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3 years ago

If the earth’s tilt on it axis were to increase by 20 degrees, what would happen to earth?

1 answer:

4 0

Each pole would become tropical during summer, but be horribly cold during winter, far colder than they are now. But the equator would be in perpetual twilight during two seasons—and summer the opposing two.

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What is meant by the phrase " a consistent method of measurement "? someone help plss

Mars2501 [29]

D.Measurements are taken in a way that is the same every time.- apex

8 0

3 years ago

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A boy throws a ball vertically up. It returns to the

NikAS [45]

Answer:

31.25 m

25m/sec

Explanation:

Given :-

Time = 5sec

V = 0 (in going up)

U = 0 (in comming down)

Find :-

H and U by which it is thrown up

Since the total time is 5 sec ,therefore half time will be taken to go up and another half will be taken to go down .

We know that ,

V = U + gt

0 = U - 10*2.5

U = 25 m/sec

Also,

V² = U² +2gs

0 = 625 - 20s

s = 625/20 = 31.25 m

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3 years ago

The department of insurance and safety is led by what official

lawyer [7]

Answer: Georgia Department of Insurance

Explanation: I hope this help :]

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3 years ago

Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

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3 years ago

A football player kicks a ball with a mass of 0.42 kg. The average acceleration of the football was 14.8 m/s².

kvasek [131]

D.6.22N. because .42kg * 14.8m/s=6.22 N[meaning newtons}.

6 0

3 years ago

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