In a single-slit experiment, the slit width is 170 times the wavelength of the light. What is the width (in mm) of the central maximum on a screen 2.4 m behind the slit?

Answer:

Gas Tungsten Arc Welding (GTAW) Resistance Spot Welding (RSW)

Explanation:

In the spot resistance welding the electrode comes into contact with the pieces of the workpiece (usually thin sheets upto 3mm) from both the sides of the workpiece and generates the heat of electric power without any spark intended. The filler material may be placed between the two joining pieces. This generates only the weld spots and not a continuous wekd seam.

In the other welding techniques the heat is generated via the non-solid energy beams such as electron beam in EBM, plasma in PAW, gas flame in Oxyacetylene welding.

Answer:

true

Explanation:

A = πr²

Differentiating in respect to r

we get

dA/dr = 2πr

since r = 4feet

so ,

2π(4) = 8π

Answer:

speed of the arrow as it leaves is 41.05 m/sec

Explanation:

We have given mass m = 0.285 kg

Average force F = 182 N

Distance traveled d = 1.32 m

We know that work done = force [tex]\times[/tex]distance

Sp work done = [tex]182\times 1.32=240.24J[/tex]

Now according to  work energy theorem work done will be equal to kinetic energy

So [tex]\frac{1}{2}mv^2=240.24[/tex]

[tex]\frac{1}{2}\times 0.285\times v^2=240.24[/tex]

[tex]v^2=1685.89[/tex]

v = 41.05 m/sec

So speed of the arrow as it leaves is 41.05 m/sec

Answer:

A combined wave of extra height will produce every other wave.

Explanation:

One wave Crest can catch up to another wave Crest if the group of waves are from different places.

Here the two group of waves have different wavelength but travelling together in the same direction so when they combine,they will produce every other wave.

The boat travels with a speed of 10m/s

Explanation:

Given-

Acceleration, a = 2m/s²

Time, t = 5s

Speed, s = ?

We know,

[tex]Acceleration = \frac{speed}{time} \\\\s = at[/tex]

[tex]s = 5 * 2m/s\\s = 10m/s[/tex]

Therefore, Boat travels with a speed of 10m/s

Answer:

0.7 secs

Explanation:

In this question, the speed does not change as the mass changes. So we can use

Δt=Δ∨x/χgμ............................equ 1

To stop, the final speed will be 0

Therefore,

Δvx=vf-vt

Δvx=0-4m/s

=     -4m/s

Now substitute the various values in equ 1

Δt=Δ∨x/χgμ

Δt=  -4m/s/(9.8m/s∧2) (0.6)

Δt=0.7 secs

Answer:

Explanation:

The box stops at zero speed.

Final Velocity = 0 ,Initial speed (s)= -4 m/s

Therefore=  change in velocity =  Vf - Vi.  ( 0 m/s- 4 m/s) =  -4 m/s

Change in velocity  = -0.4 m/s

Gravity  g = 9.8 m/s^2

Mass= 0.8 g

-4 ms divided by 9.8 ms^2 * (0.8) = 0.51 s

It takes 0.51 seconds to stop

Answer:

The jet will fly 2400 km.

Explanation:

Given the velocity of the jet flying toward the east is 1,500 kmph toward the east.

We need to find the distance covered in 1.6 hours.

In our problem we are given speed and time, we can easily determine the distance using the following formula.

[tex]Distance=Speed\times Time[/tex]

[tex]Distance=1500\times 1.6=2400\ km[/tex]

So, the supersonic jet will travel 2400 km in 1.6 hours toward the east from its starting point.

Answer:Arctic circle

Explanation:

Arctic circle is one of the circles of latitude. It is located at the northern region if the equator and passes through more than one continent. These continent includes: North Asia,

Northern America, and

Europe.

In these continent, it passes through eight known counties. These countries includes: Canada, Russia, Norway, Sweden, Finland, Denmark, Alaska and Iceland. It covers 4% of the Earth surface and it's climatic conditions are extreme.

Answer:

2.8 rad/s

Explanation:

In absence of external forces, the total angular momentum of the system must be conserved.

