Answer:
Mass is the correct answer.
Explanation:
Many drivers report a more positive handling response and a definite improvement when reducing unsprung mass. You want to keep unsprung weight to as little as possible. This minimizes the momentum and energies that your suspension has to counter. In effect, it can make your shocks more sensitive.
Answer:
C_{y} = 4.96 and θ' = 104,5º
Explanation:
To add several vectors we can decompose each one of them, perform the sum on each axis, to find the components of the resultant and then find the module and direction.
Let's start by decomposing the two vectors.
Vector A
sin θ =
/ A
cos θ = Aₓ / A
A_{y} = A sin θ
Ax = A cos θ
A_{y} = 4.9 sin 31 = 2.52
Ax = 4.9 cos 31 = 4.20
Vector B
B_{y} = B sin θ
Bx = B cos θ
B_{y} = 6 sin 156 = 2.44
Bx = 6 cos 156 = -5.48
The components of the resulting vector are
X axis
Cx = Ax + B x
Cx = 4.20 -5.48
Cx = -1.28
Axis y
C_{y} = Ay + By
C_{y} = 2.52 + 2.44
C_{y} = 4.96
Let's use the Pythagorean theorem to find modulo
C = √ (Cₙ²x2 + Cy2)
C = Ra (1.28 2 + 4.96 2)
C = 5.12
We use trigonemetry to find the angle
tan θ = C_{y} / Cₓ
θ’ = tan⁻¹ (4.96 / (1.28))
θ’ = 75.5
como el valor de Cy es positivo y Cx es negativo el angulo este en el segundo cuadrante, por lo cual el angulo medido respecto de eje x positivo es
θ’ = 180 – tes
θ‘= 180 – 75,5
θ' = 104,5º
Many ways for example they look up related things and study them or they can test it and see what happens
Answer:
The self-induced emf in this inductor is 4.68 mV.
Explanation:
The emf in the inductor is given by:

Where:
dI/dt: is the decreasing current's rate change = -18.0 mA/s (the minus sign is because the current is decreasing)
L: is the inductance = 0.260 H
So, the emf is:

Therefore, the self-induced emf in this inductor is 4.68 mV.
I hope it helps you!
Answer:
laser pointer does not make use of total internal reflection