Answer:
The coefficient of static friction between the box and floor is, μ = 0.061
Explanation:
Given data,
The mass of the box, m = 50 kg
The force exerted by the person, F = 50 N
The time period of motion, t = 10 s
The frictional force acting on the box, f = 30 N
The normal force on the box, η = mg
= 50 x 9.8
= 490 N
The coefficient of friction,
μ = f/ η
= 30 / 490
= 0.061
Hence, the coefficient of static friction between the box and floor is, μ = 0.061
Answer:
distance changing at rate of 3.94 inches/sec
Explanation:
Given data
wall decreasing at a rate = 9 inches per second
ladder L = 152 inches
distance h = 61 inches
to find out
how fast is the distance changing
solution
we know that
h² + b² = L² ..................1
h² + b² = 152²
Apply here derivative w.r.t. time
2h dh/dt + 2b db/dt = 0
h dh/dt + b db/dt = 0
db/dt = - h/b × dh/dt .............2
and
we know
h = 61
so h² + b² = L²
61² + b² = 152²
b² = 19383
so b = 139.223
and we know dh/dt = -9 inch/sec
so from equation 2
db/dt = -61/139.223 (-9)
so
db/dt = 3.94 inches/sec
distance changing at rate of 3.94 inches/sec
Answer:
A
Algebra For what values of the variables must ABCD be a parallelogram?
B
23. A 2y + 2 B 24. B
(3x + 10)
(8x + 5)º
3x + 6
54°
D
С
D
A
Зу - 9 с
ly+4
