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ioda
4 years ago
15

What is operant conditioning

Engineering
1 answer:
Degger [83]4 years ago
6 0
DescriptionOperant conditioning is a type of associative learning process through which the strength of a behavior is modified by reinforcement or punishment. It is also a procedure that is used to bring about such learning.
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g Two Standard 1/2" B18.8.2 dowel pins are to be installed in part B with an LN1 fit. The thickness of plate A is .750 +/- .005"
allochka39001 [22]

Answer:

nmuda mudaf A certain vehicle loses 3.5% of its value each year. If the vehicle has an initial value of $11,168, construct a model that represents the value of the vehicle after a certain number of years. Use your model to compute the value of the vehicle at the end of 6 years.

Explanation:

5 0
3 years ago
which of the following statements are true about client-side DNS? (Choose all that apply). a. If an APIPA address is assigned, t
Brums [2.3K]

The statements which are true about client-side DNS include all of the following;

B. Client-side DNS should be configured to point towards the DNS server that is authoritative for the domain that client wants to join.

C. Check out DNS settings using the NSLookup command.

D. Check out DNS settings using the DIG command.

E. The cache.dns file has the IP addresses of the 13 root DNS servers.

B. If a web site can be reached by IP address and not by host name, then DNS or the Hosts file would be the problem.

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In Computer technology, a DNS server can be defined as a type of server that is designed and developed to translate domain names into IP addresses, so as to allow end users access websites and other internet resources through a web browser.

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5 0
1 year ago
A coal-fired power plant equipped with a SO2 scrubber is required to achieve an overall SO2 removal efficiency of 85%. The exist
3241004551 [841]

Answer:

bypassed fraction B will be B= 0.105 (10.5%)

Explanation:

doing a mass balance of SO₂ at the exit

total mass outflow of SO₂ = remaining SO₂ from the scrubber outflow + bypass stream of SO₂

F*(1-er) =  Fs*(1-es) + Fb

where

er= required efficiency

es= scrubber efficiency

Fs and Fb = total mass inflow of  SO₂ to the scrubber and to the bypass respectively

F= total mass inflow of  SO₂

and from a mass balance at the inlet

F= Fs+ Fb

therefore the bypassed fraction B=Fb/F is

F*(1-er) =  Fs*(1-es) + Fb

1-er= (1-B)*(1-es) +B

1-er = 1-es - (1-es)*B + B

(es-er) = es*B

B= (es-er)/es = 1- er/es

replacing values

B= 1- er/es=1-0.85/0.95 = 2/19 = 0.105 (10.5%)

6 0
3 years ago
A 10 kg mass is lifted 5 m with an upward acceleration of 2 m/s^2 (Note: a process diagram is not required for this problem) a)
ohaa [14]

Answer:

a) F=20 [kgm/s^2]=20 [N]

b) W=100[kgm^2/s^2]=100[J]

c) P=44,84[kgm^2/s^3]=44,84[W]

d) W=2,778*10^-5 [kilowatt-hours]

Explanation:

a) Newton's Second Law states that <em>the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force and inversely proportional to the mass of the object.</em>

F=ma

where:

  • F, the net force [N]
  • m, mass of the object [kg]
  • a, acceleration [m/s^2]

If a 10 kg mass is lifted with an upward acceleration of 2 m/s^2, the upward force necessary is:

F=10 [kg]*2[m/s^2]=20 [kgm/s^2]=20 [N]

b) The amount of energy required to lift the box equals the magnitud of work done by the lifting force:

W=FdcosФ

where:

  • W, work executed [J]
  • F, net force [N]
  • d, displacement produced by the force [m]
  • Ф, angle between the net force and displacemt produced

Thus, the energy required to lift 5 m the mass is:

W=20[N]*5[m]cos0°=100[N.m]=100[kgm^2/s^2]=100[J]

c) To find the average power we use the formula:

P=W/t

where,

  • P, average power [W]
  • W, work executed [J]
  • t, elapsed time [s]

Thus, if the process takes 2,23 seconds the average power is:

P=100[J]/2,23[s]=44,84[J/s]=44,84[kgm^2/s^3]=44,84[W]

d) As 1 kilowatt-hours=3,6*10^6 J, then:

100 [J]*1 [kilowatt-hour]/ 3,6*10^6 [J]=2,778*10^-5 [kilowatt-hours]

6 0
3 years ago
An open rectangular tank 1 m wide and 2 m long contains gasoline to a depth of 3 m. If the height of the tank sides is 4 m. What
VARVARA [1.3K]

Answer:

ay max = 4.91 m/s²  

so here acceleration would be either right or left

Explanation:

given data

wide b = 1 m

long l  = 2 m

depth d = 3 m

height of  tank sides h = 4 m

solution

here for prevent spilling condition is

\frac{dz}{dy} ≤ - \frac{1.5 - 1 }{1}  ..........1

\frac{dz}{dy}  ≤ - 0.50

and when here

\frac{dz}{dy} =  -   \frac{ay}{g + az}    ......2  

when az is 0 ay will be

ay = -  \frac{dz}{dy} g    

and ay max will be

ay max = -( -0.50) (9.81 )

ay max = 4.91 m/s²  

so here acceleration would be either right or left

8 0
4 years ago
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