What is the maximum speed (in units of m/s) with which a car can round a
flat horizontal curve of radius (r=60 m), if the coefficient of static friction between tires and
road is (0.4)?
A) 5
B) 15.5
€) 240
D) 25.1​

Explanation:

unit is necessary to communicate values of the physical quantity for example can main to someone a particular length without using some sort of unit is impossible because a length cannot be described without a reference used to make sense of the value given

Answer:

298rad/s and 116.96 ohms

Explanation:

Given an L-R-C series circuit where

L = 0.450 H,

C=2.50×10^−5F, and resistance R= 0

In this situation we have a simple LC circuit with angular frequency

Wo = 1√LC

= 1/√(0.450)(2.50×10^-5)

= 1/√0.00001125

= 1/0.003354

= 298rad/s

B) Now we need to find the value of R such that it gives a 10% decrease in angular frequency.

Wi/W° = (100-10)/100

Wi/W° = 90/100

Wi/W° = 0.90 ............... 1

Angular frequency of oscillation

The complete aspect of the solution is attached, please check.

a. The angular frequency of the circuit when R = 0 Ohms is 294.12 rad/s.

b. The value R must have to give a decrease in angular frequency of 10.0 % compared to the initial value is equal to 116.96 Ohms.

Given the following data:

a. To determine the angular frequency of the circuit when R = 0 Ohms:

Mathematically, the angular frequency of a LC circuit is given by the formula:

[tex]\omega = \frac{1}{\sqrt{LC} } \\\\\omega =\frac{1}{\sqrt{0.450 \times 2.50\times 10^{-5}}} \\\\\omega =\frac{1}{\sqrt{1.125 \times 10^{-5}}} \\\\\omega = \frac{1}{0.0034} \\\\\omega = 294.12\;rad/s[/tex]

b. To find the value R must have to give a decrease in angular frequency of 10.0 % compared to the value calculated above:

The mathematical expression is given as follows:

[tex]\frac{\omega_f}{\omega_i} = \frac{100-10}{100} \\\\\frac{\omega_f}{\omega_i} =\frac{90}{100} \\\\\frac{\omega_f}{\omega_i} =0.9[/tex]

[tex](\frac{\omega_f}{\omega_i})^2 = 1 - \frac{R^2C}{4L} \\\\0.90^2=1 - \frac{R^2C}{4L}\\\\R=\sqrt{\frac{4L(1-0.81)}{C}} \\\\R=\sqrt{\frac{4\times 0.450 \times (0.19)}{2.50\times 10^{-5}}}\\\\R = \sqrt{\frac{0.342}{2.50\times 10^{-5}} }\\\\R =\sqrt{13680}[/tex]

R = 116.96 Ohms.

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Answer:

The electric field is  [tex]E = 636 \ V/m[/tex]

Explanation:

From the question we are told that

     The magnitude of magnetic field is [tex]B = 2.12 \mu T = 2.12*10^{-6} \ T[/tex]

      The value for speed of light is  [tex]c = 3.0 *10^8 \ m/s[/tex]

Generally the magnitude of the electric field at point P is

        [tex]E = B * c[/tex]

substituting values

         [tex]E = 2.12 *10^{-6} * 3.0 *10^{8}[/tex]

         [tex]E = 636 \ V/m[/tex]

The magnitude of electric field for the wave at point P is 636 V/m.

Given data:

The strength of magnetic field at point P is, [tex]B = 2.12 \;\rm \mu T=2.12 \times 10^{-6} \;\rm T[/tex].

The speed of light is, [tex]c = 3.0 \times 10^{8} \;\rm m/s[/tex].

The given problem is based on the concept of electric field and magnetic field. The electromagnetic wave works on the principle of oscillating magnetic field and electric field at the same region. We can find any of the two using the expression,

[tex]E = B \times c[/tex]

here,

E is the strength of electric field.

Solving as,

[tex]E = (2.12 \times 10^{-6}) \times (3 \times 10^{8})\\\\E = 636 \;\rm V/m[/tex]

Thus, we can conclude that the magnitude of electric field for the wave at point P is 636 V/m.

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Answer:

Explanation:

The e.m.f induced in the coil depend on the following :

(a) No. of turns in the coil

(b) Cross-sectional Area of the coil

(c) Magnitude of Magnetic field

(d) Angular velocity of the coil

Answer:

The correct choice is B & E.  

