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3 years ago

7

All of the following steps are included in a self-modification program except __________. A. specifying target behavior B. desig

ning a program C. focusing on one goal at a time D. gathering data about target behavior
(Psychology)

2 answers:

Y_Kistochka [10]3 years ago

8 0

Answer:

C)focusing on one goal at a time

Explanation:

Self-modification programs could be regarded as a program that help individual in managing unwanted as well as dysfunctional behavioral responses whenever they are going through a problem and try to deal with it. For instance the dysfunctional behavioral response for someone with a panic attack is avoidance. Some of the the steps that are involved in in a self-modification program are;

✓ specifying target behavior ✓designing a program

✓ gathering data about target behavior

nevsk [136]3 years ago

5 0

Answer:

c

Explanation:

on edge 2020

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If the torque required to loosen a nut that holds a wheel on a car has a magnitude of 55 n·m, what force must be exerted at the

erastova [34]

Either 175 N or 157 N depending upon how the value of 48Â° was measured from.
You didn't mention if the angle of 48Â° is from the lug wrench itself, or if it's from the normal to the lug wrench. So I'll solve for both cases and you'll need to select the desired answer.
Since we need a torque of 55 NÂ·m to loosen the nut and our lug wrench is 0.47 m long, that means that we need 55 NÂ·m / 0.47 m = 117 N of usefully applied force in order to loosen the nut. This figure will be used for both possible angles.
Ideally, the force will have a 0Â° degree difference from the normal and 100% of the force will be usefully applied. Any value greater than 0Â° will have the exerted force reduced by the cosine of the angle from the normal. Hence the term "cosine loss".
If the angle of 48Â° is from the normal to the lug wrench, the usefully applied power will be:
U = F*cos(48)
where
U = Useful force
F = Force applied
So solving for F and calculating gives:
U = F*cos(48)
U/cos(48) = F
117 N/0.669130606 = F
174.8537563 N = F
So 175 Newtons of force is required in this situation.
If the 48Â° is from the lug wrench itself, that means that the force is 90Â° - 48Â° = 42Â° from the normal. So doing the calculation again (this time from where we started plugging in values) we get
U/cos(42) = F
117/0.743144825 = F
157.4390294 = F
Or 157 Newtons is required for this case.

6 0

3 years ago

If you observe an object in the universe that is roughly 8 megaparsecs in size, what is the object most likely to be?

Dmitrij [34]

E. Galaxy Cluster

Explanation:

A galaxy cluster, or cluster of galaxies, is a structure that consists of anywhere from hundreds to thousands of galaxies that are bound together by mutual gravity.

A megaparsec is a million parsecs and there are about 3.3 light years in a mega-parsec. Parsec is rather a natural distance unit for astronomers. The standard abbreviation of a mega-parsec is Mpc.

A parsec is approximately 3.09 x 1016 meters, a megaparsec is about 3.09 x 1022 meters.

Hence, 8 megaparsecs is gigantic size and that can be only of a galaxy cluster consisting of hundreds and thousands of galaxies bounded together.

Keywords: galaxies, parsec, megaparsec, galaxy cluster

Learn more about galaxy clusters and astronomical units from:

https://brainly.in/question/4624292

brainly.com/question/14214806

brainly.com/question/13315988

#learnwithBrainly

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3 years ago

A heat pump is to be used for heating a house in winter. The house is to be maintained at 70°F at all times. When the temperatur

Anna35 [415]

Answer:

$\dot{W_{H} } = 4244.48 Btu/h$

Explanation:

Temperature of the house, $T_{H} = 70^{0} F$

Convert to rankine, $T_{H} = 70^{0}+ 460 = 530 R$

Heat is extracted at 40°F i.e $T_{L} = 40^{0}F = 40 + 460 = 500 R$

Calculate the coefficient of performance of the heat pump, COP

$COP = \frac{T_{H} }{T_{H} - T_{L} } \\COP = \frac{530 }{530 - 500 }\\ COP = \frac{530}{30} \\COP = 17.67$

The minimum power required to run the heat pump is given by the formula:

$\dot{W_{H} } = \frac{\dot{Q_{H} }}{COP} \\$ ...............(*)

Where the heat losses from the house, $\dot{Q_{H} } = 75,000 Btu/h$

Substituting these values into * above

$\dot{W_{H} } = \frac{75000}{17.67} \\ \dot{W_{H} } = 4244.48 Btu/h$

3 0

3 years ago

An 50kg car travels at 2m/s. What is the car's Kinetic energy? <br> 100J<br> 200J<br> 50J

Evgen [1.6K]

Answer:

The answer is 100J.

Explanation:

In classical mechanics, kinetic energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity squared. In this question, the mass is equals to 50kg and the velocity is 2m/s

Now,

25kg×4m/s^2 = 100kgm/s^2 or 100J

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3 years ago

A 0.200-m uniform bar has a mass of 0.795 kg and is released from rest in the vertical position, as the drawing indicates. The s

aleksklad [387]

Explanation:

Since, the rod is present in vertical position and the spring is unrestrained.

So, initial potential energy stored in the spring is $U_{s}$ = 0

And, initial potential gravitational potential energy of the rod is $U_{g} = \frac{mgL}{2}$ .

It is given that,

mass of the bar = 0.795 kg

g = 9.8 $m/s^{2}$

L = length of the rod = 0.2 m

Initial total energy T = $\frac{mgL}{2}$

Now, when the rod is in horizontal position then final total energy will be as follows.

T = $\frac{1}{2}kx^{2} + I \omega^{2}$

where, I = moment of inertia of the rod about the end = $\frac{mL^{2}}{3}$

Also, $\omega = \frac{\nu}{L}$

where, $\nu$ = speed of the tip of the rod

x = spring extension

The initial unstrained length is $x_{o}$ = 0.1 m

Therefore, final length will be calculated as follows.

x' = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m

Then, x = $x' - x_{o}$

x = $\sqrt{(0.2)^{2} + (0.1)^{2}}$ m - 0.1 m

= 0.1236 m

k = 25 N/m

So, according to the law of conservation of energy

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1 \times mL^{2}}{2 \times 3}(\frac{\nu}{L})^{2}$

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

Putting the given values into the above formula as follows.

$\frac{mgL}{2} = \frac{1}{2}kx^{2} + \frac{1}{6}mv^{2}$

$\frac{0.795 kg \times 9.8 \times 0.2 m}{2} = \frac{1}{2} \times 27 N/m \times (0.1236)^{2} + \frac{1}{6} \times 0.795 \times v^{2}$

v = 2.079 m/s

Thus, we can conclude that tangential speed with which end A strikes the horizontal surface is 2.079 m/s.

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3 years ago