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madam [21]
3 years ago
5

Briefly discuss:

Physics
1 answer:
IrinaK [193]3 years ago
8 0

Answer:

(i) Si device are coated with B or Li for neutron detection

(ii) Fast neutrons are normally first slowed down  to thermal energies before measurement

(iii) The strip detectors consists of p⁺ and n implants in the region of a very thin (300 to 400 μm) depletion zone  though which the neutrons traverse producing a readout based on the generated charges being directed to the cathode, p⁺, material  which then transmits the charge to the device electronics

Explanation:

(i) Detection of neurons with only an Si device in not possible due to the large neutrons path in Si such that the silicon needs to be coated with B or Li which readily interact with neutrons. The neutron interaction with the reactive coating produces an alpha particle which can be detected by the semi conductor and a nucleus

(ii) Neutrons having a kinetic energy that is more than 1 MeV which as such has a velocity of more than 15,000 km/s is known as fast neutrons or fission neutrons. The fast neutrons are slowed in a nuclear reactor by neutron moderation to thermal energies

Due their high speed, fast neutrons are normally slowed down which however results in the loss of some vector properties of the neutron

Techniques for fast neutron detection includes,

1) Recoil detectors which are capable of fast neutron detection without moderation

2) Bonner spheres detector first converts the fast moving electron to slow down before detection

3) Scintillation counter are widely used but require the conversion of the neutron to a charged particles before detection

(iii) Strip detectors provide high precision measurement of a particle's crossing point which can be further improved by use of low noise electronics

Applications of silicon strip detectors include

1) Particle tracking in research in particle physics

2) Particle tracking in researches in x-spectroscopy nuclear research

3) Imaging in x-talography

4) Medical research imaging

5) Particle tracking and imaging in astrophysics.

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A 2000-kg car moving with a speed of 20 m/s collides with and sticks to a 1500-kg car at rest at a stop sign. Show that because
amid [387]

Answer:

13.33m/s

Explanation:

Given data

m1= 2000kg

u1= 20m/s

m2= 1500kg

u2= 0m/s

v1= 10m/s

Required

The speed of the sticks

We know that  from the expression for the conservation of momentum

m1u1+m2u2= m1v1+m2v2

2000*20+1500*0=2000*10+1500*v2

40000=20000+1500v2

collect like terms

40000-20000= 1500v2

20000= 1500v2

v2= 20000/1500

v2= 13.33 m/s

Hence the velocity of the sticks is 13.33m/s

8 0
3 years ago
The acceleration due to gravity on the moon is 1.6 m/s2. What is the gravitational potential energy of a 1200-kg lander resting
faltersainse [42]

Gravitational potential energy is energy an object possesses because of its position in a gravitational field. The equation for gravitational potential energy is GPE = mgh. 


GPE = 1200(1.6)(350) = 672000 J


Hope this answers the question. Have a nice day.

5 0
3 years ago
Read 2 more answers
How does the medium which a sound wave travels, affect a sound
arlik [135]

The density of <u>a medium</u> is the second factor that <u>affects the speed of sound.</u>

Please vote <u>Brainliest</u> (:

6 0
3 years ago
Please help me will mark u brainiest
Vesnalui [34]

Answer:

C. 2.5 N

Explanation:

The sum of downwards moments equals sum of moments upwards

<u>Downwards moments </u>

Moments= Force*distance

               = [10/100 * 2 ] +[20/100*4]

               =[0.1*2] +[0.2*4]

               =[0.2+0.8]

               = 1.0 N

Equate this answer to upward force ;

1.0 N = 0.4 *U

1/0.4 =U

2.5 N =U

8 0
3 years ago
A pilot heads his jet due east. The jet has a speed of 475 mi/h relative to the air. The wind is blowing due north with a speed
oksian1 [2.3K]

Explanation:

a. The velocity of the wind as a vector in component form will be represented as v vector:

    v=30j

b.The velocity of the jet relative to the air as a vector in component form will be represented as u vector

    u=475i

c. The true velocity of the jet as a vector will be represented as w:

  w=u+v

  w=475i+30j

d.  The true speed of the jet will be calculated as:

    IwI=\sqrt{(475)^2+(30)^2}

    IwI=\sqrt{225625+900}

    IwI=\sqrt{226525}

    IwI=476 mi/h

e. The direction of the jet will be:

tita=tan^{-1}\frac{30}{475}

tita=tan^{-1}(0.0632)

tita=3.62degrees,or,N86.38degreesS

7 0
3 years ago
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