Answer:
Explanation:
First, It's important to remember F = ma, and in this problem m = 13.3 kg
This can be reduced to a simple system of equations problem. Now if they are both going the same way then we add them, while if they are going the opposite way we subtract them. So let's call them F1 and F2, with F1 arger than F2. Now, When we add them together F1+F2 = (.723 m/s^2)*13.3kg and then when we subtract them, and have the larger one pushing toward the east, let's call F1 the larger one, F1-F2 = (.493 m/s^2)*13.3kg.
Can you solve this system of equations seeing them like this, or do you need more help?
Answer:
a= 92. 13 m/s²
Explanation:
Given that
Amplitude ,A= 0.165 m
The maximum speed ,V(max) = 3.9 m/s
We know that maximum velocity in the SHM given as
V(max) = ω A
ω=Angular speed
A=Amplitude
ω=23.63 rad/s
The maximum acceleration given as
a = ω² A
a= (23.63)² x 0.165 m/s²
a= 92. 13 m/s²
Therefore the maximum magnitude of the acceleration will be 92. 13 m/s².
the awnser is yes because when I did this assignment I got an A
Answer:
2.10L
Explanation:
Given data
V1= 2.5L
T1= 275K
P1= 2.1atm
P2= 2.7 atm
T2= 298K
V2= ???
Let us apply the gas equation
P1V1/T1= P2V2/T2
substitute into the expression we have
2.1*2.5/275= 2.7*V2/298
5.25/275= 2.7*V2/298
Cross multiply
275*2.7V2= 298*5.25
742.5V2= 1564.5
V2= 1564.5/742.5
V2= 2.10L
Hence the final volume is 2.10L
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