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schepotkina [342]
3 years ago
6

Which of the following can be an effect of a volcanic winter? select the two correct answers

Physics
2 answers:
Marysya12 [62]3 years ago
8 0

it is B. Famine and C. Climate cooling

Umnica [9.8K]3 years ago
8 0

Answer: B. famine C. climate cooling

Volcanic winter, is a natural condition that occurs after a volcanic eruption. The volcanic ash and sulfur aerosols are released in the massive amount which covers the stratosphere layer of the atmosphere. The sulfur aerosols reflects the incoming solar radiations from the sun and absorbs the terrestrial radiations from the entire atmosphere. These processes cools the temperature of the tropospheric layer of atmosphere lying below the stratosphere. This results in harsh winter conditions.

Therefore, on the basis of above information the famine and climate cooling is the effect of a volcanic winter.

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A child’s toy that is made to shoot ping pong balls consists of a tube, a spring (k = 18 N/m) and a catch for the spring that ca
UkoKoshka [18]

Answer:

The height is 3.1m

Explanation:

Here we have a conservation of energy problem, we have a conversion form eslastic potencial  energy to gravitational potencial energy, so:

E_e=\frac{1}{2}K*x^2\\E_e=\frac{1}{2}18N/m*(9.5*10^{-2}m)^2\\E_e=0.081J

then we have only gravitational potencial energy when the ball is at its maximun height.

E_g=m*g*h

because all the energy was transformed Eg=Ee

h=\frac{0.081J}{9.8m/s^2*m}

searching the web, the mass of a ping pong ball is 2.7 gr in average. so:

h=\frac{0.081J}{9.8m/s^2*(2.7*10^{-3}kg)}\\h=3.1m

6 0
3 years ago
Balance the following equations: CUCO3+H2SO4- CUSO4+H2O+CO2​
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Answer:

It is already balanced

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A piston-cylinder device initially contains 1.4 kg saturated liquid water at 200oC. Now heat is transferred to the water until t
postnew [5]

Answer:

Explanation:

Given

mass of saturated liquid water m=1.4\ kg

at 200^{\circ} specific volume is \nu =0.001157\ m^3\kg(From Table A-4,Saturated water Temperature table)

V_1=m\nu _1

V_1=1.4\times 0.001157

V_1=1.6198\times 10^{-3}\ m^3

Final Volume V_2=4V_1

V_2=4\times (1.6198\times 10^{-3})

V_2=6.4792\times 10^{-3}\ m^3

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\nu _2=\frac{6.4792\times 10^{-3}}{1.4}

\nu _2=0.004628\ m^3/kg

Now we see the value and find the temperature it corresponds to specific volume at vapor stage in the table.

T_2=T_1^{*}+\frac{T_2^{*}-T_1^{*}}{\alpha _2^{*}-\alpha _1^{*}}\times (\alpha _2-\alpha _1^{*})

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T_2=370.7^{\circ} C

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