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ArbitrLikvidat [17]

3 years ago

10

You have been asked to recommend a alternative source of energy that your town could use to generate electricity. Describe the r

enewable resource that you believe would be best for your town. Give at least two advantages and two disadvantages of the resource you have recommended.

1 answer:

snow_tiger [21]3 years ago

3 0

I believe that solar energy would be the best energy for my town. Solar energy is advantageous because the sun is up every day, and because it takes up much less space than than some other sources of renewable energy. Some disadvantages of solar energy are that if there is cloud cover, hardly any energy will be gained at all. Another disadvantage is that it is a long term investment that can take a while for it to pay for itself.

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Find a numerical value for rhoearth, the average density of the earth in kilograms per cubic meter. Use 6378km for the radius of

mylen [45]

Answer:

density = 5520 kg/m^3

Explanation:

given that

radius of earth = 6378 km

G = 6.67 x 10⁻¹¹ m³/kg.s²

g = 9.80 m/s²

we know,

$g = \dfrac{GM}{r^2}$

mass of earth

$M = \dfrac{gr^2}{G}$

$M = \dfrac{9.8 \times (6378 \times 10^3)^2}{6.67 \times 10^{-11}}$

M = 5.972 x 10²⁴ kg

density = $\dfrac{mass}{volume}$

V = volume of the earth = 4/3πr³

V = 4/3 x 3.14 x (6378 x 10³)³

V = 1.08 x 10²¹ m³

density = $\dfrac{5.972\times 10^{24}}{1.08\times 10^{21}}$

density = 5.52 x 10³ kg/m^3

density = 5520 kg/m^3

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2 years ago

Hey guys i need song requests dbl

Andrew [12]

Answer:

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8 0

3 years ago

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If two objects have the same volume but one has a greater mass, the one with greater mass

11Alexandr11 [23.1K]

Explain more. I can't understand

7 0

3 years ago

A Carnot heat engine has an efficiency of 0.400. If it operates between a deep lake with a constant temperature of 298.0k and a

tatuchka [14]

Answer:

496.7 K

Explanation:

The efficiency of a Carnot engine is given by the equation:

$\eta = 1 - \frac{T_H}{T_L}$

where:

$T_H$ is the temperature of the hot reservoir

$T_C$ is the temperature of the cold reservoir

For the engine in the problem, we know that

$\eta = 0.400$ is the efficiency

$T_C = 298.0 K$ is the temperature of the cold reservoir

Solving for $T_H$ , we find:

$\frac{T_C}{T_H}=1-\eta\\T_H = \frac{T_C}{1-\eta} =\frac{298.0}{1-0.400}=496.7 K$

6 0

3 years ago

Suppose that a wind is blowing in the direction S45°E at a speed of 30 km/h. A pilot is steering a plane in the direction N60°E

Kay [80]

Answer:

The true course: $40.29^\circ$ north of east

The ground speed of the plane: 96.68 m/s

Explanation:

Given:

$V_w$ = velocity of wind = $30\ km/h\ S45^\circ E = (30\cos 45^\circ\ \hat{i}-30\sin 45^\circ\ \hat{j})\ km/h = (21.21\ \hat{i}-21.21\ \hat{j})\ km/h$

$V_p$ = velocity of plane in still air = $100\ km/h\ N60^\circ E = (100\cos 60^\circ\ \hat{i}+100\sin 60^\circ\ \hat{j})\ km/h = (50\ \hat{i}+86.60\ \hat{j})\ km/h$

Assume:

$V_r$ = resultant velocity of the plane

$\theta$ = direction of the plane with the east

Since the resultant is the vector addition of all the vectors. So, the resultant velocity of the plane will be the vector sum of the wind velocity and the plane velocity in still air.

$\therefore V_r = V_p+V_w\\\Rightarrow V_r = (50\ \hat{i}+86.60\ \hat{j})\ km/h+(21.21\ \hat{i}-21.21\ \hat{j})\ km/h\\\Rightarrow V_r = (71.21\ \hat{i}+65.39\ \hat{j})\ km/h$

Let us find the direction of this resultant velocity with respect to east direction:

$\theta = \tan^{-1}(\dfrac{65.39}{71.21})\\\Rightarrow \theta = 40.29^\circ$

This means the the true course of the plane is in the direction of $40.29^\circ$ north of east.

The ground speed will be the magnitude of the resultant velocity of the plane.

$\therefore Magnitude = \sqrt{71.21^2+65.39^2} = 96.68\ km/h$

Hence, the ground speed of the plane is 96.68 km/h.

5 0

3 years ago