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Juliette [100K]
4 years ago
8

Two balloons are charged with an identical quantity and type of charge: -4 nc They are held apart at a separation distance of 70

cm. Determine the magnitude of the electrical force of repulsion between them.
Physics
1 answer:
blagie [28]4 years ago
8 0

Answer:

29.4 uN

Explanation:

The electric force between two charges can be calculated using Coulomb's Law. According to this law the force between two point charges is given as:

F=k\frac{q_{1} q_{2} }{r^{2}}

where k is a proportionality constant known as the Coulomb's law constant. Its value is 9 \times 10^{9} Nm²/C²

r = distance between charges = 70 cm = 0.7 m

q1 = q2 = 4nC = 4 \times 10^{-9} C

The negative sign indicates that the charges are negative. In the formula we will only use the magnitude of the charges.

Using these values in the formula, we get:

F=9 \times 10^{9} \times \frac{4 \times 10^{-9} \times 4 \times 10^{-9}}{0.7^{2}}\\\\ F=2.94\times10^{-7} N\\\\ F=29.4 \times 10^{-6} N\\\\ F=29.4 \mu N

Therefore, the magnitude of repulsive force between the given charges will be 29.4 uN

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barxatty [35]
An ice cube would transfer heat to another object whose temperature
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If you could find a block of solid nitrogen, its temperature would be
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6 0
4 years ago
Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.39 m and exactly 8 cycles are co
aivan3 [116]

Answer:

1.195 m

2.8375 s

2.21433 rad/s

Explanation:

d = Distance = 2.39 m

N = Number of cycles = 8

t = Time to complete 8 cycles = 22.7 s

Radius would be equal to the distance divided by 2

r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m

The radius is 1.195 m

Time period would be given by

T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s

Time period of the motion is 2.8375 s

Angular speed is given by

\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s

The angular speed of the motion is 2.21433 rad/s

4 0
4 years ago
A uniform horizontal beam 4.0 m long and weighing 200 N is attached to the wall by a pin connection that allows the beam to rota
wariber [46]

Answer:

The magnitude of the tension in the cable, T is 1,064.315 N

Explanation:

Here we have

Length of beam = 4.0 m

Weight = 200 N

Center of mass of uniform beam = mid-span = 2.0 m

Point of attachment of cable = Beam end = 4.0 m

Angle of cable = 53° with the horizontal

Tension in cable = T

Point at which person stands = 1.50 m from wall

Weight of person = 350 N

Therefore,

Taking moment about the wall, we have

∑Clockwise moments = ∑Anticlockwise moments

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T = 850/sin(53)  = ‭1,064.315 N.

4 0
3 years ago
A horizontal pipe contains water at a pressure of 110 kPa flowing with a speed of 1.4 m/s. When the pipe narrows to one half its
Pavel [41]

Answer:

a

  v_2 =  5.6 \  m/s

b

   P_2 = 80600 \  Pa

Explanation:

From the question we are told that  

     The pressure of the water in the pipe is  P_1= 110 \  kPa  =  110 *10^{3 } \  Pa

      The speed of the water  is v_1 =  1.4 \  m/s

       The original area of the pipe is  A_1 =  \pi \frac{d^2 }{4}

       The  new area of the pipe is  A_2 = \pi *  \frac{[\frac{d}{2} ]^2}{4}  =  \pi *  \frac{\frac{d^2}{4} }{4} = \pi \frac{d^2}{16}

         

Generally the continuity equation is mathematically represented as

       A_1 *  v_1 =  A_2 * v_2

Here v_2 is the new velocity  

So

        \pi * \frac{d^2}{4}   *  1.4  = \pi * \frac{d^2}{16}   * v_2

=>     \frac{d^2}{4}   *  1.4  =  \frac{d^2}{16}   * v_2

=>    d^2    *  1.4  =  \frac{d^2}{4}   * v_2

=>    1.4  = 0.25    * v_2

=>     v_2 =  5.6 \  m/s

Generally given that the height of the original pipe and the narrower pipe are the same , then we will b making use of the  Bernoulli's equation for constant height to calculate the pressure

This is mathematically represented as

       

             P_1 + \frac{1}{2}  *  \rho *  v_1 ^2  =  P_2 + \frac{1}{2}  *  \rho *  v_2 ^2

Here \rho is the density of water with value  \rho =  1000  \  kg /m^3

             P_2 =  P_1 + \frac{1}{2} *  \rho [ v_1^2 - v_2^2 ]

=>          P_2 =  110 *10^{3} + \frac{1}{2} *  1000 *  [ 1.4 ^2 - 5.6 ^2 ]

=>          P_2 = 80600 \  Pa

4 0
3 years ago
What is the speed of a wave that has a frequency of 25Hz and a wavelength of 3m
Arada [10]
The answer to the question is:

75m/s

Just do 25*3
4 0
3 years ago
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