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Dahasolnce [82]

3 years ago

7

Both segments of the wire are made of the same metal. Current I1 flows into segment 1 from the left. How does current I1 in segm

ent 1 compare to current 12 in segment 27 A) h 12 B) l12 C) licl D) Not enough information is provided

1 answer:

Andrei [34K]3 years ago

8 0

Answer:

current I1 = current I2

Explanation:

since the wire is made up of the same material, from Kirchoff's current law sum of currents entering a particular node or segment of wire is equals to the sum of currents leaving that particular node or segment of wire

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Should a flat grassland or a hillside have deeper soil? Explain your answer.

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A hillside of course my friend

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which of the following locations has the largest electric field? A)0.4 cm from a +3.0 uC point charge B)0.2 cm frm a +1.5 uC poi

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3 years ago

What are the features of image formed by convex mirror.

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3 years ago

Read 2 more answers

Two small nonconducting spheres have a total charge of 90.0 C.

valentina_108 [34]

Answer: (a) Smaller charge is $2.7 \times 10^{-5} C$ and larger charge is $11.7 \times 10^{-5} C$ .

(b) Smaller charge is $-11.4 \times 10^{-5}$ and larger charge is $9.1 \times 10^{-5}$ .

Explanation:

(a) When both the spheres have same charge then force is repulsive in nature as like charges tend to repel each other.

Therefore, total charge on the two non-conducting spheres will be calculated as follows.

$Q_{1} + Q_{2} = 90 \mu \times \frac{10^{-6}C}{1 \muC}$

= $9 \times 10^{-5} C$

Therefore, force between the two spheres will be calculated as follows.

F = $k\frac{Q_{1}Q_{2}}{r^{2}}$

12 N = $\frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}$

$Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}$

or, $Q_{1}(9 \times 10^{-5} - Q_{1}) = 0.104 \times 10^{-9} C^{2}$

$9 \times 10^{-5}Q_{1} - Q^{2}_{1} = 0.104 \times 10^{-9} C^{2}$

$Q^{2}_{1} - 9 \times 10^{-5}Q_{1} + 0.104 \times 10^{-9} = 0$

$Q_{1} = 11.7 \times 10^{-5} C, 2.7 \times 10^{-5} C$

This means that smaller charge is $2.7 \times 10^{-5} C$ and larger charge is $11.7 \times 10^{-5} C$ .

(b) When force is attractive in nature then it means both the charges are of opposite sign.

Hence, total charge on the non-conducting sphere is as follows.

$Q_{1} + (-Q_{2}) = 90 \mu \times \frac{10^{-6}C}{1 \muC}$

$Q_{1} - Q_{2} = 9 \times 10^{-5} C$

Now, force between the two spheres is calculated as follows.

F = $k\frac{Q_{1}Q_{2}}{r^{2}}$

12 N = $\frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}$

$Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}$

$Q_{1}(Q_{1} - 9 \times 10^{-5}) = 0.104 \times 10^{-9} C^{2}$

$Q^{2}_{1} - 9 \times 10^{-5}Q_{1} = 0.104 \times 10^{-9} C^{2}$

$Q_{1} = -11.4 \times 10^{-5}, 9.1 \times 10^{-5}$

Hence, smaller charge is $-11.4 \times 10^{-5}$ and larger charge is $9.1 \times 10^{-5}$ .

8 0

3 years ago

How long does it take electrons to get from

OlgaM077 [116]

We need to find the time it takes an electron to move in the given circuit.

The time taken for electrons to reach the starting motor from the battery is 60.65 minutes.

I = Current = 134 A

$N_A$ = Avogadro's number = $6.022\times 10^{23}\ \text{mol}^{-1}$

A = Area = $38.9\ \text{mm}^2$

L = Length = 92.2 cm

$\rho$ = Density of copper = $8960\ \text{kg/m}^3$

M = Molar mass of copper = 63.5 g/mol

$n_v$ = Number of valence electrons of copper = 1

e = Charge of electron = $1.6\times 10^{-19}\ \text{C}$

Number of charge carriers per unit volume is given by

$n=\dfrac{\rho N_An_v}{M}\\\Rightarrow n=\dfrac{8960\times 6.022\times 10^{23}\times 1}{63.5\times 10^{-3}}\\\Rightarrow n=8.497\times 10^{28}\ \text{m}^{-3}$

Time taken is given by

$t=\dfrac{LAne}{I}\\\Rightarrow t=\dfrac{92.2\times 10^{-2}\times 38.9\times 10^{-6}\times 8.497\times 10^{28}\times 1.6\times 10^{-19}}{134}=3638.83\ \text{s}\\\Rightarrow t=\dfrac{3638.83}{60}=60.65\ \text{minutes}$

The time taken for electrons to reach the starting motor from the battery is 60.65 minutes.

Learn more:

brainly.com/question/1426683

brainly.com/question/170663

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2 years ago