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ivanzaharov [21]
3 years ago
8

When the northern hemisphere is tilted toward the sun that part of the earth is closer to the sun and thus hotter? True or false

Physics
1 answer:
Ratling [72]3 years ago
8 0

<u>False</u>

<u>Explanation:</u>

Most of us have a belief that Earth is very close to the sun in summer season and so it is hotter. Also, since Earth is far away from the sun in winter season, it is colder.  Though this idea looks true, it is incorrect.

The orbit of the Earth isn't a perfect circle. It is quite tilted. During some months of the year, Earth is very close to the sun than at other times. We have winter when the Earth is very close to the sun and summer when it is far away in the Northern Hemisphere. So when we compare the distance of the Sun from the Earth, this change in Earth's distance throughout the year does not affect our weather much.

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(So your answer would be A)
6 0
3 years ago
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The free-body diagram of a crate is shown. A free body diagram with 4 force vectors. The first vector is pointing downward, labe
satela [25.4K]

The net force acting on the crate is determined as 176 N to the left.

<h3>Net force acting on the crate</h3>

The net force acting on the crate is calculated as follows;

∑F = F1 + F2 + F3 + F4

F(net) = -440y + 176x + 440y - 352x

F(net) = -176 x

The resultant force is pointing in negative x direction.

Thus, the net force acting on the crate is determined as 176 N to the left.

Learn more about net force here: brainly.com/question/14361879

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8 0
1 year ago
According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about
NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, r=4\ AU=1.496\times 10^{11}\ m

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

T^2=\dfrac{4\pi^2}{GM}r^3

M is the mass of the sun

T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3    

T^2=\sqrt{9.96\times 10^{14}}\ s

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0
3 years ago
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zepelin [54]

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8 0
3 years ago
2. A 0.8 kg tetherball hangs on the end of a cord. It is hit by a child and rises 2.1 m above the ground. a. What is the maximum
skad [1K]

Answer:

E = 16.464 J

Explanation:

Given that,

Mass of tetherball, m = 0.8 kg

It is hit by a child and rises 2.1 m above the ground, h = 21. m

We need to find the maximum gravitational potential energy of the ball. The formula for the gravitational potential energy is given by :

E = mgh

g is acceleration due to gravity

E = 0.8 kg × 9.8 m/s² × 2.1 m

= 16.464 J

So, the maximum potential energy of the ball is 16.464 J.

3 0
3 years ago
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