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3 years ago

What is the original source of energy for tidal power schemes?

2 answers:

7 0

Using the power of the tides, energy is produced from the gravitational pull from both the moon and the sun, which pulls water upwards, while the Earth's rotational and gravitational power pulls water down, thus creating high and low tides.

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Likurg_2 [28]3 years ago

7 0

Answer:

Tidal energy is a renewable energy powered by the natural rise and fall of ocean tides and currents. Some of these technologies include turbines and paddles. Tidal energy is produced by the surge of ocean waters during the rise and fall of tides

Explanation:

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A charming friend of yours who has been reading a little bit about astronomy accompanies you to the campus observatory and asks

lidiya [134]

Answer:

Explanation:

3 0

3 years ago

vA child is in danger of drowning in the Merimac river. The Merimac river has a current of 3.1 km/hr to the east. The child is 0

ikadub [295]

Answer:

$d = 2.26 km$

Explanation:

Let the child is moving with speed same as the speed of water flow

So here the position of child with respect to flow must be zero

And if the boat start at an angle with the vertical

so its relative speed with flow of water is given as

$v_x = 24.8 sin\theta$

$v_y = 24.8 cos\theta$

now the time to reach the child is given as

$\frac{0.6}{24.8 cos\theta} = \frac{2.5}{24.8 sin\theta }$

so now we have

$\theta = 76.5 degree$

So the time to catch the child is given as

$t = \frac{0.6}{24.8 cos78.2}$

$t = 0.104 h$

So distance moved by it in 0.104 h

distance moved by the boat in upstream direction given as

$x = (24.8 sin 74.8 - 3.1)(0.104)$

$x = 2.18 km$

In y direction the displacement of boat is

$y = 0.6 km$

net displacement of the ball is given as

$d = \sqrt{x^2 + y^2}$

$d = \sqrt{2.18^2 + 0.6^2}$

$d = 2.26 km$

4 0

3 years ago

Which of the following best describes how heat is transferred by conduction?

Verizon [17]

a. Particles in warmer objects make particles in colder objects move faster.

7 0

3 years ago

football player A has a mass of 110 kg is running on the field is velocity of 2 m/s football player B has a mass of 120 kg and a

valkas [14]

Complete Question:

Football player A has a mass of 110 kg, and he is running down the field with a velocity of 2 m/s. Football player B has a mass of 120 kg and is stationary. What is the total momentum after the collision?

Answer:

Total momentum = 220 Kgm/s.

Explanation:

Given the following data;

For footballer A

Mass, M1 = 110kg

Velocity, V1 = 2m/s

For footballer B

Mass, M1 = 120kg

Velocity, V1 = 0m/s since he's stationary.

To find the total momentum;

Momentum can be defined as the multiplication (product) of the mass possessed by an object and its velocity. Momentum is considered to be a vector quantity because it has both magnitude and direction.

Mathematically, momentum is given by the formula;

$Momentum = mass * velocity$

a. To find the momentum of A;

$Momentum \; A = 110 * 2$

Momentum A = 220 Kgm/s.

b. To find the momentum of B;

$Momentum \; B = 120 * 0$

Momentum B = 0 Kgm/s.

c. To find the total momentum of the two persons;

$Total \; momentum = Momentum \; A + Momentum \; B$

Substituting into the equation, we have;

$Total \; momentum = 220 + 0$

Total momentum = 220 Kgm/s. 

7 0

3 years ago

The rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3 × 10-11 e-250/T and 2

Vlada [557]

Answer:

Calculate the ratio of the rates of ozone destruction by these catalysts at 20 km, given that at this altitude the average concentration of OH is about 100 times that of Cl and that the temperature is about -50 °C

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x $10^{-11} e^{-255/T}$ and 2x $10^{-12} e^{-940/T}$

T = -50 °C = 223 K

The reaction rate will be given by [Cl] [O3] 3x $10^{-11} e^{-255/223} = 9.78^{-12} [Cl] [O3]$

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x $10^{-12} e^{-940/223} = 2.95^{-14} [OH] [O3]$

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 330 * [Cl] / [OH]

Than, the concentration of OH is approximately 100 times of Cl, and the result will be that the reaction with Cl is 3.3 times faster than the reaction with OH

Calculate the rate constant for ozone destruction by chlorine under conditions in the Antarctic ozone hole, when the temperature is about -80 °C and the concentration of atomic chlorine increases by a factor of one hundred to about 4 × 105 molecules cm-3

Knowing

Rate constants for the reactions of atomic chlorine and of hydroxyl radical with ozone are given by 3x $10^{-11} e^{-255/T}$ and 2x $10^{-12} e^{-940/T}$

T = -80 °C = 193 K

The reaction rate will be given by [Cl] [O3] 3x $10^{-11} e^{-255/193} = 8.21^{-12} [Cl] [O3]$

Than, the reaction rate of OH with O3 is

Rate = [OH] [O3] 2x $10^{-12} e^{-940/193} = 1.53^{-14} [OH] [O3]$

Considering these 2 rates we can realize the ratio of the reaction with Cl to the reaction with OH is 535 * [Cl] / [OH]

Than, considering the concentration of Cl increases by a factor of 100 to about 4 × $10^{5}$ molecules $cm^{-3}$ , the result will be that the reaction with OH will be 535 + (100 to about 4 × $10^{5}$ molecules $cm^{-3}$ ) times faster than the reaction with Cl

Explanation:

4 0

3 years ago