To solve this problem we will start by considering how to calculate the apparent weight. On the sphere this will then be given that the real weight is the sum of the apparent weight and the Buoyant Force. Therefore we will have to
Here
= True Weight
= Apparent Weight
= Buoyant Force
If we seek to find the apparent weight we will have to,
Remember that
V = Volume (Volume Sphere)
= Density (At this case water density)
g = Gravitational acceleration
Therefore the apparent weight will be 0.1526N
The Bermoulli's equation allows us to find the pressure in the narrow part of the pipe through which water circulates is:
P = 500 Pa
Bernoulli's equation is the work-energy relationship for fluids that are liquids and gases.
Where the subscripts 1 and 2 represent points of interest, P is the pressure, ρ the density of the fluid, v the velocity and y the height.
They indicate that the pipe is horizontal, that the pressure in the wide part P₁ = 200 kPa and the velocity is v₁ = 5 m / s and in the narrow part v₂=8.00 m/s, see attached.
Since the pipe is horizontal y₁ = y₂
P₁ + ½ ρ v₁² = P₂ + ½ ρ v₂²
P₂ = P₁ + ½ ρ (v₁² - v₂²)
Let's calculate
P₂ = 200 10² + ½ ρ (5² - 8²)
P₂ = 2 10⁴ - 19.5 ρ
For a specific calculation the value of the density of the fluid is needed, suppose that the fluid is water ρ = 1000 kg / m³
P₂ = 2 10² - 19.5 1000
P₂ = 500 Pa
In conclusion using the Bermoulli equation we can find the pressure in the narrow part of the pipe through which water circulates is:
P = 500 Pa
Learn more here: brainly.com/question/9506577
Answer:
a) T = (2,375 ± 0.008) s
, b) When comparing this interval with the experimental value we see that it is within the possible theoretical values.
Explanation:
a) The period of a simple pendulum is
T = 2π √ L / g
Let's calculate
T = 2π √1.40 / 9.8
T = 2.3748 s
The uncertainty of the period is
ΔT = dT / dL ΔL
ΔT = 2π ½ √g/L 1/g ΔL
ΔT = π/g √g/L ΔL
ΔT = π/9.8 √9.8/1.4 0.01
ΔT = 0.008 s
The result for the period is
T = (2,375 ± 0.008) s
b) the experimental measure was T = 2.39 s ± 0.01 s
The theoretical value is comprised in a range of [2,367, 2,387] when we approximate this measure according to the significant figures the interval remains [2,37, 2,39].
When comparing this interval with the experimental value we see that it is within the possible theoretical values.
(a) The magnitude of force that the positive charge exerts on the negative charge is - 2.209 N.
(b) If the negative charge is doubled, then the force will also get doubled.
The new force will be F = -4.418 N.
Explanation:
The force acting between two charged particles separated by a distance is termed as Coloumb's force or electrostatic force. It can be termed as electrostatic force of attraction if the the force acting between the charges are oppositely charged. And it can be termed as electrostatic force of repulsion if the charges are similar or like charges.
In the present case, there is a positive and negative charge, so electrostatic force of attraction will be acting between them. As per Coloumb's law, the electrostatic force of attraction is directly proportional to the product of charges and inversely proportional to the square of distance of separation.
Here, k is the constant of proportionality which is equal to 9 ×
and Q, q are the two charges, d is the distance of separation.
So here Q = 5.5 ×
and q = - 3.5 × and d = 0.28 m
Then, 
So the magnitude of force that the positive charge exerts on the negative charge is - 2.209 N.
(b) If the negative charge is doubled, then the force will also get doubled.
The new force will be F = -4.418 N.
