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3 years ago

At what point is the northern hemisphere pointed farthest away from the sun?

1 answer:

7 0

<span>Well, It is the aphelion point, When the Earth is farthest away from the Sun, when the Northern Hemisphere is warm. the Earth is closest to the Sun, or at the perihelion, 2 weeks after the June Solstice, when the Northern Hemisphere is enjoying warm summer months. Well this kind of weather is very nice.</span>

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What is 60 kilometers to meters (1 km = 1000 m)

mestny [16]

60,000 meters. no explanation

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3 years ago

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Suppose a man pushes a crate with a force of 20 N north. What is the magnitude and direction of the reaction force?

Goryan [66]

20 N north because He is pushing it 20 newtons north.

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2 years ago

Does the image above correctly illustrate how the coriolis effect impacts global winds?

Sedbober [7]

Yes the winds are moving in a straight line

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3 years ago

Relativistic velocity is of the order of _____ of velocity of light A- 1/15th of the velocity of light B-1/20th of the velocity

ratelena [41]

Answer:

Relativistic velocity is of the order of 1/10th of the velocity of light

Explanation:

We define relativistic speed (or velocity) as a speed that is a significant fraction of the speed of light: c = 3*10^8 m/s

Such that for these speeds, the special relativity theory starts to apply (the relativity effects starts to apply).

Usually, we define relativistic speeds as those that are of the order (or larger) of c/10, which is one-tenth of the speed of light.

Then the correct option is C:

Relativistic velocity is of the order of 1/10th of the velocity of light

4 0

3 years ago

Two friends, Al and Jo, have a combined mass of 195 kg. At the ice skating rink, they stand close together on skates, at rest an

ddd [48]

Answer:

Al's mass is 102.92 kg

Explanation:

As there are no external forces in the horizontal direction, the horizontal net force must be zero:

$F_{net} = 0$

As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:

$F_{net} = \frac{dp_{horizontal}}{dt} = 0$

$p_{horizontal_i }= p_{horizontal_f}$

where the suffix i and f means initial and final respectively.

The initial momentum will be:

$p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}$

But, as they are at rest, initially

$p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0$

$p_{horizontal_i} = 0$

So, this means:

$p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0$

We know that the have an combined mass of 195 kg:

$m_{total} = m_{Al} + m_{Jo} = 195 \ kg$ .

so:

$m_{Jo} = 195 \ kg - m_{Al}$ .

$m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0$

$m_{Al} \ v_{Al_f} - m_{Al} \ v_{Jo_f}= - 195 \ kg \ v_{Jo_f}$

$m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f}$

$m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } { v_{Al_f} - v_{Jo_f} }$

$m_{Al} = \frac{195 \ kg \ v_{Jo_f} } { v_{Jo_f} - v_{Al_f} }$

Now, we can use the values:

$v_{Al_f}= 10.2 \frac{m}{s}$

$v_{Jo_f}= - 11.4 \frac{m}{s}$

where the minus sign appears as they are moving at opposite directions

$m_{Al} = \frac{195 \ kg ( - 11.4 \frac{m}{s} ) } { (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} }$

$m_{Al} = 102.92 \ kg$

and this is the Al's mass.

5 0

3 years ago