Consider a pan of water being heated (a) by placing it on an electric range and (b) by placing a heating element in the water. Which method is a more efficient way of heating water? Explain.

Answer:

P1+1/2pv2/1+pgh1=P2+1/2pv2/2+pgh2

Answer:

D. Both "Materials" and "Packaging"

Explanation:

Product liability may refer to the manufacturer or the seller being held responsible or liable for providing any defective product into the hands of the consumer or the customer. Responsibility or liability for a defective product which causes injuries lies with all the sellers of the product from the manufacturer to the distributor to the seller.

There are majorly three product defects. They are :

1. Manufacturing defect

2. Design defect

3. Marketing defect

Answer:

a) Real Power (kW) = 1.5 kW

b) Reactive Power (kvar) is 1.1663 KVAR

c) Apparent Power (kVA) is 1.9 KVA

d) the Power Factor cos∅ is 0.7894

e) the impedance Z in polar and rectangular form is 76 ∠ 37.87° Ω

Explanation:

Given that;

V = 380v

i = 5A

P = 1500 W

determine;

a) Real Power (kW)

P = 1500W = 1.5 kW

therefore Real Power (kW) = 1.5 kW

b) Reactive Power (kvar)

p = V×i×cos∅

cos∅ = p / Vi

cos∅ = 1500 / ( 380 × 5 ) = 0.7894

∅ = cos⁻¹ (0.7894)

∅ = 37.87°

Q = VIsin∅

Q = 380 × 5 × sin( 37.87° )

Q = 1.1663 KVAR

Therefore Reactive Power (kvar) is 1.1663 KVAR

c) Apparent Power (kVA)

S = P + jQ

= ( 1500 + J 1166.3 ) VA

S = 1900 ∠ 37.87° VA

S = 1.9 KVA

Therefore Apparent Power (kVA) is 1.9 KVA

d) Power Factor

p = V×i×cos∅

cos∅ = p / Vi

cos∅ = 1500 / ( 380 × 5 ) = 0.7894

Therefore the Power Factor cos∅ is 0.7894

e) The impedance Z in polar and rectangular form

Z = 380 / ( S∠-37.87) = V/I

Z = ( 60 + j 46.647) Ω

Z = 76 ∠ 37.87° Ω

Therefore the impedance Z in polar and rectangular form is 76 ∠ 37.87° Ω

Answer:

0.75 kg

Explanation:

c = Damping coefficient = 3 kg/s

x = Displacement of spring = 0.5 m

F = Force = 1.5 N

From Hooke's law we get

[tex]F=kx\\\Rightarrow k=\dfrac{F}{x}\\\Rightarrow k=\dfrac{1.5}{0.5}\\\Rightarrow k=3\ \text{N/m}[/tex]

In the case of critical damping we have the relation

[tex]c^2-4mk=0\\\Rightarrow m=\dfrac{c^2}{4k}\\\Rightarrow m=\dfrac{3^2}{4\times 3}\\\Rightarrow m=0.75\ \text{kg}[/tex]

Mass that would produce critical damping is 0.75 kg.

0.75 kg is the mass that would produce critical damping. As spring with an m-kg mass and a damping constant 3 (kg/s) can be held stretched 0.5 meters beyond its natural length by a force of 1.5 newtons.

A change in time and position is referred to as an object's velocity. When there is no movement of the object, the velocity of the object is said to be 0.

For any body in planar motion, the velocity is always instantaneously 0 at some point in the plane of motion (if it were rigidly connected to the body). This place is known as the instantaneous center of zero velocity, or IC.

Example: The gravitational pull of the earth pushes the ball away from the thrower when a ball is thrown upwards on Earth at a constant speed. The speed of the ball increases until it reaches its maximum, at which point it starts to plummet.

Thus, it is 0.75 kg.

For more information about zero velocity, click here:

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dutile is the correct answer

Answer:

Velocity of 5 cm diameter pipe is 1.38 m/s

Explanation:

Use following equation of Relation between the Reynolds numbers of both pipes

[tex]Re_{5}[/tex] = [tex]Re_{12}[/tex]

[tex]\sqrt{\frac{V_{5}XD_{5} }{v_{5}}}[/tex]= [tex]\sqrt{\frac{V_{12}XD_{12} }{v_{12}}}[/tex]

[tex]Re_{5}[/tex] = Reynold number of water pipe

[tex]Re_{12}[/tex] = Reynold number of oil pipe

[tex]V_{5}[/tex] = Velocity of water 5 diameter pipe = ?

