Ask question

Ask question

All categories

3 years ago

13

You are working at a company that manufactures electri- cal wire. Gold is the most ductile of all metals: it can be stretched in

to incredibly long, thin wires. The company has developed a new technique that will stretch 1.00 g of gold into a wire of length L 5 2.40 km and uniform diameter. Your supervisor gives you the task of determining the resis- tance of such a wire at 20.08C.(the density of gold is 19.3 ✕ 103 kg/m3.)

1 answer:

Alona [7]3 years ago

4 0

Explanation:

We know that the relation between volume and density is as follows.

Volume = $\frac{\text{mass}}{\text{density}}$

So, V = $\frac{10^{-3}}{19.3 \times 10^{3} kg/m^{3}}$

= $5.181 \times 10^{-8} m^{3}$

Now, we will calculate the area as follows.

Area = $\frac{\text{volume}}{\text{length}}$

= $\frac{5.181 \times 10^{-8} m^{3}}{2.4 \times 10^{3}}$

= $2.15 \times 10^{-11} m^{2}$

Formula to calculate the resistance is as follows.

R = $\rho \frac{l}{A}$

= $\frac{2.44 \times 10^{-8} \times 2400}{}2.15 \times 10^{-11}}$

= $2.71 \times 10^{6} ohm$

Thus, we can conclude that the resistance of given wire is $2.71 \times 10^{6} ohm$ .

You might be interested in

Match each item with the clean water regulation it describes. (2 points)

Leona [35]

The correct match of each item to the clean water regulation it describes is as follows:

Regulates pollutants discharged into surface waters: Clean water act
Covers both surface and ground waters: Safe drinking water act
Authorizes the EPA to establish minimum standards for tap water: Safe drinking water act
Funds sewage treatment plants: Clean water act

<h3>What are the functions of clean water regulation?</h3>

Clean Water Act (CWA) is a regulatory body that establishes the basic structure for the regulation of pollutants discharge and maintenance of quality standards of the surface waters.

On the other hand, the Safe Drinking Water Act was founded to oversee the protection of the quality drinking water. The regulatory body is primarily concerned with potable water all waters, whether from above ground or underground sources.

Therefore, the correct match of each item to the clean water regulation it describes is as follows:

Regulates pollutants discharged into surface waters: Clean water act
Covers both surface and ground waters: Safe drinking water act
Authorizes the EPA to establish minimum standards for tap water: Safe drinking water act
Funds sewage treatment plants: Clean water act

Learn more about clean water regulation at: brainly.com/question/2142268

#SPJ1

6 0

2 years ago

A mortar is like a small cannon that launches shells at steep angles. A mortar crew is positioned near the top of a steep hill.

Elena-2011 [213]

1) Distance down the hill: 1752 ft (534 m)

2) Time of flight of the shell: 12.9 s

3) Final speed: 326.8 ft/s (99.6 m/s)

Explanation:

1)

The motion of the shell is a projectile motion, so we can analyze separately its vertical motion and its horizontal motion.

The vertical motion of the shell is a uniformly accelerated motion, so the vertical position is given by the following equation:

$y=(u sin \theta)t-\frac{1}{2}gt^2$ (1)

where:

$u sin \theta$ is the initial vertical velocity of the shell, with $u=156 ft/s$ and $\theta=49.0^{\circ}$

$g=32 ft/s^2$ is the acceleration of gravity

At the same time, the horizontal motion of the shell is a uniform motion, so the horizontal position of the shell at time t is given by the equation

$x=(ucos \theta)t$

where $u cos \theta$ is the initial horizontal velocity of the shell.

We can re-write this last equation as

$t=\frac{x}{u cos \theta}$ (1b)

And substituting into (1),

$y=xtan\theta -\frac{1}{2}gt^2$ (2)

where we have choosen the top of the hill (starting position of the shell) as origin (0,0).

