If the ration supplementary angle is 11:7,find the measure of the larger angle larger angle?
Answer:
A. To find the mass flow rate.
We use= 220 x 0.355/ 60
= 1.3kg/s
B. Volume flowrate is = mass flowrate / density
But density is 1000kg/m³
= 1.3kg/s/ 1000kg/m³
= 0.0013m³/s
C. Flow speead at 1
= 0.0013m³/s / (2 x 10-2m)²
= 6.5m/s
D.flow speed at 2
0.0013m³/s / (8x 10-2m)²
=1.63m/s
E. Gauge pressure at point 1
= 152+ 1/1000 ( 1.63)²- 6.5² + 1000( 9.8) ( 0-1.35)
= 119kpa
Answer:
r = 0.0548 m
Explanation:
Given that,
Singly charged uranium-238 ions are accelerated through a potential difference of 2.20 kV and enter a uniform magnetic field of 1.90 T directed perpendicular to their velocities.
We need to find the radius of their circular path. The formula for the radius of path is given by :
![r=\dfrac{1}{B}\sqrt{\dfrac{2mV}{q}}](https://tex.z-dn.net/?f=r%3D%5Cdfrac%7B1%7D%7BB%7D%5Csqrt%7B%5Cdfrac%7B2mV%7D%7Bq%7D%7D)
m is mass of Singly charged uranium-238 ion, ![m=3.95\times 10^{-25}\ kg](https://tex.z-dn.net/?f=m%3D3.95%5Ctimes%2010%5E%7B-25%7D%5C%20kg)
q is charge
So,
![r=\dfrac{1}{1.9}\times \sqrt{\dfrac{2\times 3.95\times 10^{-25}\times 2.2\times 10^3}{1.6\times 10^{-19}}}\\\\r=0.0548\ m](https://tex.z-dn.net/?f=r%3D%5Cdfrac%7B1%7D%7B1.9%7D%5Ctimes%20%5Csqrt%7B%5Cdfrac%7B2%5Ctimes%203.95%5Ctimes%2010%5E%7B-25%7D%5Ctimes%202.2%5Ctimes%2010%5E3%7D%7B1.6%5Ctimes%2010%5E%7B-19%7D%7D%7D%5C%5C%5C%5Cr%3D0.0548%5C%20m)
So, the radius of their circular path is equal to 0.0548 m.
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Answer:
At that time, she was in the southern hemisphere
Explanation:
She is witnessing the southeastern trade breeze because the cloud is passing from east to west and she is also witnessing a storm from the southeast.
Because of the Coriolis effect, which deviates the atmosphere in the left-hand side of the southern hemisphere and generates a southeastern wind, this wind did not blow directly from south to east.