Answer:
K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x
Explanation:
To calculate the kinetic energy variation, we can utilize the work-energy theorem.
W = ΔK
∫ F .dx = K - K₀
If the object starts from rest, then K₀ = 0.
So, ∫ F dx cos θ = K.
As the force and displacement directions align, the angle is zero, and hence the cosine is 1.
Now we can substitute and perform integration:
α ∫ x³ dx + β ∫ dx = K.
Thus, α x⁴ / 4 + β x = K.
Next, we evaluate from the limits F = 0 to F:
α (x⁴ / 4 - 0) + β (x - 0) = K.
Consequently, K = αX⁴ / 4 + β x.
This results in K = 1.525 10⁻⁹ x⁴ + 4.1 10⁶ x.
To finalize the computation, we need to ascertain the displacement.
To determine the average net force, we can calculate acceleration using:
x = 0.5*a*t^2
v = a*t
where x=3.6m and v=185 m/s.
Thus,
t=v/a and therefore x = 0.5*a*(v/a)^2 = 0.5 * (v^2)/a
which gives us a= (0.5*v^2)/x
Since we have the known values of v and x, we can compute a by substituting these numbers.
The average net force is then given as:
F = m*a,
with m=7.5kg.
The change in momentum (i.e., impulse) from the car during the collision remains constant regardless of whether an airbag is present. This is because the car's mass is unchanged, and the velocity change remains the same. Therefore, if the force is constant as F and reduced by a factor of 110, it follows that the collision duration must increase by the same factor when the airbag is utilized.
J(r) = Br. We know that the area of a small segment, dA, is represented as 2 π dr. Thus, I = J A and dI = J dA. Plugging in the values gives us dI = B r. 2 π dr which simplifies to dI= 2π Br² dr. Now, integrating the above equation: Given that B= 2.35 x 10⁵ A/m³, with r₁ = 2 mm and r₂ equal to 2 + 0.0115 mm, or 2.0115 mm.
Answer:
a. Angle= 28.82°
b. Approved. Although he might feel cold, he should be able to cross.
Explanation:
Velocity Vector
Velocity is a measure of how quickly something is moving in a specific direction. It is represented as a vector that has both magnitude and direction. If an object can only move in one direction, then speed can serve as the scalar equivalent of that velocity (only focusing on magnitude).
a.
The explorer aims to swim across a river to reach his campsite, as depicted in the image below. The river's velocity is vr and the explorer's swimming speed in still water is ve. If he were to swim straight towards the campsite, he would end up downstream due to the river's current. Therefore, he must swim at an angle that allows him to overcome the current while still moving towards his goal. This angle relative to the shore is what we need to determine. The explorer's speed can be broken down into its horizontal (vx) and vertical (vy) components. In order to counteract the river's flow:

We can calculate the vertical component of the explorer's swimming speed as

Thus

Finding the value of 


Then the angle is given by

b.
The component of the explorer's velocity that goes horizontally is


This represents the actual velocity directed towards the campsite
Considering that

To find t

Calculating the duration for the explorer to cross the river


As this time is under the hypothermia threshold (300 seconds), the conclusion is
Approved. Although he will feel cold, he should manage to cross successfully.