A proton is moving in a circular orbit of radius 12 cm in a uniform 0.31-T magnetic field perpendicular to the velocity of the p
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Answer:
Force=7.71KN in the opposite direction
Explanation:
Given the mass of the car(M)=1000Kg
The initial speed of the car(u)=100km/hr
we know that 1km/hr=5/18 m/sec
100km/hr=100*5/18 m/sec=27.77m/sec
the distance travelled before it stops (s)=50m
let the acceleration of the car be a
The final velocity of the car is 0.
we know that
a=7.712m/
We know that F=Ma
F=1000* 7.712=7712N=7.71KN
The velocity when function p(t)=11 is 8 .
According to the question
The position of a car at time t represented by function :
Now,
When function p(t) = 11 , t will be
11 = t²+2t-4
0 = t² + 2t - 15
or
t² +2t-15 = 0
t² +(5-3)t-15 = 0
t² +5t-3t-15 = 0
t(t+5)-3(t+5) = 0
(t-3)(t+5) = 0
t = 3 , -5
as t cannot be -ve as given ( t≥0)
so,
t = 3
Now,
the velocity when p(t)=11
As we know velocity =
therefore to get the value of velocity from function p(t)
we have to differentiate the function with respect to time
v(t) = 2t + 2
where v(t) = velocity at that time
as t = 3 for p(t)=11
so ,
v(t) = 2t + 2
v(t) = 2*3 + 2
v(t) = 8
Hence, the velocity when function p(t)=11 is 8 .
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