Look at the graph. What is the slope of the line?

During a normal reaction to a stressful event, muscles are moved to their maximum capacity, and sensitivity is

Aleonysh [2.5K]

Answer:

The paper focuses on the biology of stress and resilience and their biomarkers in humans from the system science perspective. A stressor pushes the physiological system away from its baseline state toward a lower utility state. The physiological system may return toward the original state in one attractor basin but may be shifted to a state in another, lower utility attractor basin. While some physiological changes induced by stressors may benefit health, there is often a chronic wear and tear cost due to implementing changes to enable the return of the system to its baseline state and maintain itself in the high utility baseline attractor basin following repeated perturbations. This cost, also called allostatic load, is the utility reduction associated with both a change in state and with alterations in the attractor basin that affect system responses following future perturbations. This added cost can increase the time course of the return to baseline or the likelihood of moving into a different attractor basin following a perturbation. Opposite to this is the system's resilience which influences its ability to return to the high utility attractor basin following a perturbation by increasing the likelihood and/or speed of returning to the baseline state following a stressor. This review paper is a qualitative systematic review; it covers areas most relevant for moving the stress and resilience field forward from a more quantitative and neuroscientific perspective.

Explanation:

8 0

2 years ago

A 2kg object initially going 4m/s to the right is later going 8m/s. whats the change in velocity?

Nezavi [6.7K]

The change in velocity is +4 m/s to the right (or -4 m/s to the left).

The object's mass is irrelevant.

6 0

3 years ago

Look at the advertisement. A poster that reads, "There's more to resistors than resistance. If you're really serious about cost,

ozzi

Options:

universal appeal

flattery

association

bandwagon

Answer: Association

Explanation: Advertisement is a marketing technique through which business organisations utilize the opportunities provided by both the print, electronic and other channels of communication to market a product to the target audience.

Association Advertising is a type of Advertising where certain attributes which have been known to be associated with good and quality products are used to market the products to the target audience.

7 0

3 years ago

Read 2 more answers

Two 51 g blocks are held 30 cm above a table. As shown in the figure, one of them is just touching a 30-long spring. The blocks

vivado [14]

The concept of this question can be well understood by listing out the parameters given.

The mass of the block = 51 g = 51 × 10⁻³ kg
The distance of the block from the table = 30 cm
Length of the spring = 30 cm

The purpose is to determine the spring constant.

Let us assume that the two blocks are Block A and Block B.

At point A on block A, the initial velocity on the block is zero

i.e. u = 0

We want to determine the time it requires for Block A to reach the table. The can be achieved by using the second equation of motion which can be expressed by using the formula.

$\mathsf{S = ut + \dfrac{1}{2}gt^2}$

From the above formula,

The distance (S) = 30 cm; we need to convert the unit to meter (m).

Since 1 cm = 0.01 m
Then, 30cm = 0.3 m

The acceleration (g) due to gravity = 9.8 m/s²

∴

inputting the values into the equation above, we have;

$\mathsf{0.3 = (0)t + \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 = \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 =4.9*(t^2)}$

By dividing both sides by 4.9, we have:

$\mathsf{t^2 = \dfrac{0.3}{4.9}}$

$\mathsf{t^2 = 0.0612}$

$\mathsf{t = \sqrt{0.0612}}$

$\mathsf{t =0.247 \ seconds}$

However, block B comes to an instantaneous rest on point C. This is achieved by the dropping of the block on the spring. During this process, the spring is compressed and it bounces back to oscillate in that manner. The required time needed to get to this point C is half the period, this will eventually lead to the bouncing back of the block with another half of the period, thereby completing a movement of one period.

By applying the equation of the time period of a simple harmonic motion.

$\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}$

where the relation between time (t) and period (T) is:

$\mathsf{t = \dfrac{T}{2}}$

T = 2t

T = 2(0.247)

T = 0.494 seconds

$\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}$

By making the spring constant k the subject of the formula:

$\mathsf{\dfrac{T}{2 \pi } = \sqrt{ \dfrac{m}{k}}}$

$\Big(\dfrac{T}{2 \pi }\Big)^2 = { \dfrac{m}{k}$

$\dfrac{T^2}{(2 \pi)^2 }= { \dfrac{m}{k}$

$\mathsf{ T^2 *k = 2 \pi^2*m} \\ \\ \mathsf{ k = \dfrac{2 \pi^2*m}{T^2}}$

$\mathsf{ k =\Big( \dfrac{(2 \pi)^2*(51 \times 10^{-3})}{(0.494)^2} \Big) N/m}$

$\mathbf{ k =8.25 \ N/m}$

Therefore, we conclude that the spring constant as a result of instantaneous rest caused by the compression of the spring is 8.25 N/m.

Learn more about simple harmonic motion here:

brainly.com/question/17315536?referrer=searchResults

6 0

3 years ago

An uncharged metal sphere, A, is on an insulating base. A second sphere, B, of the same size, shape, and material carrying charg

solong [7]

Answer:

0

Explanation:

If we bring the charged sphere B close to, but not touching it , to the uncharged sphere A, as charges can move freely on the conductor, a charge -Q will be built on the outer surface of the sphere A, facing to sphere B.
As the sphere A must remain neutral, a charge Q will be built on the surface, on the side farther to the sphere B, as the following condition must be met:

Q +(-Q) =0.

If we now remove sphere B, and place it far away, there will be a charge redistribution within sphere A, making to disappear the separation between Q and -Q.
The total charge on sphere A must be 0, as there is no charge transfer from sphere B to sphere A.

3 0

3 years ago