[tex]V_{12}[/tex] = Velocity of oil 12 diameter pipe = 2.30

[tex]v_{5}[/tex] = Kinetic Viscosity of water = 1 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]/s

[tex]v_{12}[/tex] = Kinetic Viscosity of oil =  4 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]/s

[tex]D_{5}[/tex] = Diameter of pipe used for water = 0.05 m

[tex]D_{12}[/tex] = Diameter of pipe used for oil = 0.12 m

Use the formula

[tex]\sqrt{\frac{V_{5}XD_{5} }{v_{5}}}[/tex]= [tex]\sqrt{\frac{V_{12}XD_{12} }{v_{12}}}[/tex]

By Removing square rots on both sides

[tex]{\frac{V_{5}XD_{5} }{v_{5}}}[/tex]= [tex]{\frac{V_{12}XD_{12} }{v_{12}}}[/tex]

[tex]{V_{5}[/tex]= [tex]{\frac{V_{12}XD_{12} }{v_{12}XD_{5}\\}}[/tex]x[tex]v_{5}[/tex]

[tex]{V_{5}[/tex]= [ (0.23 x 0.12m ) / (4 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]/s) x 0.05 ] 1 x [tex]10^{-6}[/tex] [tex]m^{2}[/tex]/s

[tex]{V_{5}[/tex] = 1.38 m/s

Answer:

Hello the needed data given is not properly arranged attached below is the properly arranged data

Answer:

b) 62.5 * 10^3 MPa

c) ≈ 285 MPa

d)  370Mpa

e)  16%

Explanation:

Given Data:

cylindrical aluminum diameter = 12.8 mm

Gauge length = 50.8 mm

A) plot of engineering stress vs engineering strain

attached below

B ) calculate Modulus of elasticity

Modulus of elasticity = Δб / Δ ε

                                   = ( 200 - 0 ) / (0.0032 - 0 ) = 62.5 * 10^3 MPa

C) Determine the yield strength

at strain offset = 0.002

hence yield strength ≈ 285 MPa

D) Determine tensile strength of the alloy

The tensile strength can be approximated at 370Mpa because that is where it corresponds to the maximum stress on the stress  vs strain ( complete plot )

E) Determine approximate ductility in percent elongation

ductility in percent elongation = plastic strain at fracture * 100

total strain = 0.165 , plastic strain = 0.16

therefore Ductility in percent elongation = 0.16 * 100 = 16%

Answer:

C. The meter will display a negative sign.

Explanation:

If you use an analog voltmeter and you measure voltage with reverse polarity you will damage it. But in this case we are using a digital multimeter. This kind of multimeter is designed to be able to deal with positive and negative voltages

Answer:

a. 4[tex]\mu m[/tex]

b. 1 m

Explanation:

According to the question, the data is as follows

The Density of water at 20 degrees celcius is 1000 kg/m^3

Viscosity is 0.001kg/m/.s

Velocity V = 25 cm/s

V = 0.25 m/s

Now

a. The creeping motion is

As we know that

Reynold Number = (Density of water × V × d) ÷ (Viscosity)

1 = (1,000 × 0.25 × d) ÷ 0.0001

d = (1 × 0.001) ÷ (1,000 × 0.25)

= 4E - 06^m

= 4[tex]\mu m[/tex]

b. Now the sphere diameter is

Reynold Number = (Density of water × V × d) ÷ (Viscosity)

250,000 = (1,000 × 0.25 × d) ÷ 0.0001

d = (250,000 × 0.001) ÷ (1,000 × 0.25)

= 1 m

Answer:

Computer programming is where you learn and see how computers work. People do this for a living as a job, if you get really good at it you will soon be able to program/ create a computer.

Explanation:

Hope dis helps! :)

Answer:

Yes, they can all be present in the same malware because each of them perform slightly differing functions.

Explanation:

Backdoor is a software which when placed into your computer will permit hackers to easily gain reentry into your computer. This can happen even after you have already patched the flaw that they have used to hack your system before.

A bot is a program that does the same task in a continuous manner akin to when you use a blender by pressing the button to blend what you have put into it.

A keylogger is a part of a hidden software that monitors and records everything you type on your computer keyboard after which it writes it onto a file, with the hopes of capturing relevant information such as your bank account number and even passwords and other sensitive means of identification.

