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3 years ago

8

After sunset, we have about half an hour of twilight. This is because

1 answer:

Evgen [1.6K]3 years ago

6 0

During this period the sky is still bright enough, even if you cannot see the sun.

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Oxygen-18 is a naturally-occuring, stable isotope and is commonly used is scientific studies as a tracer. Using the periodic tab

Viefleur [7K]

Oxygen has Atomic number 8 so all isotopes have 8 protons and 8 electrons.

So the number of neutrons in Oxygen-18 = 18 - 8 = 10.

Option B is the correct one.

5 0

4 years ago

Read 2 more answers

What is the most important factor affecting the movement of water in an ocean?

arlik [135]

Probably ocean currents since these use heat to move large amounts of water throughout the ocean, and can you make this the brainliest answer

6 0

3 years ago

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A small water pump is used in an irrigation system. The pump takes water in from a river at 10oC, 100 kPa at a rate of 5 kg/s. T

sergij07 [2.7K]

Answer:

0.98kW

Explanation:

The conservation of energy is given by the following equation,

$\Delta U = Q-W$

$\dot{m}(h_1+\frac{1}{2}V_1^2+gz_1)-\dot{W} = \dot{m}(h_2+\frac{1}{2}V_2^2+gz_)$

Where

$\dot{m} =$ Mass flow

$h_1 =$ Specific Enthalpy (IN)

$h_2 =$ Specific Enthalpy (OUT)

$g =$ Gravity

$z_{1,2} =$ Heigth state (In, OUT)

$V_{1,2} =$ Velocity (In, Out)

Our values are given by,

$T_i = 10\°C$

$P_1 = 100kPa$

$\dot{m} = 5kg/s$

$z_2 = 20m$

For this problem we know that as pressure, temperature as velocity remains constant, then

$h_1 = h_2$

$V_1 = V_2$

Then we have that our equation now is,

$\dot{m}(gz_1) = \dot{m}(gz_2)+\dot{W}$

$\dot{W} = \frac{(5)(9.81)(0-20)}{1000}$

$\dot{W} = -0.98kW$

8 0

3 years ago

A projectile is shot directly away from Earth's surface. Neglect the rotation of the Earth. What multiple of Earth's radius RE g

7nadin3 [17]

Answer:

(a) r = 1.062·R $_E$ = $\frac{531}{500} R_E$

(b) r = $\frac{33}{25} R_E$

(c) Zero

Explanation:

Here we have escape velocity v $_e$ given by

$v_e =\sqrt{\frac{2GM}{R_E} }$ and the maximum height given by

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$

Therefore, when the initial speed is 0.241v $_e$ we have

v = $0.241\times \sqrt{\frac{2GM}{R_E} }$ so that;

v² = $0.058081\times {\frac{2GM}{R_E} }$

v² = ${\frac{0.116162\times GM}{R_E} }$

$\frac{1}{2} v^2-\frac{GM}{R_E} = -\frac{GM}{r}$ is then

$\frac{1}{2} {\frac{0.116162\times GM}{R_E} }-\frac{GM}{R_E} = -\frac{GM}{r}$

Which gives

$-\frac{0.941919}{R_E} = -\frac{1}{r}$ or

r = 1.062·R $_E$

(b) Here we have

$K_i = 0.241\times \frac{1}{2} \times m \times v_e^2 = 0.241\times \frac{1}{2} \times m \times \frac{2GM}{R_E} = \frac{0.241mGM}{R_E}$

Therefore we put $\frac{0.241GM}{R_E}$ in the maximum height equation to get

$\frac{0.241}{R_E} -\frac{1}{R_E} =-\frac{1}{r}$

From which we get

r = 1.32·R $_E$

(c) The we have the least initial mechanical energy, ME given by

ME = KE - PE

Where the KE = PE required to leave the earth we have

ME = KE - KE = 0

The least initial mechanical energy to leave the earth is zero.

3 0

3 years ago

Read 2 more answers

The luxury liner Queen Elizabeth 2 has a diesel-electric powerplant with a maximum power of 90 MW at a cruising speed of 31.5 kn

AlexFokin [52]

Answer:

5558643.69 N

Explanation:

F = Force

v = Velocity = 31.5 knots

Converting to m/s

$1\ knot=0.514\ m/s$

$31.5\ knot=31.5\times 0.514\ m/s=16.191\ m/s$

Power is given by

$P=Fv\\\Rightarrow F=\frac{P}{v}\\\Rightarrow F=\frac{90\times 10^6}{16.191}\\\Rightarrow F=5558643.69\ N$

The forward force is exerted on the ship at this highest attainable speed is 5558643.69 N

5 0

3 years ago