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A TMS (transcranial magnetic stimulation) device creates very rapidly changing magnetic fields. The field near a typical pulsed-

GrogVix [38]

Explanation:

It is given that, the field near a typical pulsed-field machine rises from 0 T to 2.5 T in 200 μs

Change in magnetic field, $dB=2.5\ T$

Change in time, $dt=200\ \mu s=200\times 10^{-6}\ s=2\times 10^{-4}\ s$

Diameter, d = 2.3 cm

Radius, r = 0.0115 m

Emf is induced in the ring as the field changes. It is given by :

$E=\dfrac{d\phi}{dt}$

$E=\dfrac{d(B.A\ cos(0))}{dt}$

$E=A\dfrac{d(B)}{dt}$

$E=\pi (0.0115)^2\dfrac{2.5}{2\times 10^{-4}}$

E = 5.19 volts

So, the emf induced in the ring is 5.19 volts. Hence, this is the required solution.

5 0

3 years ago

What is the basic difference between ultraviolet, visible, and infrared radiation?

N76 [4]

To me the basic difference is their frequency and wavelength.

For frequency :
Infrared < visible light < ultraviolet

For wavelength :
Infrared > visible light > ultraviolet

3 0

4 years ago

A parallel-plate vacuum capacitor has 7.72 J of energy stored in it. The separation between the plates is 3.30 mm. If the separa

maxonik [38]

Answer

3.340J

Explanation;

Using the relation. Energy stored in capacitor = U = 7.72 J

U =(1/2)CV^2

C =(eo)A/d

C*d=(eo)A=constant

C2d2=C1d1

C2=C1d1/d2

The separation between the plates is 3.30mm . The separation is decreased to 1.45 mm.

Initial separation between the plates =d1= 3.30mm .

Final separation = d2 = 1.45 mm

(A) if the capacitor was disconnected from the potential source before the separation of the plates was changed, charge 'q' remains same

Energy=U =(1/2)q^2/C

U2C2 = U1C1

U2 =U1C1 /C2

U2 =U1d2/d1

Final energy = Uf = initial energy *d2/d1

Final energy = Uf =7.72*1.45/3.30

(A) Final energy = Uf = 3.340J

4 0

3 years ago

Two pieces of clay are moving directly toward each other. When they collide, they stick together and move as one piece. One piec

Wittaler [7]

Answer:

The fraction of the total initial kinetic energy is lost during the collision is $\dfrac{11}{17}\ J$

Explanation:

Given that,

Mass of one piece = 300 g

Speed of one piece = 1 m/s

Mass of other piece = 600 g

Speed of other piece = 0.75 m/s

We need to calculate the final velocity

Using conservation of energy

$m_{1}v_{1}+m_{2}v_{2}=(m_{1}+m_{2})v$

Put the value intro the formula

$300\times10^{-3}\times1+600\times10^{-3}\times(0.75)=(300\times10^{-3}+600\times10^{-3})v$

$v=\dfrac{00\times10^{-3}\times1+600\times10^{-3}\times(-0.75)}{(300\times10^{-3}+600\times10^{-3})}$

$v=-0.5\ m/s$

We need to calculate the total initial kinetic energy

Using formula of kinetic energy

$K.E_{i}=\dfrac{1}{2}m_{1}v_{1}^2+\dfrac{1}{2}m_{2}v_{2}^2$

Put the value into the formula

$K.E_{i}=\dfrac{1}{2}\times300\times10^{-3}\times1^2+\dfrac{1}{2}\times600\times10^{-3}\times(0.75)^2$

$K.E_{i}=0.31875\ J$

We need to calculate the total final kinetic energy

Using formula of kinetic energy

$K.E_{f}=\dfrac{1}{2}(m_{1}+m_{2})v^2$

Put the value into the formula

$K.E_{f}=\dfrac{1}{2}\times(300\times10^{-3}+600\times10^{-3})\times(-0.5)^2$

$K.E_{f}=0.1125\ J$

We need to calculate the energy lost during the collision

Using formula of energy lost

$energy\ lost=\dfrac{0.31875-0.1125}{0.31875}$

$energy\ lost=\dfrac{11}{17}\ J$

Hence, The fraction of the total initial kinetic energy is lost during the collision is $\dfrac{11}{17}\ J$

3 0

3 years ago

A laser beam strikes a plane's surface with an angle of52.

Vadim26 [7]

Answer:

Angle between incident ray and reflected ray will be 104°

Explanation:

We have given angle of incidence = 52 °

From law of reflection angle of incidence will be equal to angle of reflection

So angle of reflection will be also 52°

We have to find the angle between incident ray and reflected ray

As the incidence angle and reflected angle both is from normal of the surface and opposite to each other

So angle between incident ray and reflected ray will be 52°+52° = 104°

7 0

4 years ago