Answer:
113.7
Explanation:
maximum distance (s) = 8.9 km
reference intensity (I0) = 1 x 10^{-12} W/m^{2}
power of a juvenile howler monkey (p) = 63 x 10^{-6} W
distance (r) = 210 m
intensity (I) = power/area
where we assume the area of a sphere due to the uniformity of the output in all directions
area = 4π = 4π x
= 554,176.9 m^{2}
intensity (I) =
therefore the desired ratio I/I0 = = 113.7
Answer:
e. Not enough information to determine
Explanation:
This question is incomplete. Here is the complete question with my solution afterwards;
A ladybug sits at the outer edge of a turntable, and a gentleman bug sits halfway between her and the axis of rotation. The turntable (initially at rest) begins to rotate with its rate of rotation constantly increasing.
What is the first event that will occur?(Assume non-zero frictional force and the same coefficients of friction for both bugs.)
a. The ladybug begins to slide
b. The gentleman bug begins to slide
c. Both bugs begin to slide at the same time
d. Nothing ever happens
e. Not enough information to determine
The centripetal force acting on a rotating body or bugs can be written as,
mass of the corresponding bugs
corresponding radial distance of each bug
angular speed of the turntable
The centripetal force tries to slide the bugs in an outward direction and it is directly proportional to the products of its mass and radial distance from the axis of rotation of the turntable
∝
Since the radial distance from the axis of rotation of the turntable for each bug is given, but the mass is not given, the given information is therefore not enough to determine which bugs will slide first.
Option "e" is correct.
Answer:
x = 0.0685 m
Explanation:
In this exercise we can use the relationship between work and energy conservation
W = ΔEm
Where the work is
W = F x
The energy can be found in two points
Initial. Just when the block with its spring spring touches the other spring
Em₀ = K = ½ m v²
Final. When the system is at rest
=
= ½ k₁ x² + ½ k₂ x²
We can find strength with Newton's second law
∑ F = F - fr
Axis y
N- W = 0
N = W
The friction force has the equation
fr = μ N
fr = μ W
The job
W = (F – μ W) x
We substitute in the equation
(F - μ W) x = ½ m v² - (½ k₁ x² + ½ k₂ x²)
½ x² (k₁ + k₂) + (F - μ W) x - ½ m v² = 0
We substitute values and solve
½ x² (20 + 40) + (15 -0.2 2) x - ½ (2/32) 6² = 0
x² 30 + 14.4 x - 1,125 = 0
x² + 0.48 x - 0.0375 = 0
We solve the second degree equation
x = [-0.48 ±√(0.48 2 + 4 0.0375)] / 2
x = [-0.48 ± 0.617] / 2
x₁ = 0.0685 m
x₂ = -0.549 m
The first result results from compression of the spring and the second torque elongation.
The result of the problem is x = 0.0685 m