The angular momentum when the arms of the student are extended horizontally is given by:

[tex]L_1 = (I_0 + 2I)\omega_1[/tex]

where:

[tex]I_0=3.8 kg m^2[/tex] is the moment of inertia of the student+stool

[tex]I=mr^2[/tex] is the moment of inertia of each mass, where:

m = 2.6 kg is one mass

r = 0.71 m is the distance of each mass from the rotation axis

[tex]\omega_1=1.8 rad/s[/tex] is the initial angular velocity

So we have

[tex]L_1=(I_0+2mr^2)\omega_1[/tex]

When the student pulls the weights to a distance of r' = 0.23 m, the angular momentum is:

[tex]L_2=(I_0+2I')\omega_2[/tex]

where:

[tex]I'=mr'^2[/tex] is the new moment of inertia of each mass, with

r' = 0.23 m

Since the angular momentum must be constant, we have:

[tex]L_1=L_2\\(I_0+2mr^2)\omega_1 = (I_0+2mr'^2)\omega_2[/tex]

and solving for [tex]\omega_2[/tex], we find the final angular velocity:

[tex]\omega_2 = \frac{I_o+2mr^2}{I_0+2mr'^2}\omega_1=\frac{3.8+2(2.6)(0.71)^2}{3.8+2(2.6)(0.23)^2}(1.8)=2.8 rad/s[/tex]

Answer:

4.28405172412 eV

Explanation:

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]

[tex]\lambda[/tex] = Wavelength = 290 nm

Energy is given by

[tex]E=h\dfrac{c}{\lambda}\\\Rightarrow E=6.626\times 10^{-34}\times \dfrac{3\times 10^{8}}{290\times 10^{-9}}\\\Rightarrow E=6.8544827586\times 10^{-19}\ J[/tex]

Converting to eV

[tex]\dfrac{6.8544827586\times 10^{-19}}{1.6\times 10^{-19}}=4.28405172412\ eV[/tex]

The threshold energy is 4.28405172412 eV

Final answer:

The threshold photon energy for DNA damage is approximately 4.29 eV, calculated using the relationship E = h * c /
, where h is Planck's constant, c is the speed of light, and
is the given threshold wavelength for UV damage to DNA (290 nm).

Explanation:

To calculate the threshold photon energy in electron volts (eV) necessary to cause damage to DNA, we can use the relationship between a photon's energy (E), its wavelength (
), and Planck's constant (h). The equation is E = h * c /
, where c is the speed of light. Using the given wavelength of 290 nm (which is the threshold for significant DNA damage), we can convert this value into meters (290 nm = 290 x 10-9 m). Planck's constant (h) is approximately 6.626 x 10-34 J*s, and the speed of light (c) is about 3.00 x 108 m/s.

First, we find the energy in joules:
E = (6.626 x 10-34 J*s) * (3.00 x 108 m/s) / (290 x 10-9 m) = 6.863 x 10-19 J

Next, we convert joules to electron volts using the conversion factor 1 eV = 1.602 x 10-19 J:

E = (6.863 x 10-19 J) / (1.602 x 10-19 J/eV)
4.29 eV

Therefore, the threshold photon energy for DNA damage is approximately 4.29 eV.

Answer:

Answer is 717 N . m

Refer below for the explanation.

Explanation:

As per the question,

38 mm diameter shaft,

Yield strength 250 mpa,

P 240kn.

Refer to the picture for complete explanation.

The maximum torque that the steel shaft can withstand before yielding, according to the maximum-shearing-stress criterion, is 8224 N.m.

The problem can be solved by first calculating the maximum shear stress that the steel shaft can withstand before yielding. The shear strength is given by the yield strength divided by the square root of 3, as the maximum-shearing-stress criterion indicates that failure occurs when the maximum shear stress equals half the yield stress. Therefore, the shear strength τ of the steel shaft is τ = 250 MPa/√3 = 144.34 MPa.  

Next, we can find the maximum torque T that the shaft can handle before yielding by using the formula T=τ*J/r, where J is the polar moment of inertia and r is the radius. Because the shaft is a circular cross section, its polar moment of inertia J = π*r⁴/2. Substituting the given diameter of the shaft d=38mm, we find r = d/2 = 19mm, so J = π*(19mm)⁴/2 = 1075.21 mm⁴.  

Thus, the critical torque T causing yield by the maximum-shearing-stress criterion can be calculated: T = τ*J/r = (144.34 MPa)*(1075.21 mm⁴)/(19 mm)=8224 N.m

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Answer:

0.00970 s

Explanation:

The centripetal force that causes the charge to move in a circular motion = The force exerted on the charge due to magnetic field

Force due to magnetic field = qvB sin θ

q = charge on the particle = 5.4 μC

v = velocity of the charge

B = magnetic field strength = 2.7 T

θ = angle between the velocity of the charge and the magnetic field = 90°, sin 90° = 1

F = qvB

Centripetal force responsible for circular motion = mv²/r = mvw

where w = angular velocity.

