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2 years ago

In operant conditioning, generalization occurs when

1 answer:

5 0

Typically occurs when we associate things to other things that look alike. We see that in many experiments, specifically “Little Albert” who was conditioned to be afraid of rats but later was afraid of anything that resembled that of a rat.
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An investigator places a sample 1.0 cm from a wire carrying a large current; the strength of the magnetic field has a particular

Sonja [21]

Answer: she will have to increase the factor of current by 11

Explanation: The mathematical relationship between the strength of the magnetic field (B) created by a current carrying conductor with current (I) is given by the Bio-Savart law given below

B= $\frac{u_{0}I }{2\pi r}$

B=strength of magnetic field

I = current on conductor

r = distance on any point of the conductor from it center

u $_{0}$ = permeability of magnetic field in space

from the question, the investigator is trying to keep a constant magnetic field meaning B has a fixed value such as the constants in the formulae, the only variables here are current (I) and distance (r). We can get this a mathematical function.

by cross multipying, we have

B* 2πr= $u_{0}$ I 

by dividing through to make I subject of formulae, we have that

I = $\frac{B*2\pi r}{u_{0} }$

B, 2π and $u_{0}$ are all constants, thus

$\frac{B*2\pi r}{u_{0} }$ = k(constant)

thus we have that

I =kr (current is proportional to distance assuming magnetic field strength and other parameters are constant)

thus we have that

$\frac{I_{1} }{r_{1} }$ = $\frac{I_{2} }{r_{2} }$

$r_{1}$ =1cm and $r_{2}$ =11cm

$\frac{1_{1} }{1}$ = $\frac{I_{2} }{11}$

thus $I_{2}$ =11* $I_{1}$

which means the second current is 11 times the first current

8 0

2 years ago

A human subject with mass 96 kg, body specific heat 3500 Jkg^-1K^-1 skin temperature 34.5 degrees Celsius, surface area 1.5m^2 a

AlekseyPX

rate of heat radiation by the body is given by

$\frac{dQ}{dt} = \sigma e A(T^4 - T_s^4)$

here we know that

$\sigma = 5.67 \times 10^{-8}$

e = 0.7

$A = 1.5 m^2$

$T = 34.5 + 273 = 307.5 k$

$T_s = 21 + 273 = 294 k$

now from above formula rate of heat dissipation

$\frac{dQ}{dt} = (5.67 \times 10^{-8}}(0.7)(1.5)(307.5^4 - 294^4)$

$\frac{dQ}{dt} = 87.5$

now we know that

$Q = ms \Delta T$

from above equation

$\frac{dQ}{dt} = ms\frac{dT}{dt}$

now we have

$m s \frac{dT}{dt} = 87.5$

here we have

m = 96 kg

s = 3500

$96 \times 3500 \times \frac{dT}{dt} = 87.5$

$\frac{dT}{dt} = (2.6 \times 10^{-4}) \: ^0C/s$

$\frac{dT}{dt} = 0.94 ^0 C/hour$

7 0

2 years ago

At what time t1 does the block come back to its original equilibrium position (x=0) for the first time? Express your answer in t

Furkat [3]

Answer:

t_1 = 0.5*pi*sqrt( m / k )

Explanation:

Given:

- The block of mass m undergoes simple harmonic motion. With the displacement of x from mean position is given by:

x(t) = A*cos(w*t)

Find:

- At what time t1 does the block come back to its original equilibrium position (x=0) for the first time?

Solution:

- The first time the block moves from maximum position to its mean position constitutes of 1/4 th of one complete cycle. So, the required time t_1 is:

t_1 = 0.25*T

- Where, T : Time period of SHM.

- The time period for SHM is given by:

T = 2*pi*sqrt ( m / k )

Hence,

t_1 = 0.25 * 2 * pi * sqrt( m / k )

t_1 = 0.5*pi * sqrt( m / k )

6 0

3 years ago

How is the grouping of states in a constellation different from the grouping of planets in the solar system?

Diano4ka-milaya [45]

Stars in a constellation only appear to be grouped together when viewed from Earth, Planets really are grouped together.