Answer:
P = 251, 3 W
Explanation:
The intensity is defined as the power emitted per unit area
I = P / A
Since sound is distributed in all directions spherical shape, the area of a sphere is
A = 4π r²
let's clear the power and replace
P = I A
P = I (4π r²)
let's calculate
P = 5.00 (4π 2²)
P = 251, 3 W
I think is a high-pressure system because it is only in one particular area.
Answer:
D) True. This is what creates the body weight
Explanation:
Let's write Newton's second law for this case. For inclined planes the reference system takes one axis parallel to the plane (x axis) and the other perpendicular to the plane (y axis)
X axis
Wx -fr = ma
Y Axis
N - Wy = 0
With trigonometry we can find the components of weight
sin θ = Wₓ / W
cos θ = 
/ W
Wₓ = W sin θ
= W cos θ
W sin θ - fr = ma
From this expression as it indicates that the body is descending the force greater is the gravity that create the weight of the body
Let's examine the answers
A False This force does not apply because it is not a spring
B) False. It is balanced at all times with the component (Wy) of the weight
C) False. For there to be a rope, if it exists you should be less than the weight component for the block to lower
D) True. This is what creates the body weight
E) False. The cutting force occurs for force applied at a single point and gravity is applied at all points
Greenhouse gases
What is Global Warming? Global warming is the unusually rapid increase in Earth's average surface temperature over the past century primarily due to the greenhouse gases released by people burning fossil fuels.Jun 3, 2010
Answer:
1.19 m/s²
Explanation:
The frequency of the wave generated in the string in the first experiment is f = n/2l√T/μ were T = tension in string = mg were m = 1.30 kg weight = 1300 g , μ = mass per unit length of string = 1.01 g/m. l = length of string to pulley = l₀/2 were l₀ = lent of string. Since f is the second harmonic, n = 2, so
f = 2/2(l₀/2)√mg/μ = 2(√mg/μ)/l₀ (1)
Also, for the second experiment, the period of the wave in the string is T = 2π√l₀/g. From (1) l₀ = 2(√mg/μ)/f and from (2) l₀ = T²g/4π²
Equating (1) and (2) we ave
2(√mg/μ)/f = T²g/4π²
Making g subject of the formula
g = 2π√(2√(m/μ)/f)/T
The period T = 316 s/100 = 3.16 s
Substituting the other values into , we have
g = 2π√(2√(1300 g/1.01 g/m)/200 Hz)/3.16
g = 2π√(2 × 35.877/200 Hz)/3.16
g = 2π√(71.753/200 Hz)/3.16
g = 2π√(0.358)/3.16
g = 2π × 0.599/3.16
g = 1.19 m/s²
