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Marrrta [24]
3 years ago
11

"Two identical positive charges exert a repulsive force of 6.7 × 10−9 N when separated by a distance 3.5 × 10−10 m. Calculate th

e charge of each. The Coulomb constant is 8.98755 × 109 N · m2 /C 2 . Answer in units of C."
Physics
1 answer:
Tamiku [17]3 years ago
8 0

Answer:

The charge of each charge is 3.02\times10^{-19} C

Explanation:

When yo have two charged particles they interact exerting an electrostatic force in the other particles, the magnitude of the electrostatic force between two particles is:

F_{e}=k\frac{\mid q_{1}q_{2}\mid}{r^{2}} (1)

with q1 and q2 the charges, r the distance between them and k the Coulomb's constant (k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}})

Because the charges we're dealing are identical positive q1=q2, then (1) is:

F_{e}=k\frac{\mid q^2 \mid}{r^{2}}

Using the values the problem give us:

6.7\times10^{-9}=8.98755\times10^{9}\frac{\mid q^2 \mid}{(3.5\times10^{-10})^{2}}

solving for q:

q=\sqrt{\frac{(3.5\times10^{-10})^{2}6.7\times10^{-9}}{8.98755\times10^{9}}}

q= 3.02\times10^{-19} C

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Answer:

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2 years ago
If an object suspended by a scale shows a weight of 3 N in air, and 2 N when submerged in water, the buoyant force on the submer
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Answer:

1 N

Explanation:

Buoyant Force: This is also called upthrust, It can be defined as the force which act upward exerted by a fluid when an object is placed in it.

The S.I unit is Newton.

From the question,

Buoyant force = Weight of the object in air- weight of the object when submerged in water.

U = W-W'.......................... Equation 1

Where U = upthrust, W = weight in air, W' = weight when submerged in water.

Given: W = 3 N, W' = 2 N

Substitute into equation 1

U = 3-2

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3 years ago
A 13 kg hanging sculpture is suspended by a 95-cm-long, 5.0 g steel wire. When the wind blows hard, the wire hums at its fundame
Artyom0805 [142]

Answer:

f=81.96 \ Hz

Explanation:

Givens

L=95cm

m_{sculpture} =13kg

m_{wire}=5g

The frequency is defined by

f=\frac{v}{\lambda}

Where v is the speed of the wave in the string and \lambda is its wave length.

The wave length is defined as \lambda = 2L = 2(0.95m)=1.9m

Now, to find the speed, we need the tension of the wire and its linear mass density

v=\sqrt{\frac{T}{\mu} }

Where \mu=\frac{0.005kg}{0.95m}= 5.26 \times 10^{-3} and the tension is defined as T=m_{sculpture} g=13kg(9.81 m/s^{2} )=127.53N

Replacing this value, the speed is

v=\sqrt{\frac{127.53N}{5.26 \times 10^{-3} } }=155.71 m/s

Then, we replace the speed and the wave length in the first equation

f=\frac{v}{\lambda}\\f=\frac{155.71 m/s}{1.9m}\\ f=81.96Hz

Therefore, the frequency is f=81.96 \ Hz

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3 years ago
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Answer:

The bonds can shift because valence electrons are held loosely and more freely

Explanation:

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An object moving at a constant velocity will always have a
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It will always have a zero acceleration

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