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Savatey [412]
2 years ago
7

The 10-kg block slides down 2 m on the rough surface with kinetic friction coefficient μk = 0.2. What is the work done by the fr

iction force?

Engineering
1 answer:
Rashid [163]2 years ago
7 0

Answer:

153.2 J

Explanation:

Let's first list our given parameters;

mass (m) of the block = 10 kg

which slides down ( i.e displacement) = 2 m

kinetic coefficient of friction (μk) = 0.2

In the diagram shown below;  if we take an integral look at the component of force in the direction of the displacement; we have

F_x= Fcos 40°

F_x= 100 (cos 40°)

F_x= 76.60 N

Workdone by the friction force can now be determined as:

W = F_x × displacement

W = 76.60 × 2

W = 153.2 J

∴  the work done by the friction force = 153.2 J

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The function below takes a single string parameter: input_string. If the input contains the lowercase letter z, return the strin
Whitepunk [10]

Answer:

# string_contains function is defined with input_string

# as arguments

def string_contains(input_string):

   # if-statement that check if letter z is in input_string

   if 'z' in input_string:

       # it print"has the letter z" if it has z

       print("has the letter z")

   else:

       # else it print "not worthwhile"

       print("not worthwhile")

       

# string_contains function is called

string_contains("The animal is zebra")        

string_contains("learning is fun")

Explanation:

The code is written in Python and well commented.

Image of the output when the function is called is attached.

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2 years ago
How does Accenture generate value for clients through Agile and DevOps?
Nana76 [90]
By permanently locking in stakeholder requirements during a project's planning phase. -through highly detailed process documentation that is updated following every work cycle.
7 0
1 year ago
Read 2 more answers
A bankrupt chemical firm has been taken over by new management. On the property they found a 20,000-m3 brine pond containing 25,
Liula [17]

Answer:

Flow-rate = 0.0025 m^3/s

Explanation:

We need to assume that the flow-rate of pure water entering the pond is the same as the flow-rate of brine leaving the pond, in other words, the volume of liquid in the pond stays constant at 20,000 m^3. Using the previous assumption we can calculate the flow rate entering or leaving the tank (they are the same) building a separable differential equation dQ/dt, where Q is the milligrams (mg) of salt in a given time t, to find a solution to our problem we build a differential equation as follow:

dQ/dt = -(Q/20,000)*r  where r is the flow rate in m^3/s

what we pose with this equation is that the variable rate at which the salt leaves the pond (salt leaving over time) is equal to the concentration (amount of salt per unit of volume of liquid at a given time) times the constant rate at which the liquid leaves the tank, the minus sign in the equation is because this is the rate at which salt leaves the pond.

Rearranging the equation we get dQ/Q = -(r/20000) dt then integrating in both sides ∫dQ/Q = -∫(r/20000) dt and solving ln(Q) = -(r/20000)*t + C where C is a constant (initial value) result of solving the integrals. Please note that the integral of dQ/Q is ln(Q) and r/20000 is a constant, therefore, the integral of dt is t.

To find the initial value (C) we evaluate the integrated equation for t = 0, therefore, ln(Q) = C, because at time zero we have a concentration of 25000 mg/L = 250000000 mg/m^3 and Q is equal to the concentration of salt (mg/m^3) by the amount of liquid (always 20000 m^3) -> Q = 250000000 mg/m^3 * 20000 m^3 = 5*10^11 mg -> C = ln(5*10^11) = 26.9378. Now the equation is ln(Q) = -(r/20000)*t + 26.9378, the only thing missing is to find the constant flow rate (r) required to reduce the salt concentration in the pond to 500 mg/L = 500000 mg/m^3 within one year (equivalent to 31536000 seconds), to do so we need to find the Q we want in one year, that is Q = 500000 mg/m^3 * 20000 m^3 = 1*10^10 mg, therefore, ln(1*10^10) = -(r/20000)*31536000 + 26.9378 solving for r -> r = 0.002481 m^3/s that is approximately 0.0025 m^3/s.

Note:

  • ln() refer to natural logarithm
  • The amount of liquid in the tank never changes because the flow-rate-in is the same as the flow-rate-out
  • When solving the differential equation we calculated the flow-rate-out and we were asked for the flow-rate-in but because they are the same we could solve the problem
  • During the solving process, we always converted units to m^3 and seconds because we were asked to give the answer in m^3/seg
7 0
2 years ago
A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EF at E is 46 lb, determine the mome
Vikki [24]

Answer:

M_AD = 1359.17 lb-in

Explanation:

Given:

- T_ef = 46 lb

Find:

- Moment of that force T_ef about the line joining points A and D.

Solution:

- Find the position of point E:

                           mag(BC) = sqrt ( 48^2 + 36^2) = 60 in

                           BE / BC = 45 / 60 = 0.75

Hence,                E = < 0.75*48 , 96 , 36*0.75> = < 36 , 96 , 27 > in

- Find unit vector EF:

                           mag(EF) = sqrt ( (21-36)^2 + (96+14)^2 + (57-27)^2 ) = 115 in

                           vec(EF) = < -15 , -110 , 30 >

                           unit(EF) = (1/115) * < -15 , -110 , 30 >

- Tension            T_EF = (46/115) * < -15 , -110 , 30 > = < -6 , -44 , 12 > lb

- Find unit vector AD:

                           mag(AD) = sqrt ( (48)^2 + (-12)^2 + (36)^2 ) = 12*sqrt(26) in

                           vec(AD) = < 48 , -12 , 36 >

                           unit(AD) = (1/12*sqrt(26)) * < 48 , -12 , 36 >

                           unit (AD) = <0.7845 , -0.19612 , 0.58835 >

Next:

                           M_AD = unit(AD) . ( E x T_EF)

                           M_d = \left[\begin{array}{ccc}0.7845&-0.19612&0.58835\\36&96&27\\-6&-44&12\end{array}\right]

                            M_AD = 1835.73 + 116.49528 - 593.0568

                            M_AD = 1359.17 lb-in

3 0
2 years ago
if the mine winch drum diameter is 6 m, calculate how far the cage will drop for each single rotation of the drum
vesna_86 [32]

Answer:

The depth to which the cage will drop for each single rotation of the drum is approximately 18.85 m.

Explanation:

The mine winch drum consists of a drum around around which the haulage rope is attached

Therefore, given that the diameter of the winch drum = 6 m

The length of rope unwound by each rotation = How far the cage will drop for each single rotation of the drum

The length of the rope unwound by each rotation = The rope that goes round the circumference of the winch drum, once

∴ Since the rope that goes round the circumference of the winch drum, once = The circumference of the winch drum, we have;

The length of the rope unwound by each rotation = The circumference of the winch drum = π × The diameter of the winch drum

The length of the rope unwound by each rotation = π × 6 m = 6·π m

The length of rope unwound by each rotation = How far the cage will drop for each single rotation of the drum = 6·π m

How far the cage will drop for each single rotation of the drum = 6·π m ≈ 18.85 m

The depth to which the cage will drop for each single rotation of the drum ≈ 18.85 m.

3 0
2 years ago
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