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Rom4ik [11]
3 years ago
7

What forms of technology are scientists using to study El Niño? need help fast!?!?!?!?

Physics
2 answers:
Natali5045456 [20]3 years ago
6 0
Noaa. The buoys transmit some of the data on a daily basis to NOAA through a satellite in space.
Luda [366]3 years ago
5 0
<h2>Answer:</h2>

<u>There are several means used for El Niño detection such as </u><u>satellites, moored ATLAS and PROTEUS buoys, drifting buoys, sea level analysis, and XBT's. </u>

<u></u>

<h2>Explanation:</h2>

El Niño is the warm phase of the El Niño Southern Oscillation and is associated with a band of warm ocean water that develops in the central and east-central equatorial Pacific. Since El Niño influences global weather patterns and affects human lives and ecosystems, prediction of an El Niño event is becoming increasingly important. For short term prediction (up to 1 year) of climate variations, current observations in the Tropical Pacific are vital. Numerical models are used in many places for El Niño prediction and research.

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Its just Calcium Flouride because as you can see it is an Ionic bond (Ca is  metal and fl is a gas usually)
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A convex mirror, like the passenger-side rearview mirror on a car, has a focal length of -3.0 m . An object is 6.0 m from the mi
Artist 52 [7]

A) -2.0 m

Look at the ray diagram attached in the picture, where:

p identifies the location of the object

q identifies the location of the image

F identifies the focus of the mirror

Each tick represents 1 m

We have

p = 6.0 m is the distance of the object from the mirror

f = -3.0 m is the focal length

From the ray diagram, we see that q has a distance of 2.0 m from the mirror, and it's on the other side of the mirror compared to the object, so

q = -2.0 m

This can also be verified by using the mirror equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{-3.0 m}-\frac{1}{6.0 m}=-\frac{3}{6.0 cm}\\q = \frac{-6.0 cm}{3}=-2.0 cm

B) Upright and virtual

As we see from the picture, the image is upright, since it has same orientation as the object.

Also, we notice that the image is on the other side of the mirror, compared to the object. For a mirror,

- An image is said to be real if it is on the same side of the object

- An image is said to be virtual if it is on the opposite side of the mirror

Therefore, this means that the image is virtual.

8 0
3 years ago
Which of the following is required for work to be done on an object?
ad-work [718]
I think that in order for work to be done, the object must move in the direction of the force and move over a distance.
7 0
3 years ago
©
Andru [333]

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Two positively charged particles

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I said two positively charged particles because if I say c or d what ever it is for you guy it can be wrong so just pick the one that says Two positively charged particles

8 0
2 years ago
[High Dive) above a pool of water. According to the announcer, the divers enter the water at a speed of 56 mi/h (25 m/s). (Air r
blsea [12.9K]

Answer:

(a) The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)  it’s possible for a diver to enter the water with the velocity of 25 m/s if he has initial velocity of 14.4 m/s. The upward initial velocity can’t be physically attained

Explanation:

(a)

To find the final velocity V_{f} for an object traveling distance h taking the initial vertical component of velocity as V_{i} the kinematics equation is written as

V_{f}^{2}=V_{i}^{2}+2ah where a is acceleration

Substituting g for a where g is gravitational force value taken as 9.81

V_{f}^{2}=V_{i}^{2}+2gh

Since the initial velocity is zero, we can solve for final velocity by substituting figures, note that 70 ft is 21.3 m for h

V_{f}=\sqrt {(2gh)}= V_{f}=\sqrt {(2*9.81*21.3)}= 20.44275

Therefore, the divers enter with a speed of 20.4 m/s

The announcer's claim is incorrect because the divers enter at a speed of 20.4 and not 25 m/s as announced

(b)

The divers can enter water with a velocity of 25 m/s only if they have some initial velocity. Using the kinematic equation

V_{f}^{2}=V_{i}^{2}+2gh

Since we have final velocity of 25 m/s

V_{i}^{2}=2gh-V_{f}^{2}

V_{i}=\sqrt{(V_{f}^{2}-2gh)}

V_{i}=\sqrt{(25^{2}-2*9.81*21.3)}= 14.390761 m/s

Therefore, it’s possible for a diver to enter the water with the velocity of 25 m/5 if he has initial velocity of 14.4 m/s

In conclusion, the upward initial velocity can’t be physically attained

3 0
3 years ago
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