Answer:
- Fx = -9.15 N
- Fy = 1.72 N
- F∠γ ≈ 9.31∠-10.6°
Explanation:
You apparently want the sum of forces ...
F = 8.80∠-56° +7.00∠52.8°
Your angle reference is a bit unconventional, so we'll compute the components of the forces as ...
f∠α = (-f·cos(α), -f·sin(α))
This way, the 2nd quadrant angle that has a negative angle measure will have a positive y component.
= -8.80(cos(-56°), sin(-56°)) -7.00(cos(52.8°), sin(52.8°))
≈ (-4.92090, 7.29553) +(-4.23219, -5.57571)
≈ (-9.15309, 1.71982)
The resultant component forces are ...
Then the magnitude and direction of the resultant are
F∠γ = (√(9.15309² +1.71982²))∠arctan(-1.71982/9.15309)
F∠γ ≈ 9.31∠-10.6°
The 7.5 is a 100 times stronger than the 5.5 . The 100 comes from the increase in scale from 5.5 to 7.5 which is an increase of two so you multiply the strength of the weaker earthquake 10^7-5=10^2=100 and that gives you the strength of the stronger earthquake!
Answer:
Explanation:
Given that
T₁ = 290 K
P₁ = 100 KPa
Power P =5.5 KW
mass flow rate
Lets take the exit temperature = T₂
We know that
If we assume that process inside the compressor is adiabatic then we can say that
That is why the exit pressure will be 4091 KPa.
ANOTHER RUNNING DOG
Explanation:
In the given question it is to find a suitable reference point to describe the motion of dog. Here I could suggest that it is better to compare the dog with another running dog to create the relative speed difference to get a reliable motion variation.
Because the motion of dog is in the linear with respect to the another dog and to the acceleration produced by the dog in the required interval is easy to calculate with respect to another dog which is already in motion.
Hence, I suggest that Motion of dog can be analysed better by analyse the motion variation of dog with another dog running.
