All categories

3 years ago

What current flows through a 2.54cm diameter rod of pure silicon that is 20cm long when 1000V is applied?

Physics

1 answer:

vfiekz [6]3 years ago

3 0

Answer: $0.0039\ A$

Explanation:

Given

Diameter of the rod $d=2.54\ cm$

length of rod is $l=20\ cm$

Resistivity of silicon is $\rho=6.4\times 10^2\ \Omega-m$

cross-section of the rod $A$

$\Rightarrow A=\dfrac{\pi d^2}{4}\\\\\Rightarrow A=\dfrac{3.142\times 2.54^2\times 10^{-4}}{4}\\\\\Rightarrow A=5.067\times 10^{-4}\ m^2$

Resistance of rod is R

$\Rightarrow R=\dfrac{\rho l}{A}$

$\Rightarrow R=\dfrac{640\times 0.20}{5.067\times 10^{-4}}\\\\\Rightarrow R=25.26\times 10^4\ \Omega$

Current is given by

$\Rightarrow I=\dfrac{V}{R}\\\\\Rightarrow I=\dfrac{1000}{25.26\times 10^4}\\\\\Rightarrow I=0.0039\ A$

You might be interested in

- Explain why an aircraft will tend to lose height

olganol [36]

Answer:

the planes wings are lifting at an angle to gravity so the plane isn't lifting as much against gravity when it banks. some of the wing lift is going into turning the plane. :) so it needs more lift to bank and stay up

Explanation:

8 0

3 years ago

mario62 [17]

Answer:

An increase in air temperature because of its compression.

Explanation:

The Gay-Lussac's Law states that a gas pressure is directly proportional to its temperature in an enclosed system to constant volume.

$P = kT$

where P: is the gas pressure, T: is the gas temperature and k: is a constant.

Therefore, due to Gay-Lussac's Law, when the plunger is pushed down very rapidly, the pressure of the air increase, which leads to its temperature increase. That is why cotton flashes and burns.

I hope it helps you!

3 0

3 years ago

Light travels _______ in a material with a higher index of refraction

exis [7]

Light will travel more slowly in a material with a higher index of refraction

4 0

3 years ago

Read 2 more answers

A block of 250-mm length and 54 × 40-mm cross section is to support a centric compressive load P. The material to be used is a b

musickatia [10]

Answer:

P = 17.28*10⁶ N

Explanation:

Given

L = 250 mm = 0.25 m

a = 0.54 m

b = 0.40 m

E = 95 GPa = 95*10⁹ Pa

σmax = 80 MPa = 80*10⁶ Pa

ΔL = 0.12%*L = 0.0012*0.25 m = 3*10⁻⁴ m

We get A as follows:

A = a*b = (0.54 m)*(0.40 m) = 0.216 m²

then, we apply the formula

ΔL = P*L/(A*E) ⇒ P = ΔL*A*E/L

⇒ P = (3*10⁻⁴ m)*(0.216 m²)*(95*10⁹ Pa)/(0.25 m)

⇒ P = 24624000 N = 24.624*10⁶ N

Now we can use the equation

σ = P/A

⇒ σ = (24624000 N)/(0.216 m²) = 114000000 Pa = 114 MPa > 80 MPa

So σ > σmax we use σmax

⇒ P = σmax*A = (80*10⁶ Pa)*(0.216 m²) = 17280000 N = 17.28*10⁶ N

7 0

3 years ago

The barricade at the end of a subway line has a large spring designed to compress 2.00 m when stopping a 1.10 ✕ 105 kg train mov

Mrac [35]

Answer:

(a) k = 1684.38 N/m = 1.684 KN/m

(b) Vi = 0.105 m/s

Explanation:

(a)

First, we find the deceleration of the car. For that purpose we use 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = ?

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = 0.35 m/s

s = distance covered by train before stopping = 2 m

Therefore,

a = [(0 m/s)² - (0.35 m/s)²]/(2)(2 m)

a = 0.0306 m/s²

Now, we calculate the force applied on spring by train:

F = ma

F = (1.1 x 10⁵ kg)(0.0306 m/s²)

F = 3368.75 N

Now, for force constant, we use Hooke's Law:

F = kΔx

where,

k = Force Constant = ?

Δx = Compression = 2 m

Therefore.

3368.75 N = k(2 m)

k = (3368.75 N)/(2 m)

k = 1684.38 N/m = 1.684 KN/m

(c)

Applying Hooke's Law with:

Δx = 0.6 m

F = (1684.38 N/m)(0.6 m)

F = 1010.62 N = 1.01 KN

(b)

Now, the acceleration required for this force is:

F = ma

1010.62 N = (1.1 kg)a

a = 1010.62 N/1.1 x 10⁵ kg

a = 0.0092 m/s²

Now, we find initial velocity of train by using 3rd equation of motion:

2as = Vf² - Vi²

a = (Vf² - Vi²)/2s

where,

a = deceleration = -0.0092 m/s² (negative sign due to deceleration)

Vf = final velocity = 0 m/s (since, train finally stops)

Vi = Initial Velocity = ?

s = distance covered by train before stopping = 0.6 m

Therefore,

-0.0092 m/s² = [(0 m/s)² - Vi²]/(2)(0.6 m)

Vi = √(0.0092 m/s²)(1.2 m)

Vi = 0.105 m/s

4 0

4 years ago