This question is incomplete, the complete question is;
All of the questions for this quiz involve a drone with a mass of 1.2 kg.
The drone hovers for 29 seconds at a height of 53 meters above the ground while taking a picture of a crowd. What is the vertical impulse by the lift force (combined force of the four rotors) during this time (in Newton-seconds)?
Answer: the vertical impulse is 341.388 Ns
Explanation:
Given that;
mass m = 1.2 kg
t = 29 sec
h = 53 m
vertical impulse = ?
Vertical impulse by lift force is given as;
Vertical Impulse = F × t
= mg × t
so we substitute
Vertical impulse = 1.2 ×9.81 × 29
= 341.388 Ns
therefore the vertical impulse is 341.388 Ns
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Answer:
The Answer is C!
Explanation:
The structure of PF3(C6H5)2 is trigonal bipyramidal, with one equatorial and two axial F atoms which interchange positions when heated. Describe the low- and high- temperature 31P and 19F NMR spectra.
Answer:
For 31 P NMR spectra
low temperature
there is two types of 19f seen in low temperature and they are
one at equitorial one at axialtherefore in low temperature the 31p couples with the two types of 19F seen ( [tex]b_{f} and c_{f}[/tex]to form a triplet and this couples more with [tex]a_{f}[/tex] to form a doublet. i.e. one (1) peak
High temperature
At High temperature The exchange is fast here therefore the 31p spectra sees all 19p at once and in the same environment leading to the formation of one (1) peak
For 19 P NMR spectra
low temperature
In low temperature [tex]a_{f}, b_{f} , c_{f}[/tex] is fixed and the environment where [tex]b_{f} and c _{f}[/tex] is the same hence a peak is formed and another peak is formed by [tex]a_{f}[/tex] that makes the number of peaks = 2 peaks
High temperature
In high temperature [tex]a_{f}, b_{f} , c_{f}[/tex] exchange very fast therefore one peak is formed for all, since the fast exchanges makes NMR machine to take an average and produce just one peak for all
Explanation:
For 31 P NMR spectra
low temperature
there is two types of 19f seen in low temperature and they are
one at equitorial one at axialtherefore in low temperature the 31p couples with the two types of 19F seen ( [tex]b_{f} and c_{f}[/tex]to form a triplet and this couples more with [tex]a_{f}[/tex] to form a doublet. i.e. one (1) peak
High temperature
At High temperature The exchange is fast here therefore the 31p spectra sees all 19p at once and in the same environment leading to the formation of one (1) peak
For 19 P NMR spectra
low temperature
In low temperature [tex]a_{f}, b_{f} , c_{f}[/tex] is fixed and the environment where [tex]b_{f} and c _{f}[/tex] is the same hence a peak is formed and another peak is formed by [tex]a_{f}[/tex] that makes the number of peaks = 2 peaks
High temperature
In high temperature [tex]a_{f}, b_{f} , c_{f}[/tex] exchange very fast therefore one peak is formed for all, since the fast exchanges makes NMR machine to take an average and produce just one peak for all
Un mol de gas ideal realiza un trabajo de 3000 J sobre su entorno, cuando se expande de manera isotermica a una temperatura de 58°C, cuando su volumen inicial es de 25 L. Determinar el volumen final
Answer:
74,4 litros
Explanation:
Dado que
W = nRT ln (Vf / Vi)
W = 3000J
R = 8,314 JK-1mol-1
T = 58 + 273 = 331 K
Vf = desconocido
Vi = 25 L
W / nRT = ln (Vf / Vi)
W / nRT = 2.303 log (Vf / Vi)
W / nRT * 1 / 2.303 = log (Vf / Vi)
Vf / Vi = Antilog (W / nRT * 1 / 2.303)
Vf = Antilog (W / nRT * 1 / 2.303) * Vi
Vf = Antilog (3000/1 * 8,314 * 331 * 1 / 2,303) * 25
Vf = 74,4 litros
g A smooth pipeline with a diameter of 5 cm carries glycerin at 20 degrees Celsius. The flow rate in the pipe is 0.01 m3/s. What is the friction factor
Answer:
The friction factor is 0.303.
Explanation:
The flow velocity ([tex]v[/tex]), measured in meters per second, is determined by the following expression:
[tex]v = \frac{4\cdot \dot V}{\pi \cdot D^{2}}[/tex] (1)
Where:
[tex]\dot V[/tex] - Flow rate, measured in cubic meters per second.
[tex]D[/tex] - Diameter, measured in meters.
If we know that [tex]\dot V = 0.01\,\frac{m^{3}}{s}[/tex] and [tex]D = 0.05\,m[/tex], then the flow velocity is:
[tex]v = \frac{4\cdot \left(0.01\,\frac{m^{3}}{s} \right)}{\pi\cdot (0.05\,m)^{2}}[/tex]
[tex]v \approx 5.093\,\frac{m}{s}[/tex]
The density and dinamic viscosity of the glycerin at 20 ºC are [tex]\rho = 1260\,\frac{kg}{m^{3}}[/tex] and [tex]\mu = 1.5\,\frac{kg}{m\cdot s}[/tex], then the Reynolds number ([tex]Re[/tex]), dimensionless, which is used to define the flow regime of the fluid, is used:
[tex]Re = \frac{\rho\cdot v \cdot D}{\mu}[/tex] (2)
If we know that [tex]\rho = 1260\,\frac{kg}{m^{3}}[/tex], [tex]\mu = 1.519\,\frac{kg}{m\cdot s}[/tex], [tex]v \approx 5.093\,\frac{m}{s}[/tex] and [tex]D = 0.05\,m[/tex], then the Reynolds number is:
[tex]Re = \frac{\left(1260\,\frac{kg}{m^{3}} \right)\cdot \left(5.093\,\frac{m}{s} \right)\cdot (0.05\,m)}{1.519 \frac{kg}{m\cdot s} }[/tex]
[tex]Re = 211.230[/tex]
A pipeline is in turbulent flow when [tex]Re > 4000[/tex], otherwise it is in laminar flow. Given that flow has a laminar regime, the friction factor ([tex]f[/tex]), dimensionless, is determined by the following expression:
[tex]f = \frac{64}{Re}[/tex]
If we get that [tex]Re = 211.230[/tex], then the friction factor is:
[tex]f = \frac{64}{211.230}[/tex]
[tex]f = 0.303[/tex]
The friction factor is 0.303.