Which of the following is an example of a conductor?

A bare 4 AWG copper conductor Installed horizontally near the bottom or vertically, and within that portion of a concrete foun d

neonofarm [45]

Answer:

20 ft

Explanation:

This is the length where the direct contact could be used as ground electrode

4 0

4 years ago

Maceo is making rock candy. Which best describes the steps she should take?

Andru [333]

Answer:

The answer is heat a saturated sugar water solution, dissolve more sugar, then let the solution cool

Explanation:

3 0

3 years ago

Read 2 more answers

A video game includes an asteroid that is programmed to move in a straight line across a 17-inch monitor according to the equati

Ainat [17]

Answer:

The asteroid's acceleration at this point is $2.71\ m/s^2$

Explanation:

The equation that governs the trajectory of asteroid is given by :

$x=6.5t-2.3t^3$

The velocity of asteroid is given by :

$v=\dfrac{dx}{dt}\\\\v=\dfrac{d(6.5t-2.3t^3)}{dt}\\\\v=6.5-6.9t^2$

At some point during the trip across the screen, the asteroid is at rest. It means, v = 0

So,

$6.5-6.9t^2=0\\\\t=0.971\ s$

Acceleration,

$a=\dfrac{dv}{dt}\\\\a=\dfrac{d(6.5-6.9t^2)}{dt}\\\\a=-13.8t$

Put t = 0.971 s

$a=-13.8\times 0.197\\\\a=-2.71\ m/s^2$

So, the asteroid's acceleration at this point is $2.71\ m/s^2$ and it is decelerating.

6 0

3 years ago

What’s an example of contact friction

kotegsom [21]

Answer:

a nightstand on a lamp table

Explanation:

5 0

3 years ago

What speed would a proton need to achieve in order to circle Earth 1790 km above the magnetic equator, where the Earth's mag- ne

irinina [24]

Answer:

The velocity is 31.25 m/s and direction is toward west.

Explanation:

Given that,

Distance $h= 1790 km = 1.790\times10^{6}\ m$

Magnetic field $B=4\times10^{-8}\ T$

Mass of proton $m=1.673\times10^{-21}\ Kg$

Radius of earth $R =6.38\times10^{6}\ m$

Radius of orbit $r=R+h$

$r=6.38\times10^{6}+1.790\times10^{6}$

$r=8170000\ m$

We need to calculate the speed

Using formula of magnetic field

$Bvq=\dfrac{mv^2}{r}$

$v=\dfrac{Bqr}{m}$

Put the value into the formula

$v=\dfrac{4\times10^{-8}\times1.6\times10^{-19}\times8170000}{1.673\times10^{-21}}$

$v=31.25\ m/s$

Hence, The velocity is 31.25 m/s and direction is toward west.

6 0

3 years ago