Answer:
Electric flux,
Explanation:
It is given that,
Electric field,
We need to find the electric flux through a circular area of radius 2.66 m that lies in the xy-plane.
The electric flux is given by :
Since, k.k=i.i=j.j = 1
So,
So, the electric flux through a circular area is .Hence, this is the required solution.
Answer:
a. The component of the net force which make up the apparent weight are added to each other at the bottom and subtracted (the centripetal force from the weight) at the top)
b. Apparent weight at the top is approximately 519.06 N
Apparent weight at the bottom is approximately 558.94 N
Explanation:
a. The apparent weight at the top is different from the apparent weight at the bottom of a moving Ferris wheel because of the opposite direction in which the centripetal force acts at the top and the bottom, which are upwards and downwards respectively, while the weight acts downwards constantly
b. The given parameters are
The radius of the Ferris wheel, r = 7.2 m
The period for one complete revolution, t = 28 seconds
The angle covered in one revolution, θ = 2·π radian
The mass of the person riding on the Ferris wheel, the passenger = 55 kg
Therefore, we have;
The angular speed, ω = Δθ/Δt = 2·π/(28)
From which we have;
Centripetal force, = m × ω² × r
Substituting the known values, we have = 55 kg × (2·π/(28 s))² × 7.2 m ≈ 19.94 N
The centripetal force, = 19.94 N always acting outward from the center
Weight = Mass × Acceleration due to gravity
The weight of the passenger = 55 kg × 9.8 m/s² = 539 N
The weight of the passenger = 539 N always acting downwards
At the top of the Ferris wheel the the centripetal force is acting upwards and the weight is acting downwards
Therefore;
The net force, which is the apparent weight of the passenger at the top = 539 N - 19.94 N ≈ 519.06 N
Apparent weight at the top ≈ 519.06 N
At the bottom of the Ferris wheel the weight is acting downwards and the centripetal force is also acting downwards
Therefore;
The net force at the bottom, which is the apparent weight of the passenger at the bottom = 539 N + 19.94 N ≈ 558.94 N
Apparent weight at the bottom ≈ 558.94 N.
Answer:
Explanation:
We apply Newton's second law at the crate :
∑F = m*a (Formula 1)
∑F : algebraic sum of the forces in Newton (N)
m : mass in kilograms (kg)
a : acceleration in meters over second square (m/s²)
Data:
m=90kg : crate mass
F= 282 N
μk =0.351 :coefficient of kinetic friction
g = 9.8 m/s² : acceleration due to gravity
Crate weight (W)
W= m*g
W= 90kg*9.8 m/s²
W= 882 N
Friction force : Ff
Ff= μk*N Formula (2)
μk: coefficient of kinetic friction
N : Normal force (N)
Problem development
We apply the formula (1)
∑Fy = m*ay , ay=0
N-W = 0
N = W
N = 882 N
We replace the data in the formula (2)
Ff= μk*N = 0.351* 882 N
Ff= 309.58 N
We apply the formula (1) in x direction:
∑Fx = m*ax , ax=0
282 N - 309.58 N = 90*a
a= (282 N - 309.58 N ) / (90)
a= - 0.306 m/s²
Kinematics of the crate
Because the crate moves with uniformly accelerated movement we apply the following formula :
vf²=v₀²+2*a*d Formula (3)
Where:
d:displacement in meters (m)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
v₀ = 0.850 m/s
d = 0.75 m
a= - 0.306 m/s²
We replace the data in the formula (3)
vf²=(0.850)²+(2)( - 0.306 )(0.75 )