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3 years ago

9

Which of these is the most likely impact of solar prominence on Earth?

1 answer:

svlad2 [7]3 years ago

8 0

<span>Solar prominences themselves are of no concern because they are visible in the Hydrogen Alpha wavelength. They are anchored in place by magnetic fields. When these fields break or reconnect, it can send the plasma that makes up the prominence away from the sun. If one of these clouds impacts Earth, they are called CMEs or coronal mass ejections. Depending on the magnetic orientation of the cloud with respect to Earth's the CME can break down our magnetic field resulting in geomagnetic storms, aurorae, power grid fluctuations, and particle radiation near the poles, satellite single upset events, and radio blackouts. </span>

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<span>Thus, letter a is the answer. </span>

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What's three pieces of evidence that Wegener used to support his theory ( Alfred Wegener’s hypothesis of continental drift state

Elis [28]

Answer:

In the early part of the 20th century, scientists began to put together evidence that the continents could move around on Earth's surface. The evidence for continental drift included the fit of the continents; the distribution of ancient fossils, rocks, and mountain ranges; and the locations of ancient climatic zones.

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3 years ago

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timurjin [86]

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A force of 70 N is inclined at an angle

Bad White [126]

Answer:

The horizontal component of the 70 N force = 35·√3 N

The vertical component of the 70 N force = 35 N

Explanation:

The magnitude of the given force, F = 70 N

The angle of inclination of the given force to the horizontal, θ = 30°

By the resolution of forces, we can resolve the given force, F, into its horizontal component, Fₓ, and vertical components, $F_y$ , as follows;

The horizontal component of the given force = Fₓ = F × cos(θ)

Substituting the values, we have;

The horizontal component of the 70 N force, Fₓ = 70 × cos(30°) = 35·√3 N

The vertical component of the given force = $F_y$ = F × sin(θ)

Substituting the values, we have;

The vertical component of the 70 N force, $F_y$ = 70 × sin(30°) = 35 N.

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The expanding gases that leave the muzzle of a rifle also contribute to the recoil. A .30 caliber bullet has mass 7.20×10−3 kg a

katen-ka-za [31]

Answer:

Mg= 1.3418 kg*m/s

Explanation:

Mass of rifle

$m_{r}=2.90 kg$

Recoil velocity of rifle

$v_{r}=1.95 m/s$

Momentum of rifle

$m*v_{r}=3.0kg*1.95\frac{m}{s}==5.85 \frac{kg*m}{s}$

mass of bullet

mb=7.2x10^{-3} kg=0.0072 kg

velocity of bullet relative to muzzel

v_{b} = 601 m/s

velocity of bullet relative to earth= 601 - 1.95=599.05 m/s

momentum of bullet

$m_{b}*m_{b}=7.2x10^{-3}kg*599.05\frac{m}{s}=4.3132 \frac{kg*m}{s}$

the momentum of the propellant gases = momentum of rifle - momentum of bullet

the momentum of the propellant gases

Mg=5.655 -4.3132

Mg= 1.3418 kg*m/s

the momentum of the propellant gases in a coordinate system attached to the earth as they leave the muzzle of the rifle is 1.3418 kgm/s

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3 years ago