\frac{v_0}{g}$

The vertical distance it would have traveled in that time is

$y=H+v_0(\dfrac{v_0}{g} )-\dfrac{1}{2} g(\dfrac{v_0}{g} )^2$

$y_{max}=\dfrac{2gH+v_0^2}{2g}$

This is the maximum height the rock reaches, and after it has reached this height the rock the starts moving downwards and eventually reaches the ground. The distance it would have traveled then would be:

$y_{down}=\dfrac{2gH+v_0^2}{2g}+H$

Therefore, the total displacement throughout the rock's journey is

$y_{tot}=y_{max}+y_{down}$

$y_{tot} =\dfrac{2gH+v_0^2}{2g}+\dfrac{2gH+v_0^2}{2g}+H$

$\boxed{y_{tot} =\dfrac{2gH+v_0^2}{g}+H}$

Now wee need to figure out the time of the journey.

We already know that the rock reaches the maximum height at

$t=\dfrac{v_0}{g},$

and it should take the rock the same amount of time to return to the roof, and it takes another $t_0$ to go from the roof of the building to the ground; therefore,

$t_{tot}=2\dfrac{v_0}{g}+t_0$

where $t_0$ is the time it takes the rock to go from the roof of the building to the ground, and it is given by

$H=v_0t_0+\dfrac{1}{2}gt_0^2$

we solve for $t_0$ using the quadratic formula and take the positive value to get:

$t_0=\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

Therefore the total time is

$t_{tot}= 2\dfrac{v_0}{g}+\dfrac{-v_0+\sqrt{v_0^2+2gH} }{g}$

$\boxed{t_{tot}= \dfrac{v_0+\sqrt{v_0^2+2gH} }{g}}$

Now the average velocity is

$v_{avg}=\dfrac{y_{tot}}{t}$

$v_{avg}=\dfrac{\frac{2gH+v_0^2}{g}+H }{\frac{v_0+\sqrt{v_0^2+2gH} }{g} }$

$\boxed{v_{avg}=\dfrac{3gH+v_0^2}{v_0+\sqrt{v_0^2+2gH} } }$

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3 years ago

Why it is easier to lift a small stone but harder to lift heavy one on the surface of Earth

Nesterboy [21]

Answer:

since small stone has less mass so the gravitational pull of the earth is lesser in case of this but this is not for the bigger stone as the gravitational pull of the earth is greater...

PLEASE MARK BRAINLIEST!!!!!

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3 years ago

I need help with question two

zaharov [31]

I can't see numbers here, so here are all the answers:
1) the frequency is c/λ = 3e8/556e-9 = 5.39e14Hz
2) light travels at <span>299,792,458 m/s. So in nanoseconds it's 0.299792458m. This is about 1/3 of a meter which is about one foot.
3) length is L = ct = (</span>299,792,458 m/s)(6e-15) = 1.799e-6m or 1.799μm

6 0

3 years ago

2. What is the smallest particle into which an

just olya [345]

Explanation:

atommmmmmmmmmmmmmmmmmmmm

7 0

3 years ago

Read 2 more answers