Explanation:

A solenoid is a coil of wire (usually copper) which is used as an electromagnet. Solenoids are used to convert electrical energy to mechanical energy. When this type of device is created it is also called a solenoid. One can increase the energy density within the solenoid or the coil by upping the electric current in the coil.

Cheers!

Answer:

[tex]v_B/v_A=\sqrt{3}[/tex]

Explanation:

Consider the two kinematic equations for velocity and position of an object falling due to the action of gravity:

[tex]v=-g\,t\\ \\position=-\frac{1}{2} g\,t^2[/tex]

Therefore, if we consider [tex]t_A[/tex] the time for the object to reach point A, and [tex]t_B[/tex] the time for it to reach point B, then:

[tex]v_A=-g\,t_A\\v_B=-g\,t_B\\\frac{v_B}{v_A}= \frac{-g\,t_B}{-g\,t_A} =\frac{t_B}{t_A}[/tex]

Let's work in a similar way with the two different positions at those different times, and for which we have some information;

[tex]x_A=-x=-\frac{1}{2}\, g\,t_A^2\\x_B=-3\,x=-\frac{1}{2}\, g\,t_B^2\\ \\\frac{x_B}{x_A} =\frac{t_B^2}{t_A^2} \\\frac{t_B^2}{t_A^2}=\frac{-3\,x}{-x} \\\frac{t_B^2}{t_A^2}=3\\(\frac{t_B}{t_A})^2=3[/tex]

Notice that this quotient is exactly the square of the quotient of velocities we are looking for, therefore:

[tex](\frac{t_B}{t_A})^2=3\\(\frac{v_B}{v_A})^2=3\\ \frac{v_B}{v_A}=\sqrt{3}[/tex]

0.6764*10^-10m

Explanation:

Using E= hc/wavelength

(4.14x10^-15)x(3.0x10^8)/(65x10^-12)=0.1911x10^5 eV=19.1 keV

So subtract the calculated energy from the given energy of scattered photons

9.11-0.75=18.36 keV

To find wavelength

Wavelength= hc/ E

[(4.14x 10^-15)x (3.0x10^8)]/(18.36*10^3) =0.6764^-10 m

Answer:

θ = 0.14°

Explanation:

Here we will use the grating equation. The grating equation is as follows:

mλ = d Sin θ

where,

θ = angle = ?

m = order number = 3

λ = wavelength of light = 577 nm = 5.77 x 10⁻⁷ m

d = spacing between slits = 1/(1420 lines/cm) = 7.042 x 10⁻⁴ m

Therefore, using these values, we get:

(3)(5.77 x 10⁻⁷ m) = (7.042 x 10⁻⁴ m)Sin θ

Sin θ = (3)(5.77 x 10⁻⁷ m)/(7.042 x 10⁻⁴ m)

Sin θ = 2.46 x 10⁻³

θ = Sin⁻¹(2.46 x 10⁻³)

θ = 0.14°

Answer:

the voltage drop across this same diode will be 760 mV

Explanation:

Given that:

Temperature T = 300°K

current [tex]I_1[/tex] = 100 μA

current [tex]I_2[/tex] = 1 mA

forward voltage [tex]V_r[/tex] = 700 mV = 0.7 V

To objective is to find the voltage drop across this same diode  if the bias current is increased to 1mA.

Using the formula:

[tex]I = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}[/tex]

[tex]I_1 = I_o \begin {pmatrix} e^{\dfrac{V_r}{nv_T}-1} \end {pmatrix}[/tex]

where;

[tex]V_r[/tex] = 0.7

[tex]I_1 = I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix}[/tex]

[tex]I_2 = I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix}[/tex]

[tex]\dfrac{I_1}{I_2} = \dfrac{ I_o \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ I_o \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }[/tex]

[tex]\dfrac{100 \ \mu A}{1 \ mA} = \dfrac{ \begin {pmatrix} e^{\dfrac{0.7}{nv_T}-1} \end {pmatrix} }{ \begin {pmatrix} e^{\dfrac{V_r'}{nv_T}-1} \end {pmatrix} }[/tex]

