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3 years ago

Explain the greenhouse effect and its connection to global warming?

Engineering

1 answer:

worty [1.4K]3 years ago

6 0

Answer:

The 'greenhouse effect' is the warming of climate that results when the atmosphere traps heat radiating from Earth toward space.

Explanation:

Certain gases in the atmosphere resemble glass in a greenhouse, allowing sunlight to pass into the 'greenhouse,' but blocking Earth's heat from escaping into space. The gases that contribute to the greenhouse effect include water vapor, carbon dioxide (CO2), methane, nitrous oxides, and chlorofluorocarbons (CFCs).

On Earth, human activities are changing the natural greenhouse. Over the last century the burning of fossil fuels like coal and oil has increased the concentration of atmospheric CO2. This happens because the coal or oil burning process combines carbon (C) with oxygen (O2) in the air to make CO2. To a lesser extent, the clearing of land for agriculture, industry, and other human activities have increased the concentrations of other greenhouse gases like methane (CH4), and further increased (CO2).

The consequences of changing the natural atmospheric greenhouse are difficult to predict, but certain effects seem likely:

- On average, Earth will become warmer. Some regions may welcome warmer temperatures, but others may not.

- Warmer conditions will probably lead to more evaporation and precipitation overall, but individual regions will vary, some becoming wetter and others dryer.

- A stronger greenhouse effect will probably warm the oceans and partially melt glaciers and other ice, increasing sea level. Ocean water also will expand if it warms, contributing to further sea level rise.

- Meanwhile, some crops and other plants may respond favorably to increased atmospheric CO2, growing more vigorously and using water more efficiently. At the same time, higher temperatures and shifting climate patterns may change the areas where crops grow best and affect the makeup of natural plant communities.

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Determine the angular acceleration of the uniform disk if (a) the rotational inertia of the disk is ignored and (b) the inertia

lukranit [14]

Answer:

α = 7.848 rad/s^2 ... Without disk inertia

α = 6.278 rad/s^2 .... With disk inertia

Explanation:

Given:-

- The mass of the disk, M = 5 kg

- The right hanging mass, mb = 4 kg

- The left hanging mass, ma = 6 kg

- The radius of the disk, r = 0.25 m

Find:-

Determine the angular acceleration of the uniform disk without and with considering the inertia of disk

Solution:-

- Assuming the inertia of the disk is negligible. The two masses ( A & B ) are hung over the disk in a pulley system. The disk is supported by a fixed support with hinge at the center of the disk.

- We will make a Free body diagram for each end of the rope/string ties to the masses A and B.

- The tension in the left and right string is considered to be ( T ).

- Apply newton's second law of motion for mass A and mass B.

ma*g - T = ma*a

T - mb*g = mb*a

Where,

* The tangential linear acceleration ( a ) with which the system of two masses assumed to be particles move with combined constant acceleration.

- g: The gravitational acceleration constant = 9.81 m/s^2

- Sum the two equations for both masses A and B:

g* ( ma - mb ) = ( ma + mb )*a

a = g* ( ma - mb ) / ( ma + mb )

a = 9.81* ( 6 - 4 ) / ( 6 + 4 ) = 9.81 * ( 2 / 10 )

a = 1.962 m/s^2

- The rope/string moves with linear acceleration of ( a ) which rotates the disk counter-clockwise in the direction of massive object A.

- The linear acceleration always acts tangent to the disk at a distance radius ( r ).

- For no slip conditions, the linear acceleration can be equated to tangential acceleration ( at ). The correlation between linear-rotational kinematics is given below :

a = at = 1.962 m/s^2

at = r*α

Where,

α: The angular acceleration of the object ( disk )

α = at / r

α = 1.962 / 0.25

α = 7.848 rad/s^2

- Take moments about the pivot O of the disk. Apply rotational dynamics conditions:

Sum of moments ∑M = Iα

( Ta - Tb )*r = Iα

- The moment about the pivots are due to masses A and B.

Ta: The force in string due to mass A

Tb: The force in string due to mass B

I: The moment of inertia of disk = 0.5*M*r^2

( ma*a - mb*a )*r = 0.5*M*r^2*α

α = ( ma*a - mb*a ) / ( 0.5*M*r )

α = ( 6*1.962 - 4*1.962 ) / ( 0.5*5*0.25 )

α = ( 3.924 ) / ( 0.625 )

α = 6.278 rad/s^2

6 0

3 years ago

A Geostationary satellite has an 8kW RF transmission pointed at the earth. How much force does that induce on the spacecraft? (N

soldier1979 [14.2K]

Answer:

The force induced on the aircraft is 2.60 N

Solution:

As per the question:

Power transmitted, $P_{t} = 8 kW = 8000 W$

Now, the force, F is given by:

$P_{t} = Force(F)\times velocity(v) = Fv$ (1)

where

v = velocity

Now,

For a geo-stationary satellite, the centripetal force, $F_{c}$ is provided by the gravitational force, $F_{G}$ :

$F_{c} = F_{G}$

$\frac{mv^{2}}{R} = \frac{GM_{e}m{R^{2}}$

Thus from the above, velocity comes out to be:

$v = \sqrt{\frac{GM_{e}}{R}}$

$v = \sqrt{\frac{6.67\times 10^{- 11}\times 5.979\times 10^{24}}{42166\times 10^{3}}} = 3075.36 m/ s$

where

R = $R_{e} + H$

R = $\sqrt{GM_{e}(\frac{T}{2\pi})^{2}}$

where

G = Gravitational constant

T = Time period of rotation of Earth

R is calculated as 42166 km

Now, from eqn (1):

$8000 = F\times 3075.36$

F = 2.60 N

6 0

3 years ago

Suppose that a class CalendarDate has been defined for storing a calendar date with month, day and year components. (In our sect

laiz [17]

Answer:

note:

find the attachment

7 0

3 years ago

A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h

hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

$D_{1}$ = 150 mm

$A_{1}= \frac{\pi }{4}\times 150^{2}$

= 17671.45 $mm^{2}$

$D_{2}$ = 250 mm

$A_{2}= \frac{\pi }{4}\times 250^{2}$

= 49087.78 $mm^{2}$

The centrifugal force acting on the flywheel is fiven by

F = M ( $R_{2}$ - $R_{1}$ ) x $w^{2}$ ------------(1)

Here F = ( -UTS x $A_{1}$ + UCS x $A_{2}$ )

Since density, $\rho = \frac{M}{V}$

$\rho = \frac{M}{A\times t}$

$M = \rho \times A\times t$ $M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t$

$M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37$

$M = 8252963901$

∴ $R_{2}$ - $R_{1}$ = 50 mm

∴ F = $763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}$

F = 33618968.38 N --------(2)

Now comparing (1) and (2)

$33618968.38 = 8252963901\times 50\times \omega ^{2}$

∴ ω = 4036.61

We know

$\omega = \frac{2\pi N}{60}$

$4036.61 = \frac{2\pi N}{60}$

∴ N = 38546.82 rpm

7 0

3 years ago

Set up the following characteristic equations in the form suited to Evanss root-locus method. Give L(s), a(s), and b(s) and the

Sunny_sXe [5.5K]

Answer:

attached below is the detailed solution and answers

Explanation:

Attached below is the detailed solution

C(iii) : versus the parameter C

The parameter C is centered in a nonlinear equation, therefore the standard locus will not apply hence when you use a polynomial solver the roots gotten would be plotted against C

4 0

3 years ago