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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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In Milgram's experiment, compliance, or doing what the experimenter asked,
Orlov [11]

Answer:

In Milgram's experiment, compliance, or doing what the experimenter asked,

the teacher and the learner were in the same room. -C.

5 0
3 years ago
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An electric filament lamp is connected to a power supply and switched on.
ladessa [460]
This is because of of the heating effect of a current. The glow is a result of current passing through the filament. The current experiences resistance as a result heat is generated. When resistance is at zero, there potential differences is not needed hence temperature generated will be at a constant.
7 0
3 years ago
During a testing process, a worker in a factory mounts a bicycle wheel on a stationary stand and applies a tangential resistive
Katyanochek1 [597]

Answer:

The force is F = 1041.7N

Explanation:

The moment of Inertia I is mathematically evaluated as

               I = MR_A^2

Substituting  1.9kg for M(Mass of the wheel) and \frac{66cm}{2} * \frac{1m}{100cm} = 0.33m for R_A(Radius of wheel)

              I = 1.9 * 0.33^2

                = 0.207kgm^2

The torque on the wheel due to net force is mathematically represented as

                      \tau = FR_B  - F_rR_A

Substituting  135 N for F_r (Force acting on sprocket),\frac{8.7cm}{2} * \frac{1m}{100cm} = 0.0435m for R_B (radius of the chain) and F is the force acting on the sprocket due to the chain which is unknown for now

                     \tau = F (0.0435) - 135 (0.33)

This same torque due to the net force is the also the torque that is required to rotate the wheel to have an angular acceleration of \alpha  = 3.70 rad/s^2 and this torque can also be represented mathematically as

                   \tau = \alpha I

Now equating the two equation for torque

                                F (0.0435) - 135 (0.33) = \alpha I    

Making F the subject

                     F = \frac{\alpha I + (135*0.33) }{0.0435}

Substituting values

                  F = \frac{(3.70 * 0.207)  + (135*0.33)}{0.0435}

                       = 1041.7N

4 0
3 years ago
In a mail-sorting facility, a 2.50-kg package slides down an inclined plane that makes an angle of 20.0° with the horizontal. Th
lawyer [7]

Answer:

The coefficient of kinetic friction is 0.382.

Explanation:

Given:

Angle of inclination is, \theta=20.0°

Mass of package is, m=2.50\ kg

Initial speed of package is, u=2.00\ m/s

Final speed of the package at the bottom is, v=0\ m/s

Distance of travel along the incline is, d=12.0\ m

Acceleration due to gravity is, g=9.8\ m/s^2

Let the coefficient of kinetic friction be \mu.

Now, the frictional force will be acting along the incline but in the direction opposite to the direction of motion.

So, the net acceleration acting on the package will be up the incline and is equal to:

a=\mu g\cos\theta-g\sin\theta ----------------- 1

Now, using equation of motion, we have:

v^2-u^2=2ad\\\\0-(2.00)^2=2a(12.0)

Solving for 'a', we get:

-4.00=24.0a\\\\a=-\frac{4}{24}=-\frac{1}{6}\ m/s^2

Now, plug in the value of 'a' in equation (1). This gives,

\mu g\cos\theta-g\sin\theta=\frac{1}{6} ( Neglecting negative sign)

Plug in all the given values and solve for \mu. This gives,

9.8(-sin(20)+\mu cos(20))=\frac{1}{6}\\\\-0.342+\mu\times 0.94=0.017\\\\0.94\mu=0.342+0.017\\\\0.94\mu=0.359\\\\\mu=\frac{0.359}{0.94}=0.382

Therefore, the coefficient of kinetic friction is 0.382.

5 0
4 years ago
Jupiter’s Great Red Spot rotates completely every six days. If the spot is circular (not quite true, but a reasonable approximat
artcher [175]

Answer:

v = 567.2 km/h

Explanation:

As we know that if Jupiter Rotate one complete rotation then it means that the it will turn by 360 degree angle

so here the distance covered by the point on its surface in one complete rotation is given by

distance = 2\pi r

distance = \pi D

now we will have the time to complete the rotation given as

t = 6 days

t = 6 (24 h) = 144 h

now the speed is given by

speed = \frac{distance}{time}

speed = \frac{\pi D}{t}

speed = \frac{\pi(26000 km)}{144}

v = 567.2 km/h

5 0
3 years ago
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