Suppose n = 1

[tex]V_T = \dfrac{T}{11600} \\ \\ V_T = \dfrac{300}{11600} \\ \\ V_T = 25. 86 \ mV[/tex]

Then;

[tex]e^{\dfrac{V_r'}{nv_T}-1} = 10 \begin {pmatrix} e ^{\dfrac{ 0.7} { nV_T} -1} \end {pmatrix}[/tex]

[tex]e^{\dfrac{V_r'}{nv_T}-1} = 10 \begin {pmatrix} e ^{\dfrac{ 0.7} { 25.86} -1} \end {pmatrix}[/tex]

[tex]e^{\dfrac{V_r'}{nv_T}-1} = 5.699 \times 10^{12}[/tex]

[tex]{e^\dfrac{V_r'}{nv_T}} = 5.7 \times 10^{12}[/tex]

[tex]{\dfrac{V_r'}{nv_T}} =log_{e ^{5.7 \times 10^{12}}}[/tex]

[tex]{\dfrac{V_r'}{nv_T}} =29.37[/tex]

[tex]V_r'=29.37 \times nV_T[/tex]

[tex]V_r'=29.37 \times 25.86[/tex]

[tex]V_r'=759.5 \ mV[/tex]

[tex]Vr' \simeq[/tex] 760 mV

Thus, the voltage drop across this same diode will be 760 mV

Answer:

   θ = 10.28º

Explanation:

To find the angle of refraction use the equation of refraction

        n₁ sin θ₁ = n₂ sin θ₂

where index 1 is for incident light and index 2 is for refracted light.

         sin θ₂ = n₁ / n₂ sin θ

let's calculate

         sin = 1 / 1.3 sin 0.23

         sin = 0.175

        θ= 0.17528 rad

let's reduce to degrees

       θ = 0.17528 rad (180ª / pi rad)

       θ = 10.28º

Answer:

a)  Q = 397.57 pC , Q = 3.18 104 pC , b) C = 1.157 10⁻¹⁰ F ,  V = 3.4375 V ,

c)  U = 54.7 nJ ,  d) ΔU = 54 nJ,

Explanation:

a) The capacity of a capacitor is defined

        C = Q / V

        Q = C V

can also be calculated using geometry consideration

        C = e or A / d

we reduce to the SI system

       A = 25.0 cm² (1 m / 10² cm) 2 = 25.0 10⁻⁴ m²

       d = 1.53 cm = 1.53 10⁻² m

we substitute

         Q = eo A / d V

         Q = 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻² 275

         Q = 3.9757 10⁻¹⁰ C

let's reduce to pC

         Q = 3.9757 10⁻¹⁰ C (10¹² pC / 1 C)

          Q = 397.57 pC

when the capacitor is introduced into the water the dielectric constant is different

           Q = k Q₀

           Q = 80 397.57

           Q = 3.18 104 pC

b) Find capacitance and voltage after submerged in water

           C = k C₀

           C = 80 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²

           C = 1.157 10⁻¹⁰ F

           V = Vo / k

            V = 275/80

            V = 3.4375 V

c) The stored energy is

             U = ½ C V²

              U = ½, 85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²     275²

             U = 5.47 10⁻⁸ J

let's reduce to nJ

              109 nJ = 1 J

               U = 54.7 nJ

d) energy after submerging

             U = ½ (kCo) (Vo / k) 2

             U = ½ Co Vo2 / k

             U = U₀ / k

             U = 54.7 / 80 nJ

              U = 0.68375 nJ

the energy change is

         ΔU = U₀ -U

          ΔU = 54.7 - 0.687375

(a) Charge on the plate before immersion, Qi is 5.258 x 10⁻³ pC and the charge after, Qf is 0.421 pC.

(b) The capacitance and potential difference after immersion is 1.157 x 10⁻¹⁰ F and 3.44 V respectively.

(c) The change in energy of the capacitor is 54.02 nJ.

