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iragen [17]
2 years ago
10

Identify and explain one way in which we try to deal with infiltration? What are the benefits and cost of that particular soluti

on?
Physics
1 answer:
tigry1 [53]2 years ago
5 0

Answer:

i vigilate

Explanation:

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According to the Standard Model of Particle Physics, a neutrino is a type of
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The standard model of particle physics classifies all known particles and documents three of the fundamental forces. A neutrino is an almost massless sub-atomic particle with no charge that only interacts with matter very weakly. Neutrinos are classified as fermions which means they have half-integer intrinsic spin.
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A4) A straight wire that is 0.60 m long is carrying a current of 2.0 A. It is placed in a uniform magnetic field of strength 0.3
Klio2033 [76]

Answer:

\theta=30^{\circ}

Explanation:

It is given that,

Length of the wire, L = 0.6 m

Current flowing inside the wire, I = 2 A

Uniform magnetic field, B = 0.3 T

Force experienced by the wire in the magnetic field, F = 0.18 N

To find,

The angle made by the wire with the magnetic field.

Solve,

We know that the magnetic force acting on the wire inside the magnetic field is given by :

F=ILB\ sin\theta

sin\theta=\dfrac{F}{ILB}

sin\theta=\dfrac{0.18}{2\times 0.6\times 0.3}

\theta=30^{\circ}

Therefore, the wire makes an angle of 30 degrees with respect to magnetic field.

5 0
3 years ago
The graph shows the progress of a car which traveled 300 km in 5 hrs. Assuming that its speed remained constant, how far had the
lisov135 [29]
If the car travels
300 km.........................5 hrs
?km ..............................2h
to find out we do (300*2)/5=600/5=120km
also
300km/5h=60km/h
in 2 hours
120km
6 0
3 years ago
A commuter train passes a passenger platform at a constant speed of 39.6 m/s. The train horn is sounded at its characteristic fr
Licemer1 [7]

Complete Question

A commuter train passes a passenger platform at a constant speed of 39.6 m/s. The train horn is sounded at its characteristic frequency of 350 Hz.

(a)

What overall change in frequency is detected by a person on the platform as the train moves from approaching to receding

(b) What wavelength is detected by a person on the platform as the train approaches?

 

Answer:

a

  \Delta  f  =  81.93 \ Hz

b

  \lambda_1 =  0.867 \ m

Explanation:

From the question we are told that

      The speed of the train is  v_t  =  39.6 m/s

      The frequency of the train horn is  f_t =  350 \ Hz

Generally the speed of sound has a constant values of  v_s  =  343 m/s

  Now  according to dopplers equation when the train(source) approaches a person on the platform(observe) then the frequency on the sound observed by the observer can be mathematically represented as  

        f_1 =  f *   \frac{v_s}{v_s - v_t}

substituting values

        f_1 =  350 *  \frac{343 }{343-39.6}

       f_1 =  395.7 \ Hz

  Now  according to dopplers equation when the train(source) moves away from  the  person on the platform(observe) then the frequency on the sound observed by the observer can be mathematically represented as  

           f_2 =  f *   \frac{v_s}{v_s +v_t}

substituting values

        f_2 =  350 *   \frac{343}{343  + 39.6}

       f_2 =  313.77 \ Hz

The overall change in frequency is detected by a person on the platform as the train moves from approaching to receding is mathematically evaluated as

        \Delta  f  =  f_1 - f_2

        \Delta  f  =  395.7 - 313.77

        \Delta  f  =  81.93 \ Hz

Generally the wavelength detected by the person as the train approaches  is mathematically represented  as

          \lambda_1 =  \frac{v}{f_1 }

          \lambda_1 =  \frac{343}{395.7 }

         \lambda_1 =  0.867 \ m

4 0
3 years ago
Kathy tests her new sports car by racing with Stan, an experienced racer. Both start from rest, but Kathy leaves the starting li
Fofino [41]

Answer:

(a) t=3.87 s :time at which Kathy overtakes Stan

(b) d=37.36 m

(c) vf₁ = 15.097 m/s : Stan's final speed

    vf₂ = 19.31 m/s : Kathy's final speed

Explanation:

kinematic analysis

Because Kathy and Stan move with uniformly accelerated movement we apply the following formulas:

vf= v₀+at Formula (1)

vf²=v₀²+2*a*d Formula (2)

d= v₀t+ (1/2)*a*t² Formula (3)

Where:  

d:displacement in meters (m)  

t : time in seconds (s)

v₀: initial speed in m/s  

vf: final speed in m/s  

a: acceleration in m/s²

Nomenclature

d₁: Stan displacement   

t₁ :  Stan time

v₀₁: Stan initial speed

vf₁: Stan final speed

a₁:  Stan acceleration

d₂: car displacement   

t₂ : Kathy time

v₀₂: Kathy initial speed

vf₂: Kathy final speed

a₂:  Kathy acceleration

Data

v₀₁ = 0

v₀₂ = 0

a₁ = 3.1 m/s²

a₂= 4.99 m/s²

t₁ = (t₂ +1) s

Problem development

By the time Kathy overtakes Stan, the two will have traveled the same distance:

d₁ = d₂

t₁ = (t₂ +1)

We aplpy the Formula (3)

d₁ = v₀₁t₁ + (1/2)*a₁*t₁²

d₁ = 0 + (1/2)*(3.1)*t₁²

d₁ =  1.55*t₁² ; Stan's cinematic equation 1

d₂ = v₀₂t₂ + (1/2)*a₂*t₂²

d₂ = 0 + (1/2)*(4.99)*t₂²

d₂ = 2.495* t₂² : Kathy's cinematic equation 2

d₁ = d₂

equation 1 = equation 2

1.55*t₁²  =  2.495* t₂²  , We replace t₁ = (t₂ +1)

1.55* (t₂ +1) ² = 2.495* t₂²

1.55* (t₂² +2t₂+1) = 2.495* t₂²

1.55*t₂²+1.55*2t₂+1.55 = 2.495* t₂²

1.55t₂²+3.1t₂+1.55=2.495t₂²

(2.495-1.55)t₂² - 3.1t₂ - 1.55 = 0

0.905t₂² - 3.1t₂ - 1.55 = 0  Quadratic equation

Solving the quadratic equation we have:

(a) t₂ = 3.87 s : time at which Kathy overtakes Stan

(b) Distance in which Kathy catches Stan

we replace t₂ = 3.87 s in equation 2

d₂ = 2.495*( 3.87)²

d₂ = 37.36 m

(c) Speeds of both cars at the instant  Kathy overtakes Stan

We apply the Formula (1)

vf₁= v₀₁+a₁t₁    t₁ =( t₂+1 ) s=( 3.87 + 1 ) s = 4.87 s

vf₁= 0+3.1* 4.87

vf₁ = 15.097 m/s : Stan's final speed

vf₂ = v₀₂+a₂ t₂  

vf₂ =0+4.99* 3.87

vf₂ = 19.31 m/s : Kathy's final speed

8 0
3 years ago
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