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blagie [28]
4 years ago
6

Consider a river flowing toward a lake at an average velocity of 3 m/s at a rate of 500 m3/s at a location 90 m above the lake s

urface. Determine the total mechanical energy of the river water per unit mass and the power generation potential of the entire river at that location.
Physics
1 answer:
WINSTONCH [101]4 years ago
7 0

Answer: a) 5 x 10^5 kg/s b) 444 MW

Explanation:

Kinetic energy per unit mass Ke is

Ke = V^2 / 2

Ke = 3^2 / 2 = 4.5 J/kg = 0.0045 kJ/kg

Now potential energy per unit mass Pe is

Pe = g x z = 9.8 x 90 = 882.9 J/kg = 0.8829 kJ/kg

The total mechanical energy of the River per unit mass e = Ke + Pe = 0.0045 + 0.8829 = 0.88744 J/kg

M = P x V = 1000 x 500 = 5 x 10^5 kg/s

b) power generation potential of the entire river at the location Wmax = Emech = M x Emech = 500,000 x 0.88744 = 444,000kW = 444MW

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A 1.10 μF capacitor is connected in series with a 1.92 μF capacitor. The 1.10 μF capacitor carries a charge of +10.1 μC on one p
miss Akunina [59]

Answer:

(a). The potential on the negative plate is 42.32 V.

(b). The equivalent capacitance of the two capacitors is 0.69 μF.

Explanation:

Given that,

Charge = 10.1 μC

Capacitor C₁ = 1.10 μF

Capacitor C₂ = 1.92 μF

Capacitor C₃ = 1.10 μF

Potential V₁ = 51.5 V

Let V₁ and V₂ be the potentials on the two plates of the capacitor.

(a). We need to calculate the potential on the negative plate of the 1.10 μF capacitor

Using formula of potential difference

V_{1}=\dfrac{Q}{C_{1}}

Put the value into the formula

V_{1}=\dfrac{10.1 \times10^{-6}}{1.10\times10^{-6}}

V_{1}=9.18\ V

The potential on the second plate

V_{2}=V-V_{1}

V_{2}=51.5 -9.18

V_{2}=42.32\ v

(b). We need to calculate the equivalent capacitance of the two capacitors

Using formula of equivalent capacitance

C=\dfrac{C_{1}\timesC_{2}}{C_{1}+C_{2}}

Put the value into the formula

C=\dfrac{1.10\times10^{-6}\times1.92\times10^{-6}}{(1.10+1.92)\times10^{-6}}

C=6.99\times10^{-7}\ F

C=0.69\ \mu F

Hence, (a). The potential on the negative plate is 42.32 V.

(b). The equivalent capacitance of the two capacitors is 0.69 μF.

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3 years ago
Which of the following is a type of object that might contain billions of stars?
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A galaxy because that is the largest group and contains many things.

6 0
4 years ago
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The greatest pull of a magnet is near its poles.<br><br> True<br> False
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An object falls freely from rest on a planet where the acceleration due to gravity is it is on Earth. In the first 5 seconds it
weqwewe [10]

Answer:

A. The object falls a distance of 250 m

Explanation:

Hi there!

In the question, you have forgotten the acceleration due to gravity. However, looking on the web I´ve found a very similar problem in which the acceleration due to gravity was as twice as much as it is on Earth.

The equation of height of a falling object is the following:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the object after a time t.

y0 = initial height.

v0 = initial velocity.

t = time.

g = acceleration due to gravity (on Earth: ≅ -10 m/s² considering the upward direction as positive).

Let´s place the origin of the system of reference at the point where the object is released so that y0 = 0. Since the object falls from rest, v0 = 0.

Then, the height of the object after 5 s will be :

y = 1/2 · 2 · g · t²    (notice that the acceleration due to gravity is 2 · g)

y = g · t²

y = -10 m/s² · (5 s)²

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The object falls a distance of 250 m.

8 0
3 years ago
The moon Phobos orbits Mars
shepuryov [24]

27.9816 \times 10^{3} s is the period of orbit.

<u>Explanation: </u>

The equation that is useful in describing satellites motion is Newton form after Kepler's Third Law. The period of the satellite (T) and the average distance to the central body (R) are related as the following equation:

                  \frac{T^{2}}{R^{3}}=\frac{4 \times \pi^{2}}{G \times M_{c e n t r a l}}

Where,

T is the period of the orbit

R is the average radius of orbit

G is gravitational constant  6.673 \times 10^{-11} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2}

Here, given data

M=6.23 \times 10^{23} \mathrm{kg}

R=9.38 \times 10^{6} \mathrm{m}

Substitute the given values, we get T as

      \frac{T^{2}}{\left(9.38 \times 10^{6}\right)^{3}}=\frac{4 \times(3.14)^{2}}{\left(6.673 \times 10^{-11}\right) \times 6.23 \times 10^{23}}

      T^{2}=\frac{4 \times 9.8596 \times 825.29 \times 10^{18}}{41.57 \times 10^{12}}

      T^{2}=\frac{32548.12 \times 10^{18-12}}{41.57}=782.97 \times 10^{6}

Taking square root, we get

       T=27.9816 \times 10^{3} s

4 0
3 years ago
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