Answer:
a) a= 8.33 m/s², T = 12.495 N
, b) a = 2.45 m / s²
Explanation:
a) this is an exercise of Newton's second law. As the upper load is secured by a cable, it cannot be moved, so the lower load is determined by the maximum acceleration.
We apply Newton's second law to the lower charge
fr₁ + fr₂ = ma
The equation for the force of friction is
fr = μ N
Y Axis
N - W₁ –W₂ = 0
N = W₁ + W₂
N = (m₁ + m₂) g
Since the beams are the same, it has the same mass
N = 2 m g
We replace
μ₁ 2mg + μ₂ mg = m a
a = (2μ₁ + μ₂) g
a = (2 0.30 + 0.25) 9.8
a= 8.33 m/s²
Let's look for cable tension with beam 2
T = m₂ a
T = 1500 8.33
T = 12.495 N
b) For maximum deceleration the cable loses tension (T = 0 N), so as this beam has less friction is the one that will move first, we are assuming that the rope is horizontal
fr = m₂ a₂
N- w₂ = 0
N = W₂ = mg
μ₂ mg = m a₂
a = μ₂ g
a = 0.25 9.8
a = 2.45 m / s²
Formula:
F = ma
m = F÷A
m: mass F:force A: accelerarion
m = 450N ÷ 15m/s^2
= 30kg
Hope this helps :)
Answer:
Curves around objects
Explanation:
Diffraction is a property of light described by bending of light around an object. This ability of light to bend around edges has facilitated optical effects of light where there is interference of light waves. Other properties of light are: reflection, refraction, polarization, scattering of light, and interference of light.
Answer:
the magnitude of the displacement after 5s is 137.31 m.
Explanation:
Given;
initial velocity of the projectile, u = 60 m/s
angle of projection, θ = 60°
time of motion, t = 5s
the vertical component of the velocity, 
The magnitude of the displacement after 5s is calculated as;
Therefore, the magnitude of the displacement after 5s is 137.31 m.
