list 8 countries which have launched satellites into space and also list the name of the satellite which was launched by the cou
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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
Answer:
a) F_net = 30.47 N , θ = 10.6º
b) Fₓ = 29.95 N
Explanation:
For this exercise we use coulomb's law
F₁₂ = k
the direction of the force is on the line between the two charges and the sense is repulsive if the charges are equal and attractive if the charges are different.
As we have several charges, the easiest way to solve the problem is to add the components of the force in each axis, see attached for a diagram of the forces
X axis
Fₓ =
Y axis
Fy =
let's find the magnitude of each force
= 9 10⁹ 2.76 10⁻⁴ 1.02 10⁻⁴ / 3²
F_{ab} = 2.82 10¹ N
F_{ab} = 28.2 N
= 9 10⁹ 6.54 10⁻⁴ 1.02 10⁻⁴ / 4²
F_{bc} = 3.75 10¹ N
F_{bc} = 37.5 N
let's use trigonometry to decompose this force
tan θ = y / x
θ = tan⁻¹ and x
θ= tan⁻¹ ¾
θ = 37º
let's break down the force
sin 37 = F_{bcy} / F_{bc}
F_{bcy} = F_{bc} sin 37
F_{bcy} = 37.5 sin 37
F_{bcy} = 22.57 N
cos 37 = F_{bcx} /F_{bc}
F_{bcx} = F_{bc} cos 37
F_{bcx} = 37.5 cos 37
F_{bcx} = 29.95 N
let's do the sum to find the net force
X axis
Fₓ = 29.95 N
Axis y
Fy = 28.2 -22.57
Fy = 5.63 N
we can give the result in two ways
a) F_net = Fₓ i ^ + j ^
F_net = 29.95 i ^ + 5.63 j ^
b) in the form of module and angle
let's use the Pythagorean theorem
F_net =
F_net = √(29.95² + 5.63²)
F_net = 30.47 N
we use trigonometry for the direction
tan θ=
θ = tan⁻¹ \frac{ F_{y} }{ F_{x} }
θ = tan⁻¹ (5.63 / 29.95)
θ = 10.6º
