Conflict resolution is conceptualized as the methods and processes involved in facilitating the peaceful ending of conflict and retribution.
<span>When reading a buret, the initial reading should be taken from the top of the glassware and the final volume should still taken at the top. If the buret is completely, the initial volume for most buret would be zero. though, there are some where their initial starts at 50 decreasing to zero.</span>
Hello!
Which nuclei is NOT radioactive?
A) Am-241 B) Mg-24 C) Pu-241 D) U-238
Solving:
It is noteworthy that chemical elements located on the periodic table in the lanthanide and actinide groups are radioactive.
Am-241 (americium) belongs to the group of actinides and is a heavy and radioactive metal.
Mg-24 (magnesium) is an essential element for the body, mainly for the nervous system, in addition to synthesizing proteins and serves for hormonal control, belongs to the group of alkaline earth metals and is a non-radioactive nucleus.
Pu-241 (plutonium) is an element that is isotope of fission by plutonium, belongs to the group of actinides and is a heavy and radioactive metal.
U-238 (uranium) is an element that is isotope of non-fission uranium, belongs to the group of actinides and is a heavy and radioactive metal.
Answer:
B) Mg-24
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Answer:
Tension= 21,900N
Components of Normal force
Fnx= 17900N
Fny= 22700N
FN= 28900N
Explanation:
Tension in the cable is calculated by:
Etorque= -FBcostheta(1/2L)+FT(3/4L)-FWcostheta(L)= I&=0 static equilibrium
FTorque(3/4L)= FBcostheta(1/2L)+ FWcostheta(L)
Ftorque=(Fcostheta(1/2L)+FWcosL)/(3/4L)
Ftorque= 2/3FBcostheta+ 4/3FWcostheta
Ftorque=2/3(1350)(9.81)cos55° + 2/3(2250)(9.81)cos 55°
Ftorque= 21900N
b) components of Normal force
Efx=FNx-FTcos(90-theta)=0 static equilibrium
Fnx=21900cos(90-55)=17900N
Fy=FNy+ FTsin(90-theta)-FB-FW=0
FNy= -FTsin(90-55)+FB+FW
FNy= -21900sin(35)+(1350+2250)×9.81=22700N
The Normal force
FN=sqrt(17900^2+22700^2)
FN= 28.900N
Answer:
A blackbody, or Planckian radiator, is a cavity within a heated material from which heat cannot escape. No matter what the material, the walls of the cavity exhibit a characteristic spectral emission, which is a function of its temperature.
Example:
Emission from a blackbody is temperature dependent and at high temperature, a blackbody will emit a spectrum of photon energies that span the visible range, and therefore it will appear white. The Sun is an example of a high-temperature blackbody.