Explanation:
The given data is as follows.
= 0, 
= 0, 
= 0
= 17.0 m, 
= 2.10 sec
As the force P is constant and the mass "m" of the tool is constant then it means that the acceleration "a" will also be constant.
Now,
= 
17.0 = 2.205a
a = 
Also, we know that
F = 
m = 
So, m = 
= 1.66 kg
Since, the tool is subject to its weight W and is in free fall. Hence,
10.0 m = 
g = 2.411 
Hence, weight of tool in Newtonia is as follows.
W = mg
= 
= 4.00 N
Hence, weight of the tool on Newtonia is 4.00 N.
And, weight of the tool on the Earth is as follows.
W = 
= 23.62 N
Hence, weight of the tool on Earth is 23.62 N.
Answer:
1210 N/m between the inner and outer edges (repulsive)
252.59 N/m between inner and line charges (attractive)
1210+252.59=1462.59 N/m effective force/metre on the inner edge
Explanation:
f=k*Qq/r^2 was applied, the charges are concentrated at the edges of a ring both inner and the outer were shape edges with uniform charge densities 6.6 micro C/m of distance 4.5-2.7=1.8 cm
Answer:
Hey ! I think I can help with this question.
In the diagram you were given the speed which is 5.25m/s and you were also given distance which is 950metres.
So the formula which will be suitable in my own way will be speed=distance/time....Since you weren't given the time,then you can place the numbers in their respectively places which will be 5.25=950/time....Then you will cross multiply which will be 5.25t=950....Then 950÷5.25=180.952381 which is approximately 181. then don't forget to add your unit, time is measured in seconds so the answer is going to be 181seconds.This is what I know...I hope this helps you.....Thank you for the question
Answer:
Explanation:
ake is 1.2 x 10^12 J. Only 6.2 x 10^11 J of electrical energy is generated when the water is released. Calculate the efficiency of this energy
