Force applied on the car due to engine is given as
towards right
Also there is a force on the car towards left due to air drag
towards left
now the net force on the car will be given as
now we can say that since the two forces are here opposite in direction so here the vector sum of two forces will be the algebraic difference of the two forces.
So we can say
So here net force on the car will be 150 N towards right and hence it will accelerate due to same force.
s = displacement; u = initial velocity; t = time of motion
Frost will disturb the smooth flow of air over the wing, unpleasantly
distressing its lifting competence. In other words, this spoils the even flow
of air over the wings, by this means decreasing lifting capability. Also, frost
may avoid the airplane from becoming flying at normal departure speed.
Answer:
a) Acceleration is zero
, c) Speed is cero
Explanation:
a) the equation that governs the simple harmonic motion is
x = A cos (wt +φφ)
Where A is the amplitude of the movement, w is the angular velocity and φ the initial phase determined by the initial condition
Body acceleration is
a = d²x / dt²
Let's look for the derivatives
dx / dt = - A w sin (wt + φ)
a = d²x / dt² = - A w² cos (wt + φ)
In the instant when it is not stretched x = 0
As the spring is released at maximum elongation, φ = 0
0 = A cos wt
Cos wt = 0 wt = π / 2
Acceleration is valid for this angle
a = -A w² cos π/2 = 0
Acceleration is zero
b)
c) When the spring is compressed x = A
Speed is
v = dx / dt
v = - A w sin wt
We look for time
A = A cos wt
cos wt = 1 wt = 0, π
For this time the speedy vouchers
v = -A w sin 0 = 0
Speed is cero
Answer:
B = 8.0487mT
Explanation:
To solve the exercise it is necessary to take into account the considerations of the Magnetic Force described by Faraday,
The magnetic force is given by the formula
Where,
B = Magnetic Field
I = Current
L = Length
Angle between the magnetic field and the velocity, for this case are perpendicular, then is 90 degrees
According to our data we have that
I = 16.4A
F = 0.132N/m
As we know our equation must be modificated to Force per length unit, that is
Replacing the values we have that
Solving for B,
