Which illustration represents the arrangement of particles in a gas?

A 1.0-m-long copper wire of diameter 0.10 cm carries a current of 50.0 A to the east. Suppose we apply to this wire a magnetic f

Setler79 [48]

Answer:

The classification of that same issue in question is characterized below.

Explanation:

The given values are:

Current, I = 50.0 A

Diameter, d = 0.10 cm

(a)...

As we know,

⇒ Magnetic force = Copper wire's weight

So,

⇒ $B\times I\times L=M\times g$

On putting the estimated values, we get

⇒ $B\times 50\times 1=7.037\times 10^{-3}\times 9.81$

⇒ $50B=69.03297\times 10^{-3}$

⇒ $B=1.38\times 10^{-3} \ T$

(b)...

As we know,

⇒ $m=\delta\times L\times \frac{\pi \ d^2}{4}$

⇒ $=8960\times 1\times \frac{\pi \ (0.001)^2}{4}$

⇒ $=2240\times \pi \ 0.000001$

⇒ $=7.037\times 10^{-3} \ kg$

7 0

1 year ago

A biophysics experiment uses a very sensitive magnetic field probe to determine the current associated with a nerve impulse trav

fenix001 [56]

Answer:

The peak current carried by the axon is 5.85 x 10⁻⁸ A

Explanation:

Given;

distance of the field from the axon, r = 1.3 mm

peak magnetic field strength, B = 9 x 10⁻¹² T

To determine the peak current carried by the axon, apply the following equation;

$B = \frac{\mu I}{2\pi r}$

where;

B is the peak magnetic field

r is the distance of the magnetic field from axon

μ is permeability of free space = 4π x 10⁻⁷

I is the peak current

Re-arrange the equation and solve for "I"

$B = \frac{\mu I}{2\pi r} \\\\I = \frac{B*2\pi r}{\mu} \\\\I = \frac{9*10^{-12}*2*\pi *1.3*10^{-3}}{4\pi *10^{-7}} \\\\I = 5.85 *10^{-8} \ A$

Therefore, the peak current carried by the axon is 5.85 x 10⁻⁸ A

7 0

1 year ago

A nonrelativistic electron is accelerated from rest through a potential difference. After acceleration the electron has a de Bro

aleksley [76]

Answer:

Potential difference though which the electron was accelerated is $2.67\times 10^{-6}\ uV\ .$

Explanation:

Given :

De Broglie wavelength , $\lambda=750\ nm.$

Plank's constant , $h=6.626\times 10^{-34}\ J.s \ .$

Charge of electron , $e=-1.6\times 10^{-19}\ C.$

Mass of electron , m=9.11\times 10^{-31}\ kg. $m=9.11\times 10^{-31}\ kg.$

We know , according to de broglie equation :

$\lambda=\dfrac{h}{mv}\\\\ v=\dfrac{h}{m\lambda}\\\\v=\dfrac{6.626\times 10^{-34}\ J.s \ }{9.11\times 10^{-31}\ kg\times 750\times 10^{-9}\ m }= 969.78\ m/s .$

Now , we know potential energy applied on electron will be equal to its kinetic energy .

Therefore ,

$qV=\dfrac{mv^2}{2}\\\\ V=\dfrac{mv^2}{2q}$

Putting all values in above equation we get ,

$V=2.67\times 10^{-6}\ uV .$

Hence , this is the required solution.

5 0

1 year ago

Snowboarder Jump—Energy and Momentum

soldier1979 [14.2K]

Answer:

THE ANSWER TERMS ARE DEFINED BLOW:-

Explanation:

MOMENTUM- IT IS THE ABILITY TO INCREASE OR DEVELOP CONSTANT FORCE.

KINETIC ENERGY:- IT IS THE ENERGY THAT A PRTICLE POSSES WHEN IT IS ACTUALLY IN MOTION.

POTENTIAL ENERGY:- IT IS THE ENERGY THAT A PARTICLE POSSES WHEN IT ACTUALLY IS IN RESTING STATE.

IN THIS ACIVITY THE SNOWBOARDER IS IN THE MOTION STATE THEREFORE HE POSSES KINETIC ENERGY AND TO MAINTAIN THAT KINEITC ENERG FOR A PERIOD OF TIME,MOMENTUM PLAYS IT'S ROLE.

4 0

1 year ago

A large crate is at rest on a ramp at a loading dock

VARVARA [1.3K]

Answer:

yuhhh

Explanation:

4 0

1 year ago