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Ipatiy [6.2K]
3 years ago
8

What do we have seasons on earth

Physics
1 answer:
Goryan [66]3 years ago
8 0

Because the Earth's axis is not "straight up and down" as we move
around the sun. 

So when we're on one side of the sun, the top pole leans slightly toward
the sun.  During that time the sun shines more directly on the top half
of the Earth, and less directly on the bottom half.  The people on the
top half see the sun higher in the sky, and their weather is warmer,
while the people on the bottom half see the sun lower in the sky, and
their weather is cooler.

Then, when we're on the other side of the sun, the top pole leans slightly
away from the sun.  During that time the sun shines more directly on the
bottom half of the Earth, and less directly on the top half.  The people on
the bottom half see the sun higher in the sky, and their weather is warmer,
while the people on the top half see the sun lower in the sky, and their
weather is cooler.

The Earth makes the complete trip around the sun in one year, so the
people on the Earth go through this cycle of higher/lower sun and
warmer/cooler weather every year.

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Water at 25°C and 1 atm is flowing over a long flat plate with a velocity of 14 m/s. Determine the distance from the leading edg
Gre4nikov [31]

Answer:

L=31.9 mm

δ = 0.22 mm

Explanation:

Given that

v= 14 m/s

ρ=997 kg/m³

μ= 0.891 × 10⁻3 kg/m·s

As we know that when Reynolds number grater than 5 x 10⁵ then flow will become turbulent.

Re=\dfrac{\rho vL}{\mu}

L=\dfrac{Re\mu}{\rho v}

L=\dfrac{5\times 10^5\times 0.891\times 10^{-3}}{ 14 \times 997}\ m

L=0.0319 m

L=31.9 mm

The  thickness of the boundary layer at that location L given as

\delta =\dfrac{5L}{\sqrt{Re}}

\delta =\dfrac{5\times0.0319}{\sqrt{5\times 10^5}}

δ = 0.00022 m

δ = 0.22 mm

7 0
3 years ago
The diagram below shows the movement of matter in a portion of the water cycle.
Alisiya [41]

Answer:

d) precipitation

Hope it helps you

And if you want to, pls mark it as the brainliest answer

7 0
3 years ago
Two horizontal rods are each held up by vertical strings tied to their ends. Rod 1 has length L and mass M; rod 2 has length 2L
antiseptic1488 [7]

Answer:

Rod 1 has greater initial angular acceleration; The initial angular acceleration for rod 1 is greater than for rod 2.

Explanation:

For the rod 1 the angular acceleration is

\tau_1 = I_1\alpha _1 \\\\\alpha_1 = \dfrac{\tau_1}{I_1}

Similarly, for rod 2

\alpha_2 = \dfrac{\tau_2}{I_2}.

Now, the moment of inertia for rod 1 is

I_1 = \dfrac{1}{3}ML^2,

and the torque acting on it is (about the center of mass)

\tau_1 = Mg\dfrac{L}{2};

therefore, the angular acceleration of rod 1 is  

\alpha_1 = \dfrac{Mg\dfrac{L}{2}}{\dfrac{1}{3}ML^2},

\boxed{\alpha_1 = \dfrac{3g}{2L} }

Now, for rod 2 the moment of inertia is

I_2 = \dfrac{1}{3}(2M)(2L)^2

I_2 = \dfrac{8}{3} ML^2,

and the torque acting is (about the center of mass)

\tau _2 = (2M)g \dfrac{(2L)}{2}

\tau _2 = 2MgL;

therefore, the angular acceleration \alpha_2 is

\alpha_2 = \dfrac{2MgL;}{\dfrac{8}{3} ML^2,}.

\boxed{\alpha_2 = \dfrac{3g}{4L}}

We see here that

\dfrac{3g}{2L} > \dfrac{3g}{4L}

therefore

\boxed{\alpha_1 > \alpha_2.}

In other words , the initial angular acceleration for rod 1 is greater than for rod 2.

7 0
3 years ago
What time that take if wavelength 1/10^8 and 3*10^8 m/s?
asambeis [7]

Answer:3.33x10^(-17)

Explanation:

Period=wavelength ➗ velocity

Period=1/10^8 ➗ (3x10^8)

Period=3.33x10^(-17)

5 0
3 years ago
A bomb falls with velocity v(t)=C(1−e−kt), where C and k are constants. What is the terminal velocity of the bomb? That is, what
irinina [24]

Answer:

The terminal velocity is equal to C.

Explanation:

Making the assumption that k and t are positive, we then have that -kt is negative. The value of e^(-kt) will be equal to 1/(e^(kt))

If t increases, e^(kt) will increase exponentially and its reciprocal 1/(e^(kt)) will approach zero.

So, we have:

v(t) = C*(1-0)

v(t) = C*(1)

v(t) = C

Therefore, C is the terminal velocity.

6 0
3 years ago
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