The charge on the plate is calculated as follows;

[tex]Q =\frac{\varepsilon _o A}{Vd} \\\\Q_i = \frac{8.85 \times 10^{-12} \times (25 \times 10^{-4}) }{275\times 0.0153} \\\\Q_i = 5.258 \times 10^{-15} \ C\\\\Q_i = 5.258 \times 10^{-3} pC[/tex]

[tex]Q_f = k Q_i\\\\Q_f = 80 \times 5.258 \times 10^{-3} \ pC= 0.421 \ pC[/tex]

[tex]C = \frac{k\varepsilon _o A}{d} \\\\C = \frac{80 \times 8.85 \times 10^{-12} \times (25\times 10^{-4} )}{0.0153} \\\\C = 1.157 \times 10^{-10} \ F[/tex]

[tex]V = \frac{V_0}{k}\\\\V = \frac{275}{80} \\\\V = 3.44 \ V[/tex]

The initial energy of the capacitor is calculated as follows;

[tex]U_i = \frac{1}{2} CV^2\\\\U_ i = \frac{1}{2} \times (\frac{\varepsilon _o A}{d} )V^2\\\\U_i = \frac{1}{2} \times (\frac{8.85\times 10^{-12} \times 25 \times 10^{-4}}{0.0153} )\times 275^2\\\\U_i = 5.47 \times 10^{-8} \ J\\\\U_i = 54.7 \ nJ[/tex]

The final energy of the capacitor is calculated as follows;

[tex]U_f = \frac{1}{2} (kC) \times (\frac{V}{k} )^2\\\\U_f = \frac{1}{2} C\times \frac{V^2}{k} \\\\U_f = \frac{1}{k} (\frac{1}{2} CV^2)\\\\U_f = \frac{U_i}{k} \\\\U_f = \frac{54.7 \ nJ}{80} \\\\U_f = 0.68 \ nJ[/tex]

Change in energy is calculated as follows;

[tex]\Delta U = U_i - U_f \\\\\Delta U = 54.7 \ nJ \ - \ 0.68 \ nJ\\\\\Delta U = 54.02 \ nJ[/tex]

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Answer:

The direction of net magnetic force on the circular section of the loop is in the positive y-axis

The net magnetic force on the circular section of the loop is 3.74 N

Explanation:

The magnetic field strength [tex]B[/tex] = 1.7 T

the current [tex]I[/tex] = 0.7 A

The diameter of the loop = 2 m

the length of the circular section of the semi-circular loop [tex]l[/tex] = πd/2

==> [tex]l[/tex] = (3.142 x 2)/2 = 3.142 m

The force on the semi-circular is given as

F = [tex]BIl[/tex] sin ∅

but the loop is perpendicular to the field, therefore

sin ∅ = sin 90° = 1

F = 1.7 x 0.7 x 3.142 x 1 = 3.74 N

The right hand rule states that "if the fingers of the right hand are held parallel to each other in the direction of the magnetic field, and the thumb is held at right angle to the other fingers in the direction of the flow of current. The palm will push in the direction of the magnetic force on the conductor".

According to the right hand rule, the direction of net magnetic force on the circular section of the loop is in the positive y-axis

Answer:

D. A convex lens in air

Explanation:

This is because the air tight plastic under water will reflect light rays in the same manner as a convex lens

Answer:

2.2nC

Explanation:

Call the amount by which the spring’s unstretched length L,

the amount it stretches while hanging x1

and the amount it stretches while on the table x2.

Combining Hooke’s law with Newton’s second law, given that the stretched spring is not accelerating,

we have mg−kx1 =0, or k = mg /x1 , where k is the spring constant. On the other hand,

applying Coulomb’s law to the second part tells us ke q2/ (L+x2)2 − kx2 = 0 or q2 = kx2(L+x2)2/ke,

where ke is the Coulomb constant. Combining these,

we get q = √(mgx2(L+x2)²/x1ke =2.2nC

Answer:

1988.05 rad/s^2

Explanation:

The angular speed of the optical disk ω = 998.0 rad/s

the time taken to come to rest t = 0.502 s

The magnitude of the average angular acceleration ∝ = ω/t

∝ = 998.0/0.502 = 1988.05 rad/s^2

Answer:

1,860 g

Explanation:

In a system, the coefficient of static friction is usually higher than the coefficient of kinetic friction. This means that the kinetic friction is usually less than the static friction. From the question, since the book is already sliding, it means that kinetic friction is the friction in play. This means that before the reading on the scale that could have been possible at the moment the student overcame the static friction must be greater than the reading on the scale during sliding. The only option above 1580 g is 1860 g

Answer:

Approximately 5 x 10^18 electrons are delivered to the smart phone charger.

