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Arlecino [84]
3 years ago
11

What is Ohm's Law, and how does it work in real life.

Physics
1 answer:
Firdavs [7]3 years ago
8 0

Explanation:

Ohms law states that the electrical current present in a metallic conductor is directly proportional to the potential difference between the metallic conductor and inversely proportional to the resistance therefore if the voltage is increased resistance also increases provided that temperature and other physical properties remains constant V=IR

You might be interested in
Newton's third law says that for every action force there is a reaction force which is
olga nikolaevna [1]
A. Equal and opposite For every action, there is an equal and opposite reaction
4 0
3 years ago
g You shine orange laser light that has a wavelength of 600 nm through a narrow slit. The slit forms a diffraction pattern on a
zimovet [89]

Answer:

 λ = 3 10⁻⁷ m,   UV laser

Explanation:

The diffraction phenomenon is described by the expression

         a sin θ = m λ

let's use trigonometry

         tan θ = y / L

as in this phenomenon the angles are small

        tan θ = \frac{sin \ \theta}{cos \ \theta} = sin θ

        sin θ = y / L

we substitute

      a y / L = m  λ

let's apply this equation to the initial data

       a  0.04 / L = 1 600 10⁻⁹

       a / L = 1.5 10⁻⁵

now they tell us that we change the laser and we have y = 0.04 m for m = 2

      a 0.04 / L = 2  λ

       a / L = 50  λ

we solve the two expression is

         1.5 10⁻⁵ = 50  λ

          λ = 1.5 10⁻⁵ / 50

          λ = 3 10⁻⁷ m

    UV laser

3 0
2 years ago
In each of the parts of this question, a nucleus undergoes a nuclear decay. Determine the resulting nucleus in each case.
vaieri [72.5K]

Answer:

(A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

Explanation:

Given that,

A nucleus undergoes a nuclear decay.

(A). In alpha decay,

We know that,

When the nucleus emit alpha particle then atomic mass of particle reduce by 4 and atomic number reduce by 2.

We need to calculate the resulting nucleus

Using given data

^{227}_{89}Ac\Rightarrow ^{227-4}_{89-2}X

^{227}_{89}Ac\Rightarrow ^{223}_{87}Fr

The resulting nucleus is Fr.

(B). In beta-minus decay,

We know that,

When the nucleus emit beta- minus particle then atomic mass of particle is same and atomic number increase by 1.

We need to calculate the resulting nucleus

Using given data

^{211}_{83}Bi\Rightarrow ^{211}_{83+1}X

^{211}_{83}Bi\Rightarrow ^{211}_{84}Po

The resulting nucleus is Po.

(C). In beta-plus decay,

We know that,

When the nucleus emit beta- plus particle then atomic mass of particle is same and atomic number decrease by 1.

We need to calculate the resulting nucleus

Using given data

^{22}_{11}Na\Rightarrow ^{22}_{11-1}X

^{22}_{11}Na\Rightarrow ^{22}_{10}Ne

The resulting nucleus is Ne.

(D). In gamma decay,

We know that,

When the nucleus emit gamma particle then atomic mass and atomic number of particle is same.

We need to calculate the resulting nucleus

Using given data

^{98}_{43}Tc\Rightarrow ^{98}_{43}Tc

The resulting nucleus is Tc.

Hence, (A). The resulting nucleus is Fr.

(B). The resulting nucleus is Po.

(C). The resulting nucleus is Ne

(D). The resulting nucleus is Tc.

4 0
3 years ago
3. The pressure at the bottom of the ocean is great enough to crush submarines with steel walls that are 10 centimeters thick. S
Elden [556K]

Answer:

just awnsered this one your awnser is the the second option

3 0
2 years ago
Read 2 more answers
A 100-turn, 3.0-cm-diameter coil is at rest in a horizontal plane. A uniform magnetic field 60%u2218 away from vertical increase
Xelga [282]

Answer:

Induced emf in the coil, E = 0.157 volts

Explanation:

It is given that,

Number of turns, N = 100

Diameter of the coil, d = 3 cm = 0.03 m

Radius of the coil, r = 0.015 m

A uniform magnetic field increases from 0.5 T to 2.5 T in 0.9 s.

Due to this change in magnetic field, an emf is induced in the coil which is given by :

E=-NA\dfrac{\Delta B}{\Delta t}

E=-100\times \pi (0.015)^2\times \dfrac{2.5-0.5}{0.9}

E = -0.157 volts

Minus sign shows the direction of induced emf in the coil. Hence, the induced emf in the coil is 0.157 volts.

8 0
3 years ago
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