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MissTica
1 year ago
10

5 What is the angular displacement at the end of the 25-mm-diameter shaft and the linear displacement of point A of Figure P5.5

Physics
1 answer:
Anvisha [2.4K]1 year ago
3 0

5 What is the angular displacement at the end of the 25-mm-diameter shaft and the linear displacement of point A of Figure P5.5

<h3>What is displacement ?</h3>

A displacement is a vector in geometry and mechanics that has a length equal to the shortest distance between a point P's initial and final positions. It calculates the length and angle of the net motion, or total motion, in a straight line from the starting point to the destination of the point trajectory. The translation that links the starting point and the ending point can be used to spot a displacement.

The final location xf of a point relative to its beginning position xi, or a relative position (derived from the motion), is another way to express a displacement. The difference between the end and beginning positions can be used to define the equivalent displacement vector

To learn more about displacement  from the given link:

brainly.com/question/321442

#SPJ4

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Explanation:

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It is in helical shape and is used in mattress to give structure and support. Spring have elastic nature and follows spring forces, F = k * x

where is the applied force, k is the spring constant and x is the amount of extension.

When a heavier person sits on a mattress, more weight is applied on springs and they form coils, as weight is removed they regains its shape again.

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6. Dentists use a soft, clay-like material to make a filling for a cavity. The filling stays soft under the
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3 years ago
An object is thrown upward from the top of a 144​-foot building with an initial velocity of 128 feet per second. The height h of
ankoles [38]

Answer:

After 9 seconds the object reaches ground.

Step-by-step explanation:

We equation of motion given as h = -16t²+128t+144,

We need to find in how many seconds will the object hit the ground,

That is we need to find time when h = 0

                        0 = -16t²+128t+144

                       16t²-128t-144= 0

                       t=\frac{-(-128)\pm \sqrt{(-128)^2-4\times 16\times (-144)}}{2\times 16}\\\\t=\frac{128\pm \sqrt{25600}}{32}\\\\t=\frac{128\pm 160}{32}\\\\t=9s\texttt{ or }t=-1s

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3 years ago
Sound enters the ear, travels through the auditory canal, and reaches the eardrum. The auditory canal is approximately a tube op
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Answer:

The fundamental frequency of can is 2.7 kHz.                          

Explanation:

Given that,

A typical length for the auditory canal in an adult is about 3.1 cm, l = 3.1 cm

The speed of sound is, v = 336 m/s

We need to find the fundamental frequency of the canal. For a tube open at only one end, the fundamental frequency is given by :

f=\dfrac{v}{4l}\\\\f=\dfrac{336}{4\times 3.1\times 10^{-2}}\\\\f=2709.67\ Hz\\\\f=2.7\ kHz

So, the fundamental frequency of can is 2.7 kHz. Hence, this is the required solution.

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3 years ago
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