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aleksley [76]
3 years ago
6

During times of stress, Yanira notices that her heart pounds a little faster than normal. What part of the brain is responsible

for her increased heart rate?
Physics
2 answers:
Paul [167]3 years ago
8 0
The Medulla is the part of our brain that helps regulate our involuntary life sustaining functions like breathing, swallowing, and controlling our heart rate!! 
STALIN [3.7K]3 years ago
7 0

Answer:

Medulla

Explanation:

It is given that, During times of stress, Yanira notices that her heart pounds a little faster than normal.  

The medulla is responsible for her increased heart rate. The medulla controls involuntary actions. For example, blinking of eyes, heartbeat, etc. It is is a part of hindbrain.

Human brain consists of three part of brain i.e. forebrain, midbrain and hindbrain.

The largest part of the brain is cerebrum. It controls the voluntary action i.e. emotions, control of movement.

Others parts of the human brain are pons, cerebellum etc.

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Answer:

Gravity, Weak, Electromagnetic and Strong.

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An 84.0 kg sprinter starts a race with an acceleration of 1.76 m/s2. If the sprinter accelerates at that rate for 11 m, and then
gulaghasi [49]

Answer:

t=17.838s

Explanation:

The displacement is divided in two sections, the first is a section with constant acceleration, and the second one with constant velocity. Let's consider the first:

The acceleration is, by definition:

a=\frac{dv}{dt}=1.76

So, the velocity can be obtained by integrating this expression:

v=1.76t

The velocity is, by definition: v=\frac{dx}{dt}, so

dx=1.76tdt\\x=1.76\frac{t^{2}}{2}.

Do x=11 in order to find the time spent.

11=1.76\frac{t^2}{2}\\ t^2=\frac{2*11}{76} \\t=\sqrt{12.5}=3.5355s

At this time the velocity is: v=1.76t=1.76*3.5355s=6.2225\frac{m}{s}

This velocity remains constant in the section 2, so for that section the movement equation is:

x=v*t\\t=\frac{x}{v}

The left distance is 89 meters, and the velocity is 6.2225\frac{m}{s}, so:

t=\frac{89}{6.2225}=14.303s

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7 0
3 years ago
A vertical spring with spring stiffness constant 305 N/m oscillates with an amplitude of 28.0 cm when 0.235 kg hangs from it. Th
den301095 [7]

Answer:

The function that describe the motion in the time

y (t) = 0.28m * sin ( 36.025 * t)

Explanation:

The angular frequency of oscillation of the spring

w = √k/m

w = √305 N/m / 0.235 kg

w = 36.025 rad / s

To determine the function of the motion knowing as a motion oscillation in a amplitude a frequency

y(t) = A * sin (w t )

So

A = 28.0 cm * 1 m / 100 cm = 0.28 m

So replacing to determine the function of the motion in the time

y (t) = A sin (w t)

y (t) = 0.28m * sin ( 36.025 * t)

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