Displacement stabilizes over time. It is known that exponentials raised to infinity approach zero, hence the system model will yield as time approaches infinity, resulting in 4x'' + e−0.1tx = 0. As time approaches infinity, we deduce that 4x'' equals zero. Consequently, upon integrating, we derive 4x' = c, and further integration leads to the conclusion 4x = cx + d.
This can be determined using the principle of energy conservation. The ski lift begins with a velocity of v= 15.5 m/s, and all of its kinetic energy Ek converts into potential energy Ep, thus we set Ep equal to Ek.
Because Ek is given by (1/2)*m*v², where m denotes mass and v represents speed, while Ep equals m*g*h, where m is mass, g is 9.81 m/s², and h is height. Now:
Ek=Ep
(1/2)*m*v²=m*g*h, canceling out the mass,
(1/2)*v²=g*h, rearranging for height by dividing by g,
(1/2*g)*v²=h and substituting the values:
h=12.245 m. The hill's height rounded to the nearest tenth is h=12.25 m.
Provided Information:
Length of inclined plane = 8 m
Height of inclined plane = 2 m
Weight of the ice block = 300 N
Required Information:
Force needed to push ice block = F =?
Answer:
Force needed to push the ice block = 75 N
Explanation:
The required force to push this ice block up an inclined plane is given by
F = Wsinθ
where W is the weight of the ice block and θ is the angle indicated in the attached image.
Using trigonometric ratios,
sinθ = opposite/hypotenuse
where the opposite side is the height of the inclined plane and the hypotenuse is the length of the inclined plane.
Thus, sinθ = 2/8
θ = sin⁻¹(2/8)
which leads to θ = 14.48°
Therefore, F = 300*sin(14.48)
results in F = 75 N
This indicates that a force of 75 N is necessary to push the ice block on the specified inclined plane.
Answer:
Every option provided is accurate
Explanation:
The electrical power dissipated by a single resistor linked to a battery can be expressed as:
where
V signifies the voltage
I denotes the current
R represents the resistance
Now, let's evaluate each scenario:
A) When the voltage is doubled (V'=2V) while the current is halved (I'=I/2), the resulting power dissipation turns out to be:
--> the power remains the same
B) When the voltage is increased to double (V'=2V) and the resistance quadruples (R'=4R), the new power dissipation becomes:
--> the power is unchanged
C) If the current is doubled (I'=2I) while the resistance diminishes to one-fourth (R'=R/4), the new power dissipation is:
--> the power is unchanged
