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Lisa [10]
2 years ago
10

A straight, cylindrical wire lying along the x axis has a length L and a diameter d . It is made of a material described by Ohm'

s law with a resistivity rho . Assume potential V is maintained at the left end of the wire at x = 0. Also assume the potential is zero at x=L . In terms of L, d , V, rho, and physical constants, derive expressions for (a) the magnitude and direction of the electric field in the wire,
Physics
1 answer:
Zepler [3.9K]2 years ago
7 0

The magnitude and direction of the electric field in the wire are mathematically given as

L &=[(v / L) v / m] \hat{i}

<h3>What is the magnitude and direction of the electric field in the wire?</h3>

Generally, the equation for is  mathematically given as

A cylindrical wire that is straight and parallel to the x-axis has the following dimensions: length L, diameter d, resistivity p, diameter d, potential v, and z length. combining elements from both sides  

E d x=\int d v.

\begin{aligned}&-E \int_0^L d x=\int_v^0 d v \\\therefore E \cdot L &=v \\L &=[(v / L) v / m] \hat{i}\end{aligned}

In conclusion, the magnitude and direction of the electric field in the wire are given as

L &=[(v / L) v / m]

Read more about electric fields

brainly.com/question/15800304

#SPJ4

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16. A 95kg fullback, running at 8.2m/s, collided in midair with a 128 kg defensive tackle moving in the opposite direction. Both
Daniel [21]

a) 779 kg m/s

The momentum of an object is given by:

p = mv

where

m is the mass of the object

v is its velocity

For the fullback before the collision,

m = 95 kg

v = 8.2 m/s

Therefore, his momentum was:

p=mv=(95)(8.2)=779 kg m/s

b) -779 kg m/s

After the collision, both the fullback and the tackle come to a stop: this means that their momentum after the collision is zero,

p' = 0

The initial momentum of the fullback was

p = 779 kg m/s

Therefore, his change in momentum is

\Delta p = p' -p =0-779  = -779 kg m/s

where the negative sign indicates that the direction is opposite to the initial direction of motion.

c) -779 kg m/s

Here we can apply the law of conservation of momentum. In fact, the total momentum before and after the collision must be conserved. So we can write:

p_f + p_t = p'

where

p_f is the initial momentum of the fullback

p_t is the initial momentum of the tackle

p' is the final combined momentum after the collision

We already know that

p_f = 779 kg m/s\\p' = 0

Therefore, we can find the tackle's original momentum:

p_t = p'-p_f = 0-(779) = -779 kg m/s

where the negative sign indicates that the direction is opposite to the initial direction of motion of the fullback.

e) -6.1 m/s

To find the velocity of the tackle, we can use again the equation of the momentum:

p = mv

where here we have

p=-779 kg m/s is the original momentum of the tackle

m = 128 kg is his mass

Solving the equation for v, we find the tackle's original velocity:

v=\frac{p}{m}=\frac{-779}{128}=-6.1 m/s

So, he was moving at 6.1 m/s in the direction opposite to the fullback.

4 0
4 years ago
An airplane travels at 300 mi/h south for 2.00 h and then at 250 mi/h north for 750 miles. What is the average speed for the tri
Anettt [7]

Answer:

270 mi/h

Explanation:

Given that,

To the south,

v₁ = 300 mi/h, t₁ = 2 h

We can find distance, d₁

d_1=v_1\times t_1\\\\d_1=300\times 2\\\\d_1=600\ \text{miles}

To the north,

v₂ = 250 mi/h, d₂ = 750 miles

We can find time, t₂

t_2=\dfrac{d_2}{v_2}\\\\t_2=\dfrac{750\ \text{miles}}{250\ \text{mi/h}}\\\\t_2=3\ h

Now,

Average speed = total distance/total time

V=\dfrac{d_1+d_2}{t_1+t_2}\\\\V=\dfrac{600+750}{2+3}\\\\V=270\ \text{mi/h}

Hence, the average speed for the trip is 270 mi/h.

3 0
3 years ago
A clothes dryer uses about 30 amps of current from a 240 volt line. How much power does it use?
Aneli [31]
Not sure if this answer is write or wrong but i think its <span>4500 watts

</span>
8 0
3 years ago
Read 2 more answers
A student, along with her backpack on the floor next to her, are in an elevator that is accelerating upward with acceleration a.
Anna007 [38]

Answer:

\mu_k = \frac{2(vt - L)}{(g + a) t^2}

Explanation:

As we know that backpack is kicked on the rough floor with speed "v"

So here as per force equation in vertical direction we know that

N - mg = ma

so normal force on the block is given as

N = mg + ma

now the magnitude of kinetic friction on the block is given as

F_f = \mu N

F_f = \mu (mg + ma)

now when bag is sliding on the floor then net deceleration of the block due to friction is given as

a = - \frac{F_f}{m}

a = -\mu_k(g + a)

now we know that bag hits the opposite wall at L distance away in time t

so we have

d = v t + \frac{1}{2}at^2

L = vt - \frac{1}{2}(\mu_k)(g + a) t^2

\mu_k = \frac{2(vt - L)}{(g + a) t^2}

8 0
3 years ago
When the mass is removed, the length of the cable is found to be l0 = 4.66 m. After the mass is added, the length is remeasured
zaharov [31]

Answer:

\gamma=6.07*10^5\frac{N}{m^2}

Explanation:

For a linear elastic material Young's modulus is a constant that is given by:

\gamma=\frac{F/A}{\Delta L/L_0}

Here, F is the force exerted on an object under tensio, A is the area of the cross-section perpendicular to the applied force, \Delta L is the amount by which the length of the object changes and  L_0 is the original length of the object. In this case the force is the weight of the mass:

F=mg\\F=55kg(9.8\frac{m}{s^2})\\F=539N

Replacing the given values in Young's modulus formula:

\gamma=\frac{F/\pi r^2}{(L_1-L_0)/L_0}\\\gamma=\frac{539N/\pi(0.045m)^2}{(5.31m-4.66m)/4.66m}\\\gamma=6.07*10^5\frac{N}{m^2}

6 0
3 years ago
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