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Irina-Kira [14]
3 years ago
10

Transcranial magnetic stimulation (TMS) is a noninvasive method for studying brain function, and possibly for treatment as well.

In this technique, a rapidly changing magnetic field is applied to a person's brain. This can cause a finger to twitch, a bright spot to appear in the visual field, or a feeling of complete happiness to overwhelm a person. If the magnetic field changes at the previously mentioned rate over an area of 1.75 10-2 m2, what is the induced emf?
Physics
1 answer:
lubasha [3.4K]3 years ago
3 0

Answer:

525 V

Explanation:

A = Area = 1.75\times 10^{-2}\ m^2

\dfrac{dB}{dt} = Rate of change of magnetic field = 3\times 10^4\ T/s (assumed)

Induced electromotive force is given by

E=A\dfrac{dB}{dt}\\\Rightarrow E=1.75\times 10^{-2}\times 3\times 10^{4}\\\Rightarrow E=525\ V

The induced electromotive force is 525 V

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3. A roller-coaster car is moving over the top of a hill. At the instant it is on the hill top, it has a kinetic energy of E and
aleksandrvk [35]

Answer:

The correct option is (b).

Explanation:

Given that,

At the top of the hill, the kinetic energy is E and the gravitational potential energy is 3E.

We need to find the kinetic energy of the car on the ground.

We know that,

Mechanical energy = kinetic energy + potential energy

According to the law of conservation of energy, the total mechanical energy is conserved.

It means, when it coasts down to ground level, the kinetic energy is same as that on the top of the hill. Hence, the required kinetic energy on the ground is equal to 3E.

8 0
3 years ago
If each of the three rotor helicopter blades is 3.50 m long and has a mass of 120 kg , calculate the moment of inertia of the th
devlian [24]

Answer:

1470kgm²

Explanation:

The formula for expressing the moment of inertial is expressed as;

I = 1/3mr²

m is the mass of the body

r is the radius

Since there are three rotor blades, the moment of inertia will be;

I = 3(1/3mr²)

I = mr²

Given

m = 120kg

r = 3.50m

Required

Moment of inertia

Substitute the given values and get I

I = 120(3.50)²

I = 120(12.25)

I = 1470kgm²

Hence the moment of inertial of the three rotor blades about the axis of rotation is 1470kgm²

7 0
3 years ago
A roofer needs to get shingles onto a roof. Pulling the shingles up manually uses 1,549 N or force. The roofer decides to use a
joja [24]

Answer:

Mechanical advantage of pulleys = 3.47 (Approx)

Explanation:

Given:

Manual force = 1,549 N

Pulleys force = 446 N

Find:

Mechanical advantage of pulleys

Computation:

Mechanical advantage of pulleys = Manual force / Pulleys force

Mechanical advantage of pulleys = 1,549 / 446

Mechanical advantage of pulleys = 3.4730

Mechanical advantage of pulleys = 3.47 (Approx)

3 0
4 years ago
The following table lists the work functions of a few common metals, measured in electron volts. Metal Φ(eV) Cesium 1.9 Potassiu
Citrus2011 [14]

A. Lithium

The equation for the photoelectric effect is:

E=\phi + K

where

E=\frac{hc}{\lambda} is the energy of the incident light, with h being the Planck constant, c being the speed of light, and \lambda being the wavelength

\phi is the work function of the metal (the minimum energy needed to extract one photoelectron from the surface of the metal)

K is the maximum kinetic energy of the photoelectron

In this problem, we have

\lambda=190 nm=1.9\cdot 10^{-7}m, so the energy of the incident light is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{1.9\cdot 10^{-7} m}=1.05\cdot 10^{-18}J

Converting in electronvolts,

E=\frac{1.05\cdot 10^{-18}J}{1.6\cdot 10^{-19} J/eV}=6.5 eV

Since the electrons are emitted from the surface with a maximum kinetic energy of

K = 4.0 eV

The work function of this metal is

\phi = E-K=6.5 eV-4.0 eV=2.5 eV

So, the metal is Lithium.

B. cesium, potassium, sodium

The wavelength of green light is

\lambda=510 nm=5.1\cdot 10^{-7} m

So its energy is

E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5.1\cdot 10^{-7} m}=3.9\cdot 10^{-19}J

Converting in electronvolts,

E=\frac{3.9\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.4 eV

So, all the metals that have work function smaller than this value will be able to emit photoelectrons, so:

Cesium

Potassium

Sodium

C. 4.9 eV

In this case, we have

- Copper work function: \phi = 4.5 eV

- Maximum kinetic energy of the emitted electrons: K = 2.7 eV

So, the energy of the incident light is

E=\phi+K=4.5 eV+2.7 eV=7.2 eV

Then the copper is replaced with sodium, which has work function of

\phi = 2.3 eV

So, if the same light shine on sodium, then the maximum kinetic energy of the emitted electrons will be

K=E-\phi = 7.2 eV-2.3 eV=4.9 eV

7 0
4 years ago
At a particular instant, a stationary observer on the ground sees a package falling with speed v1 at an angle to the vertical. t
Lerok [7]
V1 * sin(θ) where θ is the angle v1 makes with the vertical.
7 0
4 years ago
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