Ill say it would be 30 degrees F
Answer:
177.65
Explanation:
Work done by the force field,F along path C is given by:
Given that:
![F(r(t))=(t-sin \ t)i+(3-cos \ t)j\\\\\\W=\int\limits_C F.dr\\=\int\limits^{6\pi}_0(t-sin \ t)i+(3-cos \ t)j).((1-cos \ t)i+sin \ t \ j)dt\\\\=\int\limits^{6\pi}_0(t-sin \ t)(1-cos \ t)+(3-cos \ t)sin \ t \ dt\\\\=\int\limits^{6\pi}_0t-tcos \ t+2sin\ t\ dt\\\\=\int\limits^{6\pi}_0-tcos \ t\ dt+[\frac{t^2}{2}-2cos \ t]\limits^{6\pi}_0\\\\=-I+177.65](https://tex.z-dn.net/?f=F%28r%28t%29%29%3D%28t-sin%20%5C%20t%29i%2B%283-cos%20%5C%20t%29j%5C%5C%5C%5C%5C%5CW%3D%5Cint%5Climits_C%20F.dr%5C%5C%3D%5Cint%5Climits%5E%7B6%5Cpi%7D_0%28t-sin%20%5C%20t%29i%2B%283-cos%20%5C%20t%29j%29.%28%281-cos%20%5C%20t%29i%2Bsin%20%5C%20t%20%5C%20j%29dt%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7B6%5Cpi%7D_0%28t-sin%20%5C%20t%29%281-cos%20%5C%20t%29%2B%283-cos%20%5C%20t%29sin%20%5C%20t%20%5C%20dt%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7B6%5Cpi%7D_0t-tcos%20%5C%20t%2B2sin%5C%20t%5C%20dt%5C%5C%5C%5C%3D%5Cint%5Climits%5E%7B6%5Cpi%7D_0-tcos%20%5C%20t%5C%20dt%2B%5B%5Cfrac%7Bt%5E2%7D%7B2%7D-2cos%20%5C%20t%5D%5Climits%5E%7B6%5Cpi%7D_0%5C%5C%5C%5C%3D-I%2B177.65)
#Integrating I by parts:
![I=\int\limits^{6\pi}_0 tcos \ t \ dt\\\\=[\int \ tcos \ t \ dt]\limits^{6\pi}_0\\\\=[t\int cos \ t \ dt-\int (\frac{dt}{dt}\intcos \ t \ dt)dt]\limits^{6\pi}_0\\\\=[tsin\ t -\int sin\ t \ dt]\limits^{6\pi}_0\\\\=0](https://tex.z-dn.net/?f=I%3D%5Cint%5Climits%5E%7B6%5Cpi%7D_0%20tcos%20%5C%20t%20%5C%20dt%5C%5C%5C%5C%3D%5B%5Cint%20%5C%20tcos%20%5C%20t%20%5C%20dt%5D%5Climits%5E%7B6%5Cpi%7D_0%5C%5C%5C%5C%3D%5Bt%5Cint%20cos%20%5C%20t%20%5C%20dt-%5Cint%20%28%5Cfrac%7Bdt%7D%7Bdt%7D%5Cintcos%20%5C%20t%20%5C%20dt%29dt%5D%5Climits%5E%7B6%5Cpi%7D_0%5C%5C%5C%5C%3D%5Btsin%5C%20t%20-%5Cint%20sin%5C%20t%20%5C%20dt%5D%5Climits%5E%7B6%5Cpi%7D_0%5C%5C%5C%5C%3D0)
Hence, work done is 177.65
Answer:
-47 °C
Explanation:
Speed of sound in air at sea level is:
v = 331 m/s + 0.6 m/s/°C T
where T is the temperature in Celsius.
The speed of sound is also the frequency times the wavelength:
v = fλ
Therefore:
fλ = 331 m/s + 0.6 m/s/°C T
(658 Hz) (0.46 m) = 331 m/s + 0.6 m/s/°C T
302.68 m/s = 331 m/s + 0.6 m/s/°C T
-28.32 m/s = 0.6 m/s/°C T
T = -47.2 °C
Rounding to two significant figures, the temperature is -47 °C.
Answer:
C
Explanation:
The Position Versus Time Graph Of This Object Is A Parabolic Curve.
Answer:
10×2=20
10×2÷64=??
I am not sure what you are trying to say...
