The tension in the first and second rope are; 147 Newton and 98 Newton respectively.
Given the data in the question
- Mass of first block;

- Mass of second block,

- Tension on first rope;

- Tension on second rope;
To find the Tension in each of the ropes, we make use of the equation from Newton's Second Laws of Motion:

Where F is the force, m is the mass of the object and a is the acceleration ( In this case the block is under gravity. Hence ''a" becomes acceleration due to gravity
)
For the First Rope
Total mass hanging on it; 
So Tension of the rope;

Therefore, the tension in the first rope is 147 Newton
For the Second Rope
Since only the block of mass 10kg is hang from the second, the tension in the second rope will be;

Therefore, the tension in the second rope is 98 Newton
Learn More, brainly.com/question/18288215
According to the Jefferson lab, "The scientific definition of work is: using a force to move an object a distance (when both the force and the motion of the object are in the same direction.)"
Answer:
The correct answer is: waxing gibbous, 3 days
Explanation:
Waning quarter moon: hair removal time and bangs cuts.
The growing quarter as a moment of growth, development and evolution. On the contrary, the waning moon is associated with a time of completion, debugging or liquidation of pending issues.
We must take advantage of the influence of the lunar cycle in our favor according to the action we are going to take. If you have trouble growing your hair, try to go to the hairdresser in a crescent moon: it will grow faster. It is no nonsense. Since I cut my bangs to the Cleopatra, the touch-ups last me for another 1-1.5 weeks. As I reviewed the bangs in a growing room, in just a couple of weeks I was returning to the hairdresser.
That affects hair removal. There are many people who take appointments to the beautician to shave by consulting the lunar calendar. The hair removal done as soon as the dwindling is the best because it lasts longer, lasts for another week until the next appointment.
Answer:
time will elapse before it return to its staring point is 23.6 ns
Explanation:
given data
speed u = 2.45 ×
m/s
uniform electric field E = 1.18 ×
N/C
to find out
How much time will elapse before it returns to its starting point
solution
we find acceleration first by electrostatic force that is
F = Eq
here
F = ma by newton law
so
ma = Eq
here m is mass , a is acceleration and E is uniform electric field and q is charge of electron
so
put here all value
9.11 ×
kg ×a = 1.18 ×
× 1.602 ×
a = 20.75 ×
m/s²
so acceleration is 20.75 ×
m/s²
and
time required by electron before come rest is
use equation of motion
v = u + at
here v is zero and u is speed given and t is time so put all value
2.45 ×
= 0 + 20.75 ×
(t)
t = 11.80 ×
s
so time will elapse before it return to its staring point is
time = 2t
time = 2 ×11.80 ×
time is 23.6 ×
s
time will elapse before it return to its staring point is 23.6 ns
The equation for work (W) done by an electric field is:
W = qΔV
where q is the magnitude of the charge and ΔV is the potential difference. The question gives you W and q, so plug n' play to find ΔV:
10 = 2ΔV
ΔV = 5