If an object has more protons than electrons, then the net charge on the object is positive. If there are more electrons than protons, then the net charge on the object is negative. If there are equal numbers of protons and electrons, then the object is electrically neutral.
Source: https://www.khanacademy.org/science/ap-physics-1/ap-electric-charge-electric-force-and-voltage/electric-charge-ap/a/electric-charge-ap1
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Answer:
1.6 x 10⁻¹⁹ C
Explanation:
Let us arrange the charges in the ascending order and round them off as follows :-
1.53 x 10⁻¹⁹ C → 1.6x 10⁻¹⁹ C
3.26 x 10⁻¹⁹C → 3.2 x 10⁻¹⁹ C
4.66 x 10⁻¹⁹C → 4.8 x 10⁻¹⁹ C
5.09 x 10⁻¹⁹C → 4.8 x 10⁻¹⁹ C
6.39 x 10⁻¹⁹C → 6.4 x 10⁻¹⁹ C
The rounding off has been made to facilitate easy calculation to come to a conclusion and to accommodate error in measurement.
Here we observe that
2 nd charge is almost twice the first charge
3 rd and 4 th charges are almost 3 times the first charge
5 th charge is almost 4 times the first charge.
This result implies that 2 nd to 5 th charges are made by combination of the first charge ie if we take e as first charge , 2nd to 5 th charges can be written as 2e, 3e ,3e and 4e. Hence e is the minimum charge existing in nature and on electron this minimum charge of 1.6 x 10⁻¹⁹ C exists.
Answer:
The strength of the magnetic field that the line produces is
.
Explanation:
From Biot-Savart law, the equation to determine the strength of the magnetic field for any straight wire can be deduced:
(1)
Where
is the permiability constant, I is the current and r is the distance from the wire.
Notice that it is necessary to express the current, I, from kiloampere to ampere.
⇒ 
Finally, equation 1 can be used:
Hence, the strength of the magnetic field that the line produces is
.

Actually Welcome to the concept of Efficiency.
Here we can see that, the Input work is given as 2.2 x 10^7 J and the efficiency is given as 22%
The efficiency is => 22% => 22/100.
so we get as,
E = W(output) /W(input)
hence, W(output) = E x W(input)
so we get as,
W(output) = (22/100) x 2.2 x 10^7
=> W(output) = 0.22 x 2.2 x 10^7 => 0.484 x 10^7
hence, W(output) = 4.84 x 10^6 J
The useful work done on the mass is 4.84 x 10^6 J
The total work done on the car is 784Joule.
<h3>What's the acceleration of the car?</h3>
- As per Newton's equation of motion, V= U+at
- U= initial velocity= 0 m/s
V= vinal velocity= 20m/s
t= time = 10s
a= acceleration
=> a= 20/10= 2m/s²
<h3>What's the distance covered by the car in 10 seconds?</h3>
- As per Newton's equation of motion,
V²-U² = 2aS
- S= distance covered by the car
- So, 20²-0=2×2×S=4S
=> 400= 4S
=> S= 400/4= 100m
<h3>What's the work done on the car due to frictional force?</h3>
Work done by frictional force= frictional force × distance
= (0.2×4×9.8)×100
= 784Joule
Thus, we can conclude that the work done on the car is 784Joule.
Learn more about the work done here:
brainly.com/question/25573309
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