The centripetal force that causes the charge to move in a circular motion = The force exerted on the charge due to magnetic field

mvw = qvB

mw = qB

w = (qB/m) = (5.4 × 10⁻⁶ × 2.7)/(4.5 × 10⁻⁸)

w = 3.24 × 10² rad/s

w = 324 rad/s

w = (angular displacement)/time

Time = (angular displacement)/w

Angular displacement = π rads (half of a circle; 2π/2)

Time = (π/324) = 0.00970 s

Hope this Helps!!!

Given, Two boxes(Let's say A and B) are connected horizontally by a light rope. Coefficient of kinetic friction between each box is 0.30. Mass of Box B is 5.00 kg and Box A is 'm'. The force is 40.0 N in the direction [tex]53.1^{o}[/tex].

To find the tension in the rope and the mass of one of the boxes, Newton's Second Law is applied separately to each box, accounting for tension, friction, and the applied force. By resolving the applied force and calculating the friction, we can solve for tension and then for the unknown mass.

To solve for these, we must apply Newton's Second Law of Motion (F = ma) to each box separately.

Part A: Tension T in the Rope

For Box B (5.00 kg), which has the force applied:

Part B: Mass m of Box A

For Box A (with unknown mass m):

Appropriate use of Newton's laws and kinetic friction calculations will yield answers to both parts of the question.

Answer:

The correct option is

(e)either (c) or (d) could be correct.

Explanation:

The electric field of a charge radiates out in all directions and the intensity of the electric field strength given by E = F/q₀, diminishes as the lines of force moves further away from the source. The direction of F and E is in the line of potential motion of the source charge in the field.

Equipotential surfaces are locations in the radiated electric that have the same field strength or electric potential. The work done in moving within an equipotential surface is zero and as such since

Work = Force × distance = 0 where distance ≠ 0.

The force acting between two points on an equipotential surface is also zero or the component of the force within an equipotential surface is zero and since there is a force in the electric field, it is acting at right angles to the equipotential surface which could be horizontally to the left or right directions where the equipotential surfaces due to the charge distribution are in the vertical plane.

Therefore it is either horizontally to the left, or horizontally to the right.

Answer:

[tex]W=1.22*10^{-23}J[/tex]

Explanation:

Torque and energy of an electric dipole in an electric field we find:

[tex]W=U(\alpha_{o}+\pi )-U(\alpha_{o} )=-pE(cos(\alpha_{o}+\pi )-cos(\alpha_{o} ))\\W=2pECos\alpha_{o}\\ W=2(3.02*10^{-25}C.m )(46.oN/C)Cos64\\W=1.22*10^{-23}J[/tex]

The work required to turn an electric dipole 180° in a uniform electric field is -4.89 × 10^-24 J.

To calculate the work required to turn an electric dipole 180° in a uniform electric field, we can use the formula:

W = -pE(1 - cosθ)

where W is the work done, p is the dipole moment, E is the electric field strength, and θ is the angle between the dipole moment and the electric field.

Plugging in the given values:

W = - (3.02 × 10-25 C·m)(46.0 N/C)(1 - cos(180° - 64°))

Simplifying the equation gives the work done to be -4.89 × 10-24 J.

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Answer:

1920 J

Explanation:

Answer:

19.6 L

Explanation:

In this process we have an isothermal process, a transformation in which the temperature of the gas remains constant.

In an isothermal process, the work done by the ideal gas on the surrounding is:

[tex]W=nRT ln\frac{V_f}{V_i}[/tex]

where:

n is the number of moles of the gas

R is the gas constant

T is the absolute temperature of the gas

[tex]V_i[/tex] is the initial volume of the gas

[tex]V_f[/tex] is the final volume

In this problem, we have:

n = 1.44 mol

T = 258 K is the gas temperature

[tex]V_i=14.5 L[/tex] is the initial volume of the gas

[tex]V_f=27.0 L[/tex] is the final volume

Solving the equation for W, we find the work done by the gas:

[tex]W=(1.44)(8.314)(258)ln\frac{27.0}{14.5}=1920 J[/tex]

The average speed taken by Sam to travel 100 miles in 3 hours is 33.33 miles per hour.

