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3 years ago

Why do surface winds tend to cross the isobars?

2 answers:

6 0

Geostrophic wind blows parallel to the isobars because the Coriolis force and pressure gradient force are in balance. However it should be realized that the actual wind is not always geostrophic. Especially near the surface. The surface of the Earth exerts a frictional drag on the air blowing just above it.

scZoUnD [109]3 years ago

3 0

Geostrophic winds blows parallel to the isobars. That is because the Coriolis force and pressure gradient force ( PGF ) are in balance. But near the surface the friction can act to change the direction of the wind and to slow it down. Coriolis force decreases at the surface and PGF stays the same. The difference in terrain conditions affects how much friction is exerted. Hills and forests force the wind to change direction more than flat areas. Answer: Friction reduces the speed so Coriolis is weakened.

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lilavasa [31]

Number 5 is A
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3 years ago

To make sure you understand how to use the equation, suppose that there are 1000 habitable planets in our galaxy, that 1 in 10 h

krok68 [10]

Answer:

Explanation:

Number of habitable planets = 1000

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3 years ago

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oksian1 [2.3K]

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3 years ago

Read 2 more answers

Describe what happens to the moving boat when the oars are out of the water and the forward thrust is zero

Tatiana [17]

Answer:

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6 0

3 years ago

A 5.92 g object moving to the right at 17.1 cm/s makes an elastic head-on collision with an 11.84 g object that is initially at

Lelechka [254]

Answer:

v₁f = -5.7 cm/s

Explanation:

Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

$m_{1} * v_{10} = m_{1} *v_{1f} + m_{2} * v_{2f} (1)$

Rearranging terms, we have:

$m_{1} * (v_{10} - *v_{1f} ) = m_{2} * v_{2f} (2)$

We also know that the collision is elastic, so total kinetic energy must be conserved , as follows:

$\Delta K = 0 \\ \\ \frac{1}{2} * m_{1} *v_{10} ^{2} = \frac{1}{2}* m_{1} *v_{1f} ^{2} + \frac{1}{2}* m_{2} *v_{2f} ^{2} (3)$

Rearranging , and simplifying common terms, we have:

$m_{1}* (v_{10} ^{2} -v_{1f} ^{2} ) = m_{2} *v_{2f} ^{2} (4)$

Replacing by the givens, doing some algebra and dividing (4) by (2), we find the following relationship:

$v_{10} + v_{1f} = v_{2f}$

Replacing the expression above in (1), as m₂ = 2*m₁, we can find the value of v₁f, as follows:

$m_{1} * v_{10} = m_{1} * v_{1f} +2*m_{1} * (v_{10} + v_{1f})\\ \\ -(m_{1} * v_{10}) = 3* m_{1} *v_{1f} \\ \\ v_{1f} = - \frac{v_{10} }{3} = \frac{-17.1cm/s}{3} = -5.7 cm/s$

7 0

3 years ago