Answer:
a) a = 2.383 m / s², b) T₂ = 120,617 N
, c) T₃ = 72,957 N
Explanation:
This is an exercise of Newton's second law let's fix a horizontal frame of reference
in this case the mass of the sleds is 30, 20 10 kg from the last to the first, in the first the horizontal force is applied.
a) request the acceleration of the system
we can take the sledges together and write Newton's second law
T = (m₁ + m₂ + m₃) a
a = T / (m₁ + m₂ + m₃)
a = 143 / (10 +20 +30)
a = 2.383 m / s²
b) the tension of the cables we think through cable A between the sledges of 1 and 20 kg
on the sled of m₁ = 10 kg
T - T₂ = m₁ a
in this case T₂ is the cable tension
T₂ = T - m₁ a
T₂ = 143 - 10 2,383
T₂ = 120,617 N
c) The cable tension between the masses of 20 and 30 kg
T₂ - T₃ = m₂ a
T₃ = T₂ -m₂ a
T₃ = 120,617 - 20 2,383
T₃ = 72,957 N
Answer:C) Both have rocky composition
The force applied would be 1.05*9.8 = 10.3 N
the pressure is equal to F/a
area will be πr^2 = 0.002826
thus pressure will be = 10.3/0.002826= 3644.72 N/m^2
Answer:
y = 67.6 feet, y = 114.4/ (22 - 3t)
Explanation:
For this exercise let's use that light travels in a straight line and some trigonometric relationships, the symbols are in the attached diagram
Large triangle Projector up to the screen
tan θ = y / L
For the small triangle. Projector up to the person
tan θ = y₀ / (L-d)
The angle is the same, so we equate the two equations
y₀ / (L -d) = y / L
y = y₀ L / (L-d)
The distance from the screen (d), we look for it with kinematics
v = d / t
d = v t
we replace
y = y₀ L / (L - v t)
y = 5.2 22 / (22 - 3 t)
y = 114.4 (22 - 3t)⁻¹
This is the equation of the shadow height change as a function of time
For the suggested distance the shadow has a height of
y = 114.4 / (22-13)
y = 67.6 feet