Explanation:

Step 1 :

Distance traveled by Sam  = 100 miles

Time taken by Sam to travel the distance = 3 hours

We need to calculate the Sam's average speed in traveling 100 miles for three hours.

Step 2 :

Total distance traveled  =  speed × total time taken

Hence we have speed = total distance divided by total time taken

Speed = 100/3 = 33.33 miles per hour

Step 3 :

Answer :

The average speed taken by Sam to travel 100 miles in 3 hours is 33.33 miles per hour.

Answer:

never, current is not induced

Explanation:

The induced emf in the ring equals the rate of change of magnetic flux in the ring.

E = -dФ/dt = -dAB/dt = -BdA/dt since B the magnetic field is constant.

E = -BdA/dt

Now dA/dt = dA/dy × dy/dt where dA/dy = rate of change of area with vertical distance as the ring is falling. dy/dt = speed of ring.

Since the ring is falling freely before being caught, its speed v is gotten from v = u + at where u = 0 and a = -g

v = 0 - gt = -gt

v = -gt  

So, dA/dt = dA/dy × dy/dt = vdA/dy = -gtdA/dy

So E = -BdA/dt = -B × -gtdA/dy = BgtdA/dy

Since dA/dy = 0 since the area of the ring does not change with vertical distance. So,

E = BgtdA/dy = Bgt × 0 = 0

E = 0

So, emf is never induced because the flux through the ring remains constant

Final answer:

Bending the knees when landing from a jump increases the impact time and decreases the force experienced by the body, as per the impulse-momentum theorem. This technique prevents injuries by distributing the force over a longer period, thus reducing the force's intensity on the bones.

Explanation:

When we jump from a height to the ground and bend our knees upon landing, we are effectively reducing the impact force experienced by our body. This is explained by the impulse-momentum theorem, which states that the change in momentum (or impulse) of an object is equal to the force applied multiplied by the time the force is applied. The momentum of a person jumping from a height is quite large due to their mass and the speed at which they are moving toward the ground due to gravity. If we land with straight legs, that momentum is transferred very quickly to our legs and the ground, which results in a very high force over a short time. This can lead to injuries as bones can fracture if the force on them is too large.

By bending our knees, we increase the time over which our momentum is brought to zero, which means the force exerted on our legs and subsequently on the bones is lessened, as force is inversely proportional to the time over which the momentum changes. This principle applies to various real-world situations like rolling on the ground after a jump, using a parachute, or even the crumple zones in cars, all designed to extend the impact time and reduce the force felt by the occupants.

The gravitational attraction between Earth and the person causes the person to accelerate towards the Earth. However, due to Earth's significantly greater mass, Earth's movement is negligible compared to that of the person. This results in the person experiencing a much greater acceleration and subsequently a larger change in momentum when they hit the ground after a jump.

Answer:

The answer is A, primary/primary.

Refer below for the explanation.

Explanation:

The multipoint grounded neutral is intended to reduce the primary neutral voltage drop, assist in clearing utility line-to-neutral faults, and reduce elevated voltage caused by line-to-ground faults.

Answer:

the ball takes 1.53 s to reach the top of its trajectory.

Explanation:

given information:

the speed, v = - 15 m/s (moving upward)

(a) How much time does it take for the ball to reach the top of its trajectory?

we know that the speed for the vertical motion is

v = v₀ - gt, v₀ = 0

where

v = speed (m/s)

g = gravitational force (9.8 m/s²)

t = time (s)

thus

v = - gt

-15 = - 9.8 t

t = 15/9.8

 = 1.53 s

so, the time that is needed by the ball to reach the top its trajectory is 1.53 s

Answer:

The ball takes 1.53 seconds to reach its top trajectory

Explanation:

The velocity of the ball will keep pushing it upwards until the velocity becomes zero. Therefore, the ball will reach the top of its trajectory when velocity i.e. V=0,

Fundamental principal of velocity is V = Vo + g*t

where, V=0  

Vo = 15 m/s

g = -9.8 m/s^2 (since ball is going upwards against the gravity)

t = ?  

