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3 years ago

What do we have seasons on earth

1 answer:

8 0

Because the Earth's axis is not "straight up and down" as we move
around the sun.

So when we're on one side of the sun, the top pole leans slightly toward
the sun. During that time the sun shines more directly on the top half
of the Earth, and less directly on the bottom half. The people on the
top half see the sun higher in the sky, and their weather is warmer,
while the people on the bottom half see the sun lower in the sky, and
their weather is cooler.

Then, when we're on the other side of the sun, the top pole leans slightly
away from the sun. During that time the sun shines more directly on the
bottom half of the Earth, and less directly on the top half. The people on
the bottom half see the sun higher in the sky, and their weather is warmer,
while the people on the top half see the sun lower in the sky, and their
weather is cooler.

The Earth makes the complete trip around the sun in one year, so the
people on the Earth go through this cycle of higher/lower sun and
warmer/cooler weather every year.

You might be interested in

What is the ratio of the sun’s gravitational pull on Mercury to the sun’s gravitational pull on the earth?

Marta_Voda [28]

Answer:

The answer is $\frac{F_{Sun-Mercury} }{F_{Sun-Earth} } =0,3709$ . Let's learn why.

Explanation:

Newton's law of universal gravitation says;

$F_{g} =G.\frac{m_{1}.m_{2}}{r^{2}}$

Here G is a universal gravitational constant and is measured experimentally.

Sun's gravitational pull on mercury is:

$F_{Sun-Mercury} =G.\frac{m_{sun}.3,30.10^{23}}{(5,79.10^{10})^{2} }$

Therefore $F_{Sun-Mercury} = Gm_{sun} 98,4366$

Sun's gravitational pull on Earth is:

$F_{Sun-Earth} =G.\frac{m_{sun} 5,97.10^{24} }{(1,50.10^{11}) ^{2}}$

Therefore $F_{Sun-Earth} =Gm_{sun} 265,33$

As a result;

$\frac{F_{Sun-Mercury}}{F_{Sun-Earth} }=\frac{Gm_{sun}98,4366}{Gm_{sun}265,33 } =0,3709$

4 0

3 years ago

Substitution solve for x & y -4x - 2y equals 14 -10x+7y=-25

den301095 [7]

Answer:

-4×-2y=14 (1)

-10×+7y=-25 (2)

multiplying eq 1 by 7 and eq 2 by 2 and add eq. 1 and 2

-28×-14y=98

-20×+14y=-50

___________

-28×=48

×=48/-28

×=-12/7

now

-4×-2y=14

-4*-12/7-2y=14

48/7-2y=14

-2y=14-48/7

-2y=(98-48)/7

-2y=50/7

y=-50/14

y=-25/7

8 0

3 years ago

You shoot an arrow into the air. Two seconds later (2.00 s) the arrow has gone straight upward to a height of 35.0 m above its l

sdas [7]

This question can be solved by using the equations of motion.

a) The initial speed of the arrow is was "9.81 m/s".

b) It took the arrow "1.13 s" to reach a height of 17.5 m.

We will use the second equation of motion to find out the initial speed of the arrow.

$h= v_it + \frac{1}{2}gt^2\\$

where,

vi = initial speed = ?

h = height = 35 m

t = time interval = 2 s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

$35\ m = (v_i)(2\ s)+\frac{1}{2}(9.81\ m/s^2)(2\ s)^2\\\\v_i(2\ s)=19.62\ m\\\\v_i = \frac{19.62\ m}{2\ s}$

vi = 9.81 m/s

To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.

$h= v_it + \frac{1}{2}gt^2\\$

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 17.5 m

vi = initial speed = 9.81 m/s

t = time = ?

Therefore,

$17.5 = (9.81)t+\frac{1}{2}(9.81)t^2\\4.905t^2+9.81t-17.5=0$

solving this quadratic equation using the quadratic formula, we get:

t = -3.13 s (OR) t = 1.13 s

Since time can not have a negative value.

Therefore,

t = 1.13 s

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

4 0

2 years ago

Who wrote the universal gravitation

ira [324]

I think it was Isaac Newton

3 0

3 years ago

When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the terminal speed. Once he has reache

Masteriza [31]

Answer:

b) True. the force of air drag on him is equal to his weight.

Explanation:

Let us propose the solution of the problem in order to analyze the given statements.

The problem must be solved with Newton's second law.

When he jumps off the plane

fr - w = ma

Where the friction force has some form of type.

fr = G v + H v²

Let's replace

(G v + H v²) - mg = m dv / dt

We can see that the friction force increases as the speed increases

At the equilibrium point

fr - w = 0

fr = mg

(G v + H v2) = mg

For low speeds the quadratic depended is not important, so we can reduce the equation to

G v = mg

v = mg / G

This is the terminal speed.

Now let's analyze the claims

a) False is g between the friction force constant

b) True.

c) False. It is equal to the weight

d) False. In the terminal speed the acceleration is zero

e) False. The friction force is equal to the weight

3 0

3 years ago