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SashulF [63]
3 years ago
10

PSYCHOLOGY QUESTION

Physics
2 answers:
Anna71 [15]3 years ago
7 0

Answer:

Some anxiety disorders have different symptoms and it is important to figure out how those symptoms develop. Symptoms may develop from physiological, psychological, or environmental influences. Physiological factors may include heredity, chemical and hormonal imbalances, gender (females are more prone to anxiety than males), and anxiety sensitivity, which has to do with how one’s body reacts to chemical changes. Psychological influences focus on how one’s mind reacts to fearful situations. Environmental factors also influence how individuals deal with fearful situations.  Environmental stress levels, how we think or interpret situations, and our observations of how others react to similar situations all impact how we manage anxiety.

Explanation:

Julli [10]3 years ago
3 0

Answer:

Not sure what yours looking for but When a disease is widespread, epidemiological studies investigate what associated factors, such as location, swx, exposure to chemicals and many others, make a population more or less likely to have an illness, condition, or disease thus helping determine its etiology.

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how many days does it take a team of drivers to drive a truck california and back to lansing if the drivers average 15m/s includ
Scilla [17]

Answer:

t = 2.2 [days] and is there is a round trip, it will be double time t = 4.4 [days]

Explanation:

First, we need to arrange the problem to work in the same unit system (SI).

We need to convert the 1800 [miles] to meters, therefore:

1800[miles] * \frac{1609.34[m]}{1[mile]} }=2896812[m] = 2896.8[km]

Now using the following equation of kinematics, for the avarage velocity  we have:

v=\frac{x}{t} \\where \\v=velocity [m/s]\\t = time [s]\\x=distance traveled [m]\\

therefore:

t=\frac{x}{v} \\t=\frac{2896812}{15}\\ t=193120.8[s]

Now we can convert from seconds into days.

193120.8[s]*\frac{1[hr]}{3600[s]}*\frac{1[day]}{24[hr]}\\  t = 2.2[days]

Now if the truck has the need to come back, the team will spend double time.

t= 4.4 [days]

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3 years ago
What is a normal resting heart rate for adults, according to the American
kicyunya [14]
The answer of this question is
No B
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3 years ago
Complete the following statement: When a net torque is applied to a rigid object, it always produces a:______.
Tatiana [17]

When a net torque is applied to a rigid object, it always produces a <em>change in angular velocity</em>. <em>(e.)</em>

7 0
3 years ago
Selma made an error while creating a chart to describe the technological design process for a new vacuum. Which best describes h
Makovka662 [10]
It would be helpful if you can provide the chart or any additional information about the what Selma made. Anyhow, as I've searched on the internet, the answer is - Switch the stages "Evaluate the solution" and "Design a solution."
6 0
3 years ago
Read 2 more answers
An object has the acceleration graph shown in (Figure 1). Its velocity at t=0s is vx=2.0m/s. Draw the object's velocity graph fo
timama [110]

Answer:

Explanation:

We may notice that change in velocity can be obtained by calculating areas between acceleration lines and horizontal axis ("Time"). Mathematically, we know that:

v_{b}-v_{a} = \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt

v_{b} = v_{a}+ \int\limits^{t_{b}}_{t_{a}} {a(t)} \, dt

Where:

v_{a}, v_{b} - Initial and final velocities, measured in meters per second.

t_{a}, t_{b} - Initial and final times, measured in seconds.

a(t) - Acceleration, measured in meters per square second.

Acceleration is the slope of velocity, as we know that each line is an horizontal one, then, velocity curves are lines with slopes different of zero. There are three region where velocities should be found:

Region I (t = 0 s to t = 4 s)

v_{4} = 2\,\frac{m}{s}  +\int\limits^{4\,s}_{0\,s} {\left(-2\,\frac{m}{s^{2}} \right)} \, dt

v_{4} = 2\,\frac{m}{s}+\left(-2\,\frac{m}{s^{2}} \right) \cdot (4\,s-0\,s)

v_{4} = -6\,\frac{m}{s}

Region II (t = 4 s to t = 6 s)

v_{6} = -6\,\frac{m}{s}  +\int\limits^{6\,s}_{4\,s} {\left(1\,\frac{m}{s^{2}} \right)} \, dt

v_{6} = -6\,\frac{m}{s}+\left(1\,\frac{m}{s^{2}} \right) \cdot (6\,s-4\,s)

v_{6} = -4\,\frac{m}{s}

Region III (t = 6 s to t = 10 s)

v_{10} = -4\,\frac{m}{s}  +\int\limits^{10\,s}_{6\,s} {\left(2\,\frac{m}{s^{2}} \right)} \, dt

v_{10} = -4\,\frac{m}{s}+\left(2\,\frac{m}{s^{2}} \right) \cdot (10\,s-6\,s)

v_{10} = 4\,\frac{m}{s}

Finally, we draw the object's velocity graph as follows. Graphic is attached below.

3 0
4 years ago
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