I think it's mass...
The mass of an object refers to the amount of matter that is contained by the object; the weight of an object is the force of gravity acting upon that object. Mass is related to how much stuff is there and weight is related to the pull of the Earth (or any other planet) upon that stuff.
Answer:
a) x₀ = - 2 m , b) y = 4.47 m
Explanation:
A wave travels in the middle with constant speed, let's use the equation of uniform motion
v = d / t
t = d / v
The distance to the first listeners, see attached
d₁ = x₀-x
t = (x₀ +7) / v
The distance to the second listener
d₂ = x - x₀
t = (+ 3- x₀) / v
As the wave arrives at the same time, we can equal the two equations
(x₀ +7) / v = (3 -x₀) / v
x₀ + 7 = 3 - x₀
2 x₀ = 3 - 7
x₀ = -4/2
x₀ = - 2 m
b) The time it takes for the wave to reach the listeners of the x-axis, where the speed of sound is 340 m / s
t = 5/340
t = 0.0147 s
Let's look for the distance the wave travels for the listener axis and
v = d₃ / t
d₃ = v.t
d₃ = 340 * 0.0147
d₃ = 5 m
For the distance component we use the Pythagorean triangle
d₃² = x₀² + y²
y² = d₃² - x₀²
y = √ (d₃² -4)
y = √ (5² -4)
y = 4.47 m
Answer:
IV because the process of water is equal to 5,8 to 783253.23 to the hendroxagram of 4.
Explanation:
Work = Force times Distance
W = Fd
Given W = 750J, F = 125N;
750 = 125d
Solving for d:
d = 750/125
d = 6
The box moved a distance of 6 meters.
Answer:
F = GMmx/[√(a² + x²)]³
Explanation:
The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is
dF = GmdM/L²
Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.
So, the horizontal components add from two symmetrically opposite mass elements dM,
Thus, the horizontal component of the force is
dF' = dFcosФ where Ф is the angle between L and the x axis
dF' = GmdMcosФ/L²
L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.
L = √(a² + x²)
cosФ = x/L
dF' = GmdMcosФ/L²
dF' = GmdMx/L³
dF' = GmdMx/[√(a² + x²)]³
Integrating both sides we have
∫dF' = ∫GmdMx/[√(a² + x²)]³
∫dF' = Gm∫dMx/[√(a² + x²)]³ ∫dM = M
F = GmMx/[√(a² + x²)]³
F = GMmx/[√(a² + x²)]³
So, the force due to the sphere of mass m is
F = GMmx/[√(a² + x²)]³