A Spyware is somehow similar to a keylogger just that it steals information from your computer and sends it to someone else.

A root kit is a bad software that is capable of modifying the operating system or other privileged access devices in order to gain continuous access into your system for the purpose of gathering of information and/or reducing the system’s functionality.

Yes, they can all be present in the same malware because each of them perform slightly differing functions.

Solution :

Sieve Size (in)                   Weight retain(g)

3                                         1.62

2                                         2.17

[tex]$1\frac{1}{2}$[/tex]                                       3.62

[tex]$\frac{3}{4}$[/tex]                                        2.27

[tex]$\frac{3}{8}$[/tex]                                        1.38

PAN                                    0.21

Given :

Sieve       weight       % wt. retain    % cumulative       % finer

size        retained                               wt. retain

No. 4        59.5            10.225%          10.225%            89.775%

No. 8        86.5            14.865%          25.090%           74.91%

No. 16       138              23.7154%        48.8054%         51.2%

No. 30      127.8           21.91%              70.7154%          29.2850%

No. 50      97               16.6695%         87.3849%         12.62%

No. 100     66.8            11.4796%         98.92%              1.08%

Pan            6.3               1.08%              100%                   0%

                581.9 gram

Effective size = percentage finer 10% ([tex]$$D_{20}[/tex])

0.149 mm, N 100, % finer 1.08

0.297, N 50 , % finer 12.62%

x  ,   10%

[tex]$y-1.08 = \frac{12.62 - 1.08}{0.297 - 0.149}(x-0.149)$[/tex]

[tex]$(10-1.08) \times \frac{0.297 - 0.149}{12.62 - 1.08}+ 0.149=x$[/tex]

x = 0.2634 mm

Effective size, [tex]$D_{10} = 0.2643 \ mm$[/tex]

Now, N 16 (1.19 mm)  ,  51.2%

N 8 (2.38 mm)  ,  74.91%

x,  60%

[tex]$60-51.2 = \frac{74.91-51.2}{2.38-1.19}(x-1.19)$[/tex]

x = 1.6317 mm

[tex]$\therefore D_{60} = 1.6317 \ mm$[/tex]

Uniformity co-efficient = [tex]$\frac{D_{60}}{D_{10}}$[/tex]

   [tex]$Cu= \frac{1.6317}{0.2643}$[/tex]

Cu = 6.17

Now, fineness modulus = [tex]$\frac{\Sigma \text{\ cumulative retain on all sieve }}{100}$[/tex]

[tex]$=\frac{\Sigma (10.225+25.09+48.8054+70.7165+87.39+98.92+100)}{100}$[/tex]

= 4.41

which lies between No. 4  and No. 5 sieve [4.76 to 4.00]

So, fineness modulus = 4.38 mm

Answer:

Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3/s and at a velocity of 7 m/s, and leaves in the normal direction along the pump casing, as shown in Fig. PI3-39. Determine the force acting on the shaft (which is also the force acting on the bearing of the shaft) in the axial direction.

Step-by-step solution:

Step 1 of 5

Given data:-

The velocity of water is .

The water flow rate is.

Answer:

51112.5 ft^3

Explanation:

Determine the volume needed along the road when we assume a 6% shrinkage

shrinkage factor = 1 - shrinkage  = 1 - 0.06 =  0.94

first we have to calculate the volume between the cross sectional areas (i.e. A1 ---- A5 ) using average end area method

Volume between A1 - A2

= (125 ft - 0 ft) * [(130 ft^2 + 140 ft^2) / 2]

 = 125 ft * 135 ft^2

= 16875 ft^3

Volume between A2 - A3

= (250 ft - 125 ft) * [(140 ft^2 + 60 ft^2) / 2]

= 125 ft * (200 ft^2 / 2)

= 12500 ft^3

Volume between A3 - A4

= (375 ft - 250 ft) * [(60 ft^2 + 110 ft^2) / 2]

= 125 ft * (170 ft^2 / 2)

= 10625 ft^3

Volume between A4 - A5

(500 ft - 375 ft) * [(110 ft^2 + 120 ft^2) / 2]

 = 125 ft * 115 ft^2

= 14375 ft3

Hence the total volume along the 500 ft road

= ∑ volumes between cross sectional areas

=  16875 ft^3 + 12500 ft^3 + 10625 ft^3 + 14375 ft^3 = 54375 ft^3

Finally the volume needed along this road is calculated as

Total volume * shrinkage factor

= 54375 * 0.94  = 51112.5 ft^3

Answer:

