Answer:
Calcium chloride is very effective, working at temperatures below most products, and is significantly more effective than sodium chloride because of its ability to extract moisture from its surroundings and to cause exothermic or heat generating reactions.
Explanation:
Hope this helps
I had to look for the options and here is my answer:
Based on the actual options attached to this question, the statements that are considered true about electromagnetic induction are the following:
-It is possible to induce a current in a closed loop of wire without the aid of a power supply or battery.
-It is possible to induce a current in a closed loop of wire by changing the strength of a magnetic field enclosed by the wire.
-It is possible to induce a current in a closed loop of wire by change the orientation of a magnetic field enclosed by the wire.
-It is possible to induce a current in a closed loop of wire located in a uniform magnetic field by either increasing or decreasing the <span>area enclosed by the loop.</span>
The answer is electron :)
Answer:
The angular acceleration of the blades is
.
Explanation:
Given that,
Initial angular velocity of the blade of fan, ![\omega_i=270\ rpm=28.27\ rad/s](https://tex.z-dn.net/?f=%5Comega_i%3D270%5C%20rpm%3D28.27%5C%20rad%2Fs)
Final angular velocity of the blade of a fan, ![\omega_f=440\ rpm=46.07\ rad/s](https://tex.z-dn.net/?f=%5Comega_f%3D440%5C%20rpm%3D46.07%5C%20rad%2Fs)
Time, t = 5.95 s
The angular acceleration of the blades is equal to the rate of change of its angular velocity. It is given by :
![\alpha =\dfrac{\omega_f-\omega_i}{t}\\\\\alpha =\dfrac{46.07-28.27}{5.95}\\\\\alpha =2.99\ rad/s^2\\\\\alpha =3\ rad/s^2](https://tex.z-dn.net/?f=%5Calpha%20%3D%5Cdfrac%7B%5Comega_f-%5Comega_i%7D%7Bt%7D%5C%5C%5C%5C%5Calpha%20%3D%5Cdfrac%7B46.07-28.27%7D%7B5.95%7D%5C%5C%5C%5C%5Calpha%20%3D2.99%5C%20rad%2Fs%5E2%5C%5C%5C%5C%5Calpha%20%3D3%5C%20rad%2Fs%5E2)
So, the magnitude of the angular acceleration of the blades is
.
Answer:
(a) ![a_{y}= 1.931m/s^{2}](https://tex.z-dn.net/?f=a_%7By%7D%3D%201.931m%2Fs%5E%7B2%7D)
(b) ![t=10.478s](https://tex.z-dn.net/?f=t%3D10.478s)
Explanation:
For Part(a)
Apply ∑Fy=ma
![T-Mg=Ma_{y} \\](https://tex.z-dn.net/?f=T-Mg%3DMa_%7By%7D%20%5C%5C)
Solve for a
![a_{y}=\frac{T-Mg}{M}=\frac{2.80mg-Mg}{M} \\a_{y}=\frac{(2.80*575kg-1345kg)}{1345kg}*(9.80m/s^{2} )\\a_{y}= 1.931m/s^{2}](https://tex.z-dn.net/?f=a_%7By%7D%3D%5Cfrac%7BT-Mg%7D%7BM%7D%3D%5Cfrac%7B2.80mg-Mg%7D%7BM%7D%20%5C%5Ca_%7By%7D%3D%5Cfrac%7B%282.80%2A575kg-1345kg%29%7D%7B1345kg%7D%2A%289.80m%2Fs%5E%7B2%7D%20%29%5C%5Ca_%7By%7D%3D%201.931m%2Fs%5E%7B2%7D)
For Part(b)
Assume the maximum acceleration (maximum tension upwards is greater than the weight of the chain+boulder)
As the given data
![a_{y}=1.931m/s^{2}\\ y-y_{o}=106m\\v_{oy}=0](https://tex.z-dn.net/?f=a_%7By%7D%3D1.931m%2Fs%5E%7B2%7D%5C%5C%20%20y-y_%7Bo%7D%3D106m%5C%5Cv_%7Boy%7D%3D0)
Using equation of simple motion