The net charge on an atom is equal to the overall difference between the number of protons in the nucleus versus the number of electrons around the nucleus, where a negative sign represents less protons and a positive sign represents more protons (than electrons).
Answer:
The correct answer is "6666.67 N".
Explanation:
The given values are:
Mass,
m = 0.100
Relative speed,
v = 4.00 x 10³
time,
t = 6.00 x 10⁻⁸
As we know,
⇒ 
On substituting the given values, we get
⇒ 
⇒ 
Answer:
Super-critical mass
Explanation:
This term refers to the mass, in which the amount of fission processes per unit of time increases to the point, where some intrinsic feedback mechanism causes the reactor to reach an equilibrium point at a high temperature or power, that is, It becomes critical again, or it is destroyed due to the amount of processes.
Answer:
0.8712 m/s²
Explanation:
We are given;
Velocity of first car; v1 = 33 m/s
Distance; d = 2.5 km = 2500 m
Acceleration of first car; a1 = 0 m/s² (constant acceleration)
Velocity of second car; v2 = 0 m/s (since the second car starts from rest)
From Newton's equation of motion, we know that;
d = ut + ½at²
Thus,for first car, we have;
d = v1•t + ½(a1)t²
Plugging in the relevant values, we have;
d = 33t + 0
d = 33t
For second car, we have;
d = v2•t + ½(a2)•t²
Plugging in the relevant values, we have;
d = 0 + ½(a2)t²
d = ½(a2)t²
Since they meet at the next exit, then;
33t = ½(a2)t²
simplifying to get;
33 = ½(a2)t
Now, we also know that;
t = distance/speed = d/v1 = 2500/33
Thus;
33 = ½ × (a2) × (2500/33)
Rearranging, we have;
a2 = (33 × 33 × 2)/2500
a2 = 0.8712 m/s²
Answer:
500km
Explanation:
Given parameters:
Speed = 200km/hr
Time taken = 2.5hrs
Unknown:
Distance = ?
Solution:
To solve this problem, we use the speed, time and distance equation.
Therefore;
Distance = Speed x time
So;
Distance = 200 x 2.5 = 500km
