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Tom [10]
3 years ago
11

How would you design an experiment to condition a rabbit to salivate to the ringing of a cell phone?​

Physics
1 answer:
Klio2033 [76]3 years ago
5 0
Food causes the response (salivation). Pair the food each time with a cell phone ringing. Continue this process with repetition.
• Eventually, the rabbit will learn to salivate at the ringing
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The largest tendon in the body, the Achilles tendon, connects the calf muscle to the heel bone of the foot. This tendon is typic
tia_tia [17]

Answer:

Tension on tendon = 1669800N

Explanation:

Detailed explanation and calculation is shown in the image below

7 0
3 years ago
Forces that act in equal and opposite directions on an object
Akimi4 [234]
These are known as balanced forces because they will not change the motion of the object, and it will remain at rest unless forces become unbalanced- meaning they would be unequal and not opposing. 
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3 years ago
The blades in a blender rotate at a rate of 7700 rpm. when the motor is turned off during operation, the blades slow to rest in
Tpy6a [65]

Angular acceleration = (change in angular speed) / (time for the change)

Change in angular speed = (speed at the end) - (speed at the beginning)

For this fan, speed at the end = 7700 rpm, speed at the end = 0 .

Change in angular speed = -7700 rpm

Angular acceleration = (-7700 rpm) / (2.5 sec)

<em>Angular acceleration = -3,080 rev per minute / sec</em>

That's a perfectly good and true answer to the question, but the units are ugly.  We really need to fix the units, and convert them into something prettier before we hand in this assignment.

1 rev = 2π radians, and

1 minute = 60 seconds .

So

Angular acceleration =

(-3,080 rev/min-sec) · (2π rad/rev) · (1 min/60 sec)

AngAccel = (-3,080 · 2π · 1 / 60) · (rev·rad·min / min·sec·rev·sec)

AngAccel = ( -102 and 2/3 · π) · (rad/s²)

<em>AngAccel = -322.5 radian/s²</em>

7 0
3 years ago
Question 5 of 10
muminat
The correct answer to this question is D
5 0
3 years ago
A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.
kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

A)

In order to calculate the image position, we can use the lens equation:

\frac{1}{p}+\frac{1}{q}=\frac{1}{f}

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm

The negative sign means that the image is virtual.

B)

In order to calculate the image height, we use the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm

And the positive sign means the image is upright.

6 0
3 years ago
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