Umm I think the ocean floor but I don't think so
Excellent work!
Your calculations are correct, but near the end, you have forgotten to cancel u. If you cancel the u, the linear term becomes a constant, and the resultant equation becomes a simple quadratic which is much easier to solve.
I get u=7.468 m/s using g=9.81 (as you did)
You will rework and should get 7.468 m/s as well.
Congrats for the good work!
Answer:
Answer is option b) 2.97m
Explanation:
With the relationship between the force exerted by the runner and the mass that it has, I can determine the acceleration it will have:
F= m × a ⇒ a= (650 kg ×(m/s^2)) / (70kg)= 9.286 (m/s^2)
With the acceleration that prints the force exerted and the time I can determine the distance traveled in the interval:
Distance= (1/2) × a × t^2 = (1/2) × 9.286 (m/s^2) × ((0.8s)^2)= 2.97m
Answer:
ρ_body = 1000 kg / m³
Explanation:
This is an exercise in fluid mechanics, specifically we must use the Archimedean principle, which states that the thrust is equal to the weight of the dislodged liquid.
In this case let's start by finding the volume of our body
oak block
v = l to h
v = 0.02 0.02 0.05
V = 2 10⁻⁵ m³
cooper block indicate that it has the same dimensions so its volume is the same, the total volume of the body is
V_total = 4 10⁻⁵ m³
as they indicate that the body is fully submerged there is a balance between weight and thrust
B - W = 0
the push is
B = ρ_fluid g V_total
the body weight is
ρ_body = M / V_total
M = ρ_body V_total
W = Mg
W = ρ_body V_total g
we substitute
ρ_fluid g V_total = ρ_body V_total g
ρ_body = ρ_fluid
in this case the body is in equilibrium in the fluid, in case the density of the body is greater than that of the fluid, the body sinks
Therefore the average density is equal to the density of the fluid, since since it is water the density is
ρ_body = 1000 kg / m³
To solve this exercise it is necessary to apply the concepts related to Work and Kinetic Energy. Work from the rotational movement is described as

In the case of rotational kinetic energy we know that

PART A)
is given in revolutions and needs to be in radians therefore


Replacing in the work equation we have to



PART B) From the torque and moment of inertia it is possible to calculate the angular acceleration and the final speed, with which the kinetic energy can be determined.

Rearrange for the angular acceleration,



From the kinematic equations of angular motion we have,




In this way the rotational kinetic energy would be given by


