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4 years ago

Which of the following is most responsible for El Niño?

Physics

2 answers:

lesantik [10]4 years ago

6 0

Unusually cold water in the Atlantic causes an increase in wind activity.

kozerog [31]4 years ago

6 0

Answer:

The calming down of trade winds in the Pacific, which leads to the buildup of warm water

Explanation:

El Niño is the climate condition that occurs when the top part of the Pacific Ocean comes in contact with the atmosphere above it. It causes the water of the Pacific Ocean to become warmer because the trade winds are becoming more calmer.

El Niño is warm stage or period of the El Niño-Southern Oscillation.

The effects of El Niño on the environment includes; flooding,severe drought, break out of fire incidents, heat stress e.t.c

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Gravity ____.

vitfil [10]

I am like 98% sure it is C

4 0

4 years ago

Read 2 more answers

A particle has a charge of q = +4.9 μC and is located at the origin. As the drawing shows, an electric field of Ex = +242 N/C ex

irina1246 [14]

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=0$

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=0$

$F_{B_y}=3.21\cdot 10^{-3}N$ (+z axis)

$F_{E_x}=1.19\cdot 10^{-3} N$ (+x axis)

$F_{B_x}=3.21\cdot 10^{-3} N$ (+y axis)

$F_{B_y}=3.21\cdot 10^{-3}N$ (-x axis)

Explanation:

The electric force exerted on a charged particle is given by

$F=qE$

where

q is the charge

E is the electric field

For a positive charge, the direction of the force is the same as the electric field.

In this problem:

$q=+4.9\mu C=+4.9\cdot 10^{-6}C$ is the charge

$E_x=+242 N/C$ is the electric field, along the x-direction

So the electric force (along the x-direction) is:

$F_{E_x}=(4.9\cdot 10^{-6})(242)=1.19\cdot 10^{-3} N$

towards positive x-direction.

The magnetic force instead is given by

$F=qvB sin \theta$

where

q is the charge

v is the velocity of the charge

B is the magnetic field

$\theta$ is the angle between the directions of v and B

Here the charge is stationary: this means $v=0$ , therefore the magnetic force due to each component of the magnetic field is zero.

In this case, the particle is moving along the +x axis.

The magnitude of the electric force does not depend on the speed: therefore, the electric force on the particle here is the same as in part a,

$F_{E_x}=1.19\cdot 10^{-3} N$ (towards positive x-direction)

Concerning the magnetic force, we have to analyze the two different fields:

- $B_x$ : this field is parallel to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=0^{\circ}$ , so the force due to this field is zero.

$- B_y$ : this field is perpendicular to the velocity of the particle, which is moving along the +x axis. Therefore, $\theta=90^{\circ}$ . Therefore, $\theta=90^{\circ}$ , so the force due to this field is:

$F_{B_y}=qvB_y$

where:

$q=+4.9\cdot 10^{-6}C$ is the charge

$v=345 m/s$ is the velocity

$B_y = +1.9 T$ is the magnetic field

Substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And the direction of this force can be found using the right-hand rule:

- Index finger: direction of the velocity (+x axis)

- Middle finger: direction of the magnetic field (+y axis)

- Thumb: direction of the force (+z axis)

As in part b), the electric force has not change, since it does not depend on the veocity of the particle:

$F_{E_x}=1.19\cdot 10^{-3}N$ (+x axis)

For the field $B_x$ , the velocity (+z axis) is now perpendicular to the magnetic field (+x axis), so the force is

$F_{B_x}=qvB_x$

And by substituting,

$F_{B_x}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+x axis)

- Thumb: force (+y axis)

For the field $B_y$ , the velocity (+z axis) is also perpendicular to the magnetic field (+y axis), so the force is

$F_{B_y}=qvB_y$

And by substituting,

$F_{B_y}=(4.9\cdot 10^{-6})(345)(1.9)=3.21\cdot 10^{-3} N$

And by using the right-hand rule:

- Index finger: velocity (+z axis)

- Middle finger: magnetic field (+y axis)

- Thumb: force (-y axis)

3 0

4 years ago

Using the ideas from this section and the periodic table, choose the more reactive nonmetal. C or F

alexira [117]

c i hope thats the answer

4 0

3 years ago

Read 2 more answers

Ship A is located 4.1 km north and 2.3 km east of ship B. Ship A has a velocity of 22 km/h toward the south and Ship B has a vel

castortr0y [4]

Answer:

a) x component = -31.25 km/hr

b) y component = 46.64 km/hr

Explanation:

Given data:

A position is 4km north and 2.5 km east to B

Ship A velocity = 22 km/hr

ship B velocity = 40 km/hr

A velocity wrt to velocity of B

$\vec{V_{AB}} =\vec{V_A} - \vec{V_B}$

$\vec{V_A} = 22 km/hr$

$\vec{V_B} = 40 cos38\hat{i} + 40sin 38 \hat{j}$

$= 31.52\hat{i} + 24.62 \hat{j}$

putting respective value to get velocity of A with respect to B

$\vec{V_{AB}} = -22 \hat {j} - (31.52\hat{i} + 24.62 \hat{j})$

$\vec{V_{AB}} = -31.52\hat{i} - 46.62\hat{j}$

a) x component = -31.25 km/hr

b) y component = 46.64 km/hr

6 0

4 years ago

Car A is parked on a hill and Car B is driving down a level highway. Which statement best describes the two cars? A) Car A has k

12345 [234]

The answer is b because kinetic energy is the energy of motion, while potential energy is the energy stored in an object because of its position.

5 0

4 years ago

Read 2 more answers