Answer: D
Explanation: :) Just took the quizz
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
Answer:
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Explanation:
For this problem let's use Newton's second law applied to each body
Body A
X axis
T = m_A a
Axis y
N- W_A = 0
Body B
Vertical axis
W_B - T = m_B a
In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension
We write the equations
T = m_A a
W_B –T = M_B a
We solve this system of equations
m_B g = (m_A + m_B) a
a = m_B / (m_A + m_B) g
In this initial case
m_A = M
m_B = M
a = M / (1 + 1) M g
a = ½ g
Let's find the tension
T = m_A a
T = M ½ g
T = ½ M g
Now we change the mass of the second block
m_B = 2M
a = 2M / (1 + 2) M g
a = 2/3 g
We seek tension for this case
T’= m_A a
T’= M 2/3 g
Let's look for the relationship between the tensions of the two cases
T’/ T = 2/3 M g / (½ M g)
T’/ T = 4/3
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
The ratio of the distance moved by the point at which the effort is applied in a simple machine to the distance moved by the point at which the load is applied, in the same time. In the case of an ideal (frictionless and weightless) machine, velocity ratio = mechanical advantage. Velocity ratio is sometimes called distance ratio.
Answer:
Explanation:
When the number of slits increases, the intensity of fringes increases.
So, the fringes appear to be more bright.
As we know that the fringe width is inversely proportional to the number of slits, so as the number of slits increases, the fringe width decreases, hence the fringes are narrower, bright and close together.
