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scZoUnD [109]
3 years ago
8

How does earth's orbit affect climate change? Contrast a circular orbit to elliptical (oval) orbit. Give two opposite extreme cl

imate change due to earth's orbital change
Physics
1 answer:
Sergio039 [100]3 years ago
3 0

Answer: extreme heat and extreme cold.

Explanation: Climate experts prove that with climate change, the planet can experience extreme weather. In particular temperature record highs out pace record lows and some types of extreme weather such as extreme heat, intense precipitation, and drought have become more frequent or severe in recent decades.

The orbit has an effect on climate by determining the amount of incoming sunlight. The cycle of ice ages are linked to changes in the earth's orbit, so they are important to the long-term climate variability of the earth. Earth's orbit around the sun is due to the gravitational attraction between the earth and the sun.

The solar radiation bombarding Earth at any given time, makes the planet warmer. So Earth's place in each of these cycles should have a measurable effect on long term climate trends — and it does. Another factor is to consider which hemisphere happens to be receiving the heaviest bombardment. This is because land warms faster than oceans do, and the Northern Hemisphere is covered by more land and less ocean than the Southern Hemisphere

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A vehicle of mass 100kg has a kinetic energy of 5000 J at an instant. The velocity at that instant is​
snow_lady [41]

Answer:

  • \bf\pink{10\:m/s}

Explanation:

<h2><u>Given</u> :-</h2>

  • \sf\red{Mass \ of \ vehicle \ (m) = 100 \ kg}
  • \sf\orange{Kinetic \ energy \ (K.E.) = 5000 \ J}

<h2><u>To Find</u> :-</h2>

  • \sf\green{Velocity\: of\: the \:vehicle \:at \:the \:instant}

<h2><u>Formula to be used</u> :-</h2>

  • \sf\blue{K.E. = \dfrac{1}{2} mv^{2}}

Where,

  • K.E. = Kinetic energy possessed by the body
  • M = Mass of the body
  • V = Velocity of the body

<h2><u>Solution</u> :-</h2>

\to\:\:\sf\red{K.E. = \dfrac{1}{2}mv^{2}}

\to\:\:\sf\orange{5000 = \dfrac{1}{2} \times 100 \times v^{2}}

\to\:\:\sf\green{5000 = 50 \times v^{2}}

\to\:\:\sf\blue{\dfrac{5000}{50} = v^{2}}

\to\:\:\sf\purple{100 = v^{2}}

\to\:\:\sf\red{\sqrt{100} = v}

\to\:\:\sf\orange{ 10 = v}

\to\:\:\bf\pink{v = 10\:m/s}

  • Velocity of the vehicle at the instant is \bf\green{10\:m/s}
7 0
2 years ago
Read 2 more answers
Three boxes in contact rest side-by-side on a smooth, horizontal floor. Their masses are 5.0-kg, 3.0-kg, and 2.0-kg, with the 3.
Ivahew [28]

Answer:

(a)Look at the attached graphic

(b)

(b)-1 Equation 1  : m1= 5kg

       50-F1= 5 *a

(b)-2 Equation 2 : m2= 3kg

        F1-F2= 3 *a

(b)-3 Equation 3 : m3= 2kg

         F2 = 2*a  

(c) F1 =25 N

(d) F2 =10 N

Explanation:

We apply Newton's second law:

∑F = m*a (Formula 1)

∑F : algebraic sum of the forces in Newton (N)

m : mass in kilograms (kg)

a : acceleration in meters over second square (m/s²)

(a) Draw the free-body diagrams for each of the boxes

Look at the attached graphic

(b) Write Newton’s equation for each mass along the horizontal direction.

Data: m1=  5.0-kg ,m2= 3.0-kg , ,m3= 2.0-kg

<em>Look</em> <em>m1 free-body diagram:</em>

∑Fx = m1*a

50-F1= 5 *a Equation 1

<em>Look</em> <em>m2 free-body diagram:</em>

∑Fx = m2*a

F1-F2= 3 *a Equation 2

<em>Look</em> <em>m3 free-body diagram:</em>

∑Fx = m3*a

F2 = 2*a     Equation 3

(c) What magnitude force does the 3.0-kg box exert on the 5.0- kg box?

<em>Look</em> <em>Free body diagram of the mass set</em>

∑Fx = m*a   m= m1+m2+m3= 5+3+2 = 10 kg

50 = 10*a

a= 50/10 = 5 m/s²

We replace a = 5 m/s² in the equation 1:

50-F1= 5 *5

50-25= F1

F1 = 25 N

<em> (d) </em><em>What magnitude force does the 3.0-kg box exert on the 2.0kg box?</em>

We replace a= 5 m/s² in the equation 3

F2 = 2*5 = 10 N

4 0
3 years ago
You're driving your new sports car at 85 mph over the top of a hill that has a radius of curvature of 525 m.
Bumek [7]

Explanation:

It is given that,

Speed of the sports car, v = 85 mph = 37.99 m/s

The radius of curvature, r = 525 m

Let W_N is the normal weight and W_A is the apparent weight of the person. Its apparent weight is given by :

W_A=mg-\dfrac{mv^2}{r}

So, \dfrac{W_A}{W_N}=\dfrac{mg-\dfrac{mv^2}{r}}{mg}

\dfrac{W_A}{W_N}=\dfrac{g-\dfrac{v^2}{r}}{g}

\dfrac{W_A}{W_N}=\dfrac{9.8-\dfrac{(37.99)^2}{525}}{9.8}

\dfrac{W_A}{W_N}=0.719

or

\dfrac{W_A}{W_N}=71.9\%

Hence, this is the required solution.

5 0
3 years ago
Which describes one event that causes an eclipse?
Burka [1]

Answer:

Earths shadow covering the moon would create a lunar eclipse.

Explanation:

because i just know

3 0
3 years ago
The ballistic pendulum is a device used to measure the speed of a projectile such as a bullet. The projectile of mass m is fired
My name is Ann [436]

Answer:

Relation between initial speed of bullet and height h is given as

v = \frac{m + M}{m}\sqrt{2gh}

Explanation:

As we know that system of block and bullet swings up to height h after collision

So we have

(m + M)gh = \frac{1}{2}(m + M)v_1^2

so we have

v_1 = \sqrt{2gh}

so speed of the block + bullet just after the impact is given by above equation

Now we also know that there is no force on the system of bullet + block in the direction of motion

So we can use momentum conservation

mv = (m + M)v_1

now we have

v = \frac{m + M}{m}\sqrt{2gh}

5 0
3 years ago
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