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Transcranial magnetic stimulation (TMS) is a noninvasive technique used to stimulate regions of the human brain. A small coil is

attashe74 [19]

Answer:

0.125 volts

Explanation:

The induced emf can be sufficient to stimulate neuronal activity.

One such device generates a magnetic field within the brain that rises from zero to 1.5 T in 120 ms.

We need to find the induced emf within a circle of tissue of radius 1.6 mm and that is perpendicular to the direction of the field. The formula for the induced emf is given by :

$\epsilon=-\dfrac{d\phi}{dt}$

Where

$\phi$ is magnetic flux

So,

$\epsilon=-\dfrac{d(BA)}{dt}\\\\=2\pi r\times \dfrac{dB}{dt}\\\\=2\pi \times 1.6\times 10^{-3}\times \dfrac{1.5-0}{120\times 10^{-3}}\\\\=0.125\ V$

So, the induced emf is equal to 0.125 volts.

7 0

3 years ago

An airplane has a momentum of 8.55 x 107 kg.m/s[S] and a velocity of 900 km/h[S]. Determine the mass of the airplane.

kow [346]

Answer:

342,000kg

Explanation:

p=mv

8.55*10^7 kg*m/s=m(900 km/h)

85,500,000 kg*m/s=m(900 km/h)

(85,500,000 kg*m/s)/(900 km/h)=m

Get same units.... 900km/h = 250m/s

m/s cancel in the division, you are left with just kg!!

85,500,000/250=342,000kg! That's it!

6 0

3 years ago

What safety feature melts to protect a circuit?

N76 [4]

A fuse melts to protect a circuit.

7 0

3 years ago

Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri

Ainat [17]

Answer:

The taken is $t_A = 19.0 \ s$

Explanation:

Frm the question we are told that

The speed of car A is $v_A = 22 \ m/s$

The speed of car B is $v_B = 29.0 \ m/s$

The distance of car B from A is $d = 300 \ m$

The acceleration of car A is $a_A = 2.40 \ m/s^2$

For A to overtake B

The distance traveled by car B = The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is

$d = v_B * t_A$

Where $t_B$ is the time taken by car B

Now this can also be represented as using equation of motion as

$d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300$

Now substituting values

$d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

Equating the both d

$v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

substituting values

$29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A =1.2 t_A^2 - 300$

$1.2 t_A^2 - 7 t_A - 300 = 0$

Solving this using quadratic formula we have that

$t_A = 19.0 \ s$

7 0

3 years ago

If we start with 1.000 g of cobalt-60, 0.675 g will remain after 3.00 yr. this means that the of is _____

Tju [1.3M]

Cobalt-60 is undergoing a radioactivity decay.

The formula of the decay is n=N(1/2)∧(T/t).
Where N ⇒ original mass of cobalt
 n ⇒ remaining mass of cobalt after 3 years
T ⇒ decaying period
t ⇒ half-life of cobalt.

So,
0.675 = 1 × 0.5∧(3/t)
log 0.675 = log 0.5∧(3/t)
3/t = log 0.675 ÷log 0.5
3/t= 0.567

t = 3÷0.567
= 5.290626524

the half-life of Cobalt-60 is 5.29 years.

8 0

3 years ago