All categories

3 years ago

Ohms law? explanation.

Physics

1 answer:

shusha [124]3 years ago

4 0

Answer:

$\boxed{ \bold{ \sf{see \: below}}}$

Explanation:

$\boxed{ \bold{ \underline{ \huge{ \boxed{ \sf{ohm's \: law}}}}}}$

The relation between current through a metallic conductor and potential difference across its ends was studied systematically by a German physicist , George Simon Ohm in 1826 AD . This relation is now known as Ohm's law. It states that the electric current passing through a conductor is directly proportional to the potential difference across its two ends at a constant physical condition [ Temperature , cross - sectional area , length , nature of material etc ]

Hope I helped!

Best regards!!

You might be interested in

Energy transformation in wound spring of a toy car? give your own answrr

BaLLatris [955]

Answer:

The work done in winding the spring gets stored in the wound up spring in the form of elastic potential energy (i.e potential energy due to change in shape). ... During this process, the potential energy stored in it gets converted to kinetic energy. This turns the wheels of the toy car.

Explanation:

7 0

3 years ago

Read 2 more answers

How to solve question #2?

Vera_Pavlovna [14]

2a for example the first one,2sec. You know that every second it moves 3metres further. So 2x3=6 but you start at 0.50m so 6+0.50=6.5

5 0

3 years ago

You push a 1.30 kg physics book 2.80 m along a horizontal tabletop with a horizontal push of 1.55 N while the opposing force of

Rzqust [24]

Answer:

Explanation:

Step one:

given data

mass m= 1.3kg

distance moved s= 2.8m

opposing frictional force= 0.34N

assume g= 9.81m/s^2

we know that work done= force *distance moved

1. work done to push the book= 1.55*2.8=4.34J

2. Work against friction = force of friction x distance

= 0.34*2.8=0.952J

Step two:

the work done on the book is the net work, which is

Network done= work done to push the book- Work against friction

Network done= 4.32-0.952=3.36J

Therefore the work of the 1.55N 3.36J

4 0

2 years ago

A spring stretches by 0.0208 m when a 3.39-kg object is suspended from its end. How much mass should be attached to this spring

sashaice [31]

Answer:

COMPLETE QUESTION

A spring stretches by 0.018 m when a 2.8-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f = 3.0 Hz?

Explanation:

Given that,

Extension of spring

x = 0.0208m

Mass attached m = 3.39kg

Additional mass to have a frequency f

Let the additional mass be m

Using Hooke's law

F= kx

Where F = W = mg = 3.39 ×9.81

F = 33.26N

Then,

F = kx

k = F/x

k = 33.26/0.0208

k = 1598.84 N/m

The frequency is given as

f = ½π√k/m

Make m subject of formula

f² = ¼π² •(k/m

4π²f² = k/m

Then, m4π²f² = k

So, m = k/(4π²f²)

So, this is the general formula,

Then let use the frequency above

f = 3Hz

m = 1598.84/(4×π²×3²)

m = 4.5 kg

4 0

3 years ago

A closed container initially holds 50 monatomic Aparticles that have a combined energy of 480 units. After 100 monatomic B parti

Molodets [167]

Answer:

"8 units" is the appropriate answer.

Explanation:

According to the question,

Throughout equilibrium all particles are of equivalent intensity, and as such the integrated platform's total energy has been uniformly divided across all individuals.

Now,

The total energy will be:

= $480+720$