The answer is in the picture.
I don't think that 4m has anything to do with the problem.
anyway. here.
A___________________B_______C
where A is the point that the train was released.
B is where the wheel started to stick
C is where it stopped
From A to B, v=2.5m/s, it takes 2s to go A to B so t=2
AB= v*t = 2.5 * 2 = 5m
The train comes to a stop 7.7 m from the point at which it was released so AC=7.7m
then BC= AC-AB = 7.7-5 = 2.7m
now consider BC
v^2=u^2+2as
where u is initial speed, in this case is 2.5m/s
v is final speed, train stop at C so final speed=0, so v=0
a is acceleration
s is displacement, which is BC=2.7m
substitute all the number into equation, we have
0^2 = 2.5^2 + 2*a*2.7
0 = 6.25 + 5.4a
a = -6.25/5.4 = -1.157
so acceleration is -1.157m/(s^2)
He is discussing Bargaining Power of Buyers competitive force.
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Answer:
F = 7.68 10¹¹ N, θ = 45º
Explanation:
In this exercise we ask for the net electric force. Let's start by writing the configuration of the charges, the charges of the same sign must be on the diagonal of the cube so that the net force is directed towards the interior of the cube, see in the attached numbering and sign of the charges
The net force is
F_ {net} = F₂₁ + F₂₃ + F₂₄
bold letters indicate vectors. The easiest method to solve this exercise is by using the components of each force.
let's use trigonometry
cos 45 = F₂₄ₓ / F₂₄
sin 45 = F_{24y) / F₂₄
F₂₄ₓ = F₂₄ cos 45
F_{24y} = F₂₄ sin 45
let's do the sum on each axis
X axis
Fₓ = -F₂₁ + F₂₄ₓ
Fₓ = -F₂₁₁ + F₂₄ cos 45
Y axis
F_y = - F₂₃ + F_{24y}
F_y = -F₂₃ + F₂₄ sin 45
They indicate that the magnitude of all charges is the same, therefore
F₂₁ = F₂₃
Let's use Coulomb's law
F₂₁ = k q₁ q₂ / r₁₂²
the distance between the two charges is
r = a
F₂₁ = k q² / a²
we calculate F₂₄
F₂₄ = k q₂ q₄ / r₂₄²
the distance is
r² = a² + a²
r² = 2 a²
we substitute
F₂₄ = k q² / 2 a²
we substitute in the components of the forces
Fx =
Fx =
( -1 + ½ cos 45)
F_y = k \frac{q^2}{a^2} ( -1 + ½ sin 45)
We calculate
F₀ = 9 10⁹ 4.25² / 0.440²
F₀ = 8.40 10¹¹ N
Fₓ = 8.40 10¹¹ (½ 0.707 - 1)
Fₓ = -5.43 10¹¹ N
remember cos 45 = sin 45
F_y = - 5.43 10¹¹ N
We can give the resultant force in two ways
a) F = Fₓ î + F_y ^j
F = -5.43 10¹¹ (i + j) N
b) In the form of module and angle.
For the module we use the Pythagorean theorem
F =
F = 5.43 10¹¹ √2
F = 7.68 10¹¹ N
in angle is
θ = 45º
Answer:
I gotchu fam. 0.0435 kg
Explanation:
m1v1+m2v2=m1v1f+m2v2f
(0.5)(23)+(m2)(0)=(0.5)(17)+(m2)(69)
11.5=8.5+69m2
3=69m2
0.0435 kg