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Mars2501 [29]
3 years ago
9

When a voltage difference is applied to a piece of metal wire, a 5.0 mA current flows through it. If this metal wire is now repl

aced with a silver wire having twice the diameter of the original wire, how much current will flow through the silver wire? The lengths of both wires are the same, and the voltage difference remains unchanged. (The resistivity of the original metal is 1.68 � 10-8 ? ? m, and the resistivity of silver is 1.59 � 10-8 ? ? m.)
Answer
5.3 mA
11 mA
19 mA
21 mA
Physics
1 answer:
zheka24 [161]3 years ago
5 0

Answer:

I = 21.13 mA ≈ 21 mA

Explanation:

If

I₁ = 5 mA

L₁ = L₂ = L

V₁ = V₂ = V

ρ₁ = 1.68*10⁻⁸ Ohm-m

ρ₂ = 1.59*10⁻⁸ Ohm-m

D₁ = D

D₂ = 2D

S₁ = 0.25*π*D²

S₂ = 0.25*π*(2*D)² = π*D²

If we apply the equation

R = ρ*L / S

where (using Ohm's Law):

R = V / I

we have

V / I = ρ*L / S

If V and L are the same

V / L =  ρ*I / S

then

(V / L)₁ = (V / L)₂  ⇒     ρ₁*I₁ / S₁ = ρ₂*I₂ / S₂

If

S₁ = 0.25*π*D²   and

S₂ = 0.25*π*(2*D)² = π*D²

we have

ρ₁*I₁ / (0.25*π*D²) = ρ₂*I₂ / (π*D²)

⇒    I₂ = 4*ρ₁*I₁ / ρ₂

⇒     I₂ = 4*1.68*10⁻⁸ Ohm-m*5 mA / 1.59*10⁻⁸ Ohm-m

⇒     I₂ = 21.13 mA

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Answer:

Hello your question is incomplete hence I will give you a general answer on how A van de Graaff generator works

answer :

If the electrons falls through a PD of 150mV the electron will gain energy of   150MeV

Explanation:

when a Van de Graff generator is used to accelerate an electron through a PD ( potential difference ) of any value the particle ( electron )  the electron will gain energy  ( eV )  which is is equivalent in value of the PD it accelerated through

hence if the electrons falls through a PD of 150mV the electron will gain energy of   150MeV

4 0
3 years ago
A mechanic pushes a 3540 kg car from rest to a speed of v, doing 4864 J of work in the process. Find the speed v. Neglect fricti
topjm [15]

Answer:

1.66 m/s

Explanation:

Work or kinetic energy = \frac{1}{2} mv^{2}

4864=\frac{1}{2} (3540)v^{2}

v = 1.66 m/s

6 0
3 years ago
What is the kinetic energy of a 1130 kg truck that is moving with a velocity of 40 m/s?
Makovka662 [10]

Answer:

The answer is 904,000.

Kinetic energy=1/2mv^2.

1/2×1130×40^2.

1/2×1808000=904,000Joules.

8 0
3 years ago
A jet plane is launched from a catapult on an aircraft carrier. In 2.0 s it reaches a speed of 42 m/s at the end of the catapult
Dvinal [7]

Answer:

Explanation:

Given

time taken t=2\ s

Speed acquired in 2 sec v=42\ m/s

Here initial velocity is zero u=0

acceleration is the rate of change of velocity in a given time

a=\frac{v-u}{t}

a=\frac{42-0}{2}=21\ m/s^2

Distance travel in this time

s=ut+0.5at^2

where  

s=displacement

u=initial velocity

a=acceleration

t=time

s=0+\0.5\times 21\times (2)^2

s=42\ m

so Jet Plane travels a distance of 42 m in 2 s                                

5 0
3 years ago
A transformer has 18 turns of wire in its primary coil and 90 turns in its secondary coil. An alternating voltage with an effect
Fantom [35]

Answer:

I_s=5.8A

Explanation:

Not considering any type of losses in the transformer, the input power in the primary is equal to the output power in the secondary:

P_p=P_s

So:

V_p*I_p=V_s*I_s

Where:

V_p=Voltage\hspace{3}in\hspace{3}the\hspace{3}primary\hspace{3}coil\\V_s=Voltage\hspace{3}in\hspace{3}the\hspace{3}secondary\hspace{3}coil\\I_p=Current\hspace{3}in\hspace{3}the\hspace{3}primary\hspace{3}coil\\I_s=Current\hspace{3}in\hspace{3}the\hspace{3}secondary\hspace{3}coil

Solving for I_s

I_s=\frac{V_p*I_p}{V_s}

Replacing the data provided:

I_s=\frac{110*29}{550} =5.8A

4 0
3 years ago
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