Answer:
Maximum acceleration in the simple harmonic motion will be
Explanation:
We have given amplitude of simple harmonic motion is A = 0.43 m
Time period of the oscillation is T = 3.9 sec
We have to find the maximum acceleration
For this we have to find the angular frequency
Angular frequency will be equal to 
Maximum acceleration is given by 
So maximum acceleration in the simple harmonic motion will be
Answer:
temperature on left side is 1.48 times the temperature on right
Explanation:
GIVEN DATA:

T1 = 525 K
T2 = 275 K
We know that


n and v remain same at both side. so we have

..............1
let final pressure is P and temp 

..................2
similarly
.............3
divide 2 equation by 3rd equation
![\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}](https://tex.z-dn.net/?f=%5Cfrac%7B21%7D%7B11%7D%5E%7B-2%2F3%7D%20%5Cfrac%7B21%7D%7B11%7D%5E%7B5%2F3%7D%20%3D%20%5B%5Cfrac%7BT_1%20%7Bf%7D%7D%7BT_2%20%7Bf%7D%7D%5D%5E%7B5%2F3%7D)

thus, temperature on left side is 1.48 times the temperature on right
Answer:
a = 2.22 [m/s^2]
Explanation:
First we have to convert from kilometers per hour to meters per second
![40 [\frac{km}{h}]*[\frac{1h}{3600s}]*[\frac{1000m}{1km}] = 11.11 [m/s]](https://tex.z-dn.net/?f=40%20%5B%5Cfrac%7Bkm%7D%7Bh%7D%5D%2A%5B%5Cfrac%7B1h%7D%7B3600s%7D%5D%2A%5B%5Cfrac%7B1000m%7D%7B1km%7D%5D%20%3D%2011.11%20%5Bm%2Fs%5D)
We have to use the following kinematics equation:

where:
Vf = final velocity = 11.11 [m/s]
Vi = initial velocity = 0
a = acceleration [m/s^2]
t = time = 5 [s]
The initial speed is taken as zero, as the car starts from zero.
11.11 = 0 + (a*5)
a = 2.22 [m/s^2]
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