Stars that are closer look— A. brighter B. dimmer C. white

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

4 years ago

Calculate the average Power in kilowatts required to pull a car up a ramp in the amount of work is 250 KJ over a period of 45 se

kherson [118]

For this case we have that by definition, the average power is given by:

$P = \frac {W} {t}$

Where:

W: It is the work done

t: It's time

According to the data we have to:

$W = 250 \ KJ = 250,000 \ J\\t = 45 \ s$

So:

$P = \frac {250,000} {45}\\P = 5555.55 \ watts = 5.56 \ kw$

Answer:

$5.56 \ kw$

5 0

4 years ago

A spaceship, at rest in some inertial frame in space, suddenly needs to accelerate. The ship forcibly expels 103 kg of fuel from

Cerrena [4.2K]

Answer:

$V_s = 1.8*10^5m/s$

Explanation:

There is no external force applied, therefore there is a moment's preservation throughout the trajectory.

Initial momentum = Final momentum.

The total mass is equal to

$m_T= m_1 +m_2$

Where,

$m_1 =$ mass of ship

$m_2 =$ mass of fuell expeled.

As the moment is conserved we have,

$0=V_fm_2+V_sm_1$

Where,

$V_f =$ Velocity of fuel

$V_s =$ Velocity of Space Ship

Solving and re-arrange to $V_s$ we have,

$V_s = \frac{V_f m_2 }{m_1}$

$V_s = \frac{3/5c}{10^6}$

$V_s = 3.5*10^{-3}c$

Where c is the speed of light.

Therefore the ship be moving with speed

$V_s = \frac{3}{5}*10^{-3}*3*10^8m/s$

$V_s = 1.8*10^5m/s$

5 0

3 years ago

What are two pieces of information you must know about velocity on an object

levacccp [35]

Distance traveled(in a certain direction) and time
Velocity is all about direction traveled in comparison to speed which is just distance with out direction.

Hope this helps you

3 0

3 years ago

A radar station sends out a 250000 Hz sound wave at a speed of 340 m/s. The sound wave bounces off a weather ballon and returns

Eddi Din [679]

Answers:

a)The balloon is 68 m away of the radar station

b) The direction of the balloon is towards the radar station

Explanation:

We can solve this problem with the Doppler shift equation:

$f'=\frac{V+V_{o}}{V-V_{s}} f$ (1)

Where:

$f=250,000 Hz$ is the actual frequency of the sound wave

$f'=240,000 Hz$ is the "observed" frequency

$V=340 m/s$ is the velocity of sound

$V_{o}=0 m/s$ is the velocity of the observer, which is stationary

$V_{s}$ is the velocity of the source, which is the balloon

Isolating $V_{s}$ :

$V_{s}=\frac{V(f'-f)}{f'}$ (2)

$V_{s}=\frac{340 m/s(240,000 Hz-250,000 Hz)}{240,000 Hz}$ (3)

$V_{s}=-14.16 m/s$ (4) This is the velocity of the balloon, note the negative sign indicates the direction of motion of the balloon: It is moving towards the radar station.

Now that we have the velocity of the balloon (hence its speed, the positive value) and the time ( $t=4.8 s$ ) given as data, we can find the distance:

$d=V_{s}t$ (5)

$d=(14.16 m/s)(4.8 s)$ (6)

Finally:

$d=68 m$ (8) This is the distance of the balloon from the radar station

6 0

3 years ago