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2 years ago

Can someone help answers these questions? (I attached the pictures)

Physics

1 answer:

RoseWind [281]2 years ago

3 0

Nothing works if Switch-3 is open.

-- None

-- A, E, F, G

-- None

-- A, B, E, F, G

-- A, C, D, E, F, G (everything except B)

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2. According to research, what is the most reliable form of identifying potentially effective reinforcers?

Dafna11 [192]

The most reliable form of identifying potentially effective reinforcers is known as Preference Assessment. It is an adaptive procedure vastly used in the field of reinforcers.

Explanation:

Preference Assessment identifies items that are likely to be effective as reinforcers by identifying a particular learner's preference for them.

Reinforcement is an important part of being an effective behavior technician and the preference assessments are crucial.
Plenty of research has shown that when you give the learner an opportunity to show you what they want that is a much better indication of what will work as a reinforcer rather than asking family.
Preference assessments identify attention, objects or activities that have a strong potential for serving as reinforcers for target clients. There are ethical reasons to use them.
Beyond the scope of the guide is the exhaustive review of the Preference assessment methodology and literature. However, interested readers go through Virus-Ortega et al(2014) for decision- making process and selecting an appropriate method of preference assessment for individuals with disabilities.

6 0

3 years ago

What Coulombs discovered almost 300<br> years ago

tamaranim1 [39]

Answer:

ummm hehe this is my time to shine

Explanation:

MERICIA!!!!!!!!!!!!!!!!!!!!!!!

5 0

3 years ago

Read 2 more answers

hree identical resistors are connected in series. When a certain potential difference is applied across the combination, the tot

pav-90 [236]

Answer:

The power dissipated if the three resistors were connected in parallel across the same potential difference is 405 W

Explanation:

Given;

three identical resistors connected in series

let the first resistor = R₁

let the second resistor = R₂

let the third resistor = R₃

Rt = R₁ + R₂ + R₃

Since the resistors are identical, thus, R₁ = R₂ = R₃ = R

Rt = 3R

Power is given as;

P = IV = V² / R

$P = \frac{V^2}{R_t} = \frac{V^2}{3R} \\\\3P = \frac{V^2}{R} ------equation(i)$

If the 3 identical resistor connection were changed to parallel, then the equivalent resistance in the circuit will be;

$\frac{1}{R_t} = \frac{1}{R_1} +\frac{1}{R_2} + \frac{1}{R_3} \\\\But, R_1 = R_2 = R_3\\\\\\frac{1}{R_t} = \frac{1}{R} +\frac{1}{R} + \frac{1}{R} \\\\\frac{1}{R_t} =\frac{3}{R} \\\\R_t = \frac{R}{3} \\\\P = \frac{V^2}{R_t} = \frac{3V^2}{R} \\\\P_{parallel} = \frac{3V^2}{R} ---------equation (ii)\\\\From \ equation \ (i), 3P_{series} = \frac{V^2}{R}, Substitute \ this \ into \ equation \ (ii)\\\\P = 3(\frac{V^2}{R} )\\\\P = 3(3P)\\\\P_{parallel} = 9P_{series}\\\\P_{parallel} = 9(45)\\\\$

$P_{parallel} = 405 \ W$

Therefore, the power dissipated if the three resistors were connected in parallel across the same potential difference is 405 W

4 0

3 years ago

White light, with frequencies ranging from 4.00 x 10^14 Hz to 7.90 x 10^14 Hz, is incident on a barium surface. Given that the w

REY [17]

Answer:

0.7515875 eV

$4\times 10^{14}\leq f$

Explanation:

f = Maximum frequency = $7.9\times 10^{14}\ Hz$

h = Planck's constant = $6.626\times 10^{-34}\ m^2kg/s$

W = Work function = 2.52 eV

Converting to Joules

$W=2.52\times 1.6\times 10^{-19}\\\Rightarrow W=4.032\times 10^{-19}\ J$

Maximum photon energy is given by

$E=hf\\\Rightarrow E=6.626\times 10^{-34}\times 7.9\times 10^{14}\\\Rightarrow E=5.23454\times 10^{-19}\ J$

Maximum Kinetic energy is given by

$K=E-W\\\Rightarrow K=5.23454\times 10^{-19}-4.032\times 10^{-19}\\\Rightarrow K=1.20254\times 10^{-19}\ J$

Converting to eV

$1.20254\times 10^{-19}\times \frac{1}{1.6\times 10^{-19}}=0.7515875\ eV$

The maximum kinetic energy of electrons ejected from this surface is 0.7515875 eV

$W=hf\\\Rightarrow f=\frac{W}{h}\\\Rightarrow f=\frac{4.032\times 10^{-19}}{6.626\times 10^{-34}}\\\Rightarrow f=6.08512\times 10^{14}\ Hz$

The range of frequencies for which no electrons are ejected is

$4\times 10^{14}\leq f$

5 0

2 years ago

A driver, traveling at 22.0 m/s, slows down her 2.00 x 103 kg car to stop for a red light. What work is done by the friction for

Artemon [7]

Some work will be done on friction between wheels and road but it is negligible compared to work done on friction on breaks.

W = Ek = (m*v^2)/2 = 2000*22^2/2 = 1000*22^2 = 484KJ

Because car is not changing its potential energy, there is no work to be done on while changing it which means that all goes on changing kinetic energy (energy of motion)

6 0

3 years ago