Pressure is the force applied perpendicular to a surface divided by the area of the surface. Therefore, from the given values, we can easily calculate the area by dividing the force to the pressure. We do as follows:
Area = F / P
Area = 15000 / 190000
Area = 0.0789 m^2
Hope this answers the question. Have a nice day.
Answer:
The direction of the magnetic force on a moving charge is perpendicular to the plane formed by v and B and follows right hand rule–1 (RHR-1)
Explanation:
hope this helps
In rivers that are close to sea
Answer:
a) variation of the energy is equal to the work of the friction force
b) W = Em_{f} -Em₀
, c) he conservation of mechanical energy
Explanation:
a) In an analysis of this problem we can use the energy law, where at the moment the mechanical energy is started it is totally potential, and at the lowest point it is totally kinetic, we can suppose two possibilities, that the friction is zero and therefore by equalizing the energy we set the velocity at the lowest point.
Another case is if the friction is different from zero and in this case the variation of the energy is equal to the work of the friction force, in value it will be lower than in the calculations.
b) the calluses that he would use are to hinder the worker's friction force and energy
W = Em_{f} -Em₀
N d = ½ m v² - m g (y₂-y₁)
y₂-y₁ = 35 -10 = 25m
c) if there is no friction, the physical principle is the conservation of mechanical energy
If there is friction, the principle is that the non-conservative work is equal to the variation of the energy
Answer:
a) α = 0.338 rad / s² b) θ = 21.9 rev
Explanation:
a) To solve this exercise we will use Newton's second law for rotational movement, that is, torque
τ = I α
fr r = I α
Now we write the translational Newton equation in the radial direction
N- F = 0
N = F
The friction force equation is
fr = μ N
fr = μ F
The moment of inertia of a saying is
I = ½ m r²
Let's replace in the torque equation
(μ F) r = (½ m r²) α
α = 2 μ F / (m r)
α = 2 0.2 24 / (86 0.33)
α = 0.338 rad / s²
b) let's use the relationship of rotational kinematics
w² = w₀² - 2 α θ
0 = w₀² - 2 α θ
θ = w₀² / 2 α
Let's reduce the angular velocity
w₀ = 92 rpm (2π rad / 1 rev) (1 min / 60s) = 9.634 rad / s
θ = 9.634 2 / (2 0.338)
θ = 137.3 rad
Let's reduce radians to revolutions
θ = 137.3 rad (1 rev / 2π rad)
θ = 21.9 rev
