Answer: a) 7.1 * 10^3 N; b) -880 N directed out of the curve.
Explanation: In order to solve this problem we have to use the Newton laws, then we have the following:
Pcos 15°-N=0
Psin15°-f= m*ac
from the first we obtain N, the normal force
N=750Kg*9.8* cos (15°)= 7.1 *10^3 N
Then to calculate the frictional force (f) we can use the second equation
f=P sin (15°) -m*ac where ac is the centripetal acceletarion which is equal to v^2/r
f= 750 *9.8 sin(15°)-750*(85*1000/3600)^2/150= -880 N
The bouncing person because the bounce helped him survive
Answer:
38 cm from q1(right)
Explanation:
Given, q1 = 3q2 , r = 60cm = 0.6 m
Let that point be situated at a distance of 'x' m from q1.
Electric field must be same from both sides to be in equilibrium(where EF is 0).
=> k q1/x² = k q2/(0.6 - x)²
=> q1(0.6 - x)² = q2(x)²
=> 3q2(0.6 - x)² = q2(x)²
=> 3(0.6 - x)² = x²
=> √3(0.6 - x) = ± x
=> 0.6√3 = x(1 + √3)
=> 1.03/2.73 = x
≈ 0.38 m = 38 cm = x
I hope my answer helped u under stand better
