A car of mass 750kg accelerates away from traffic lights. At the end of the first 100m it has reached a speed of 12 m/s. During
2 answers:
You might be interested in
Answer:
v = √[gR (sin θ - μcos θ)]
Explanation:
The free body diagram for the car is presented in the attached image to this answer.
The forces acting on the car include the weight of the car, the normal reaction of the plane on the car, the frictional force on the car and the net force on the car which is the centripetal force on the car keeping it in circular motion without slipping.
Resolving the weight into the axis parallel and perpendicular to the inclined plane,
N = mg cos θ
And the component parallel to the inclined plane that slides the body down the plane at rest = mg sin θ
Frictional force = Fr = μN = μmg cos θ
Centripetal force responsible for keeping the car in circular motion = (mv²/R)
So, a force balance in the plane parallel to the inclined plane shows that
Centripetal force = (mg sin θ - Fr) (since the car slides down the plane at rest, (mg sin θ) is greater than the frictional force)
(mv²/R) = (mg sin θ - μmg cos θ)
v² = R(g sin θ - μg cos θ)
v² = gR (sin θ - μcos θ)
v = √[gR (sin θ - μcos θ)]
Hope this Helps!!!
The net force on q₃ will be 17.51 N. The net force is the algebraic sum of the two forces on the pleading q₃
<h3 /><h3>What is Columb's law?</h3>The force of attraction between two charges, according to Coulomb's law, is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.
The force,by the charge q₁ on the q₃;
The force,by the charge q₂ on the q₃;
The net force is the sum of the two forces;
Hence, the net force on q₃ will be 17.51 N.
To learn more about Columb's law, refer to the link;
#SPJ1