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3 years ago

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Your teacher (175kg) and the lab (15kg) are 2m apart. What is the gravitational force between them? Draw a FBD, and label

1 answer:

Pachacha [2.7K]3 years ago

7 0

The gravitational force between two objects is given by
$F=G \frac{m_1 m_2 }{r^2}$
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is the separation between the two objects

In this problem, $m_1 = 175 kg$ , $m_2 =15 kg$ and $r=2 m$ , therefore the gravitational force between the two objects is
$F=(6.67 \cdot 10^{-11} m^3 kg^{-1} s^{-2}) \frac{(175 kg)(15 kg)}{(2m)^2}=4.38 \cdot10^{-8} N$

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A sprinter speeds up to 3 m/s during the last 2 seconds of the race with an

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Answer:

<em>The initial speed of the sprinter was 2.2 m/s</em>

Explanation:

<u>Constant Acceleration Motion</u>

It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.

The following relation applies:

$v_f=v_o+at$

Where a is the constant acceleration, vo the initial speed, vf the final speed, and t the time.

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3 years ago

There are four charges, each with a magnitude of 2.06 µC. Two are positive and two are negative. The charges are fixed to the co

mars1129 [50]

Answer:

0.208 N

Explanation:

We are given that

$q_1=q_2=2.06\mu C=2.06\times 10^{-6} C$

$q_3=q_4=-2.06\mu C=-2.06\times 10^{-6} C$

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The magnitude of the net electrostatic force experienced by any charge at point 4

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$F_{net}=\sqrt{F^2+F^2+0}-F_2$

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$F_2=\frac{Kq^2}{2d^2}$

$F_{net}=\frac{\sqrt 2kq^2}{d^2}-\frac{kq^2}{2d^2}=\frac{kq^2}{d^2}(\sqrt 2-\frac{1}{2})$

Where $k=9\times 10^9$

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3 years ago

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malfutka [58]

Answer:

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Answer:

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Explanation:

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