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amm1812
3 years ago
15

From the above diagram, which phase of the Moon will result in the greatest difference between high and low tide? A. New Moon B.

Last Quarter C. First Quarter D. Crescent Moon

Physics
2 answers:
aliya0001 [1]3 years ago
8 0
New moon, because the New Moon (A) because the sun and the moon work together most when it is there on the tides
9966 [12]3 years ago
4 0

Answer: A. New Moon

Explanation:

Spring Tides occur during the new moon or full moon phase. Spring tides are a result of greatest difference between high and low tides. This is because during new moon or full moon, the Earth, the moon and the Sun are aligned in a straight line which leads to maximum gravitational pull causing the maximum difference in high tidal waves and low tidal waves. From the given options, correct answer is option A. New Moon.  

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A material must readily accept electron flow to be a good conductor of electricity. Electrical conductors are electrical charge carriers with electrons that move with ease from atom to atom when charged with voltage. Examples of good conductors are copper, brass, steel, gold, and aluminum.

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Calculate the acceleration if you push with a 20-N horizontal force on a 2-kg block on a horizontal friction-free air table
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What happens to the temperature of a substance during a phase change?
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3 years ago
A tank, shaped like a cone has height 12 meter and base radius 1 meter. It is placed so that the circular part is upward. It is
Temka [501]

Answer:

376966.991 Joules

Explanation:

Given that :

the height = 12  m

Let assume the tank have a thickness  = dh

The radius of the tank by using the concept of similar triangle is :

\dfrac{1}{r} = \dfrac{12}{h}

r = \dfrac{h}{12}

The area of the tank = \mathbf{\pi r^2}

The area of the tank = \mathbf{\pi( \dfrac{h}{12})^2}

The area of the tank = \mathbf{ \dfrac{\pi}{144}h^2}

The volume of the tank is  = area × thickness

= \mathbf{ \dfrac{\pi}{144}h^2 \  dh}

Weight of the element = \rho_ g * volume

where;

\rho_g = density of water ; which is given as 10000 N/m³

So;

Weight of the element = \mathbf{ 10000 *\dfrac{\pi}{144}h^2 \  dh}

Weight of the element = \mathbf{69.44 \ \pi  \ h^2 \  dh}

However; the work required to pump this water = weight × height  rise

where the height rise = 12 - h

the work required to pump this water  = \mathbf{69.44 \ \pi  \ h^2 \  dh}(12 - h)

the work required to pump this water  = \mathbf{69.44 \pi (12h^2-h^3)dh}

We can determine the total workdone by integrating the work required to pump this water

SO;

Workdone = \mathbf{\int\limits^{12}_0 {69.44 \pi(12h^2-h^3)} dh}

= \mathbf{ 69.44 \pi \int\limits^{12}_0 {(12h^2-h^3)} dh}

=  \mathbf{ 69.44 \pi[ \frac{12h^3}{3}-  \frac{h^4}{4}]^{12}}_0} }

= \mathbf{69.44 \pi [ \frac{12^4}{3}-\frac{12^4}{4}]}

= \mathbf{69.44 \pi*12^4 [ \frac{4-3}{12}]}

= \mathbf{69.44 \pi*12^4 *\frac{1}{12}}

= 376966.991 Joules

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