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marissa [1.9K]
3 years ago
12

How does the solar wind affect earth

Physics
1 answer:
Vika [28.1K]3 years ago
8 0
It affects it by the intense clouds of the high energy particles that it often contains which are produced by solar storms.
~ I hope this helps you!!~
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Suppose your surface body temperature averaged 90 degrees F. How much radiant energy in W/m^2 would be emitted from your body?
Debora [2.8K]

493 \; \text{W}\cdot \text{m}^{-2}.

<h3>Explanation</h3>

The Stefan-Boltzmann Law gives the energy radiation <em>per unit area</em> of a black body:

\dfrac{P}{A} = \sigma \cdot T^{4}

where,

  • P the total power emitted,
  • A the surface area of the body,
  • \sigma the Stefan-Boltzmann Constant, and
  • T the temperature of the body in degrees Kelvins.

\sigma = 5.67 \times 10^{-8} \;\text{W}\cdot \text{m}^{-2} \cdot \text{K}^{-4}.

T = 90 \; \textdegree{}\text{F} = (\dfrac{5}{9} \cdot (90-32) + 273.15) \; \text{K} = 305.372 \; \text{K}.

\dfrac{P}{A} = \sigma \cdot T^{4} = 5.67 \times 10^{-8} \times 305.372^{4} = 493\; \text{W}\cdot \text{m}^{-2}.

Keep as many significant figures in T as possible. The error will be large when T is raised to the power of four. Also, the real value will be much smaller than 493\; \text{W}\cdot \text{m}^{-2} since the emittance of a human body is much smaller than assumed.

5 0
3 years ago
Quel est le type de rayonnement produit par la radioactivité
Dmitriy789 [7]

Il existe troi types de rayons produits lors de la désintégration des éléments radioactifs:

-- "particules alpha" . . . noyaux d'hélium, composés chacun de 2 protons et 2 neutrons

-- "rayons bêta" ou "particules bêta" . . . flux d'électrons

-- "rayons gamma" . . . rayonnement électromagnétique avec les longueurs d'onde les plus courtes connues et l'énergie la plus élevée

8 0
3 years ago
In order to identify them, animal fossils and other fossils nearby can be: A. compared B. piled C. seen
tigry1 [53]

Answer:

A.compared

Explanation:

Fossils help figure out the time that organisms lived. If you know one of the fossils, it can be used as a reference for others around.

5 0
3 years ago
A 0.274 kg block is pushed up a frictionless ramp inclined at angle of 32° above the horizon. The push force is a constant 2.55
SpyIntel [72]

Answer:

The answer is 2,416 m/s. Let's jump in.

Explanation:

We do work with the amount of energy we can transfer to objects. According to energy theory:

W = ΔE

Also as we know W = F.x

We choose our reference point as a horizontal line at the block's rest point.<u> At the rest, block doesn't have kinetic energy</u> and <u>since it is on the reference point(as we decided) it also has no potential energy.</u>

Under the force block gains;

W = F.x → W=2,55.0,71=1,8105\frac{N}{m}

In the second position block has both kinetic and potential energy. Following the law of conservation of energy;

W = ΔE = Kinetic energy + Potantial Energy

W = ΔE = \frac{1}{2} mV^{2} + mgh

Here we can find h in the triangle i draw in the picture using sine theorem;

In a triangle \frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}

In our situation

\frac{0,71}{sin90} =\frac{h}{sin32} → h=0,376

Therefore

1,8105=\frac{1}{2} 0,274V^{2} +0,274.9,81.0,376

→ V=2,416

7 0
3 years ago
A 3250-kg aircraft takes 12.5 min to achieve its cruising
scoundrel [369]

Answer:

effeciency n = = 49%

Explanation:

given data:

mass of aircraft 3250 kg

power P = 1500 hp = 1118549.81 watt

time = 12.5 min

h = 10 km = 10,000 m

v =85 km/h = 236.11 m/s

n = \frac{P_0}{P}

P_o = \frac{total\ energy}{t} = \frac{ kinetic \energy + gravitational\ energy}{t}

kinetic energy= \frac{1}{2} mv^2  =\frac{1}{2} 3250* 236 = 90590389.66 kg m^2/s^2

kinetic energy = 90590389.66 kg m^2/s^2

gravitational energy = mgh = 3250*9.8*10000 = 315500000.00  kg m^2/s^2

total energy = 90590389.66 +315500000.00 = 409091242.28 kg m^2/s^2

P_o =\frac{409091242.28}{750} = 545454.99 j/s

effeciency\ n = \frac{P_o}{P} = \frac{545454.99}{1118549.81} = 0.49

effeciency n = = 49%

8 0
3 years ago
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