On average, how many stars would we have to search before we would expect to hear a signal? assume there are 500 billion stars i
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Refer to the figure shown below.
Let m₁ and m₂ e the two masses.
Let a = the acceleration.
Let T = tension over the frictionless pulley.
Write the equations of motion.
m₂g - T = m₂a (1)
T - m₁g = m₁a (2)
Add equations (1) and (2).
m₂g - T + T - m₁g = (m₁ + m₂)a
(m₂ - m₁)g = (m₁ + m₂)a
Divide through by m₁.
(m₂/m₁ - 1)g = (1 + m₂/m₁)a
Define r = m₂/m₁ as the ratio of the two masses. Then
(r - 1)g = (1 +r)a
r(g-a) = a + g
r = (g - a)/(g + a)
With = 2 ft/s from rest, the acceleration is
a = 2/32.2 = 0.062 ft/s²
Therefore
r = (32.2 - 0.062)/(32.2 + 0.062) = 0.9962
Answer:
The ratio of masses is 0.9962 (heavier mass divided by the lighter mass).
Let m₁ and m₂ e the two masses.
Let a = the acceleration.
Let T = tension over the frictionless pulley.
Write the equations of motion.
m₂g - T = m₂a (1)
T - m₁g = m₁a (2)
Add equations (1) and (2).
m₂g - T + T - m₁g = (m₁ + m₂)a
(m₂ - m₁)g = (m₁ + m₂)a
Divide through by m₁.
(m₂/m₁ - 1)g = (1 + m₂/m₁)a
Define r = m₂/m₁ as the ratio of the two masses. Then
(r - 1)g = (1 +r)a
r(g-a) = a + g
r = (g - a)/(g + a)
With = 2 ft/s from rest, the acceleration is
a = 2/32.2 = 0.062 ft/s²
Therefore
r = (32.2 - 0.062)/(32.2 + 0.062) = 0.9962
Answer:
The ratio of masses is 0.9962 (heavier mass divided by the lighter mass).
Answer:
a) F = 2.66 10⁴ N, b) h = 1.55 m
Explanation:
For this fluid exercise we use that the pressure at the tap point is
Exterior
P₂ = P₀ = 1.01 105 Pa
inside
P₁ = P₀ + ρ g h
the liquid is water with a density of ρ=1000 km / m³
P₁ = 0.85 1.01 10⁵ + 1000 9.8 5
P₁ = 85850 + 49000
P₁ = 1.3485 10⁵ Pa
the net force is
ΔP = P₁- P₂
Δp = 1.3485 10⁵ - 1.01 10⁵
ΔP = 3.385 10⁴ Pa
Let's use the definition of pressure
P = Fe / A
F = P A
the area of a circle is
A = pi r² = [i d ^ 2/4
let's reduce the units to the SI system
d = 100 cm (1 m / 100 cm) = 1 m
F = 3.385 104 pi / 4 (1) ²
F = 2.66 10⁴ N
b) the height for which the pressures are in equilibrium is
P₁ = P₂
0.85 P₀ + ρ g h = P₀
h =
h =
h = 1.55 m