It must exist in p<span>lasma state.</span>
Answer:
The minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
Explanation:
We know by equation of motion that,

Where, v= final velocity m/sec
u=initial velocity m/sec
a=Acceleration m/
s= Distance traveled before stop m
Case 1
u= 13 m/sec, v=0, s= 57.46 m, a=?

a = -1.47 m/
(a is negative since final velocity is less then initial velocity)
Case 2
u=29 m/sec, v=0, s= ?, a=-1.47 m/
(since same friction force is applied)

s = 285.94 m
Hence the minimum stopping distance when the car is moving at
29.0 m/sec = 285.94 m
The orbital with the lowest energy is 3s.
Answer:
Explanation:
kinetic energy required = 1.80 MeV
= 1.8 x 10⁶ x 1.6 x 10⁻¹⁹ J
= 2.88 x 10⁻¹³ J
If v be the velocity of proton
1/2 x mass of proton x v² = 2.88 x 10⁻¹³
= .5 x 1.67 x 10⁻²⁷ x v² = 2.88 x 10⁻¹³
v² = 3.45 x 10¹⁴
v = 1.86 x 10⁷ m /s
If V be the potential difference required
V x e = kinetic energy . where e is charge on proton .
V x 1.6 x 10⁻¹⁹ = 2.88 x 10⁻¹³
V = 1.8 x 10⁶ volt .