74,4 litros

Explanation:

Dado que

W = nRT ln (Vf / Vi)

W = 3000J

R = 8,314 JK-1mol-1

T = 58 + 273 = 331 K

Vf = desconocido

Vi = 25 L

W / nRT = ln (Vf / Vi)

W / nRT = 2.303 log (Vf / Vi)

W / nRT * 1 / 2.303 = log (Vf / Vi)

Vf / Vi = Antilog (W / nRT * 1 / 2.303)

Vf = Antilog (W / nRT * 1 / 2.303) * Vi

Vf = Antilog (3000/1 * 8,314 * 331 * 1 / 2,303) * 25

Vf = 74,4 litros

Answer:

-9.5° C

Explanation:

See attachment for calculations.

On the concluding parts, from the attachment, we have that

√[(297000 * 4)/(1056)] = 297 - T(l), and solving further, we get

297 - T(l) = √(1188000/1056)

297 - T(l) = √1125

297 - T(l) = 33.5

T(l) = 297 - 33.5

T(l) = 263.5

When you convert back to °C, we have

263.5 - 273 = -9.5° C

Answer:

Updated question

The big ben clock tower in London has clocks on all four sides. If each clock has a minute hand that is 11.5 feet in length, how far does the tip of each hand travel in 52 minutes?

The distance traveled by the tip of the minute hand of the clock would be 62.59 ft

Explanation:

Let us assume the shape of the clock is circular.

the minute hand is equal to the radius = 11.5 ft

Diameter = radius x 2

Diameter = 11.5 x 2 = 23 ft

The distance traveled by the tip of the minute hand can be calculated thus;

the fraction of the circumference traveled by the minute hand would be;

52/60 = 0.8667

Circumference of the clock would be;

C = pi x d

where C is the circumference

pi is a constant

d is the diameter

C = 3.14 x 23

C = 72.22 ft

Therefore the fraction of the circumference covered by the minute hand would be;

72.22 ft x 0.8667 = 62.59 ft

Therefore the distance traveled by the tip of the minute hand of the clock would be 62.59 ft

Answer:

The total mass in the tank = 0.45524  kg

The amount of heat transferred = 3426.33 kJ

Explanation:

Given that:

The volume of the tank V = 0.06 m³

The pressure of the liquid and the vapor of H2O (p) = 15 bar

The initial quality of the mixture [tex]\mathbf{x_{initial} - 0.20}[/tex]

By applying the energy rate balance equation;

[tex]\dfrac{dU}{dt} = Q_{CV} - m_eh_e[/tex]

where;

[tex]m_e =- \dfrac{dm_{CV}}{dt}[/tex]

Thus, [tex]\dfrac{dU}{dt} =Q_{CV} + \dfrac{dm_{CV}}{dt}h_e[/tex]

If we integrate both sides; we have:

[tex]\Delta u_{CV} = Q_{CV} + h _e \int \limits ^2_1 \ dm_{CV}[/tex]

[tex]m_2u_2 - m_1 u_1 = Q_{CV} + h_e (M_2-m_1) \ \ \ --- (1)[/tex]

We obtain the following data from the saturated water pressure tables, at p = 15 bar.

Since:

[tex]h_e =h_g[/tex]

Then: [tex]h_g = h_e = 2792.2 \ kJ/kg[/tex]

[tex]v_f = 1.1539 \times 10^{-3} \ m^3 /kg[/tex]

[tex]v_g = 0.1318 \ m^3/kg[/tex]

Hence;

[tex]v_1 = v_f + x_{initial} ( v_g-v_f)[/tex]

[tex]v_1 = 1.1529 \times 10^{-3} + 0.2 ( 0.1318-1.159\times 10^{-3} )[/tex]

[tex]v_1 = 0.02728 \ m^3/kg[/tex]

Similarly; we obtained the data for [tex]u_f \ \& \ u_g[/tex] from water pressure tables at p = 15 bar

[tex]u_f = 843.16 \ kJ/kg\\\\ u_g = 2594.5 \ kJ/kg[/tex]

Hence;

[tex]u_1 = u_f + x_{initial } (u_g -u_f)[/tex]

[tex]u_1 =843.16 + 0.2 (2594.5 -843.16)[/tex]

[tex]u_1 = 1193.428[/tex]

However; the initial mass [tex]m_1[/tex] can be calculated by using the formula:

[tex]m_1 = \dfrac{V}{v_1}[/tex]

[tex]m_1 = \dfrac{0.06}{0.02728}[/tex]

[tex]m_1 = 2.1994 \ kg[/tex]

From the question, given that the final quality; [tex]x_2 = 1[/tex]

[tex]v_2 = v_f + x_{final } (v_g - v_f)[/tex]

[tex]v_2 = 1.1539 \times 10^{-3} + 1(0.1318 -1.1539 \times 10^{-3})[/tex]

[tex]v_2 = 0.1318 \ m^3/kg[/tex]

Also;

[tex]u_2 = u_f + x_{final} (u_g - u_f)[/tex]

[tex]u_2 = 843.16 + 1 (2594.5 - 843.16)[/tex]

[tex]u_2 = 2594.5 \ kJ/kg[/tex]

Then the final mass can be calculated by using the formula:

[tex]m_2 = \dfrac{V}{v_2}[/tex]

[tex]m_2 = \dfrac{0.06}{0.1318}[/tex]

[tex]m_2 = 0.45524 \ kg[/tex]

Thus; the total mass in the tank = 0.45524  kg

FInally; from the previous equation (1) above:

[tex]m_2u_2 - m_1 u_1 = Q_{CV} + h_e (M_2-m_1) \ \ \ --- (1)[/tex]

[tex]Q = (m_2u_2-m_1u_1) - h_e(m_2-m_1)[/tex]

Q = [(0.45524)(2594.5) -(2.1994)(1193.428)-(2792.2)(0.45524-2.1994)]

Q = [ 1181.12018 - 2624.825543 - (2792.2)(-1.74416 )]

Q = 3426.33 kJ

Thus, the amount of heat transferred = 3426.33 kJ

Answer:

2.44 mV

Explanation:

This question has to be one of analog quantization size questions and as such, we use the formula

Q = (V₂ - V₁) / 2^n

Where

n = 12

V₂ = higher voltage, 5 V

V₁ = lower voltage, -5 V

Q = is the change in voltage were looking for

On applying the formula and substitutiting the values we have

Q = (5 - -5) / 2^12

Q = 10 / 4096

Q = 0.00244 V, or we say, 2.44 mV

Answer:

mountain ranges may be

Answer:

Option B (Cold working) would be the correct alternative.

Explanation:

The other possibility isn't linked to the given scenario. Therefore the alternative above is the right one.

Answer:64.10 Btu/lbm

Explanation:

Work done in an isothermally compressed steady flow device is expressed as

Work done = P₁V₁ In { P₁/ P₂}

Work done=RT In { P₁/ P₂}

where P₁=13 psia

          P₂= 80 psia

Temperature =°F Temperature is convert to  °R

T(°R) = T(°F) + 459.67

T(°R) = 55°F+ 459.67

=514.67T(°R)

According to the properties of molar gas, gas constant and critical properties table, R  which s the gas constant of air is given as 0.06855 Btu/lbm

Work = RT In { P₁/ P₂}

0.06855 x 514.67 In { 13/ 80}

=0.06855 x 514.67 In {0.1625}

= 0.06855 x 514.67  x -1.817

=- 64.10Btu/lbm

The required work therefore for this  isothermal compression is 64.10 Btu/lbm

Answer:

A tax is a monetary payment without the right to individual consideration, which a public law imposes on all taxable persons - including both natural and legal persons - in order to generate income. This means that taxes are public-law levies that everyone must pay to cover general financial needs who meet the criteria of tax liability, whereby the generation of income should at least be an auxiliary purpose. Taxes are usually the main source of income of a modern state. Due to the financial implications for all citizens and the complex tax legislation, taxes and other charges are an ongoing political and social issue.

I dont know I asked this to

Explanation:

Answer:

a) the  rate of cooling provided is 18604.8 Btu/h

b) the rate of exergy destruction in the evaporator is 0.46 Btu/Ibm

c) the second-law efficiency of the evaporator is 37.39%

Explanation:

Given that;

Temperature of sink TL = 50°F = 510 R

Temperature at evaporator inlet TI = 10°F = 470 R

mass flow rate m" = 0.08 lbm/s

quality of refrigerant at evaporator inlet x1 = 0.3

quality of refrigerant at evaporator exit x2 = 1.0

T₀ = 77°F = 537 R

h1 = 107.5 Btu/lbm

s1 = 0.2851 Btu/lbm·R,

h2 = 172.1 Btu/ lbm,

s2 = 0.4225 Btu/lbm·R.

a) rate of cooling provided, in Btu/h.

