What are some of the benefits of learned optimism that have been found inresearch?
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A. Fewer health problems
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O
B. All of these
C. Making more money
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D. A lower divorce rate

Answers

Answer 1

Answer:

The benefits of learned optimism that have been found in research are Fewer health problems, Making more money, and a lower divorce rate. The correct option is B.

Learned optimismhas been associated with numerous benefits in research, including fewer health problems, making more money, and a lower divorce rate. Optimistic people tend to have better physical and mental health, which leads to fewer health problems. Additionally, optimistic people tend to be more successful in their careers and finances, which can lead to higher income and better financial stability. Finally, optimistic people tend to have better relationships, including lower divorce rates, as they are better able to handle conflicts and maintain positive attitudes toward their partners.

In summary, learned optimism has a range of benefitsfor individuals, including better physical and mental health, greater success in work and education, better relationships with others, and improved resilience. These benefits make learned optimism an important skill for individuals to develop in order to lead happier, healthier, and more successful lives.

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If a microwave oven produces electromagnetic waves with a frequency of 2.30 ghz, what is their wavelength?

Answers

Answer: wavelength is $<strong> 1.30 * 10^8 nm.</strong>$

The frequency of the microwave is, f = 2.30 GHz.

To Find frequency use the formula:

$c=f$ λ

Where, c is the speed of electromagnetic wave or light. f is the frequency, and λ is the wavelength of light.

Rearranging, $\lambda = (c)/(f)$

Plug in the values,

$\lambdam = (3 * 10^8 m/s)/(2.30 GHz(10^9 Hz)/(1 GHz))=0.130 m(10^9 nm)/(1 m) = 1.30 * 10^8 nm.$

$\lambda = (c)/(\nu) = (3 \cdot 10^8)/(2.3 \cdot 10^9) = (3)/(23) \approx 0.130435 \approx 0.13 \ m.$

An airplane with a speed of 92.3 m/s is climbing upward at an angle of 51.1 ° with respect to the horizontal. When the plane's altitude is 532 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth.
(b) Relative to the ground, determine the angle of the velocity vector of the package just before impact. (a) Number Units (b) Number Units

Answers

Answer:

$D = 1162.7 \ m$

$\beta =- 65.55^o$

Explanation:

From the question we are told that

The speed of the airplane is $u = 92.3 \ m/s$

The angle is $\theta = 51.1^o$

The altitude of the plane is $d = 532 \ m$

Generally the y-component of the airplanes velocity is

$u_y = v * sin (\theta )$

=> $u_y = 92.3 * sin ( 51.1 )$

=> $u_y = 71.83 \ m/s$

Generally the displacement traveled by the package in the vertical direction is

$d = (u_y)t + (1)/(2)(-g)t^2$

=> $-532 = 71.83 t + (1)/(2)(-9.8)t^2$

Here the negative sign for the distance show that the direction is along the negative y-axis

=> $4.9t^2 - 71.83t - 532 = 0$

Solving this using quadratic formula we obtain that

$t = 20.06 \ s$

Generally the x-component of the velocity is

$u_x = u * cos (\theta)$

=> $u_x = 92.3 * cos (51.1)$

=> $u_x = 57.96 \ m/s$

Generally the distance travel in the horizontal direction is

$D = u_x * t$

=> $D = 57.96 * 20.06$

=> $D = 1162.7 \ m$

Generally the angle of the velocity vector relative to the ground is mathematically represented as

$\beta = tan ^(-1)[(v_y)/(v_x ) ]$

Here $v_y$ is the final velocity of the package along the vertical axis and this is mathematically represented as

$v_y = u_y - gt$

=> $v_y = 71.83 - 9.8 * 20.06$

=> $v_y = -130.05 \ m/s$

and v_x is the final velocity of the package which is equivalent to the initial velocity $u_x$

$\beta = tan ^(-1)[-130.05}{57.96 } ]$

$\beta =- 65.55^o$

The negative direction show that it is moving towards the south east direction

A body of mass 80kg was lifted vertically through a distance of 5.0 metres. Calculate the work done on the body. ( Acceleration due to gravity g=10ms²)

Answers

Answer:

80×5×10=4000J

so therefore, work done on the body is 4000J

Sharece knows that wave peaks and valleys can add and subtract. What would be the net effect if she was able to cross Wave 1 (a large-amplitude wave in a valley phase) with Wave 2 (a wave with slightly smaller amplitude than Wave 2, in a peak phase)?Sharece knows that wave peaks and valleys can add and subtract. What would be the net effect if she was able to cross Wave 1 (a large-amplitude wave in a valley phase) with Wave 2 (a wave with slightly smaller amplitude than Wave 2, in a peak phase)?

Answers

Answer:

The two waves will add vectorially to produce a small amplitude wave in a valley phase.

Explanation:

The two waves will add vectorially to produce a small amplitude wave in a valley phase. This is because the amplitudes of the waves are slightly different and in opposite directions. When wave 1 cancels out all of wave 2, the resultant wave would be the slight difference between both waves, and it would be in the direction of wave 1 which is a valley phase.

Which describes the motion of the box based on the resulting free-body diagram?1. It is moving up with a net force of 20 N.
2. It is moving to the right with a net force of 10 N.
3. It is in dynamic equilibrium with a net force of 0 N.
4. It is in static equilibrium with a net force of 0 N.

Answers

The statement "It is in dynamic equilibrium with a net force of 0 N" describes the motion of the box based on the resulting free-body diagram. (option 3)

What is a free-body diagram?

A free-body diagram is a diagram that shows all the forces acting on an object. If the net force on an object is zero, then the object is in equilibrium. This means that the object is not accelerating and is either at rest or moving with constant velocity.

In the case of the box in the free-body diagram, there are two forces acting on it: the force of gravity and the force of the table pushing up on the box. The force of gravity is pulling the box down, but the force of the table is pushing the box up.

These two forces are equal in magnitude and opposite in direction, so they cancel each other out. This means that the net force on the box is zero and the box is in dynamic equilibrium.

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Answer:

4. It is in static equilibrium with a net force of 0 N.

Explanation:

Just got it right :)

Two large parallel metal plates are 1.6 cm apart and have charges of equal magnitude but opposite signs on their facing surfaces. Take the potential of the negative plate to be zero. If the potential halfway between the plates is then +3.8 V, what is the electric field in the region between the plates?

Answers

Answer:

475 N/C

Explanation:

As we know that, the electric field in parallel plate capacitor is same (constant) throughout. And is potential gradient.

So, Electric field is given by

Electric field = potential gradient

$Electric FIeld = (Change\: in\: Potential)/(Distance)$

Here, the potential change is 3.8V and the distance from negative plate to positive plate is 1.6 cm. The potential from negative plate to the center is (1.6/2)cm i.e., 0.8 cm.

But we have to take distance in SI units So, distance= $0.8 * 10^(-2) m$