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Cerrena [4.2K]
3 years ago
7

A baseball travels 200 metes in 6 seconds, what is the baseball’s velocity?

Physics
1 answer:
goldenfox [79]3 years ago
8 0

Answer:

33.33 m/sec

Explanation:

A baseball travels 200 metes in 6 seconds,

what is the baseball’s velocity?

use the formula: velocity = distance over time

where (d) distance = 200 m

and (t) time = 6 sec.

plugin values into the formula:

v = d / t

  = 200 m / 6 sec

  = 33.33 m/sec.

therefore, the baseball's velocity is 33.33 m/sec

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This is just testing your ability to recall that kinetic energy is given by: 

<span>k.e. = ½mv² </span>

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<span>k.e. = ½ * 9.1 * 10^-31 * (7.00 * 10^5)² </span>
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Are there any exceptions to the rule that planets rotate with small axis tilts and in the same direction as they orbit the sun?
Alona [7]
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A meteoroid is in a circular orbit 600 km above the surface of a distant planet. The planet has the same mass as Earth but has a
AVprozaik [17]
<h2>Answer:</h2><h2>The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/s^{2}</h2>

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The acceleration of the meteoroid due to the gravitational force exerted by the planet = ?

By formula, g = \frac{GM}{r^{2} }

where g is the acceleration due to the gravity

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Substituting the values, we get

g =  \frac{(6.67 * 10^{-11}) (5.972 * 10^{24})  }{5733^{2} }

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Two identical small metal spheres with q1 &gt; 0 and |q1| &gt; |q2| attract each other with a force of magnitude 72.1 mN when se
Brrunno [24]

1) +2.19\mu C

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F=k\frac{q_1 q_2}{r^2} (1)

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the charges

When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of

\frac{Q}{2}

where Q is the total charge between the two spheres.

So we can actually rewrite the force as

F=k\frac{(\frac{Q}{2})^2}{r^2}

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We can solve the equation for Q:

Q=2\sqrt{\frac{Fr^2}{k}}=2\sqrt{\frac{(0.02163)(1.41)^2}{8.98755\cdot 10^9}}}=4.37\cdot 10^{-6} C

So, the final charge on the sphere on the right is

\frac{Q}{2}=\frac{4.37\cdot 10^{-6} C}{2}=2.19\cdot 10^{-6}C=+2.19\mu C

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Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that

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q_2 = Q-q_1

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F=k \frac{q_1 (Q-q_1)}{r^2}

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Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is

q_1 = +6.70 \mu C

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3 years ago
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