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r-ruslan [8.4K]
3 years ago
9

What is magnetic resonance ???​

Physics
2 answers:
Lemur [1.5K]3 years ago
7 0

Answer:

feu vkdu qtg

gi.rls joi.n for study here i am also a girl here we study girl is allowed here boy are not allowed here we only study so co.me for study in me.et.ing gir.ls co.me for study here its boys disturb me.eting so boys not allowed girls is allowed i am also a girl its a safe me.et.ing

MrRa [10]3 years ago
4 0

Answer:

haajajjsjajajajaaaiiajaka

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A person takes a trip, driving with a constant speed of 89.5 km/h, except for a 22.0-min rest stop. If the person’s average spee
DanielleElmas [232]

Answer:

(a): It spends on the trip 2.75 hr (2 hours 45 minutes)

(b):  The person travels 213.9 km.

Explanation:

Vp= 77.8 km/h

V1= 89.5 km/h

V2= 0 km/h

t2= 0.36hr

t1= ?

Vp= (V1*t1 + V2*t2) / (t1 + t2)

clearing t1:

t1= 2.39 hr

Trip time= t1+t2

Trip time= 2.75 hr (a)

Distance traveled= V1*t1

Distace traveled= 213.9 km (b)

7 0
3 years ago
A 1-mCi source of^60 Co is placed in the center of a cylindrical water-filled tank with an inside diameter of 20 cm and depth of
LenKa [72]

To solve this problem we need the concepts of Energy fluency and Intensity from chemical elements.

The energy fluency is given by the equation

\Psi=4RcE\pi

Where

\Psi =The energy fluency

c = Activity of the source

r = distance

E = electric field

In the other hand we have the equation for current in materials, which is given by

I= I_0 e^{-\mu_{h20}X_{h2o}} e^{-\mu_{Fe}X_{Fe}}

Then replacing our values we have that

I = 1*10^{-3} * 3.3*10^{10} * e ^{-0.06*1.1} e^{-0.058*7.861}

I = 1.3*10^7 Bq

We can conclude in this part that 1.3*10^7Bq is the activity coming out of the cylinder.

Now the energy fluency would be,

\Psi = \frac{cE}{4\pir^2}

\Psi = \frac{1.3*10^7*2*1.25}{4\pi*11^2}

\Psi = 2.14*10^4 MeV/cm^2.s

The  uncollided flux density at the outer surface of the tank nearest the source is \Psi = 2.14*10^4 MeV/cm^2.s

6 0
3 years ago
What's the diameter of a dish antenna that will receive 10−20W of power from Voyager at this time? Assume that the radio transmi
Murrr4er [49]

Complete Question:

The Voyager 1 spacecraft is now beyond the outer reaches of our solar system, but earthbound scientists still receive data from the spacecraft s 20-W radio transmitter. Voyager is expected to continue transmitting until about 2025, when it will be some 25 billion km from Earth.

What s the diameter of a dish antenna that will receive 10−20W of power from Voyager at this time? Assume that the radio transmitter on Voyager transmits equally in all directions(isotropically).  In fact, the antenna on Voyager focuses the signal in a beam aimed at the earth, so this problem over-estimates the size of the receiving dish needed.

Answer:

d = 2,236 m.

Explanation:

The received power on Earth, can be calculated as the product of the intensity (or power density) times the area that intercepts the power radiated.

As we assume that  the transmitter antenna is ominidirectional, power is spreading out over a sphere with a radius equal to the distance to the source.

So, we can get the power density as follows:

I = P /A = P / 4*π*r², where P = 20 W, and r= 25 billion km = 25*10¹² m.

⇒ I = 20 W / 4*π* (25*10¹²)² m²

The received power, is just the product of this value times the area of the receiver antenna, which we assumed be a circle of diameter d:

Pr = I. Ar =( 20W / 4*π*(25*10¹²)² m²) * π * (d²/4) = 10⁻²⁰ W

Simplifying common terms, we can solve for d:

d= √(16*(25)²*10⁴/20) = 2,236 m.

3 0
3 years ago
Coherent microwaves of wavelength 5.00 cm enter a tall, narrow window in a building otherwise essentially opaque to the microwav
zzz [600]
The solution for this problem is:

For 1st minimum, let m be equal to 1. 

d = slit width

D = screen distance. 

Θ = arcsin (m * lambda/ (d))


= 0.13934 rad, 7.9836 deg 

y = D*tan (Θ)

y = 6.50 * tan (7.9836)

= 0.91161 m is the distance from the central maximum to the first-order minimum 
8 0
3 years ago
Suppose the initial position of an object is zero, the starting velocity is 3 m/s and the final velocity was 10 m/s. The object
Nookie1986 [14]

Explanation:

We have,

The initial position of an object is zero.

The starting velocity is 3 m/s and the final velocity was 10 m/s.

The object moves with constant acceleration..

The area covered under the velocity-time graph gives displacement of the object. The correct option is "the area of the rectangle plus the area of the triangle under the line".

8 0
3 years ago
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