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Pepsi [2]
3 years ago
8

Two small spheres spaced 20.0 cm apart have equal charge. how many excess electrons must be present on each sphere if the magnit

ude of the force of repulsion between them is 4.57 * 10-21 n?
Physics
1 answer:
uysha [10]3 years ago
8 0
<span>890 Coulomb's law is F = k * q1 * q2 / r^2 where F = force k = Coulomb's constant 8.99x10^9 N m^2/C^2 q1,q2 = signed charges r = distance between charges Since the charges are the same, let's simplify the equation, solve for q, then substitute the known values and calculate. F = k * q1 * q2 / r^2 F = k * q^2 / r^2 F*r^2 = k * q^2 F*r^2/k = q^2 sqrt(F*r^2/k) = q sqrt(4.57x10^-21 N * (0.2m)^2 / (8.99x10^9 N m^2/C^2)) = q sqrt((4.57x10^-21 N * 0.04m^2) / (8.99x10^9 N m^2/C^2)) = q sqrt((1.828x10^-22 N*m^2) / (8.99x10^9 N m^2/C^2)) = q sqrt(2.03337x10^-32 C^2) = q 1.42596x10^-16 C = q So each sphere has to have an excess of 1.42596x10^-16 Coulombs of electrons. A coulomb is 6.24150934x10^18 electrons, so let's do the multiplication: 1.42596x10^-16 * 6.24150934x10^18 = 8.9001426584664x10^2 = 890 So each sphere has an extra 890 electrons.</span>
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The water-ice particles forming Saturn's rings are frozen together into a thin sheet that rotates around Saturn like a solid bod
vaieri [72.5K]

Answer:

B. False

Explanation:

According to research by several scientists, Saturn's rings aren't solid, as they appear from Earth.  They are actually made up of floating chunks of water ice, rocks and dust that range in diferent sizes from specks to enormous, even house-sized pieces that orbit Saturn in a ring pattern.

4 0
3 years ago
A square wave has amplitude 0 V for the low voltage and 4 V for the high voltage. Calculate the average voltage by integrating o
Margarita [4]

Answer:

V_{average} = \frac{1}{2}  V_o  ,     V_{average} = 2 V

Explanation:

he average or effective voltage of a wave is the value of the wave in a period

            V_average = ∫ V dt

in this case the given volage is a square wave that can be described by the function

           V (t) = \left \{ {{V=V_o \ \ \  t<  \tau /2} \atop {V=0 \ \  \ \  t> \tau /2 }   } \right.

to substitute in the equation let us separate the into two pairs

             V_average = \int\limits^{1/2}_0 {V_o} \, dt + \int\limits^1_{1/2} {0} \, dt

             V_average = V_o \ \int\limits^{1/2}_0 {} \, dt

             V_{average} = \frac{1}{2}  V_o

we evaluate  V₀ = 4 V

             V_{average} = 4 / 2)

             V_{average} = 2 V

6 0
3 years ago
20 cm long 10 cm wide and 5 cm thick as a mass of 500 g determine the greatest pressure that can be exerted by block on the flat
uysha [10]

100000 Pascal

Explanation:

pressure= force/area

Max pressure= force/min area

so f=5

min area= 5×10^-5

5÷5*10^-5 = 100000pascal

8 0
2 years ago
Passengers on an airplane move from rest to 86 meters per second before the airplane takes off. If the airplane takes 100 second
VikaD [51]

Utilize the formula:  V _{f} = V _{i} + a\Delta t

V _{f} = Final Velocity (86 m/s)

V _{f} = Initial Velocity (0 m/s)

a = acceleration (m/s²)

\Deltat = Time (100 seconds)

As a result,

86 m/s = 0 + (a)(100 seconds)

Using algebra, divide 86 m/s by 100 seconds:

86 m/s = 100a

a = 0.86 m/s²

Rounded to one decimal place: 0.9 m/s²

Let me know if you have any questions!

4 0
3 years ago
A 97.6-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =
Mila [183]

Answer:

v=6.65m/sec

Explanation:

From the Question we are told that:

Mass m=97.6

Coefficient of kinetic friction  \mu k=0.555

Generally the equation for Frictional force is mathematically given by

 F=\mu mg

 F=0.555*97.6*9.8

 F=531.388N

Generally the  Newton's equation for Acceleration due to Friction force is mathematically given by

 a_f=-\mu g

 a_f=-0.555 *9.81

 a_f=-54455m/sec^2

Therefore

 v=u-at

 v=0+5.45*1.22

 v=6.65m/sec

4 0
3 years ago
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