Explain this statement: Cells are the basic unit of structure and function in organisms.

A "gauge 8" jumper cable has a diameter d of 0.326 centimeters. The cable carries a current I of 30.0 amperes. The electric fiel

AveGali [126]

Answer:

0.0979 N/c

Explanation:

Electric field, E is given as a product of resistivity and current density

E=jP where P is resistivity and j is current density

But the current density is given as

$j=\frac {I}{A}$ where I is current and A is area and $A=\pi r^{2}$

Substituting this into the first equation then $E=P\times \frac {I}{\pi r^{2}}$

Given diameter of 0.259 cm= 0.00259 m and the radius will be half of it which is 0.001295 m

$E=1.72\times 10^{-8}\times \frac {30}{\pi \times 0.001295^{2}}=9.79\times 10^{-2} N/c=0.0979 N/c$

4 0

3 years ago

Usually the force of gravity on electrons is neglected. To see why, we can compare the force of the Earth’s gravity on an electr

miv72 [106K]

Answer:

$6.4\cdot 10^{-15} N$

Explanation:

The electric force exerted on the electron is given by:

$F=qE$

where

$q=1.6\cdot 10^{-19}C$ is the magnitude of the electron charge

E = 40000 V/m is the electric field

Substituting,

$F=(1.6\cdot 10^{-19} C)(40000 V/m)=6.4\cdot 10^{-15} N$

By comparison, the gravitational force exerted on the electron is:

$F=mg$

where

$m=9.10939\cdot 10^{-31} kg$ is the mass of the electron

g = 9.8 m/s^2 is the acceleration due to gravity

Substituting,

$F=(9.10939\cdot 10^{-31} kg)(9.8 m/s^2)=8.93\cdot 10^{-30}N$

6 0

4 years ago

A block with mass of 10 kg is on a frictionless surface. One hand on the left side of the block is pushing it to the right. A se

igomit [66]

Answer:

$W_2=-12J$

Explanation:

The work of force 2 will be given by the vectorial equation $W_2=F_2.d$ . We know the value of $F_1$ and have information about its movement, which relates to the net force $F=F_1+F_2$ .

About this movement we can obtain the acceleration using the equation $v_f^2=v_i^2+2ad$ . Since it departs from rest we have $a=\frac{v_f^2}{2d}$ .

And then using Newton's 2dn Law we can obtain the net force F=ma, thus we will have $F_2=F-F_1=ma-F1=\frac{mv_f^2}{2d}-F_1$

And we had the work done by force 2 as:

$W_2=F_2.d=\frac{mv_f^2}{2}-F_1d$

(The sign will be given algebraically since we take positive the direction to the right.)

With our values:

$W_2=\frac{(10kg)(2m/s)^2}{2}-(8N)(4m)=-12J$

Another (shorter but maybe less intuitive way for someone who is learning) way of doing this would have been to say that the work done by both forces would be equal to the variation of kinetic energy:

$W_1+W_2=K_f-K_i=K_f=\frac{mv_f^2}{2}$