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Lynna [10]
3 years ago
6

Why isn't the earth the same distance from sun all year long?

Physics
2 answers:
miskamm [114]3 years ago
7 0

Answer:

Earth's orbit is elliptical

Explanation:

The earth's orbit is not a straight circle and the sun is not in the very center of it. The orbit is more of a wonky oval with the sun closer to one side of it than the other causing the earth's distance from the sun to vary throughout the year.

attashe74 [19]3 years ago
5 0

Because the planets (and everything else in the solar system) don't travel in circular orbits with the sun at the center of the circle.  

They travel in elliptical orbits, with the sun at one focus of the ellipse.

Different points on the ellipse have different distances from the focus where the sun is.

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What is a fact about a punnet square
Elena L [17]
Punnet squares were only recently added into the curriculum. They allow a more organized/ efficient way to do binomial, or polynomial multiplication. They may also be known for being used in biology to show what genes an offspring could potentially have.


Hope this helps :)
4 0
3 years ago
Read 2 more answers
In a weak acid solution, _____.
erastovalidia [21]

Answer:

A)

Explanation:

8 0
3 years ago
During your summer internship for an aerospace company, you are asked to design a small research rocket. The rocket is to be lau
Thepotemich [5.8K]

Answer:

T=6.75s

Explanation:

We must separate the motion into two parts, the first when the rocket's engines is on  and the second when the rocket's engines is off. So, we need to know the height (h_1) that the rocket reaches while its engine is on and we need to know the distance (h_2) that it travels while its engine is off.

For solving this we use the kinematic equations:

In the first part we have:

h_1=v_0T+\frac{1}{2}aT^2\\h_1=0*T+\frac{1}{2}(16\frac{m}{s^2})T^2\\h_1=8\frac{m}{s^2}T^2\\

and the final speed is:

v_f=v_0+aT\\v_f=0+16\frac{m}{s^2}T\\v_f=16\frac{m}{s^2}T

In the second part, the final speed of the first part it will be the initial speed, and the final speed is zero, since gravity slows it down the rocket.

So, we have:

v_f^2=v_0^2+2gh_2\\2gh_2=v_f^2-v_0^2\\h_2=\frac{v_f^2-v_0^2}{2g}\\h_2=\frac{0^2-(16\frac{m}{s^2}T)^2}{2(-9.8\frac{m}{s^2})}\\h_2=\frac{-256\frac{m^2}{s^4}T^2}{-19.6\frac{m}{s^2}}\\h_2=13.06\frac{m}{s^2}T^2

The sum of these heights will give us the total height, which is known:

h=h_1+h_2\\960m=8\frac{m}{s^2}T^2+13.06\frac{m}{s^2}T^2\\960m=21.06\frac{m}{s^2}T^2\\T^2=\frac{960m}{21.06\frac{m}{s^2}}\\T^2=45.58s^2\\T=\sqrt{45.58s^2}\\T=6.75s

This is the time that its needed in order for the rocket to reach the required altitude.

5 0
3 years ago
A bowling bowl is thrown off the top of a building. Determine distance the ball has fallen after falling for 3.0 s. Remember, th
Zanzabum

d = distance the bowling ball has fallen = ?

g = acceleration due to gravity acting on the ball by earth = 9.8 m/s²

t = time of fall for the ball = 3.0 s

distance the ball has fallen is given as

d = (0.5) g t²

inserting the above values in the equation above

d = (0.5) (9.8 m/s²) (3.0 s)²

d = (0.5) (9.8 m/s²) (9.0 s²)

d = (4.9 m/s²) (9.0 s²)

d = 44.1 m

hence the distance fallen by the ball comes out to be 44.1 m

7 0
4 years ago
Which technology is intended to work at a distance of about 5 to 10 centimeters with transmission speeds of 250 Kbps?
Mars2501 [29]

Answer:

NFC Near Field Communication

Explanation:

The Near Field Communication is a communication protocol, for extra short distance with a maximum of 10 centimeters, but usually used in 4 to 5 cm. Its intended to be used in contactless pay systems and in transportation card. Actually has been used to transfer multimedia from cell phones and other devices. The maximum data rate is around 424 kbit/s, with mean in 250 Kbp

5 0
4 years ago
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