Answer:
Incomplete question check attachment for diagram
Explanation:
Given that,
Mass of car
M = 1500kg
Enter curve at Point A with speed of
Va = 100km/hr = 100× 1000/3600
Va = 27.78m/s
The car was slow down at a constant rate till it gets to point C at speed of
Vc = 50km/r = 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 given in the attachment is
S=200m
We want to find the total horizontal forces at point A, B and C exerted by the road on the tire
The constant tangential acceleration can be calculated using 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 because the car was decelerating
Since the car is slowing down at a constant rate, the tangential acceleration is equal 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 calculate the normal component of acceleration(centripetal acceleration) at each point since we know the radius of curvature
The centripetal acceleration is 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 point of inflection
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.
Since tangential acceleration is constant, then, this is the tangential force at each point A, B and C
Now, normal force
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.
Then, the horizontal force acting at each point is
Using Vector of right angle triangle
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
<em>Hello, </em>
<h3><u>QUESTION)</u></h3>Assuming that the initial velocity of the jumper is zero, on Earth any freely falling object has an acceleration of 9.8 m/s².
<em>✔ We have : a = v/Δt = ⇔ Δt = v/a </em>
Answer:
a=
Explanation:
The net force, of the box is expressed as a product of acceleration and mass hence
where m is mass and a is acceleration
Making a the subject, a=
From the attached sketch,
∑ where
is frictional force and
is horizontal angle
Substituting ∑ as
in the equation where we made a the subject
a=
Since we’re given the value of F as 240N, as 41.5N,
as
and mass m as 30kg
a=
Answer:
E = 1.04*10⁻¹ N/C
Explanation:
Assuming no other forces acting on the proton than the electric field, as this is uniform, we can calculate the acceleration of the proton, with the following kinematic equation:
As the proton is coming at rest after travelling 0.200 m to the right, vf = 0, and x = 0.200 m.
Replacing this values in the equation above, we can solve for a, as follows:
According to Newton´s 2nd Law, and applying the definition of an electric field, we can say the following:
F = mp*a = q*E
For a proton, we have the following values:
mp = 1.67*10⁻²⁷ kg
q = e = 1.6*10⁻¹⁹ C
So, we can solve for E (in magnitude) , as follows:
⇒ E = 1.04*10⁻¹ N/C