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Montano1993 [528]
4 years ago
5

If earth rotated on an axis that was not tilted, how would this affect the seasons?

Physics
2 answers:
VARVARA [1.3K]4 years ago
8 0
If earth was not tilted on a axis, we wouldn't have any seasons. we wouldn't have seasons because the axis tilts us towards or away from the sun, and if you're tilted torwards the sun the sun you would have summer. if you're tilted away from the sun, you have winter. if there was no axis, the temperature would be the same all around the earth, so therefore we would not have seasons. 

hopefully this helps 
anastassius [24]4 years ago
6 0

Answer:

There would not be any seasons.

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A woman is standing in the ocean, and she notices that after a wave crest passes, five more crests pass in a time of 54.0 s. The
timama [110]

Answer:

a) f=0.1 Hz ; b) T=10s

c)λ= 36m

d)v=3.6m/s

e)amplitude, cannot be determined

Explanation:

Complete question is:

Determine, if possible, the wave's (a) frequency, (b) period, (c) wavelength, (d) speed, and (e) amplitude.

Given:

number of wave crests 'n'= 5

pass in a time't' 54.0s

distance between two successive crests 'd'= 36m

a) Frequency of the waves 'f' can be determined by dividing number of wave crests with time, so we have

f=n/t

f= 5/ 54 => 0.1Hz

b)The time period of wave 'T' is the reciprocal of the frequency

therefore,

T=1/f

T=1/0.1

T=10 sec.

c)wavelength'λ' is the distance between two successive crests i.e 36m

Therefore, λ= 36m

d) speed of the wave 'v' can be determined by the product of frequency and wavelength

v= fλ => 0.1 x 36

v=3.6m/s

e) For amplitude, no data is given in this question. So, it cannot be determined.

5 0
3 years ago
If a ball is thrown horizontally with a speed of 65 mph, how far will it fall while traveling 90 ft of horizontal distance?
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3 years ago
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Black_prince [1.1K]
<span>
Reaction rates are affected by reactant concentrations and temperature. this is accounted for by the c</span>ollision model.

-Hope this helps.
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3 years ago
Zero, a hypothetical planet, has a mass of 4.5 x 1023 kg, a radius of 3.2 x 106 m, and no atmosphere. A 10 kg space probe is to
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a.) K 2=K 1 +GmM( r 21− r 11)=2.2×10 7J

b.) ​K 2 +GmM( r 11− r 21)=6.9×10 7 J

Applying Law of  Energy conservation :

K 1+U 1

=K 2+U 2

⇒K 1− r 1GmM

=K 2− r 2 GmM

where M=5.0×10 23kg,r1

=> R=3.0×10 6m and m=10kg

(a) If K 1​

=5.0×10 7J and r 2

=4.0×10 6 m, then the above equation leads to

K 2=K 1 +GmM (r 21− r 11)=2.2×10 7J

(b) In this case, we require K 2

=0 and r2

=8.0×10 6m, and solve for K 1:K 1

​=K 2 +GmM (r 11− r 21)=6.9×10 7 J

Learn more about Kinetic energy on:

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2 years ago
During which segments are two states of matter present? the answer is D
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Answer:

2,4

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