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Goshia [24]
3 years ago
6

What is the repulsive force between two pith balls that are 8.00 cm apart and have equal charges of – 30.0 nC?

Physics
1 answer:
Nastasia [14]3 years ago
8 0

Answer:

F=1.26*10^{-3}N

Explanation:

Assuming the pith balls as point charges, we can calculate the repulsive force between them, using Coulomb's law:

F=\frac{kq_1q_2}{d^2}

We observe that the magnitude of the electric force is directly proportional to the product of the magnitude of both signed charges(q_1,q_2) and inversely proportional to the square of the distance(d) that separates them.

Replacing the given values, where k is the Coulomb constant:

F=\frac{8.99*10^{9}\frac{N\cdot m^2}{C^2}(-30*10^{-9}C)(-30*10^{-9}C)}{(8*10^{-2}m)^2}\\F=1.26*10^{-3}N

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A bug on the surface of a pond is observed to move up and down a total vertical distance of 6.5 cm , from the lowest to the high
m_a_m_a [10]

Answer:

factor that bug maximum KE change is 0.52284

Explanation:

given data

vertical distance = 6.5 cm

ripples decrease to =  4.7 cm

solution

We apply here formula for the KE of particle that executes the simple harmonic motion that is express as

KE = (0.5) × m × A² × ω²     .................1

and kinetic energy is  directly proportional to square of the amplitude.

so

\frac{KE2}{KE1} =  \frac{A2^2}{A1^2}      .............2

\frac{KE2}{KE1} = \frac{4.7^2}{6.5^2}

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5 0
3 years ago
A 20 kg wagon is rolling to the right across a floor. A person attempts to catch and stop the crate and applies a force of 70 N,
Serggg [28]

Answer:

1.736m/s²

Explanation:

According to Newton's second law;

\sum F_x = ma_x\\

Fm - Ff = ma_x\\ where;

Fm is the moving force = 70.0N

Ff is the frictional force acting on the body

Ff = \mu R\\Ff = \mu mg\\

\mu is the coefficient of friction

m is the mass of the object

g is the acceleration due to gravity

a is the acceleration/deceleration

The equation becomes;

Fm - Ff = ma_x\\Fm - \mu mg = ma\\

Substitute the given parameters

Fm - \mu mg = ma\\70 - 0.18(20)(9.8) = 20a\\70-35.28 = 20a\\34.72 = 20a\\a = \frac{34.72}{20}\\a =  1.736m/s^2\\

Hence the deceleration rate of the wagon as it is caught is 1.736m/s²

7 0
3 years ago
The Earth's Radius is 6.3710x106 m and mass is 5.9742x1024 kg. What is the acceleration due to gravity at Mount Everest (elevati
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Answer is

9.773m/s^2

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Given,

h=8848m

The value of sea level is 9.08m/s^2. So, Let g′ be the acceleration due to the gravity on Mount Everest.

g′=g(1 − 2h/h)

=9.8(1 - 6400000/17696)

=9.8(1 − 0.00276)

9.8×0.99724

=9.773m/s^2

Thus, the acceleration due to gravity on the top of Mount Everest is =9.773m/s^2

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hope this helps :)

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