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3 years ago

What is the relationship between the movement of the atmosphere and surface currents

Physics

1 answer:

maxonik [38]3 years ago

3 0

Answer:

Waves and surface currents are wind-generated. The atmosphere pushes the ocean surface along. But as the ocean moves, heat is redistributed and eventually tossed back into the atmosphere. The exchange of heat influences the pressure field in the atmosphere, which modifies the wind field.

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What is the formula for finding the velocity?

Kazeer [188]

Answer:

v = Δs/Δt

and use this calcultor"Velocity Calculator | Definition | Formula - Omni Calculator"if u have problems to slove.cos

Explanation:

8 0

3 years ago

Read 2 more answers

Which force below acts at 90 degrees and is applied by a surface/wall to an object.

Molodets [167]

Answer:

B. normal

please mark as brainliest

4 0

2 years ago

Using a scale diagram, calculate the resultant force acting on a sailing boat when an easterly wind provides 2, point, 50, k, N,

EastWind [94]

Answer:

F = 3.6 kN, direction is 9.6º to the North - East

Explanation:

The force is a vector, so one method to find the solution is to work with the components of the vector as scalars and then construct the resulting vector.

Let's use trigonometry to find the component of the forces, let's use a reference frame where the x-axis coincides with the East and the y-axis coincides with the North.

Wind

X axis

F₁ = 2.50 kN

Tide

cos 30 = F₂ₓ / F₂

sin 30 = F_{2y} / F₂

F₂ₓ = F₂ cos 30

F_{2y} = F₂ sin 30

F₂ₓ = 1.20cos 30 = 1.039 kN

F_{2y} = 1.20 sin 30 = 0.600 kN

the resultant force is

X axis

Fₓ = F₁ₓ + F₂ₓ

Fₓ = 2.50 +1.039

Fₓ = 3,539 kN

F_y = F_{2y}

F_y = 0.600

to find the vector we use the Pythagorean theorem

F = $\sqrt{F_x^2 +F_y^2}$

F = $\sqrt{ 3.539^2 + 0.600^2 }$

F = 3,589 kN

the address is

tan θ = F_y / Fₓ

θ = tan⁻¹ $\frac{F_y}{F_x}$

θ = tan⁻¹ $\frac{0.6}{3.539}$ 0.6 / 3.539

θ = 9.6º

the resultant force to two significant figures is

F = 3.6 kN

the direction is 9.6º to the North - East

7 0

3 years ago

A rocket is moving at 1/4 the speed of light relative to Earth. At the center of this rocket a light suddenly flashes. To an obs

Sedaia [141]

Answer:

B. the light will reach the front of the rocket at the same instant that it reaches the back of the rocket.

Explanation:

To an observer at rest in the rocket who can't see either sides of the rocket, the speed of the light is constant which means the distance to the front or the back is same and would appear to reach the rocket at the same time.

Although from the point of view of the person on the earth, the front of the rocket is travelling in opposite direction of the light while the back of the rocket is moving closer to the light. This means that the distance travelled by the light going forward will be longer going backwards. And since the speed of light is constant in both directions, the light will reach the back of the rocket before it reaches the front for the observer on the earth.

5 0

3 years ago

Read 2 more answers

A projectile is launched horizontally from a 20-m tall edifice with a vox of 25 m/s. How long will it take for the projectile to

NISA [10]

Answer:

a) First let's analyze the vertical problem:

When the projectile is on the air, the only vertical force acting on it is the gravitational force, then the acceleration of the projectile is the gravitational acceleration, and we can write this as:

a(t) = -9.8m/s^2

To get the vertical velocity we need to integrate over time to get:

v(t) = (-9.8m/s^2)*t + v0

where v0 is the initial vertical velocity because the object is thrown horizontally, we do not have any initial vertical velocity, then v0 = 0m/s

v(t) = (-9.8m/s^2)*t

To get the vertical position equation we need to integrate over time again, to get:

p(t) = (1/2)*(-9.8m/s^2)*t^2 + p0

where p0 is the initial position, in this case is the height of the edifice, 20m

then:

p(t) = (-4.9m/s^2)*t^2+ 20m

The projectile will hit the ground when p(t) = 0m, then we need to solve:

(-4.9m/s^2)*t^2+ 20m = 0m

20m = (4.9m/s^2)*t^2

√(20m/ (4.9m/s^2)) = t = 2.02 seconds

The correct option is a.

b) The range will be the total horizontal distance traveled by the projectile, as we do not have any horizontal force, we know that the horizontal velocity is 25 m/s constant.

Now we can use the relationship:

distance = speed*time

We know that the projectile travels for 2.02 seconds, then the total distance that it travels is:

distance = 2.02s*25m/s = 50.5m

Here the correct option is a.

c) Again, the horizontal velocity never changes, is 25m/s constantly, then here the correct option is option b. 25m/s

d) Here we need to evaluate the velocity equation in t = 2.02 seconds, this is the velocity of the projectile when it hits the ground.

v(2.02s) = (-9.8m/s^2)*2.02s = -19.796 m/s

The velocity is negative because it goes down, and it matches with option d, so I suppose that the correct option here is option d (because the sign depends on how you think the problem)

4 0

3 years ago