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NeTakaya
3 years ago
11

For the simple harmonic motion equation

Physics
1 answer:
My name is Ann [436]3 years ago
3 0
Given that : d = 5sin(pi t/4), So, maximum displacement, d = 5*(+1) = 5 Also, maximum displacement, d = 5*(-1) = -5 
You might be interested in
10 POINTS!!! Determine the pressure of your book in pascals (Pa). Show your work! (the pressure of the book in psi is 0.03 psi)
Musya8 [376]

You've listed a lot of data here, in both metric and customary units,
and I'm not even sure it's all needed.  Let me try and boil it down:

Pressure on a surface =
               (total force on a surface)
divided by (area of the surface).

The answer to the question is the pressure expressed in pascals. 
There's actually enough information here to answer the question
in 2 different ways.  We could ...

-- simply convert (0.03 pound per inch²) to pascals, or
-- go through the whole calculation of force, area, and then their quotient. 

To me, converting 0.03 psi to Pa looks easier.

-- 1 pascal = 1 newton / 1 meter²

-- On Earth, 1 kilogram of mass weighs 9.8 Newtons and 2.2 pounds.
From this, we can calculate that

                    2.2 pounds of force = 9.8 newtons of force.

                     1 pound = 4.45 newtons

(0.03 pound/inch²) x (4.45 newton/pound) x (1inch/2.54cm)² x (100cm/1m)² =

 (0.03 x 4.45 x 1² x 100²) / (2.54² x 1²)    newton/meter²  =  206.9 Pa .

7 0
4 years ago
The best way to ward of ailments later in life is to exercise at least 30 min per day.
hichkok12 [17]
True, physical excercise helps
5 0
4 years ago
A force of 150 N is exerted 22° north of east. What is the
svetlana [45]

Answer: B

Explanation:

3 0
3 years ago
Hooke's law describes an ideal spring. Many real springs are better described by the restoring force (F Sp ) s =−kΔs−q(Δs) 3 (p)
Montano1993 [528]

Answer with Explanation:

We are given that

Restoring force,(FS_p)s=-k\Delta s-q(\Delta s)^3

k=350N/m

q=750 N/m^3

We have to find the work must you do to compress this spring 15 cm.

\Delta s=15 cm=0.15 m

Using 1 m=100 cm

Work done=\int_{0}^{0.15}-Fd(\Delta s)

W=-\int_{0}^{0.15}(-k\Delta s-q(\Delta s)^3))d(\Delta s)

W=k[\frac{(\Delta s)^2}{2}]^{0.15}_{0}+q[\frac{(\Delta s)^4}{4}]^{0.15}_{0}

W=0.01125k+0.000127q=0.01125\times 350+0.000127\times 750

W=4.033 J

Ideal spring work=0.5k(\Delta s)^2=0.5\times 350\times (0.15)^2=3.938 J

Percentage increase in work=\frac{4.033-3.938}{3.928}\times 100=2.4%

6 0
4 years ago
(30 points) Air at 500 kPa and 400 K enters an adiabatic nozzle at a velocity of 30 m/s and leaves at 300 kPa and 350 K. Using v
Sophie [7]

Answer:

Explanation:

Check attachment for solution

8 0
3 years ago
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