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1 year ago

Large-scale environmental catastrophes _______.

1 answer:

5 0

I'll give just one. The eruption of Mount Tambora in 1815 was one of the volcanic eruptions with widespread damage. Mount Tambora erupted, bringing thermal waves and tsunamis that killed 38,000 people (estimated). It is not just the tsunami and thermal waves, but also the ash plume that caused a cool down on the earth's atmosphere by 5°C causing "The Year Without Summer" and caused crops worldwide to fail, or degrade causing famine. That is all I know :) have a good day.

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Planetary orbits... are spaced more closely together as they get further from the Sun. are evenly spaced throughout the solar sy

BaLLatris [955]

Answer:

E) are almost circular, with low eccentricities.

Explanation:

Kepler's laws establish that:

All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).

A planet describes equal areas in equal times (Kepler's second law).

The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).

$T^{2} = a^{3}$

Where T is the period of revolution and a is the semi-major axis.

Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.

However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)

In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).

Therefore, option A and B can not be true.

In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.

4 0

2 years ago

You catch a volleyball (mass 0.270 kg) that is moving downward at 7.50 m/s. In stopping the ball, your hands and the volleyball

sesenic [268]

Explanation:

The work done equals the change in energy.

W = ΔKE

W = 0 − ½mv²

W = -½ (0.270 kg) (-7.50 m/s)²

W = -7.59 J

Work is force times displacement.

W = Fd

-7.59 J = F (-0.150 m)

F = 50.6 N

3 0

2 years ago

A 1000.–kilogram car traveling 20.0 meters per second east experiences an impulse of 2000. newton • seconds west. What is the fi

emmainna [20.7K]

Impulse is equal to change in momentum. So if impulse is 2000 then to solve for new velocity we just set it equal to equation for momentum.

First find original momentum by p=mv
p=1000*20=20000

So then taking that value minus the impulse since it was in opposite direction of original momentum it will slow it down some. To find new velocity we just take

20000-2000=18000=mv

v=18000/1000 =18m/s

Hope this helps!! Any questions please ask!!
Thank you!

5 0

1 year ago

Consider a star that is a sphere with a radius of 6.32 108 m and an average surface temperature of 5350 K. Determine the amount

Mariulka [41]

Answer:

The value is $\Delta s = 8.537 *10^{25 } \ J/K$

Explanation:

From the we are told that

The radius of the sphere is $r = 6.32 *10^{8} \ m$

The temperature is $T_x = 5350 \ K$

The average temperature of the rest of the universe is $T_r = 2.73 \ K$

Generally the change in entropy of the entire universe per second is mathematically represented as

$\Delta s = s_r - s_x$

Here $s_r$ is the entropy of the rest of the universe which is mathematically represented as

$s_r = \frac{Q}{T_r}$

Here Q is the quantity of heat radiated by the star which is mathematically represented as

$Q = 4 \pi * r^2 * \sigma * T^4_x$

Here $\sigma$ is the Stefan-Boltzmann constant with value

$\sigma = 5.67 * 10^{-8 }W\cdot m^{-2} \cdot K^{-4}.$

=> $Q = 4 \pi * (6.32*10^{8})^2 * 5.67 * 10^{-8 } * 5350 ^4$

=> $Q = 2.332 *10^{26} \ J$

$s_r = \frac{2.332 *10^{26}}{2.73}$

=> $s_r = 8.5415 *10^{25}\ J/K$

Here $s_x$ is the entropy of the rest of the universe which is mathematically represented as

$s_x = \frac{Q}{T_x}$

=> $s_x = \frac{2.332 *10^{26} }{5350}$

=> $s_x = 4.359 *10^{22} \ J/K$

$\Delta s = 8.5415 *10^{25} - 4.359 *10^{22}$

=> $\Delta s = 8.537 *10^{25 } \ J/K$

7 0

1 year ago

Sally finds herself stranded on a frozen pond so slippery that she can't stand up or walk on it. To save herself, she throws one

8_murik_8 [283]

Answer:

a) 2.5 m/s. (In the opposite direction to the direction in which she threw the boot).

b) The centre of mass is still at the starting point for both bodies.

c) It'll take Sally 12 s to reach the shore which is 30 m from her starting point.

Explanation:

Linear momentum is conserved.

(mass of boot) × (velocity of boot) + (mass of sally) × (velocity of Sally) = 0

5×30 + 60 × v = 0

v = (-150/60) = -2.5 m/s. (Minus inicates that motion is in the opposite direction to the direction in which she threw the boot).

b) At time t = 10 s,

Sally has travelled 25 m and the boot has travelled 300 m.

Taking the starting point for both bodies as the origin, and Sally's direction as the positive direction.

Centre of mass = [(60)(25) + (5)(-300)]/(60+5)

= 0 m.

The centre of mass is still at the starting point for both bodies.

c) The shore is 30 m away.

Speed = (Distance)/(time)

Time = (Distance)/(speed) = (30/2.5)

Time = 12 s

Hope this Helps!!!

7 0

2 years ago

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