The charge of an electron is 1.6x10^'' C. How many electrons does it take to make 1 C of charge?
1 answer:
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Answer:
body position 4 is (-1,133, -1.83)
Explanation:
The concept of center of gravity is of great importance since in this all external forces are considered applied, it is defined by
x_cm = 1 /M ∑ m_{i}
y_cm = 1 /M ∑ y_{i} mi
Where M is the total mass of the body, mi is the mass of each element
give us the mass and position of this masses
body 1
m1 = 2.00 ka
x1 = 0 me
y1 = 0 me
body 2
m2 = 2.20 kg
x2 = 0m
y2 = 5 m
body 3
m3 = 3.4 kg
x3 = 2.00 m
y3 = 0
body 4
m4 = 6 kg
x4=?
y4=?
mass center position
x_cm = 0
y_cm = 0
let's apply to the equations of the initial part
X axis
M = 2.00 + 2.20 + 3.40
M = 7.6 kg
0 = 1 / 7.6 (2 0 + 2.2 0 + 3.4 2 + 6 x4)
x4 = -6.8 / 6
x4 = -1,133 m
Axis y
0 = 1 / 7.6 (2 0 + 2.20 5 +3.4 0 + 6 y4)
y4 = -11/6
y4 = -1.83 m
body position 4 is (-1,133, -1.83)
Answer:
q = 0.036 C
Explanation:
Given that,
Current passes through a defibrillator, I = 18 A
Time, t = 2 ms
We need to find the charge moved during this time. We know that,
Electric current = charge/time
Put all the values,
So, 0.036 C of charge moves during this time.
Answer:
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Hope this helps!
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Answer:
a) Final position is x = 0.90 m
b) Final position is x = 0.133 m
Explanation:
The workdone between two points is usually approximated as the area under the force-distance curve between those two points.
From the graph,
As at the initial position, x = 0.40 m and the corresponding F = 0.8 N,
The area from that point onwards up to the end of that particular bar = 0.8 (0.5 - 0.4) = 0.08 J
The next bar has force = 0.4 N and the width of the bar = (0.75 - 0.50) = 0.25 m
Work done under this bar = 0.4 × 0.25 = 0.1 J
Total work done from the starting position up to this point now = 0.08 + 0.1 = 0.18 J, still less than 0.21 J
So, the final position has to be on the last bar. Let the position be x. The force on the last bar = 0.2 N
0.21 = 0.18 + 0.2 (x - 0.75)
0.03 = 0.2x - 0.15
0.2x = 0.18
x = 0.9 m
Therefore, the final position of the object, to do 0.21 J worth of work, starting from x = 0.4 m is 0.90 m.
b) For this part, negative work is done, this means, we will move in the negative direction to try and trace this total work done.
From the starting point where the initial position is 0.40 m, the force here is 0.80 N
The workdone under this bar to the left is
The workdone = 0.8 (0.25 - 0.4) = - 0.12 J
Since we're tracing -0.19 J, the final position has to be on the last bar (on the left), Let the position be x. The force on the last bar on the left (could also be referred to as the first bar) = 0.60 N
- 0.19 = -0.12 + 0.6 (x - 0.25)
-0.07 = 0.6x - 0.15
0.6x = 0.08
x = (0.08/0.6) = 0.133 m
Therefore, the final position of the object, after doing -0.19 J worth of work, starting from x = 0.4 m is 0.133 m.
Hope this Helps!!!
