I'm pretty sure it's D. negative acceleration
<span>We can draw a free body diagram for the safe to show all the horizontal forces acting on the safe. The three horizontal forces are Clyde's force, Bonnie's force, and friction. Since the safe slides at a constant speed (no acceleration), the net force must be zero.
0 = Clyde's force + Bonnie's force - friction
friction = Clyde's force + Bonnie's force
mg mu = Clyde's force + Bonnie's force
mu = (Clyde's force + Bonnie's force) / (mg)
mu = (445 N + 350 N) / (300 kg x 9.80~m/s^2)
mu = 0.27
The coefficient of friction is 0.27</span>
Answer:
θ = 66º
Explanation:
This exercise of Newton's second law must be solved in part, let's start by finding the slowing down acceleration of the ball
a = v² / r
the radius of the circle is
sin θ = r / L
r = L sin θ
we substitute
a = v² /L sin θ
now let's write Newton's second law
vertical axis
T_y -W = 0
T_y = W
radial axis
Tₓ = m a (1)
let's use trigonometry for the components of the string tension
cos θ = T_y / T
sin θ = Tₓ / T
Tₓ = T sin θ
we substitute in 1
T sin θ =
T L sin² θ = m v²
we write our system of equations
T cos θ = m g
T L sin ² tea = m v²
we divide the two equations
L = v² / g
(1 -cos²)/ cos θ =
1 - cos² θ = cos θ
cos² θ + 0.97044 cos θ -1 = 0
we change variable cos θ = x
x² + 0.97044 x - 1 =0
x=
since the square root is imaginary there is no real solution to the problem, suppose that the radius is 1 m r = 1 m
T sin θ =
T cos θ = m g
resolved
tan θ =
θ = tan⁻¹ ( 4.75²/ 1 9.81)
θ = 66º
Answer:
The exergy is 279.77 kJ/kg
The exergy of the exhaust gases is 143.24 kJ/kg
Explanation:
the step by step is in the image