The proton would be 2 and d a part of 1 then calculate that hope this helped
All electromagnetic waves travel at the same speed in a vacuum: 3.0 x 10^5 (300,000) kilometres per second. some electromagnetic waves are part of the visible light spectrum and some do emit harmful radiation, but certainly not all. they travel fine on earth without the vacuum of space too.
The Electric field is zero at a distance 2.492 cm from the origin.
Let A be point where the charge 
C is placed which is the origin.
Let B be the point where the charge 
C is placed. Given that B is at a distance 1 cm from the origin.
Both the charges are positive. But charge at origin is greater than that of B. So we can conclude that the point on the x-axis where the electric field = 0 is after B on x - axis.
i.e., at distance 'x' from B.
Using Coulomb's law, 
,
= 
k is the Coulomb's law constant.
On substituting the values into the above equation, we get,
Taking square roots on both sides and simplifying and solving for x, we get,
1.67x = 1+x
Therefore, x = 1.492 cm
Hence the electric field is zero at a distance 1+1.492 = 2.492 cm from the origin.
Learn more about Electric fields and Coulomb's Law at brainly.com/question/506926
#SPJ4
<span>velocity is defined as the rate of change of displacement irrespective of the length of the path travelled while speed is the average rate of covering distance. but in the liming case where the instantaneous velocity is given as v=dx/dt where dx is the small displacement in a small interval dt, both the speed and velocity have the same magnitude and the direction of velocity is the direction of the tangent to the corresponding displacement-time curve.</span>
Given:-
- Time taken by the particle (t) = 6 s
- Average speed (v) = 40 m/s
To Find: Distance (s) travelled by the particle.
We know,
s = vt
where,
- s = Distance travelled,
- v = Speed &
- t = Time taken.
Putting the values,
s = (40 m/s)(6 s)
→ s = 240 m ...(Ans.)
