Answer:
The question is incomplete; refer to the attachment for the diagram
Explanation:
Provided that,
The mass of the car
M = 1500kg
Entering the curve at Point A with a speed of
Va = 100km/hr = 100× 1000/3600
Va = 27.78m/s
The car slows down at a constant rate until it reaches point C at a speed of
Vc = 50km/hr = 50×1000/3600
Vc = 13.89m/s
Radius of curvature at point A
p = 400m
Radius of curvature at point B
p = 80m
The distance from point A to point B as indicated in the attachment is
S=200m
We need to determine the total horizontal forces at points A, B, and C applied by the road on the tire
The constant tangential acceleration can be computed using the equation of motion
Vc² = Va² + 2as
13.89² = 27.78² + 2 × a × 200
192.9 = 771.6 + 400a
400a = 192.9—771.6
400a = -578.7
a = -578.7 / 400
a = —1.45 m/s²
at = —1.45m/s²
The tangential acceleration is -1.45m/s² and it is negative due to the car decelerating
Since the vehicle is decreasing speed at a uniform rate, the tangential acceleration remains the same at every point
At point A
at = -1.45m/s²
At point B
at = -1.45m/s²
At point C
at = -1.45m/s²
Now,
We can compute the normal component of acceleration (centripetal acceleration) at each point since we know the radius of curvature
The centripetal acceleration can be calculated using
ac = v²/ p
At point A ( p = 400)
an = Va²/p = 27.78² / 400
an = 1.93 m/s²
At point B (p = ∞), since point B is the inflection point
Then,
an = Vb²/p = Vb/∞ = 0
an = 0
At point C ( p = 80m)
an = Vc²/p = 13.89² / 80 = 2.41m/s²
an = 2.41 m/s²
Then,
The tangential force is
Ft = M•at
Ft = 1500 × 1.45
Ft = 2175 N.
Given that the tangential acceleration remains constant, this represents the tangential force at A, B, and C
Next, the normal force
At point A ( an = 1.93m/s²)
Fn = M•an
Fn = 1500 × 1.93
Fn = 2895 N
At point B (an=0)
Fn = M•an
Fn = 0 N
At point C (an= 2.41m/s²)
Fn = M•an
Fn = 1500 × 2.41
Fn = 3615 N.
Thus, the horizontal force acting at each point is
Using the right triangle vector method
F = √(Fn² + Ft²)
At point A
Fa = √(2895² + 2175²)
Fa = √13,111,650
Fa = 3621 N
At point B
Fb = √(0² + 2175²)
Fb = √2175²
Fb = 2175 N
At point C
Fc = √(3615² + 2175²)
Fc = √17,798,850
Fc = 4218.88 N
Fc ≈ 4219N
Answer:
Maximum emf = 5.32 V
Explanation:
Provided data includes:
Number of turns, N = 10
Radius of loop, r = 3 cm = 0.03 m
It made 60 revolutions each second
Magnetic field, B = 0.5 T
We are tasked to determine the maximum emf produced in the loop, which is founded on Faraday's law. The induced emf can be calculated by:
For the maximum emf,
Therefore,
Hence, the maximum emf generated in the loop is 5.32 V.
Response:
The new resistance is half of the original resistance.
Explanation:
Resistance in a wire is represented by:
= resistivity of the material
L and A are the physical dimensions
If a wire is exchanged for one where all linear dimensions are doubled, i.e. l' = 2l and r' = 2r
The updated resistance of the wire can be calculated as follows:
The new resistance equals half of the original resistance. Thus, this provides the solution needed.
The overall force acting on the vehicle is zero
Explanation:
Let's evaluate the situation separately for the vertical direction and the horizontal direction along the slope.
Considering the direction perpendicular to the slope, two forces are in effect:
Equilibrium exists here, indicating the net force in this direction is zero.
Now let’s examine the parallel direction to the slope. We have two forces present:
The car moves at a constant speed in this direction, indicating that its acceleration is zero.
Thus, according to Newton's second law,
implying the net force is zero:
Learn more about slopes and friction: