i. Draw the velocity diagram for the instant shown and determine the velocity of
sliding of the piston relative to the cylinder and the angular velocity of link
DP.
ii. Draw the acceleration diagram for the instant shown and determine the
acceleration of sliding of the piston relative to the cylinder and the angular
acceleration of link DP.

The invention process can be regarded as the process involving the  creation of an idea as well as development of this idea.

It should be noted that invention process involves;

Things that must consider during the invention process are;

Therefore, invention process involves creation of an idea and this idea must be unique.

Learn more about invention process at;

https://brainly.com/question/4459550

Answer:

i) Primary sensing element.

ii) Variable conversion element.

iii) Variable manipulation element.

Explanation:

Answer:

The main functional elements of a measurement system are:

i) Primary sensing element.

ii) Variable conversion element.

iii) Variable manipulation element.

v) Data transmission element.

Answer:

The golfer is at greater risk.

Explanation:

The golfer is holding a metal club. Metal is a good conductor for electricity (lightning), meaning electrons can pass through easily. Her caddy is at lesser risk because she is holding a plastic handled umbrella. Plastic is an insulator, which does not easily allow the movement of electrons to pass.

Answer:

Option A is correct

Explanation:

As we know  

Inductive Susceptance = ½(pi)*f*L

Or  Inductive Susceptance is inversely  proportional to the frequency

Likewise conductive Susceptance = 2 (pi)*f*C

Conductive Susceptance is directly proportional to the frequency

When the frequency will reach the value zero, then the Inductive Susceptance will become infinite

Hence, inductor will dominate in determining the equivalent impedance of this parallel combination

Option A

The image of the assembly is missing, so i have attached it.

Answer:

T = 210 Lb.in

τ_ab = 7818.705 lb/in²

τ_bc = 2361.036 lb/in²

Explanation:

We know that formula for shear stress due to the twisting couple is given by;

τ = Tr/J

Where;

T is torque

r is radius of shaft

J is polar moment of inertia.

Let's first find J and T

J = πd⁴/32

For section AB,

J_ab = π(0.75⁴ - 0.68⁴)/32

J_ab = 0.010072 in⁴

Similarly;

J_bc = π(1⁴ - 0.86⁴)/32

J_bc = 0.044472 in⁴

Now, let's find the torque.

From the image attached, we can see that the force acting at point 1 is 15 lb and also the force acting at point 2 is 15 lb. Distance from Point 1 to A; L1 = 6 in

Distance from point 2 to A; L2 = 8 in

Thus, taking moments about point A, we have;

T = PL1 + PL2

T = 15(6) +15(8)

T = 210 Lb.in

r_ab = 0.75/2

r_ab = 0.375 in

r_bc = 1/2

r_bc = 0.5 in

Thus;

τ_ab = 210 × 0.375/0.010072

τ_ab = 7818.705 lb/in²

τ_bc = 210 × 0.5/0.044472

τ_bc = 2361.036 lb/in²

Answer:

hahahaha

Explanation:

Answer:

148.6 mins

Explanation:

Aim: using Cr-Au galvanic cell to power a 75 watt light bulb

Calculate how long the Galvanic cell will power the light bulb

power ( W ) = electric potential * current  ---- ( 1 )

charge = current * time ------ ( 2 )

first step : determine the electric potential

E⁰cell = E⁰cathode  -  E⁰anode ( values gotten from balancing  reaction )

          = ( 1.5 - ( -0.74 )) =  2.24 V

Given that amount of AU^+3 is limited the

Ecell ( cell voltage ) ≈ E⁰cell  =  2.24 V

back to equation 1

Power = 2.24 * current

∴ current = 75 / 2.24 = 33.48 A

back to equation 2

Charge = current * time

Time = Charge / current  = ( 3 * 1.032 * 96485 ) / (33.48 * 60 )

                                         = 148.6 mins

Note : 1 mol AU^+3  will give  3 e^-

          1.032 mol AU^+3 = 3 * 1.032 mol e^-

hence charge = 3 * 1.032 * 96485

Answer:

that not even a question

Answer: As to the more sophisticated way of detecting "smoke" from an object a human may use in hotel rooms, this sensor called a Fresh Air Sensor does not just detect and, and but alerts the management about a smoking incident in a hotel room

Answer:

[tex]D=0.1160m[/tex]

Explanation:

From the question we are told that:

Output Power [tex]P=1.2kw[/tex]

Density [tex]\rho=1.29kg/m^3[/tex]

Wind speed [tex]V=17.5mph=>7.8m/s[/tex]

Efficiency [tex]\gamma=60\%=>0.60[/tex]

Let Betz Limit

  [tex]C_p=\frac{16}{27}[/tex]

Generally the equation for Turbine Efficiency is mathematically given by

  [tex]\gamma=\frac{P}{P'}[/tex]

Where

P'=input power

 [tex]P' = In\ power[/tex]

 [tex]P'=1/2*C_p*\rho u^3*A[/tex]

 [tex]P'=1/2*C_p*\rho u^3*\frac{\pi}{4}*D^2[/tex]

Therefore the blade diameter for a two-blade propeller type rotor is

 [tex]\gamma=\frac{P*2*27*4}{16*\rho u^3*\pi*D^2}[/tex]

 [tex]D^2=\frac{P*2*27*4}{16*\rho u^3*\pi*\gamma}[/tex]

 [tex]D^2=\frac{1.2*2*27*4}{16*1.29*7.91^3*\pi*0.60}[/tex]

 [tex]D^2=0.0135[/tex]

 [tex]D=\sqrt{0.0135}[/tex]

 [tex]D=0.1160m[/tex]

Answer:

0.1160

Explanation: give them brainliest they deserve it

Answer:

International Organization For Standardization

Explanation:

This is the body within Geneva, Switzerland that promotes worldwide industrial and commercial standards

The seismic response of nuclear power plant structures is often calculated using lumped parameter methods. A finite element model of the structure is coupled to the soil with a spring-dashpot system used to represent the interaction process. The parameters of the interaction model are based on analytic solutions to simple problems which are idealizations of the actual problems of interest. The objective of the work reported in this paper is to compare predicted responses using the standard lumped parameter models with experimental data. These comparisons are shown to be good for a fairly uniform soil system and for loadings that do not result in nonlinear interaction effects such as liftoff. 7 references, 7 figures.

Answer:

L = 10.32 m

Explanation:

The resistance of a copper-nickel alloy wire is given by the following formula:

[tex]R = \frac{\rho L}{A}[/tex]

where,

R = Resistance = 0.5 Ω

ρ = resistivity of copper-nickel alloy = 3.8 x 10⁻⁸ Ωm

L = Length of wire = ?

A = cross-sectional area of wire = [tex]\frac{\pi d^2}{4} = \frac{\pi(1\ x\ 10^{-3}\ m)^2}{4}[/tex] = 7.85 x 10⁻⁷ m²

Therefore,

[tex]0.5\ \Omega = \frac{(3.8\ x\ 10^{-8}\ \Omega.m)L}{7.85\ x\ 10^{-7}\ m^2}[/tex]

L = 10.32 m

Answer:

Printing press , lenses etc,

Explanation:

Almost all of the Renaissance innovations that influenced the world, printing press was one of them. The printing press is widely regarded as the Renaissance's most significant innovation. The press was invented by Johannes Gutenberg, and Germany's culture was altered forever.

Astrophysics, humanist psychology, the printed word, vernacular vocabulary of poetry, drawing, and sculpture practice are only a few of the Renaissance's main inventions.

Answer: hello some parts of your question is missing attached below is the missing information

The radiator of a car is a type of heat exchanger. Hot fluid coming from the car engine, called the coolant, flows through aluminum radiator tubes of thickness d that release heat to the outside air by conduction. The average temperature gradient between the coolant and the outside air is about 130 K/mm . The term ΔT/d  is called the temperature gradient which is the temperature difference ΔT between coolant inside and the air outside per unit thickness of tube

answer : Total surface area = 3/2 * area of old radiator

Explanation:

we will use this relation

K = [tex]\frac{Qd }{A* change in T }[/tex]

change in T =  ΔT  

therefore New Area  ( A ) = 3/2 * area of old radiator

Given that the thermal conductivity is the same in the new and old radiators

Answer:

  A. Oil changes

Explanation:

It depends on the car and its usage and environment. Usually oil is supposed to be changed every few months, more often if the car is driven a lot. Coolant changes may be indicated as seasons change, so will generally occur less frequently than oil changes.

Tire and brake replacement depend on usage and driving habits. Some owners may never have to replace either one, if they trade their car every year or two. Folks who drive with their foot on the brake pedal may have to replace brakes relatively often.

The most frequent task is generally oil changes.

Answer:

Answer:

C = $0.0032 per day

Explanation:

We are given;

Dimension of cell phone; 50 mm × 45 mm × 20 mm

Temperature of charger; T1 = 33°C = 306K

Emissivity; ε = 0.92

convection heat transfer coefficient; h = 4.5 W/m².K

Room air temperature; T∞ = 22°C = 295K

Wall temperature; T2 = 20°C = 293 K

Cost of electricity; C = $0.18/kW.h

Chargers are usually in the form of a cuboid, and thus, surface Area is;

A = (50 × 45) + 2(50 × 20) + 2(45 × 20)

A = 6050 mm²

A = 6.05 × 10^(-3) m²

Formula for total heat transfer rate is;

E_t = hA(T1 - T∞) + εσA((T1)⁴ - (T2)⁴)

Where σ is Stefan Boltzmann constant with a value of; σ = 5.67 × 10^(-8) W/m².K⁴

Thus;

E_t = 4.5 × 6.05 × 10^(-3) (306 - 295) + (0.92 × 6.05 × 10^(-3) × 5.67 × 10^(-8)(306^(4) - 293^(4)))

E_t = 0.7406 W = 0.7406 × 10^(-3) KW

Now, we know C = $0.18/kW.h

Thus daily cost which has 24 hours gives;

C = 0.18 × 0.7406 × 10^(-3) × 24

C = $0.0032 per day

Answer:

B. False

Explanation:

In laminar flow, the Reynolds number (RE) is less than 2100 while in turbulent flow, it is greater than 4000.

Now,the entrance lengths for both laminar and turbulent flow is usually dependent on Reynolds number.

Thus is because the hydrodynamic entrance length for laminar flow is given by the formula;

Le = 0.05 DRe

Where;

Le is hydrodynamic entrance length

D is diameter of tube

Re is Reynolds number

Meanwhile, for turbulent flow, entrance length is given by;

Le = 1.359D(Re)^(1/4)

In both cases, we can see that the hydrodynamic entry length is directly proportional to the Reynolds number and not independent. Thus, we can say that the statement in the question is false.

Answer:

L = Henry

C = Farad

Explanation:

The electrical parameter represented as L is the inductance whose unit is Henry(H).

The electrical parameter represented as C is the inductance whose unit is Farad

Resonance frequency occurs when the applied period force is equal to the natural frequency of the system upon which the force acts :

To obtain :

At resonance, Inductive reactance = capacitive reactance

Equate the inductive and capacitive reactance

Inductive reactance(Xl) = 2πFL

Capacitive Reactance(Xc) = 1/2πFC

Inductive reactance(Xl) = Capacitive Reactance(Xc)

2πFL = 1/2πFC

Multiplying both sides by F

F * 2πFL = F * 1/2πFC

2πF²L = 1/2πC

Isolating F²

F² = 1/2πC2πL

F² = 1/4π²LC

Take the square root of both sides to make F the subject

F = √1 / √4π²LC

F = 1 /2π√LC

Hence, the proof.

Solution :

The road gradient is given as = tan θ = 0.08

So, θ = 457 degrees, sin θ = 0.08

a). While moving downhill, the terminal velocity the forward force which acts due to the gravitation = backward force due to drag

[tex]$Mg\sin \theta = C_d \times \frac{1}{2} \rho AV^2$[/tex]

Taking the air density, [tex]$\rho = 1.2 kg/m^3$[/tex] and putting the values we get

[tex]$100 \times 9.8 \times 0.08 = C_d \times \frac{1}{2} \times 1.2 \times 0.46 \times 15^2$[/tex]

Therefore, [tex]$C_d = 1.26$[/tex]

This is approximately the same as the value of coefficient of drag given in the question.

Hence verified.

b). Drag force on level road, F = [tex]$C_d \times \frac{1}{2}\rho A V^2$[/tex]

  Hence, deceleration :

  [tex]$F/m = C_d \times \frac{1}{2m} \rho AV^2$[/tex]

  [tex]$dV/dt = C_d \times \frac{1}{2m} \rho AV^2$[/tex]

  [tex]$(dV/ds)(ds/dt) = C_d \times \frac{1}{2m} \times \rho AV^2$[/tex]

  [tex]$V(dV/ds) = C_d \times \frac{1}{2m} \times \rho AV^2$[/tex]

  [tex]$ dV/V= C_d \times \frac{1}{2m} \times \rho A \times ds$[/tex]

Integrating both the sides, we get

[tex]$\ln (V_2/V_1) = C_d \times \frac{1}{2m} \times \rho A \times (s_2-s_1)$[/tex]

Hence, [tex]$s_2-s_1 = \frac{2m}{C_d} \times \ln (V_2/V_1) / \rho A$[/tex]

Putting the values,

Distance = [tex]$2 \times \frac{100}{1.2} \times \frac{ \ln(10/15)}{(1.2 \times 0.46)}$[/tex]

              = -122.5 m

Therefore, ignoring the negative sign, we get distance = 122.5 m

Answer:

  0°

Explanation:

The resistor voltage has the same phase as the circuit current. There is no phase difference.

Answer:

Explanation:

Answer:

1. i. 20 Nm ii. 4.85 HP

2. 16.5 %

Explanation:

1) Determine the torque and the power developed at 4% slip when a reduced voltage of 340V is applied.

i. Torque

Since slip is constant at 4 %,torque, T ∝ V² where V = voltage

Now, T₂/T₁ = V₂²/V₁² where T₁ = torque at 690 V = P/2πN where P = power = 20 HP = 20 × 746 W = 14920 W, N = rotor speed = N'(1 - s) where s = slip = 4% = 0.04 and N' = synchronous speed = 120f/p where f = frequency = 60 Hz and p = number of poles = 4.

So, N' = 120 × 60/4 = 30 × 60 = 1800 rpm

So, N = N'(1 - s) = 1800 rpm(1 - 0.04) = 1800 rpm(0.96) = 1728 rpm = 1728/60 = 28.8 rps

So, T = P/2πN = 14920 W/(2π × 28.8rps) = 14920 W/180.96 = 82.45 Nm

T₂ = torque at 340 V, V₁ = 690 V and V₂ = 340 V

So, T₂/T₁ = V₂²/V₁²

T₂ = (V₂²/V₁²)T₁

T₂ = (V₂/V₁)²T₁

T₂ = (340 V/690 V)²82.45 Nm

T₂ = (0.4928)²82.45 Nm

T₂ = (0.2428)82.45 Nm

T₂ = 20.02 Nm

T₂ ≅ 20 Nm

ii. Power

P = 2πT₂N'

= 2π × 20 Nm × 28.8 rps

= 1152π W

= 3619.11 W

converting to HP

= 3619.11 W/746 W

= 4.85 HP

2) What must be the new slip for the motor to develop the same torque when the reduced voltage is applied

Since torque T ∝ sV² where s = slip and V = voltage,

T₂/T₁ = s₂V₂²/s₁V₁²

where T₁ = torque at slip, s₁ = 4% and voltage V₁ = 690 V and T₂ = torque at slip, s₂ = unknown and voltage V₂ = 340 V

If the torque is the same, T₁ = T₂ ⇒ T₂T₁ = 1

So,

T₂/T₁ = s₂V₂²/s₁V₁²

1 = s₂V₂²/s₁V₁²

s₂V₂² = s₁V₁²

s₂ = s₁V₁²/V₂²

s₂ = s₁(V₁/V₂)²

substituting the values of the variables into the equation, we have

s₂ = s₁(V₁/V₂)²

s₂ = 4%(690/340)²

s₂ = 4%(2.0294)²

s₂ = 4%(4.119)

s₂ = 16.47 %

s₂ ≅ 16.5 %

Answer:

[tex]2\ \text{N}[/tex]

Explanation:

[tex]A[/tex] = Cross sectional area of bone = [tex]0.0004\ \text{m}^2[/tex]

[tex]P[/tex] = Pressure = [tex]5000\ \text{Pa}[/tex]

[tex]F[/tex] = Force applied to the bone

Force is given by

[tex]F=PA[/tex]

[tex]\Rightarrow F=5000\times 0.0004[/tex]

[tex]\Rightarrow F=2\ \text{N}[/tex]

The force applied to the bone is [tex]2\ \text{N}[/tex].

Answer:

(2) ( (x+y)⁴)³(3)+(3)x(4x+y)

Explanation:

correct me if I'm wrong^_^

Answer:

The rate at which entropy is produced within the turbine is 22.762 kilojoules per kilogram-Kelvin.

Explanation:

By either the Principle of Mass Conservation and First and Second Laws of Thermodynamics, we model the steam turbine by the two equations described below:

Principle of Mass Conservation

[tex]\dot m_{in} - \dot m_{out} = 0[/tex] (1)

First Law of Thermodynamics

[tex]-\dot Q_{out} + \dot m \cdot \left[h_{in}-h_{out}+ \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) - w_{out} \right] = 0[/tex] (2)

Second Law of Thermodynamics

[tex]-\frac{\dot Q_{out}}{T_{out}} + \dot m\cdot (s_{in}-s_{out}) + \dot S_{gen} = 0[/tex] (3)

By dividing each each expression by [tex]\dot m[/tex], we have the following system of equations:

[tex]-q_{out} + h_{in}-h_{out} + \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2}) - w_{out} = 0[/tex] (2b)

[tex]-\frac{q_{out}}{T_{out}} + s_{in}-s_{out} + s_{gen} = 0[/tex] (3b)

Where:

[tex]\dot Q_{out}[/tex] - Heat transfer rate between the turbine and its surroundings, in kilowatts.

[tex]q_{out}[/tex] - Specific heat transfer between the turbine and its surroundings, in kilojoules per kilogram.

[tex]T_{out}[/tex] - Outer surface temperature of the turbine, in Kelvin.

[tex]\dot m[/tex] - Mass flow rate through the turbine, in kilograms per second.

[tex]h_{in}[/tex], [tex]h_{out}[/tex] - Specific enthalpy of water at inlet and outlet, in kilojoules per kilogram.

[tex]v_{in}[/tex], [tex]v_{out}[/tex] - Speed of water at inlet and outlet, in meters per second.

[tex]w_{out}[/tex] - Specific work of the turbine, in kilojoules per kilogram.

[tex]s_{in}[/tex], [tex]s_{out}[/tex] - Specific entropy of water at inlet and outlet, in kilojoules per kilogram-Kelvin.

[tex]s_{gen}[/tex] - Specific generated entropy, in kilojoules per kilogram-Kelvin.

By property charts for steam, we get the following information:

Inlet

[tex]T = 400\,^{\circ}C[/tex], [tex]p = 3000\,kPa[/tex], [tex]h = 3231.7\,\frac{kJ}{kg}[/tex], [tex]s = 6.9235\,\frac{kJ}{kg\cdot K}[/tex]

Outlet

[tex]T = 100\,^{\circ}C[/tex], [tex]p = 101.42\,kPa[/tex], [tex]h = 2675.6\,\frac{kJ}{kg}[/tex], [tex]s = 7.3542\,\frac{kJ}{kg\cdot K}[/tex]

If we know that [tex]h_{in} = 3231.7\,\frac{kJ}{kg}[/tex], [tex]h_{out} = 2675.6\,\frac{kJ}{kg}[/tex], [tex]v_{in} = 160\,\frac{m}{s}[/tex], [tex]v_{out} = 100\,\frac{m}{s}[/tex], [tex]w_{out} = 540\,\frac{kJ}{kg}[/tex], [tex]T_{out} = 350\,K[/tex], [tex]s_{in} = 6.9235\,\frac{kJ}{kg\cdot K}[/tex] and [tex]s_{out} = 7.3542\,\frac{kJ}{kg\cdot K}[/tex], then the rate at which entropy is produced withing the turbine is:

[tex]q_{out} = h_{in} - h_{out} + \frac{1}{2}\cdot (v_{in}^{2}-v_{out}^{2})-w_{out}[/tex]

[tex]q_{out} = 3231.7\,\frac{kJ}{kg} - 2675.6\,\frac{kJ}{kg} + \frac{1}{2}\cdot \left[\left(160\,\frac{m}{s} \right)^{2}-\left(100\,\frac{m}{s} \right)^{2}\right] - 540\,\frac{kJ}{kg}[/tex]

[tex]q_{out} = 7816.1\,\frac{kJ}{kg}[/tex]

[tex]s_{gen} = \frac{q_{out}}{T_{out}}+s_{out}-s_{in}[/tex]

[tex]s_{gen} = \frac{7816.1\,\frac{kJ}{kg} }{350\,K} + 7.3542\,\frac{kJ}{kg\cdot K} - 6.9235\,\frac{kJ}{kg\cdot K}[/tex]

[tex]s_{gen} = 22.762\,\frac{kJ}{kg\cdot K}[/tex]

The rate at which entropy is produced within the turbine is 22.762 kilojoules per kilogram-Kelvin.

