A charged particle moving in a magnetic field experiences a force equal to:
Thus, the magnitude of the force that the proton experiences is given by:
The magnetic field is perpendicular to the proton's velocity, therefore, we have 
. Replacing the given values, we obtain:
Answer:
Explanation:
Velocity of plane relative to ground V_pg = ?
Given the velocity in vector form ,
velocity of plane relative to air V_pw = 120 cos30 i + 120sin30j
V_wg = 60 i
V_pg = V_pw +V_wg
= 120 cos30 i + 120sin30j + 60i
= 164 i + 60 j
magnitude
=251 km / h
=
Answer:
x component 3.88 y- component 14.488
Explanation:
We have given a vector A which has a magnitude of 15 m/sec which is at 75° counter-clock wise ( anti-clock wise) from x -axis which is clearly shown in bellow figure
Now x-component will be 15 cos75°=3.8822 ( as it makes an angle of 75° with x-axis )
y- component will be 15 sin 75°=14.488
For verification the resultant of x and y component should be equal to 15
So 
Answer:
<h2> 4kg</h2>
Explanation:
Step one:
given
length of rod=2m
mass of object 1 m1=1kg
let the unknown mass be x
center of mass<em> c.m</em>= 1.6m
hence 1kg is 1.6m from the <em>c.m</em>
and x is 0.4m from the <em>c.m</em>
Taking moment about the <em>c.m</em>
<em>clockwise moment equals anticlockwise moments</em>
1*1.6=x*0.4
1.6=0.4x
divide both sides by 0.4 we have
x=1.6/0.4
x=4kg
The mass of the other object is 4kg