0 = 15 + (-9.8 * t)

-15 = -9.8t

-15 / -9.8 = t

t = 1.53 Seconds

The ball takes 1.53 seconds to reach its top trajectory

Answer:

that the angle must be increased to maintain the minimum time of discrimination due to the increase in the speed of sound in material

Explanation:

The direction of sound is detected by the difference in time of reception of each ear, the speed of the wave is

             v = d / t

             t = d / v

In air the velocity is v = 330 m / s, let's use trigonometry

           Cos 3 = d / L

           L = d / cos 3

The difference in distance is

            Δd = d - d / cos 3 = d (1- 1 / cos3)

             t = Δd / 330

When the animal is in the water the speed of sound is

              v = 1540 m / s

So time is

             t' = Δd ’/ 1540

            t ’= Dd’ / 4.67  330

So if  t = t’  is the minimum response time, the distance must be increased

            Δd ’= 4.6 Δd

            1-1 / cos θ = 4.6 (1- 1 / cos 3) = -4.6 0.00137 = -0.00631

           1 + 0.0063 = 1 / cos θ

           1.00631 = 1 / cos θ

           Cos θ = 1 / 1.00631

           Tea = 6.5

We see that the angle must be increased to maintain the minimum time of discrimination due to the increase in the speed of sound in material

Answer:

The answer to the question is spread

To change the shadow size, the spread parameter must be added to the box-shadow property, specifying the size of the shadow relative to the size of the object.

Explanation:

The spread parameter is used to contract or expand a shadow to change the shadow size. To increase the shadow size the radius is set to a positive value while to decrease the size of the shadow, the spread radius is set to a negative value and a zero value will leave the shadow size unchanged

Example

text-shadow: 3px 3px 20x 5px lightblue; color: blue.

The 'spread radius' parameter in the 'box-shadow' property in CSS is used to adjust the shadow size. It specifies how much the shadow expands or contracts.

To change the shadow size in CSS, the spread radius parameter must be added to the box-shadow property, specifying the size of the shadow relative to the size of the object. The box-shadow property in CSS is used to apply shadows to an element. It takes values for x-offset, y-offset, blur radius, spread radius (which you are asking about), and color. The spread radius specifies how much the shadow expands. Negative values can also be used to make the shadow smaller. Here's an example:

box-shadow: 10px 5px 5px 2px grey;

In this example, '2px' is the spread radius, and 'grey' is the color of the shadow.

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Answer:

(A). The tension in the rope that connects the boxes is 10.50 N.

(B). The value of m is 7 kg.

Explanation:

Given that,

Mass of box B = 5.00 kg

Mass of box A = m

Force = 40.0 N

Direction= 53.1°

Acceleration = 1.50 m/s²

Coefficient of kinetic friction = 0.30

(A). We need to calculate the tension in the rope that connects the boxes

Using balance equation

[tex]T=ma+m\cos\theta[/tex]

Put the value into the formula

[tex]T=5\times1.50+5.00\cos53.1[/tex]

[tex]T=10.50\ N[/tex]

(B). We need to calculate the value of m

Using formula of tension

[tex]T=ma[/tex]

[tex]m=\dfrac{T}{a}[/tex]

Put the value into the formula

[tex]m=\dfrac{10.50}{1.50}[/tex]

[tex]m=7\ kg[/tex]

Hence, (A). The tension in the rope that connects the boxes is 10.50 N.

(B). The value of m is 7 kg.

Answer:

Question not completed, so I analysed the question first

Tony drove to the mountains last weekend. there was heavy traffic on the way there, and the trip took 6 hours. when tony drove home, there was no traffic and the trip only took 4 hours. if his average rate was 22 miles per hour faster on the trip home, how far away does tony live from the mountains?

Explanation:

Let use variables to solve the problems

Let the first trip to be mountain take x hours

Let the trip back home take y hours

Let the speed to while going to the mountain be a miles/hour

Then, while going home it was b miles/hour faster than while going to the mountain.

Then, speed going home is (a+b)miles / hour

The formula for speed is given as

Speed=distance/time

The constant through out the journey is distance, the two journey has the same distance.

Then,

Distance =speed×time

For first journey going to the mountain

Distance = a×x=ax miles

For the second journey going home

Distance =y×(a+b)

Distance Mountain= distance home

ax=y(a+b)

Make a subject of the formula

ax=ya+yb

ax-ya=yb

a(x-y)=yb

a=yb/(x-y)

Therefore, distance from mountain is

Distance=speed ×time

Distance= a×x=ax

Now, applying the questions

So from the questions

x=6hours, y=4hours

Also, b=22miles/hour

Then,

a=yb/(x-y)

a=4×22/(6-4)

a=88/2

a=44miles/hour

Then, the house distance from the mountain is

Distance=ax

Distance =44×6

Distance =264miles

Answer:

480 miles.

