The average velocity or displacement of a particle for the first time interval is <u>Δs / Δt = 6 cm/s.</u>
Solution:
As we know that displacement is calculated in centimeters and the unit of time is second.
The average velocity for the first interval [1,2] is given
Δs / Δt = s (t2) - s (t) / t2 - t1
Δs / Δt = 2sin2 π + 3cos 2 π - ( 2sin π + 3cos π ) / 2 - 1
Δs / Δt = 2(0) + 3(1) - 2(0) - 3 (-1) / 1
Δs / Δt = 6 cm/s
Thus the average velocity or displacement of a particle for the first time interval is Δs / Δt = 6 cm/s
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The complete question is:
The displacement of a particle moving back and forth along a line is given by the following equation s(t) = 2sin π t + 3cos π t. Estimate the instantaneous velocity of the particle when t = 1
The velocity of the body is zero; option A
<h3>What is the motion of an oscillating body?</h3>
The motion of an oscillating body is known as simple harmonic motion.
Simple harmonic motion involves a periodical motion of a body whose acceleration is directed towards a fixed point.
For a body that is oscillating up and down at the end of a spring, considering when the body is at the top of its up-and-down motion, the velocity of the body at the top and down is zero since the body comes to rest at the top and down position of its motion.
In conclusion, oscillating bodies undergo simple harmonic motion.
Learn more about simple harmonic motion at: brainly.com/question/24646514
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Answer:Purpose of Cloning
A forensic image of a hard drive captures everything on the hard drive, from the physical beginning to the physical end. ... Hard drives are susceptible to failure. Having two clones gives an investigator one to examine and one to fall back on.Therapeutic cloning involves creating a cloned embryo for the sole purpose of producing embryonic stem cells with the same DNA as the donor cell. These stem cells can be used in experiments aimed at understanding disease and developing new treatments for disease.
Explanation:
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Answer:
A) a = 2.31[m/s^2]; B) t = 14.4 [s]
Explanation:
We can solve this problem using the kinematic equations, but firts we must identify the data:
Vf= final velocity = take off velocity = 120[km/h]
Vi= initial velocity = 0, because the plane starts to move from the rest.
dx= distance to run = 240 [m]
![v_{f} ^{2} =v_{i} ^{2}+2*g*dx\\where:\\v_{f}=120[\frac{km}{h} ]*\frac{1hr}{3600sg} * \frac{1000m}{1km} =33.33[m/s]\\\\Replacing\\33.33^{2}=0+2*a*(240)\\ a=\frac{11108.88}{2*240}\\ a=2.31[m/s^2]\\](https://tex.z-dn.net/?f=v_%7Bf%7D%20%5E%7B2%7D%20%3Dv_%7Bi%7D%20%5E%7B2%7D%2B2%2Ag%2Adx%5C%5Cwhere%3A%5C%5Cv_%7Bf%7D%3D120%5B%5Cfrac%7Bkm%7D%7Bh%7D%20%5D%2A%5Cfrac%7B1hr%7D%7B3600sg%7D%20%2A%20%5Cfrac%7B1000m%7D%7B1km%7D%20%3D33.33%5Bm%2Fs%5D%5C%5C%5C%5CReplacing%5C%5C33.33%5E%7B2%7D%3D0%2B2%2Aa%2A%28240%29%5C%5C%20a%3D%5Cfrac%7B11108.88%7D%7B2%2A240%7D%5C%5C%20%20a%3D2.31%5Bm%2Fs%5E2%5D%5C%5C)
To find the time we must use another kinematic equation.
![v_{f} =v_{i} +a*t\\replacing:\\33.33=0+(2.31*t)\\t=\frac{33.33}{2.31}\\ t=14.4[s]](https://tex.z-dn.net/?f=v_%7Bf%7D%20%3Dv_%7Bi%7D%20%2Ba%2At%5C%5Creplacing%3A%5C%5C33.33%3D0%2B%282.31%2At%29%5C%5Ct%3D%5Cfrac%7B33.33%7D%7B2.31%7D%5C%5C%20t%3D14.4%5Bs%5D)