QL = m"( h2 - h1)

we substitute

QL = 0.08( 172.1 - 107.5

= 0.08 × 64.6

= 5.168 Btu/s

we convert to Btu/h

5.168 × 60 × 60

QL = 18604.8 Btu/h

Therefore the  rate of cooling provided is 18604.8 Btu/h

b) The rate of exergy destruction in the evaporator

Entropy generation can be expressed as;

S_gen = m"(s2 - s1) - QL/TL

so we substitute

S_gen = 0.08( 0.4225 -  0.2851  ) - 5.168 / 510

= 0.010992 - 0.01013

S_gen = 0.00086 Btu/ibm.R

now the energy destroyed expressed as;

X_dest = T₀ × S_gen

so

X_dest =  537 × 0.00086

X_dest = 0.46 Btu/Ibm

Therefore the rate of exergy destruction in the evaporator is 0.46 Btu/Ibm

c)  The second-law efficiency of the evaporator.

Energy expended is expressed as;

X_exp = m"(h1 - h2) - m"T₀(s1 - s2)

we substitute

= 0.08( 107.5 - 172.1 ) - [0.08 × 537 ( 0.2851 - 0.4225 )

= -5.168 - [ - 5.9027)

= -5.168 + 5.9027

= 0.7347 Btu/s

Now second law efficiency is expressed as;

nH = 1 - (X_dest / X_esp)

= 1 - ( 0.46 / 0.7347 )

= 1 - 0.6261

= 0.3739

nH = 37.39%

Therefore the second-law efficiency of the evaporator is 37.39%

Answer:

a) The stresses due to bending moments is much more than the stresses from transverse shear.

c) The deformations due to bending moments is much more than the deformations from transverse shear.

Explanation:

Strain in an object suspended is a function of the stress which the suspended body passed through. The stress which is the function of the force experienced by the body over a given area helps is straining the moment. This lead to the strain energy from bending moment being greater than the strain energy from a transverse shear force.

Answer:

155fts

Explanation:

We apply the bernoulli's equation to get the depth of water.

We have the following information

P1 = pressure at top water surface = 0

V1 = velocity at too water surface = 0

X1 = height of water surface = h

Hf = friction loss = 0

P2 = pressure at exit = 0

V2 = velocity at exit if penstock = 100ft/s

X2 = height of penstock = 0

g = acceleration due to gravity = 32.2ft/s²

Applying these values to the equation

0 + 0 + h = 0 + v2²/2g +0 + 0

= h = 100²/2x32.2

= 10000/64.4

= 155.28ft

= 155

Answer:

commercial freezing

Explanation:

smaller ice crystals are formed this causes less damage to cell membranes so the quality is less effected

Answer:

[tex]10.8\ \text{lb/ft^2}[/tex]

[tex]101.96\ \text{lb/ft}^2[/tex]

Explanation:

[tex]v_1[/tex] = Velocity of car = 65 mph = [tex]65\times \dfrac{5280}{3600}=95.33\ \text{ft/s}[/tex]

[tex]\rho[/tex] = Density of air = [tex]0.00237\ \text{slug/ft}^3[/tex]

[tex]v_2=0[/tex]

[tex]P_1=0[/tex]

[tex]h_1=h_2[/tex]

From Bernoulli's law we have

[tex]P_1+\dfrac{1}{2}\rho v_1^2+h_1=P_2+\dfrac{1}{2}\rho v_2^2+h_2\\\Rightarrow P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 95.33^2\\\Rightarrow P_2=10.8\ \text{lb/ft^2}[/tex]

The maximum pressure on the girl's hand is [tex]10.8\ \text{lb/ft^2}[/tex]

Now [tex]v_1[/tex] = 200 mph = [tex]200\times \dfrac{5280}{3600}=293.33\ \text{ft/s}[/tex]

[tex]P_2=\dfrac{1}{2}\rho v_1^2\\\Rightarrow P_2=\dfrac{1}{2}\times 0.00237\times 293.33^2\\\Rightarrow P_2=101.96\ \text{lb/ft}^2[/tex]

The maximum pressure on the girl's hand is [tex]101.96\ \text{lb/ft}^2[/tex]