Ask question

Ask question

All categories

3 years ago

11

Our planet heats unevenly, explain how this affects wind patterns. Use the words sea breeze and land breeze please :)

2 answers:

slava [35]3 years ago

7 0

Answer: The heating of the Earth's surface and atmosphere by the sun drives convection within the atmosphere and ocean. This convection produces winds and ocean currents. The greater the pressure differences between a low-pressure area and a high-pressure area, the stronger the winds.

Explanation:

jeka57 [31]3 years ago

6 0

Earth heating jhgffhhgggg

You might be interested in

An electric generator transforms mechanical energy into electrical energy. This process could be done by which of these?

Dimas [21]

Answer:

its d

Explanation:

6 0

3 years ago

Read 2 more answers

there is a fish called an archer fish that shoots drops of water at insects resting on branches above the water. If the Archer f

Mariulka [41]

Answer:

=3.5 m/s

Explanation:

y = x tanθ - 1/2 g x² / (u²cos²θ )

y = 0.25 , x = 0.5, θ = 40°

.25 = .50 tan40 - .5 x 9.8x x²/ u²cos²40

.25 = .42 - 2.0875/u²

u = 3.5 m / s.

4 0

3 years ago

The cycle of the moon through its phases or the synodic month is___.

ikadub [295]

It is 29 and a half days long

3 0

3 years ago

Read 2 more answers

I just need x isolated

mamaluj [8]

Answer:

$x=\frac{-y+\sqrt{y^2+4rt} }{2r}$

$x=\frac{-y-\sqrt{y^2+4rt} }{2r}$

Explanation:

$rx+y=\frac{t}{x}\\\\x(rx+y)=(\frac{t}{x})x\\\\rx^2+yx=t\\\\rx^2+yx-t=t-t\\\\rx^2+yx-t=0$

Solve using the quadratic formula.

$x=\frac{-y+\sqrt{y^2+4rt} }{2r}$

$x=\frac{-y-\sqrt{y^2+4rt} }{2r}$

7 0

3 years ago

Read 2 more answers

A crate lies on a plane tilted at an angle θ = 22.5 ∘ to the horizontal, with μk = 0.19.

ValentinkaMS [17]

A) $2.03 m/s^2$

Let's start by writing the equation of the forces along the directions parallel and perpendicular to the incline:

Parallel:

$mg sin \theta - \mu_k R = ma$ (1)

where

m is the mass

g = 9.8 m/s^2 the acceleration of gravity

$\theta=22.5^{\circ}$

$\mu_k = 0.19$ is the coefficient of friction

R is the normal reaction

a is the acceleration

Perpendicular:

$R-mg cos \theta =0$ (2)

From (2) we find

$R=mg cos \theta$

And substituting into (1)

$mg sin \theta - \mu_k mg cos \theta = ma$

Solving for a,

$a=g sin \theta - \mu_k g cos \theta = (9.8)(sin 22.5)-(0.19)(9.8)(cos 22.5)=2.03 m/s^2$

B) 5.94 m/s

We can solve this part by using the suvat equation

$v^2-u^2=2as$

where

v is the final velocity

u is the initial velocity

a is the acceleration

s is the displacement

Here we have

v = ?

u = 0 (it starts from rest)

$a=2.03 m/s^2$

s = 8.70 m

Solving for v,

$v=\sqrt{u^2+2as}=\sqrt{0+2(2.03)(8.70)}=5.94 m/s$

6 0

3 years ago