Answer:

2901 vehicles

Explanation:

We are given;

Percentage of large trucks & buses; p_t = 7% = 0.07

Percentage of recreational vehicles; p_r = 3% = 0.03

PHF = 0.90

Driver population adjustment; f_p = 0.92

First of all, let's Calculate the heavy vehicle factor from the formula;

f_hv = 1/[1 + p_t(e_t - 1) + p_r(e_r - 1)]

Where;

e_t = passenger car equivalents for trucks and buses

e_r = passenger car equivalents for recreational vehicles

From the table attached, for a mountainous terrain, e_t = 6 and e_r = 4. Thus;

f_hv = 1/[1 + 0.07(6 - 1) + 0.03(4 - 1)]

f_hv = 1.44

Let's now calculate the initial hourly volume from the formula;

v_p = V1/(PHF × N × f_hv × f_p)

Where;

v_p = 15-minute passenger-car equivalent flow rate

V1 = hourly volume

N = number of lanes in each direction

From online tables of LOS criteria for multilane freeway segments, v_p = 1300 pc/hr/ln

Thus;

1300 = V1/(0.9 × 3 × 1.44 × 0.92)

V1 = 1300 × (0.9 × 3 × 1.44 × 0.92)

V1 = 4650 veh/hr

Now, let's Calculate the final hourly volume;

From online sources, the maximum capacity of a 6 lane highway with free-flow speed of 50 mi/h is 2000 pc/hr/ln.

We are told the online peak-hour factor increases to 0.95 and so PHF = 0.95.

Thus;

2000 = V2/(0.95 × 3 × 1.44 × 0.92)

V2 = 2000(0.95 × 3 × 1.44 × 0.92)

V2 = 7551 veh/hr

Number of vehicles added to the highway = V2 - V1 = 7551 - 4650 = 2901 vehicles

Answer:

Horizontal force = 89.2 N

Explanation:

The frictional force = coefficient of friction * magnitude of the force (weight of the body) * cos theta

Substituting the given values, we get -

Frictional Force = 0.3*300 * cos 25 = 89.2 N

Horizontal force = 89.2 N

Answer:

- 46.5171kW

Explanation:

FIrst, the value given:

P1 = 1.05 bar (Initial pressure)

P2 = 12 bar (final pressure)

Heat transfer, Q = - 3.5 kW (It is negative because the compressor losses heat to the surroundings)

Mgaseous nitrogen = Mair = 28.0134 Kg/mol (constant)

Universal gas constant, Ru = 8.3143 Kj/Kgmolk

Specific gas constant, R = 0.28699 Kj/KgK

Initial temperature, T1 = 300 K

Final temperature, T2 = 400 K

Finding the volume:

P1V1 = RT1

V1 = RT1 ÷ P1

= (0.28699 Kj/KgK X 300k) ÷ 105

Note convert bar to Kj/Nm by multiply it by 100

V1 =  0.81997 m3/Kg

To get the mass flow rate:

m = volumetric flow rate / V1

= (21 m3/min x 1/60seconds) ÷ 0.81997 m3/Kg

= 0.4268Kg/s

Using tables for the enthalpy,

hT1 = 300.19 KJ/Kg

hT2 = 400.98 KJ/Kg

The enthalpy change = hT2 - hT1

= 100.79 KJ/Kg

Power, P = Q - (m X enthalpy change)

= - 3.5 - (0.4268 X 100.79)

= - 46.5171kW

Answer:

Hydrostatic force = 41168 N

Explanation:

Complete question

A triangular plate with a base 5 ft and altitude 3 ft is submerged vertically in water  so that the top is 4 ft below the surface. If the base is in the surface of water, find the force against onr side of the plate. Express the hydrostatic force against one side of the plate as an integral and evaluate it. (Recall that the weight density of water is 62.5 lb/ft3.)

Let "x" be the side length submerged in water.

Then

w(x)/base = (4+3-x)/altitude

w(x)/5 = (4+3-x)/3

w(x) = 5* (7-x)/3

Hydrostatic force = 62.5 integration of  x * 4 * (10-x)/3 with limits from 4 to 7

HF = integration of 40x - 4x^2/3

HF = 20x^2 - 4x^3/9 with limit 4 to 7

HF = (20*7^2 - 4*7^(3/9))- (20*4^2 - 4*4^(3/9))

HF = 658.69 N *62.5 = 41168 N