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OverLord2011 [107]
3 years ago
11

When do both hemispheres receive the same amount of the sun’s energy?

Physics
2 answers:
sveta [45]3 years ago
4 0

Answer: During the months of March and September

Explanation: Hemisphere is a combination of two words; hemi meaning half and sphere referring to the earth which is a spherical celestial body.

The Hemisphere of the Earth is divided into two; Northern and Southern Hemisphere.

The Northern Hemisphere refers to the Hemisphere above the equator and Southern Hemisphere refers to the Hemisphere below the equator.

The earth completes it's revolution about the sun in 365 days. For the first 365 days, the Northern Hemisphere receives more amount of the sun energy while for the second half, the Southern Hemisphere received more amount of the sun's energy.

During equinox however, both the Northern and the Southern Hemisphere receive the same amount of the sun's energy due to the zero tilt (rotation) of the Earth's axis.

The equinox in the Northern Hemisphere is referred to as Spring equinox while that of the Southern Hemisphere is referred to as Autumnal equinox.

These equinoxes occurs in the months of March and September every year.

Juli2301 [7.4K]3 years ago
3 0
In March and September both of the hemispheres receive the same amount of energy from the sun. this is because neither end of the axis is tilted toward the sun.
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A 44-cm-diameter water tank is filled with 35 cm of water. A 3.0-mm-diameter spigot at the very bottom of the tank is opened and
cricket20 [7]

Answer:

The frequency f = 521.59 Hz

The rate at which the frequency is changing = 186.9 Hz/s

Explanation:

Given that :

Diameter of the tank = 44 cm

Radius of the tank = \frac{d}{2} =\frac{44}{2} = 22 cm

Diameter of the spigot = 3.0 mm

Radius of the spigot = \frac{d}{2} =\frac{3.0}{2} = 1.5 mm

Diameter of the cylinder = 2.0 cm

Radius of the cylinder = \frac{d}{2} = \frac{2.0}{2} = 1.0 cm

Height of the cylinder = 40 cm = 0.40 m

The height of the water in the tank from the spigot = 35 cm = 0.35 m

Velocity at the top of the tank = 0 m/s

From the question given, we need to consider that  the question talks about movement of fluid through an open-closed pipe; as such it obeys Bernoulli's Equation and the constant discharge condition.

The expression for Bernoulli's Equation is as follows:

P_1+\frac{1}{2}pv_1^2+pgy_1=P_2+\frac{1}{2}pv^2_2+pgy_2

pgy_1=\frac{1}{2}pv^2_2 +pgy_2

v_2=\sqrt{2g(y_1-y_2)}

where;

P₁ and P₂ = initial and final pressure.

v₁ and v₂ = initial and final fluid velocity

y₁ and y₂ = initial and final height

p = density

g = acceleration due to gravity

So, from our given parameters; let's replace

v₁ = 0 m/s ; y₁ = 0.35 m ; y₂ = 0 m ; g = 9.8 m/s²

∴ we have:

v₂ = \sqrt{2*9.8*(0.35-0)}

v₂ = \sqrt {6.86}

v₂ = 2.61916

v₂ ≅ 2.62 m/s

Similarly, using the expression of the continuity for water flowing through the spigot into the cylinder; we have:

v₂A₂ = v₃A₃

v₂r₂² = v₃r₃²

where;

v₂r₂ = velocity of the fluid and radius at the spigot

v₃r₃ = velocity of the fluid and radius at the cylinder

v_3 = \frac{v_2r_2^2}{v_3^2}

where;

v₂ = 2.62 m/s

r₂ = 1.5 mm

r₃ = 1.0 cm

we have;

v₃ = (2.62  m/s)* (\frac{1.5mm^2}{1.0mm^2} )

v₃ = 0.0589 m/s

∴ velocity  of the fluid in the cylinder =  0.0589 m/s

So, in an open-closed system we are dealing with; the frequency can be calculated by using the expression;

f=\frac{v_s}{4(h-v_3t)}

where;

v_s = velocity of sound

h = height of the fluid

v₃ = velocity  of the fluid in the cylinder

f=\frac{343}{4(0.40-(0.0589)(0.4)}

f= \frac{343}{0.6576}

f = 521.59 Hz

∴ The frequency f = 521.59 Hz

b)

What are the rate at which the frequency is changing (Hz/s) when the cylinder has been filling for 4.0 s?

The rate at which the frequency is changing is related to the function of time (t) and as such:

\frac{df}{dt}= \frac{d}{dt}(\frac{v_s}{4}(h-v_3t)^{-1})

\frac{df}{dt}= -\frac{v_s}{4}(h-v_3t)^2(-v_3)

\frac{df}{dt}= \frac{v_sv_3}{4(h-v_3t)^2}

where;

v_s (velocity of sound) = 343 m/s

v₃ (velocity  of the fluid in the cylinder) = 0.0589 m/s

h (height of the cylinder) = 0.40 m

t (time) = 4.0 s

Substituting our values; we have ;

\frac{df}{dt}= \frac{343*0.0589}{4(0.4-(0.0589*4.0))^2}

= 186.873

≅ 186.9 Hz/s

∴ The rate at which the frequency is changing = 186.9 Hz/s  when the cylinder has been filling for 4.0 s.

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Which of the following statements are TRUE about nuclear energy?
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Answer:

d

Explanation:

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