We introduce an electron in a homogeneous electric field. Can we have a net torque on the electron? Please explain.
1 answer:
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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 )
Answer:
Force, F = 77 N
Explanation:
A child in a wagon seem to fall backward when you give the wagon a sharp pull forward. It is due to Newton's third law of motion. The forward pull on wagon is called action force and the backward force is called reaction force. These two forces are equal in magnitude but they acts in opposite direction.
We need to calculate the force is needed to accelerate a sled. It can be calculated using the formula as :
F = m × a
Where
m = mass = 55 kg
a = acceleration = 1.4 m/s²
F = 77 N
So, the force needed to accelerate a sled is 77 N. Hence, this is the required solution.