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4 years ago

What is the repulsive force between two pith balls that are 9.00 cm apart and have equal charges of -28.0 nC?

Physics

1 answer:

Alexus [3.1K]4 years ago

7 0

Answer:

Force, $F=8.71\times 10^{-4}\ N$

Explanation:

Given that,

Charges on pith balls, $q_1=q_2=-28\ nC=-28\times 10^{-9}\ C$

Distance between balls, d = 9 cm = 0.09 m

Let F is the repulsive force between two pith balls. We know that the repulsive force between two charges is given by :

$F=k\dfrac{q_1^2}{d^2}$

$F=9\times 10^9\times \dfrac{(-28\times 10^{-9})^2}{(0.09)^2}$

F = 0.000871 N

$F=8.71\times 10^{-4}\ N$

So, the repulsive force between the pith balls is $8.71\times 10^{-4}\ N$ . Hence, this is the required solution.

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A rope exerts a force F on a 20.0-kg crate. The crate starts from rest and accelerates upward at 5.00 m/s2 near the surface of t

tia_tia [17]

Answer:

400 J

Explanation:

Given:

Δy = 4.00 m

v₀ = 0 m/s

a = 5.00 m/s²

Find: v²

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (5.00 m/s²) (4.00 m)

v² = 40.0 m²/s²

Find KE:

KE = ½ mv²

KE = ½ (20.0 kg) (40.0 m²/s²)

KE = 400 J

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3 years ago

A skateboarder travels on a horizontal surface with an initial velocity of 3.6 m/s toward the south and a constant acceleration

Dimas [21]

Answer:

a) The x-position of the skateboarder is 0.324 meters.

b) The y-position of the skateboarder is -2.16 meters.

c) The x-velocity of the skateboard is 1.08 meters per second.

d) The y-velocity of the skateboard is -3.6 meters per second.

Explanation:

a) The x-position of the skateboarder is determined by the following expression:

$x(t) = x_{o} + v_{o,x}\cdot t + \frac{1}{2}\cdot a_{x} \cdot t^{2}$ (1)

Where:

$x_{o}$ - Initial x-position, in meters.

$v_{o,x}$ - Initial x-velocity, in meters per second.

$t$ - Time, in seconds.

$a_{x}$ - x-acceleration, in meters per second.

If we know that $x_{o} = 0\,m$ , $v_{o,x} = 0\,\frac{m}{s}$ , $t = 0.60\,s$ and $a_{x} = 1.8\,\frac{m}{s^{2}}$ , then the x-position of the skateboarder is:

$x(t) = 0\,m + \left(0\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(1.8\,\frac{m}{s^{2}} \right) \cdot (0.60\,s)^{2}$

$x(t) = 0.324\,m$

The x-position of the skateboarder is 0.324 meters.

b) The y-position of the skateboarder is determined by the following expression:

$y(t) = y_{o} + v_{o,y}\cdot t + \frac{1}{2}\cdot a_{y} \cdot t^{2}$ (2)

Where:

$y_{o}$ - Initial y-position, in meters.

$v_{o,y}$ - Initial y-velocity, in meters per second.

$t$ - Time, in seconds.

$a_{y}$ - y-acceleration, in meters per second.

If we know that $y_{o} = 0\,m$ , $v_{o,y} = -3.6\,\frac{m}{s}$ , $t = 0.60\,s$ and $a_{y} = 0\,\frac{m}{s^{2}}$ , then the x-position of the skateboarder is:

$y(t) = 0\,m + \left(-3.6\,\frac{m}{s} \right)\cdot (0.60\,s) + \frac{1}{2}\cdot \left(0\,\frac{m}{s^{2}}\right)\cdot (0.60\,s)^{2}$

$y(t) = -2.16\,m$

The y-position of the skateboarder is -2.16 meters.

c) The x-velocity of the skateboarder ( $v_{x}$ ), in meters per second, is calculated by this kinematic formula:

$v_{x}(t) = v_{o,x} + a_{x}\cdot t$ (3)

If we know that $v_{o,x} = 0\,\frac{m}{s}$ , $t = 0.60\,s$ and $a_{x} = 1.8\,\frac{m}{s^{2}}$ , then the x-velocity of the skateboarder is:

$v_{x}(t) = \left(0\,\frac{m}{s} \right) + \left(1.8\,\frac{m}{s} \right)\cdot (0.60\,s)$

$v_{x}(t) = 1.08\,\frac{m}{s}$

The x-velocity of the skateboard is 1.08 meters per second.

d) As the skateboarder has a constant y-velocity, then we have the following answer:

$v_{y} = -3.6\,\frac{m}{s}$

The y-velocity of the skateboard is -3.6 meters per second.

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3 years ago

I need a detailed explanation on What is nuclear fusion is and what company’s are using it

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Answer: here are some of the places that use nuclear fusion

Commonwealth Fusion Systems (CFS)

TAE Technologies.

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2 years ago

Please help...........................

Marianna [84]

Hi!

Neutrons are neutral, which means they don't exactly have an electrical charge. It's because of this neutral charge that it is represented with a '0'.

On the other hand, protons and electrons <em>do </em>have electrical charges. Electrons flow around the outside of the nucleus, with a negative charge.

Protons are stored in the nucleus with the neutrons, holding a positive charge.

Hopefully, this helps! =)

6 0

4 years ago

Read 2 more answers

An object accelerates from 20 m/s down to 10 m/s in 5 seconds.

DedPeter [7]

Answer:

The acceleration of the object is $-2\ \frac{m}{sec^2}$