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VARVARA [1.3K]
3 years ago
15

Two tiny conducting spheres are identical and carry charges of -20μC and +50μC. They are separeted by a distance of 2.50cm. (a)

what is the magnitude of the force that each sphere each sphere experience, and is the force attractive or repulsive ? (b) The spheres are brought into contact and then separated toa distance of 2.50cm. Determine the magnitude of the force that each sphere now experiences, and state whether the force is attractive or repulsive.
Physics
1 answer:
Dahasolnce [82]3 years ago
7 0

Answer:

14400\ \text{N}, Attractive

3240\ \text{N}, Repulsive

Explanation:

q_1 = -20 μC

q_2 = 50 μC

r = Distance between the charges = 2.5 cm

k = Coulomb constant = 9\times 10^9\ \text{Nm}^2/\text{C}^2

Electrical force is given by

F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{9\times 10^9\times (-20\times 10^{-6})\times (50\times 10^{-6})}{(2.5\times10^{-2})^2}\\\Rightarrow F=-14400\ \text{N}

The magnitude of force each sphere will experience is 14400\ \text{N}

Since the charges have opposite charges they will attract each other.

Now the charges are brought into contact with each other so the resultant charge will be

q=\dfrac{q_1+q_2}{2}\\\Rightarrow q=\dfrac{-20+50}{2}\\\Rightarrow q=15\ \mu\text{C}

F=\dfrac{kq^2}{r^2}\\\Rightarrow F=\dfrac{9\times 10^9\times (15\times 10^{-6})^2}{(2.5\times 10^{-2})^2}\\\Rightarrow F=3240\ \text{N}

The magntude of the force the spheres experience will be 3240\ \text{N}

The spheres have the same charge now so they will repel each other.

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a.  4 m/s b. 0.2 V

Explanation:

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Since V = vBd

v = V/Bd    given that V = 60.0 mV = 0.060 V, substituting the values of the other variables, we have

v = 0.060 V/(0.500 T × 0.03 m)

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b. What would the Hall voltage be for the same flow rate through a 10.0-cm-diameter pipe with the same field applied?

Since the hall voltage, V = vBd and v = flow-rate = 4 m/s, B = magnetic field strength = 0.500 T and d' = diameter of pipe = 10.0 cm = 0.10 m

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V = 0.2 V

6 0
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Question 7 of 25
tiny-mole [99]

Answer:

2PBr₃ + 3Cl₂ → 2PCl₃ + 3Br₂

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Explanation:

A balanced chemical equation is a chemical equation that have an equal number of elements of each type on both sides of the equation

Among the given chemical reactions, we have;

2PBr₃ + 3Cl₂ → 2PCl₃ + 3Br₂

In the above reaction;

The number of phosphorus, P, on either side of the equation = 2

The number of bromine atoms, Br, on either side of the equation = 6

The number of chlorine atoms, Cl, on either side of the equation = 6

Therefore, the number of elements in the reactant side and products side of the reaction are equal and the reaction is balanced

The second balanced chemical reaction is 2Na + MgCl₂ → 2NaCl + Mg

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3 years ago
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995 N

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Gravitational constant, g, is taken as 9.81 hence mass, m of surface is W/g where W is weight of surface

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Using radius of orbit of 6371km

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Where G=6.67*10^{-11} and M=5.94*10^{24}

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F= 995.01142 then rounded off

F=995N

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