Explanation:

The electric current in a circuit is the flow of charges through a circuit with time.

The charges through the circuit are due to the electrons that flow through the circuit.

An individual electrons has a charge of -1.60 x 10^-19 C on it.

If the current through the circuit is -0.75 C/s, then the number of electrons that are delivered is gotten by dividing the charge per second by the charge on an electron.

==> -0.75/(-1.60 x 10^-19) = 4.67 x 10^18 electrons ≅ 5 x 10^18 electrons are delivered to the smart phone charger.

Answer:

A) Workdone = 223.57 N-m

B) 22357 J of energy

C) Number of donuts = 10.7 donuts

Explanation:

A) The work done is calculated from the formula;. Work done = Force × Distance

We are given;

Mass; m = 43 kg

Distance = 0.53 m

Force(weight) = mg = 43 × 9.81

Thus;

Work done = 43 × 9.81 × 0.53

Workdone = 223.57 N-m

B) We are told she does 23 repetitions a day.

Thus, we assume 23% efficiency.

So, Work = Energy

Thus;

At 100% efficiency;

Energy = (223.57/100%) × 23 repetitions = 5142.11 J

Now, since she is only 23% efficient, she will expend; 5142.11/0.23 J = 22357 J of energy to do 5390 J of work.

C) from conversions; 4.18 J = 1 calorie

Thus;

22357 J ÷ 4.18 J/cal = 5348.565 calories

We how many 500 calorie donuts she can eat in a day to supply that energy.

Thus;

Number of donuts = 5348.565 cal ÷ 500 cal /donut

Number of donuts = 10.7 donuts

Answer:

The glove takes 1.02s to return to the pitchers hand.

Explanation:

Given;

initial velocity the pitcher's glove, u = 5 m/s

Apply kinematic equation

s = ut - ¹/₂gt²

where;

g is acceleration due to gravity = 9.8 m/s²

t is the time takes the glove to return to the pitchers hand

s is the displacement of the glove, which will be equal to zero when the glove returns to the pitchers hand. (s = 0)

0 = ut - ¹/₂gt²

ut = ¹/₂gt²

u = ¹/₂gt

gt = 2u

t = (2u) / g

t = (2 x 5) / 9.8

t = 1.02 s

Therefore, the glove takes 1.02s to return to the pitchers hand.

Answer:

   t = t₀ / 2

Explanation:

In this exercise we must use Newton's second law

          F = m a

          a = F / m

now we can use kinematics

  as in object part of rest (v₀ = 0)

        v =a t₀

        t₀ = v / a

these results are with the first experiment

now repeat the experiment, but F = 2F₀

           a = 2F₀ / m = 2 a₀

          v = 2 a₀ t

          t = v / 2a₀

          t = t₀ / 2

The time interval that is required to reach the same final speed (V) is equal to [tex]t=\frac{\Delta t}{2}[/tex].

Given the following data:

To find the time interval that is now required to reach the same final speed (V), we would apply Newton's Second Law of Motion:

Mathematically, Newton's Second Law of Motion is given by this formula;

[tex]F = \frac{M(V-U)}{t}[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]F = \frac{M(V-0)}{\Delta t}\\\\F = \frac{MV}{\Delta t}[/tex]

When the experiment is repeated, the magnitude of the force is doubled:

[tex]F = 2F[/tex]

Now, we can find the time interval that is required to reach the same final speed (V):

[tex]F = \frac{M(V-0)}{t}\\\\t=\frac{MV}{F}[/tex]

Substituting the value of F, we have:

[tex]t=\frac{MV}{2F} \\\\t=\frac{MV}{\frac{2MV}{\Delta t}} \\\\t=MV \times \frac{\Delta t}{2MV} \\\\t=\frac{\Delta t}{2}[/tex]

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Answer: F

Out of the page.

Explanation:

For an electron with a charge of -e, the magnitude of the force on it is F = BeV

Where

F = force on the electron

e = charge ( electrons )

V = velocity

B = magnetic field

F is the force acting on all the electrons in a wire which gives rise to the F = BIL

Where

I = current

L = length of the wire

The force F is always at the right angle to the particle's velocity and its direction can be found using the left hand rule.

When the electron is moving in the plane of the page in the direction indicated by the arrow, the force on the electron is directed out of the page.

Answer:

103nm

Explanation:

Pls see attached file

Answer:

60 Ω

Explanation:

R(com) = 15 Ω

1/R(com) = 1/R1 + 1/R2 + 1/R3 ..... + 1/Rn

1/15 = 1/20 + 1/R2

1/R2 = 1/15 - 1/20

1/R2 = (4 - 3) / 60

1/R2 = 1/60

R2 = 60 Ω

así, la combinada de resistencia necesaria es 60 Ω

Answer:

C) radio waves, microwaves, infrared, visible, ultraviolet, x-rays, gamma rays

Explanation:

radio waves have lowest  energy , lowest  frequency and highest  wavelength

gamma rays  have highest  energy , highest  frequency and least  wavelength

Answer: C

Explanation:

Answer:

382Hz

Explanation:

The question lacks the required option. Find the complete question in the attachment.

The long organ pipe open at both ends is called an open pipe. The fundamental frequency for an open pipe is expressed as F0 = V/2L

Harmonics are integral multiples of the fundamental frequency. For open pipes its harmonics are 2fo, 3fo, 4fo, 5fo...

Given fundamental frequency f0 to be 85 Hz, the following frequencies will be present as a standing wave;

First overtone f1 = 2fo = 2(85) = 170Hz

Second overtone f2 = 3fo = 3(85) = 255Hz

Third overtone = 4fo = 4(85) = 340Hz

Based on the option it can be seen that the only frequency that is not present as a standing wave is 382Hz

Explanation:

Distance = speed × time

d = (18.0 m/s) (1 hr × 3600 s/hr)

d = 64,800 m

d = (18.0 m/s) (1 min × 60 s/min)

d = 1080 m

d = (18.0 m/s) (1 s)

d = 18.0 m

Answer:

reviewing the final statements, the correct one is the quarter

The nail exerts an equal force on the hammer in the opposite direction.

Explanation:

This is an action-reaction problem or Newton's third law, which states that forces in naturals occur in pairs.

This is the foregoing, the hammer exerts a force on the nail of magnitude F and it will direct downwards, if we call this action and the nail exerts a force on the hammer of equal magnitude but opposite direction bone directed upwards, each force is applied in one of the bodies.

The difference in result that each force is that the force between the nail exerts a very high pressure (relation between the force between the nail area), instead the area of ​​the hammer is much greater, therefore the pressure is small.

When reviewing the final statements, the correct one is the quarter

The nail exerts an equal force on the hammer in the opposite direction.

Answer:

a) 1.35 x 10^11 m

b) 0.53 µs

c) 8 ns

Explanation:

Radar involves the use of radio wave which has speed c = 3 x 10^8 m/s

a) for 900 s,

The distance for a round trip = v x t

==>  (3 x 10^8) x 900 =  2.7 x 10^11 m

The distance of Venus is half this round trip distance = (2.7 x 10^11)/2 = 1.35 x 10^11 m

b) for a 80.0 m distance of the car from the radar source, the radar will travel a total distance of

d = 2 x 80 = 160 m

the time taken = d/c = 160/(3 x 10^8) = 5.3 x 10^-7 s = 0.53 µs

c) accuracy in distance Δd = 11.5 m

Δt = accuracy in time = Δd/c = 11.5/(3 x 10^8) = 3.8 x 10^-8 = 38 ns

Explanation:

Pressure is the same for both plungers.

P = P

F / A = F / A

F / (¼ π d²) = F / (¼ π d²)

F / d² = F / d²

5 N / (0.05 m)² = F / (1 m)²

F = 2000 N

None of the options are correct.

Answer:

16 times.

Explanation:

The rate of the radiation dose is , R = 200 ×10^{-3} rad/hr

Time consumed, t = 40 hr

The magnitude of Q.F for the neutrons, Q.F = 2

Thus the effective radiation dose is:

[tex]R_{Eff} = Rt(Q.F) \\= 200 \times 10^{-3} \frac{rad}{hr} (40hr)(2) \\= 16 \ rad[/tex]

Thus, the effective dose allowable yearly = 16 times