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irina [24]
3 years ago
5

What is the kinetic energy of a ball with a mass of 9 kg and a velocity of 13 m/s ?

Physics
1 answer:
Nataly [62]3 years ago
5 0

Answer:

760.5J

Explanation:

Given that KE=1/2mv^2

1/2*9*13^2

=760.5J

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A ball is kicked from the origin with an initial speed of 10 = 25.0 /, at an angle with
pav-90 [236]

Answer:

2.5                  2.566335 sec

Explanation:

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2 years ago
What is a prokaryote cell/ eukaryote cell?
Wewaii [24]
A prokaryote is a microscopic single-celled organism that has neither a distinct nucleus with a membrane nor other specialized organelles. Prokaryotes include the bacteria and cyanobacteria.

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8 0
3 years ago
Newton's First and Second Laws of Motion:Question 3Kepler-10b, the first confirmed terrestrial extrasolar planet, is about 564 l
OleMash [197]

Answer:

F = 85696.5 N = 85.69 KN

Explanation:

In this scenario, we apply Newton's Second Law:

Unbalanced\ Force = ma\\Upthrust - Weight\ of\ Space\ Craft = ma\\F - W = ma\\F - mg = ma\\F = m(g + a)\\

where,

F = Upthrust = ?

m = mass of space craft = 5000 kg

g = acceleration due to gravity on surface of Kepler-10b = (1.53)(9.81 m/s²)

g = 15.0093 m/s²

a = acceleration required = 2.13 m/s²

Therefore,

F = (5000\ kg)(15.0093\ m/s^{2} + 2.13\ m/s^{2})\\

<u>F = 85696.5 N = 85.69 KN</u>

8 0
3 years ago
Two objects, one of mass m and the other of mass 2m, are dropped from the top of a building. when they hit the ground
Setler79 [48]
They will both hit the ground at the same time because gravitational acceleration for all objects is the same.
6 0
3 years ago
1) On the way to the moon, the Apollo astro-
kramer
(1) You must find the point of equilibrium between the two forces,

<span>G * <span><span><span>MT</span><span>ms / </span></span><span>(R−x)^2 </span></span>= G * <span><span><span>ML</span><span>ms / </span></span><span>x^2
MT / (R-x)^2 = ML / x^2

So,

x = R * sqrt(ML * MT) - ML / (MT - ML)
R = is the distance between Earth and Moon.

</span></span></span>The result should be,
x = 3.83 * 10^7m
from the center of the Moon, and 

R - x = 3.46*10^8 m
from the center of the Earth.


(2) As the distance from the center of the Earth is the number we found before,
d = R - x = 3.46*10^8m
The acceleration at this point is
g = G * MT / d^2
g = 3.33*10^-3 m/s^2
6 0
3 years ago
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