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svetlana [45]
3 years ago
12

Two students designed an experiment to study the effect of solar radiations on four cities of Earth. They used a globe to repres

ent Earth and a light bulb to represent the sun. The four cities were labeled A, B, C, and D on the globe at the latitudes shown in the chart below. City Latitude A 33° 53' S B 64° 03' N C 39° 56' N D 2° 09' S The students moved the globe slowly around the light bulb in an elliptical orbit. Based on the experiment, the students are most likely to conclude that the least variation in seasons will be experienced at City A City B City C City D
Physics
1 answer:
Furkat [3]3 years ago
3 0
Great experiment !  Everybody should try it if they can get the equipment. 
It demonstrates a lot of things that are very hard to explain in words. 

I hope the students remembered to tilt the axis of the globe.  If they didn't,
and instead kept it straight up and down, then each city had pretty much
the same amount of bulb-light all the way around, and there were no seasons.

If the axis of the globe was tilted, then City-D had the least variation in
seasons.  City-D is only 2° from the equator, so the sun is more direct
there all year around than it is at any of the others.
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2
Karolina [17]
I’m pretty sure the answer would be D, sorry if it’s not correct!
5 0
3 years ago
1. What does the endocrine system have in common with the muscular system?
Lostsunrise [7]

Answer:

B Both are directly related to movement.

7 0
3 years ago
Convert 200in/10s into m/s (1m = 39.37in)
stiv31 [10]
5.08




hope you got it right !
7 0
3 years ago
If the swimmer starts at rest, slides without friction, and descends through a vertical height of 2.41 m
AveGali [126]

Answer:

6.88 m/s

Explanation:

The Conservation of Energy states that:

Initial Kinetic Energy + Initial Potential Energy = Final Kinetic Energy + Final Potential Energy

So we can write

mgh_{i}+\frac{1}{2}mv_{i} ^{2}=mgh_{f}+\frac{1}{2}mv_{f} ^{2}

We can cancel the common factor of m which leaves us with

gh_{i}+\frac{1}{2}v_{i} ^{2}=gh_{f}+\frac{1}{2}v_{f} ^{2}

Lets solve for v_f

gh_{i}+\frac{v_{i} ^{2}}{2}=gh_{f}+\frac{v_{f} ^{2}}{2}

Subtract gh_f from both sides of the equation.

gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f}=\frac{v_{f} ^{2}}{2}

Multiply both sides of the equation by 2.

2(gh_{i}+\frac{v_{i} ^{2}}{2}-gh_{f})={v_{f} ^{2}

Simplify the left side.

Apply the distributive property.

2(gh_{i})+2\frac{v_{i} ^{2}}{2}+2(-gh_{f})={v_{f} ^{2}

Cancel the common factor of 2.

2gh_{i}+v_{i} ^{2}-2gh_{f}={v_{f} ^{2}

Take the square root of both sides of the equation to eliminate the exponent on the right side.

{v_{f}=\sqrt{2gh_{i}+v_{i} ^{2}-2gh_{f}}

We are given g,v_{i},h_{i},h_{f}.

We can now solve for the final velocity.

{v_{f}=\sqrt{(2*9.81*2.41)+(0^{2})-(2*9.81*0)

Anything multiplied by 0 is 0.

{v_{f}=\sqrt{2*9.81*2.41

{v_{f}=\sqrt{47.2842

v_f=6.88

7 0
1 year ago
An excited 92 kg football player celebrates a touchdown by carelessly running straight into the goalpost at 9.4 m/s. He bounces
yKpoI14uk [10]

Answer:

i. 15.6 m/s

ii. I = 1.44 KNs

Explanation:

The impulse, I, on a body is the product of force applied on it and the time it acts.

i.e I = F x t

Impulse is sometimes expressed as the change in momentum of a body. It is measured in Ns.

i. mass, m, of the player = 92 kg

initial velocity of the player, u = 9.4 m/s

final velocity of the player, v = 6.2 m/s

Since he bounces back on hitting the pole, then the sign of initial and final velocities are of opposite sign.

So that,

change in velocity of the player = final velocity - initial velocity

                                          = 6.2 - (-9.4)

                                         = 6.2 + 9.4

                                         = 15.6 m/s

change in velocity of the player is 15.6 m/s

ii. Impulse, I = m(v - u)

                    = 92 x 15.6

                    = 1435.2

Impulse on the player is 1.44 KNs.

7 0
3 years ago
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