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Step2247 [10]
2 years ago
10

The mean time between collisions for electrons in room temperature copper is 2.5 x 10-14 s. What is the electron current in a 2

mm diameter copper wire where the internal electric field strength is 0.01 V/m?
Physics
1 answer:
Darya [45]2 years ago
4 0

Answer:

1.87 A

Explanation:

τ = mean time between collisions for electrons = 2.5 x 10⁻¹⁴ s

d = diameter of copper wire = 2 mm = 2 x 10⁻³ m

Area of cross-section of copper wire is given as

A = (0.25) πd²

A = (0.25) (3.14) (2 x 10⁻³)²

A = 3.14 x 10⁻⁶ m²

E = magnitude of electric field = 0.01 V/m

e = magnitude of charge on electron = 1.6 x 10⁻¹⁹ C

m = mass of electron = 9.1 x 10⁻³¹ kg

n = number density of free electrons in copper = 8.47 x 10²² cm⁻³ = 8.47 x 10²⁸ m⁻³

i = magnitude of current

magnitude of current is given as

i = \frac{Ane^{2}\tau E}{m}

i = \frac{(3.14\times 10^{-6})(8.47\times 10^{28})(1.6\times 10^{-19})^{2}(2.5\times 10^{-14}) (0.01)}{(9.1\times 10^{-31})}

i  = 1.87 A

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Andrews [41]

Answer:

2.35 m/s²

Explanation:

Given that

Mass of the smaller crate, m₁ = 21 kg

Mass of the larger crate, m₂ = 90 kg

Tensión of the rope, T = 261 N

We know that the sum of all forces for the two objects with a force of friction F and a tension T are:

(i) m₁a₁ = F

(ii) m₂a₂ = T - F, where m and a are the masses and accelerations respectively.

1) no sliding can also mean that:

a₁ = a₂ = a

This makes us merge the two equations written above together as:

m₂a = T - m₁a

If we then solve for a, we would have something like this

a = T / (m₁+m₂)

a = 261 / (21 + 90)

a = 261 / 111

a = 2.35 m/s²

Therefore, the needed acceleration of the small crate is 2.35 m/s²

8 0
3 years ago
Help pleasseeee URGENT
Zina [86]

Answer:

The speed of the 8-ball is 2.125 m/s after the collision.

Explanation:

<u>Law Of Conservation Of Linear Momentum</u>

The total momentum of a system of masses is conserved unless an external force is applied. The momentum of a body with mass m and velocity v is calculated as follows:

P=mv

If we have a system of masses, then the total momentum is the sum of all the individual momentums:

P=m_1v_1+m_2v_2+...+m_nv_n

When a collision occurs, the velocities change to v' and the final momentum is:

P'=m_1v'_1+m_2v'_2+...+m_nv'_n

In a system of two masses, the law of conservation of linear momentum is simplified to:

m_1v_1+m_2v_2=m_1v'_1+m_2v'_2

The m1=0.16 Kg 8-ball is initially at rest v1=0. It is hit by an m2=0.17 Kg cue ball that was moving at v2=2 m/s.

After the collision, the cue ball comes to rest v2'=0. It's required to find the final speed v1' after the collision.

The above equation is solved for v1':

\displaystyle v'_1=\frac{m_1v_1+m_2v_2-m_2v'_2}{m_1}

\displaystyle v'_1=\frac{0.16*0+0.17*2-0.17*0}{0.16}

\displaystyle v'_1=\frac{0.34}{0.16}

v'_1=2.125\ m/s

The speed of the 8-ball is 2.125 m/s after the collision.

8 0
3 years ago
after a circuit has been turned off, so it is important to make sure they are discharged before you touch them. Suppose a 120 mF
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Answer:

The time is 110.16\times10^{-3}\ sec

Explanation:

Given that,

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Current = 50 mA

We need to calculate the discharge current

Using formula of discharge current

i_{0}=\dfrac{V_{0}}{R}

Put the value into the formula

i_{0}=\dfrac{150}{1.8\times10^{3}}

i_{0}=83.3\times10^{-3}\ A

We need to calculate the time

Using formula of current

i=\dfrac{V_{0}}{R}e^{\frac{-t}{RC}}

Put the value into the formula

50=\dfrac{150}{1.8\times10^{3}}e^{\frac{-t}{RC}}

\dfrac{50}{83.3}=e^{\frac{-t}{RC}}

\dfrac{-t}{RC}=ln(0.600)

t=0.51\times1.8\times10^{3}\times120\times10^{-6}

t=110.16\times10^{-3}\ sec

Hence, The time is 110.16\times10^{-3}\ sec

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magnitude
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