Explanation:

Let, S = rate on his way to the mountains.

Assume, Sgoing x time going = Sreturning x time returning

= S × 12 hours = (S + 20mph) ×8 hours

= 12 × S = 8 × S + 160.

4 × S = 160

S = 40 miles/hour

The trip took 12 hours at 40 miles per hour, so distance was:

= 12 hours × 40 mph

= 480 miles.

Answer:

Explanation:

Solution:

- Finding large moons comparable in size to their planets result from impacts of two astro-bodies. The probability of such an event occurring is very rare.

- Even at the best luck, one moon can be made from the result of giant impact. While the probability of 6 planets having moons of comparable sizes is close to impossible.

Answer:

Maximum velocity: Position C, Maximum positive acceleration: Position A.

Explanation:

Let consider that spring is compressed in the negative direction. Then, the maximum velocity occurs at position C, when spring is not compressed nor stretched. Since force in spring is of reactive nature, that is, the direction of force is opposed to the direction of movement, a maximum positive acceleration occurs at position A.

Final answer:

To find the initial velocity of a 1500 kg car that left 68-m-long skid marks, the kinetic energy (½mv²) equal to the work done by friction is used, resulting in the equation v = √(2μkgd). Plugging in the values gives the car's speed.

Explanation:

To determine how fast a 1500 kg car was traveling before it skidded to a halt on a wet road with a coefficient of kinetic friction (μk) of 0.47, leaving 68-meter-long skid marks, we can use the work-energy principle. The kinetic energy the car had before skidding must equal the work done by friction to halt the car.

Work done by friction (W) = μk × normal force (N) × distance (d)

Since the normal force (N) is equal to the weight of the car (mg) where ‘m’ is the mass of the car and ‘g’ is the acceleration due to gravity, normal force can be replaced in this equation:

W = μk × mg × d

The kinetic energy (KE) the car initially had is given by:

KE = ½ m × v2

Setting work done by friction equal to kinetic energy:

½ m × v2 = μk × mg × d

We can then solve for velocity (‘v’), which ends up being:

v = √(2 × μk × g × d)

By plugging in the values (m = 1500 kg, μk = 0.47, g = 9.8 m/s2, d = 68 m) we get the initial velocity.

Calculation:

v = √(2 × 0.47 × 9.8 m/s2 × 68 m)

After completing the calculation, we can determine the car's initial velocity that resulted in the 68-m-long skid marks on the wet road.

Answer:

The answer to the question is;

When she is 8 m from the building fast the length of her shadow on the building is decreasing at [tex]\frac{2}{9} m/s[/tex] or 0.22 m/s.

Explanation:

We have

Distance of the spotlight from the building = 20 m

Distance  of woman from the building when her speed is measured = 8 m

Height of the woman = 2 m

Actual speed of the woman = 0.8 m/s

Comparing the distance of the woman from the spotlight and the wall from the spotlight, we have when the woman is 8 m from the building she is 12 m from the spotlight    

Therefore we have

[tex]\frac{12}{20} = \frac{2}{y}[/tex] where y is the shadow cast by the woman on the building = 10/3

When the woman is x distance from the building, she is 20 - x meters from the spotlight

Therefore the above equation can be written  as

[tex]\frac{20-x}{20} = \frac{2}{y}[/tex]  which gives [tex]1 - \frac{1}{20}*x = 2* \frac{1}{y}[/tex] finding the derivative of both sides gives

[tex]-\frac{1}{20}dx =-2*\frac{1}{y^2}dy[/tex] hence we have by dividing by dt gives [tex]-\frac{1}{20}\frac{dx}{dt} =-2*\frac{1}{y^2}\frac{dy}{dt}[/tex]

However we know that [tex]\frac{dx}{dt} = 0.8 m/s[/tex]

Therefore [tex]-\frac{0.8}{20} = -0.18\frac{dy}{dt}[/tex]

The rate of decrease of her shadow [tex]\frac{dy}{dt}[/tex] is given by

[tex]\frac{dy}{dt} = \frac{0.8}{3.6} =\frac{2}{9} m/s[/tex] or 0.222 m/s.