We also know that the hill goes down with a slope of $\alpha=-41.0^{\circ}$ from the horizontal, so we can write the position (x,y) of the hill as

$y=x tan \alpha$ (3)

Therefore, the shell hits the slope of the hill when they have same x and y coordinates, so when (2)=(3):

$xtan\alpha = xtan \theta - \frac{1}{2}gt^2$

Substituting (1b) into this equation,

$xtan \alpha = x tan \theta - \frac{1}{2}g(\frac{x}{ucos \theta})^2\\x (tan \theta - tan \alpha)-\frac{g}{2u^2 cos^2 \theta} x^2=0\\x(tan \theta - tan \alpha-\frac{gx}{2u^2 cos^2 \theta})=0$

Which has 2 solutions:

x = 0 (origin)

and

$tan \theta - tan \alpha=\frac{gx}{2u^2 cos^2 \theta}=0\\x=(tan \theta - tan \alpha) \frac{2u^2 cos^2\theta}{g}=1322 ft$

So, the distance d down the hill at which the shell strikes the hill is

$d=\frac{x}{cos \alpha}=\frac{1322}{cos(-41.0^{\circ})}=1752 ft=534 m$

2)

In order to find how long the mortar shell remain in the air, we can use the equation:

$t=\frac{x}{u cos \theta}$

where:

x = 1322 ft is the final position of the shell when it strikes the hill

$u=156 ft/s$ is the initial velocity of the shell

$\theta=49.0^{\circ}$ is the angle of projection of the shell

Substituting these values into the equation, we find the time of flight of the shell:

$t=\frac{1322}{(156)(cos 49^{\circ})}=12.9 s$

3)

In order to find the final speed of the shell, we have to compute its horizontal and vertical velocity first.

The horizontal component of the velocity is constant and it is

$v_x = u cos \theta =(156)(cos 49^{\circ})=102.3 ft/s$

Instead, the vertical component of the velocity is given by

$v_y=usin \theta -gt$

And substituting at t = 12.9 s (time at which the shell strikes the hill),

$v_y=(156)(cos 49^{\circ})-(32)(12.9)=-310.4ft/s$

Therefore, the final speed of the shell is:

$v=\sqrt{v_x^2+v_y^2}=\sqrt{(102.3)^2+(-310.4)^2}=326.8 ft/s=99.6 m/s$

Learn more about projectile motion:

brainly.com/question/8751410

#LearnwithBrainly

5 0

3 years ago

A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 m/s upward.

fgiga [73]

complete question:

A child bounces a 60 g superball on the sidewalk. The velocity change of the superball is from 22 m/s downward to 15 m/s upward. If the contact time with the sidewalk is 1/800 s, what is the magnitude of the average force exerted on the superball by the sidewalk

Answer:

F = 1776 N

Explanation:

mass of ball = 60 g = 0.06 kg

velocity of downward direction = 22 m/s = v1

velocity of upward direction = 15 m/s = v2

Δt = 1/800 = 0.00125 s

Linear momentum of a particle with mass and velocity is the product of the mass and it velocity.

p = mv

When a particle move freely and interact with another system within a period of time and again move freely like in this scenario it has a definite change in momentum. This change is defined as Impulse .

I = pf − pi = ∆p

F = ∆p/∆t = I/∆t

let the upward velocity be the positive

Δp = mv2 - m(-v1)

Δp = mv2 - m(-v1)

Δp = m (v2 + v1)

Δp = 0.06( 15 + 22)

Δp = 0.06(37)

Δp = 2.22 kg m/s

∆t = 0.00125

F = ∆p/∆t

F = 2.22/0.00125

F = 1776 N

4 0

3 years ago

What is the measuring devices of a mass

erik [133]

The measuring devices for mass is scales and Balances

6 0

3 years ago

A person can see clearly up close but cannot focus on objects beyond 75.0 cm. She opts for contact lenses to correct her vision.

Pavel [41]

Answer:

(a) nearsighted

(b) diverging

(c) the lens strength in diopters is 1.33 D, and considering the convention for divergent lenses normally prescribed as: -1 33 D

Explanation:

(a) The person is nearsighted because he/she cannot see objects at distances larger than 75 cm.

(b) the type of correcting lens has to be such that it counteracts the excessive converging power of the eye of the person, so the lens has to be diverging (which by the way carries by convention a negative focal length)

(c) the absolute value of the focal length (f) is given by the formula:

$f=\frac{1}{d} =\frac{1}{0.75} = 1.33\,D$

So it would normally be written with a negative signs in front indicating a divergent lens.

7 0

3 years ago