It's called covelant bonding
20.o hope this helps if not I am sorry
Answer:
Explanation:
When beam is balanced and not rotating with Suki standing on it , let reaction force on the supports be R₁ and R₂. Then
R₁ +R₂ = 336 + 590
= 929
Now the moment beam begins to tip , reaction on distant support R₁ = 0
only R₂ will exists on the support near to Suki.
Taking torque about this support of weight of beam acting from the middle point and weight of suki of 590N ,who is x distance from the support towards the other end.
336 x 1.5 = 590 x
x = .85 m
ie , from second support , Suki can not go beyond a distance of .85 m towards the second end.
Answer:
a) 24
b) 3.3 sec
c) 29.8 m/s
d) 48.85 m
Explanation:
a)
α = angular acceleration = - 28.4 rad/s²
r = radius of the tire = 0.32 m
w₀ = initial angular velocity = 93 rad/s
w = final angular velocity = 0 rad/s
θ = angular displacement
Using the equation
w² = w₀² + 2αθ
0² = 93² + 2 (- 28.4) θ
θ = 152.3 rad
n = number of revolutions
Number of revolutions are given as
b)
t = time taken to stop
using the equation
w = w₀ + αt
0 = 93 + (- 28.4) t
t = 3.3 sec
c)
v₀ = initial velocity of the car
initial velocity of the car is given as
v₀ = r w₀ = (0.32) (93) = 29.8 m/s
d)
v = final velocity = 0 m/s
a = linear acceleration = rα = (0.32) (- 28.4) = - 9.09 m/s²
d = distance traveled by car before stopping
Using the equation
v² = v₀² + 2 a d
0² = 29.8² + 2 (- 9.09) d
d = 48.85 m
Answer:
Vf = 41.6 [m/s]
Explanation:
To solve this problem we must use the equations of kinematics.
Vf² = Vo² + (2*g*y)
where:
Vf = final velocity [m/s]
Vo = initial velocity = 0
g = gravity acceleration = 9.81 [m/s²]
y = height = 88.2 [m]
Note: The positive sign of the equation tells us that the acceleration of gravity goes in the direction of motion.
Vf² = Vo² + (2*g*y)
Vf² = 0 + (2*9.81*88.2)
Vf = (1730.48)^0.5
Vf = 41.6 [m/s]
