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Bas_tet [7]
3 years ago
9

Hola me pueden ayudar necesito los cuerpos de:

Physics
2 answers:
NNADVOKAT [17]3 years ago
7 0
I don’t speak Spanish no hablo español
kumpel [21]3 years ago
3 0
The answer is Madera
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Name at least four factors involved in performing a good throw.
atroni [7]

Answer:

You need to have a good stance, release, power, speed, body, movement, and aim. You should be able to throw an effective pass or throw by using your body and having the correct amount of balance and form.

Explanation:

7 0
3 years ago
A 1,600 kg car is traveling with a speed of 20 m/s. Find the net force that is required to bring the car to a halt in a distance
Pie
256N. I love physics so.... yerp
4 0
4 years ago
Read 2 more answers
What is the % error in using g = 10.0 m/s^2? (Take the value ofg as 9.8 m/s^2)
Alenkasestr [34]

Answer:

So percentage error will be 2 %

Explanation:

We have given initial value of acceleration due to gravity g=10m/sec^2

And final value of acceleration due to gravity g=9.8m/sec^2

We have to find the percentage error

We know that percentage error is given by percentage\ error=\frac{initial\ value-final\ value}{initial\ value}\times 100

So percentage\ error=\frac{10-9.8}{10}\times 100=2 %

4 0
3 years ago
The mean distance of an asteroid from the Sun is 2.98 times that of Earth from the Sun. From Kepler's law of periods, calculate
lutik1710 [3]

Answer:

The asteroid requires 5.14 years to make one revolution around the Sun.

Explanation:

Kepler's third law establishes that the square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit:

T^{2} = a^{3} (1)

Where T is the period of revolution and a is the semi-major axis.

In the other hand, the distance between the Earth and the Sun has a value of 1.50x10^{8} Km. That value can be known as well as an astronomical unit (1AU).

But 1 year is equivalent to 1 AU according with Kepler's third law, since 1 year is the orbital period of the Earth.

For the special case of the asteroid the distance will be:

a = 2.98(1.50x10^{8}Km)

a = 4.47x10^{8}Km

That distance will be expressed in terms of astronomical units:

4.47x10^{8}Km.\frac{1AU}{1.50x10^{8}Km} ⇒ 2.98AU

Finally, from equation 1 the period T can be isolated:

T = \sqrt{a^{3}}

T = \sqrt{(2.98)^{3}}  

T = \sqrt{26.463592}

T = 5.14AU

Then, the period can be expressed in years:

5.14AU.\frac{1yr}{1AU} ⇒ 5.14 yr

T = 5.14 yr

Hence, the asteroid requires 5.14 years to make one revolution around the Sun.

8 0
3 years ago
while doing supw at school,you found a picture of Guru rinpoche in a waste pit.what would be your course of action?​
koban [17]
Dude you gotta rewrite that I have no idea what the question is
7 0
3 years ago
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