Answer:
In the beginning, I was not familiar to assess assessments of the other students. Ifelt a little bit weird that is it possible to check assignments while having an instructor.I was also a bit frustrated, to be honest, that why do we have to assess thoseassessments. It was kind of extra burden for me. But after few weeks assessingmore assignments, my feeling had changed because I was learning lots of thingsthat were changing my perspectives. I was gaining extra knowledge from my peersin the form of assessments. Yes, I am comfortable with assessing assessments,because I got to learn many vocabularies and making structures of the sentencecorrectly by improving grammatically as I am not a native English speaker. Thus, inthis way, I was learning something new in each and every assessment.
Answer:
A.) 1372 N
B.) 1316 N
C.) 1428 N
Explanation:
Given that a 140 kg load is attached to a crane, which moves the load vertically. Calculate the tension in the cable for the following cases:
a. The load moves downward at a constant velocity
At constant velocity, acceleration = 0
T - mg = ma
T - mg = 0
T = mg
T = 140 × 9.8
T = 1372N
b. The load accelerates downward at a rate 0.4 m/s??
Mg - T = ma
140 × 9.8 - T = 140 × 0.4
1372 - T = 56
-T = 56 - 1372
- T = - 1316
T = 1316N
C. The load accelerates upward at a rate 0.4 m/s??
T - mg = ma
T - 140 × 9.8 = 140 × 0.4
T - 1372 = 56
T = 56 + 1372
T = 1428N
Answer:
0.04s
Explanation:
Specific heat of water c = 4.186J/g.C
The heat energy it would take to heat up 0.5 kg of water from 30C to 72 C is
If the power of the electric kettle is 2.2kW (or 2200W) and suppose the work efficiency is 100%, then the time it takes to transfer that power is
Answer:
The correct option is;
- 4x
Explanation:
From the inverse square law, as the distance from the source of a physical quantity increases, the intensity of the source is spread over an area proportional to the square of the distance of the object from the source
The inverse square law can be presented as follows;
As the distance, r, increases, the surface it covers also increases by the power of 2
Therefore, where the distance increases from r to 2·r, we have;
When, I, remain constant
The surface increases to 4·S by the inverse square law
Therefore, the correct option is 4 × x.
