Answer:
0.114m
Explanation:
From the general expression for the radius of the proton's resulting orbit, we have
where q is is the charge of the proton 
m is the mass of the proton 
B is the magnetic field 
and v i the speed.
to determine the speed, we use the expression
Kinetic Energy=
where <em>V </em>is the voltage value i.e 1.0kv
and v is the speed
Hence, from simple rearrangement we have the speed v to be
if we substitute value, we have
carrying out careful arithmetic we arrive at
.
using the value for the speed in the expression for the radius of the orbit as stated earlier, we have
Answer:
Explanation:
a ) It is given that bomb was at rest initially , so , its momentum before the explosion was zero.
b ) We shall apply law of conservation of momentum along x and y direction separately because no external force acts on the bomb.
If v be the velocity of the third part along a direction making angle θ
with x axis ,
x component of v = vcosθ
So momentum along x axis after explosion of third part = mv cosθ
= 10 v cosθ
Momentum along x of first part = - 5 x 42 m/s
momentum of second part along x direction =0
total momentum along x direction before explosion = total momentum along x direction after explosion
0 = - 5 x 42 + 10 v cosθ
v cosθ = 21
Similarly
total momentum along y direction before explosion = total momentum along y direction after explosion
0 = - 5 x 38 + 10 v sinθ
v sinθ= 21
squaring and and then adding the above equation
v² cos²θ +v² sin²θ = 21² +19²
v² = 441 + 361
v = 28.31 m/s
Tanθ = 21 / 19
θ = 48°
11 is the answer for this question
Answer:
a) 86 atm
b) 86 atm
c) 645 m/s
Explanation:
See attachment for calculations on how i arrived at the answer
