The correct option that can be deduced for both Object P and Q is Option b) I and II only
To solve this question correctly, we need to understand the concept of density and it relation to mass and volume.
<h3>What is Density?</h3>
Density is a physical property of an object and can be expressed by using the relation:
From the given parameters, we are being told that:
This implies that Q has a greater density that P. Since Q has a greater density than P, Q will be heavier since it will have greater mass.
However, Q will not be denser than water because if that happens, P will be have a greater density which is untrue in this scenario.
Therefore, we can conclude that:
- 1. Q is heavier than P
- II. 1cm³ of Q has a greater mass than 1cm³ of P
Learn more about density here:
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Answer:
the required revolution per hour is 28.6849
Explanation:
Given the data in the question;
we know that the expression for the linear acceleration in terms of angular velocity is;
= rω²
ω² = / r
ω = √( / r )
where r is the radius of the cylinder
ω is the angular velocity
given that; the centripetal acceleration equal to the acceleration of gravity a = g = 9.8 m/s²
so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m
Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m
so we substitute
ω = √( 9.8 m/s² / 3909.87 m )
ω = √0.002506477 s²
ω = 0.0500647 ≈ 0.05 rad/s
we know that; 1 rad/s = 9.5493 revolution per minute
ω = 0.05 × 9.5493 RPM
ω = 0.478082 RPM
1 rpm = 60 rph
so
ω = 0.478082 × 60
ω = 28.6849 revolutions per hour
Therefore, the required revolution per hour is 28.6849
Answer:
0.1
Explanation:
mass, m = 5 kg
θ = 60°
Force, F = 10 N
velocity is constant , it means the net force is zero.
So, the component of force along the surface is equal to the friction force
FCosθ = friction force
10 x cos 60 = μ x m x g
where, μ is the coefficient of friction
5 = μ x 5 x 9.8
μ = 0.1
Thus, the coefficient of friction is 0.1
Answer:
a) V = - x ( σ / 2ε₀)
c) parallel to the flat sheet of paper
Explanation:
a) For this exercise we use the relationship between the electric field and the electric potential
V = - ∫ E . dx (1)
for which we need the electric field of the sheet of paper, for this we use Gauss's law. Let us use as a Gaussian surface a cylinder with faces parallel to the sheet
Ф = ∫ E . dA = 
/ε₀
the electric field lines are perpendicular to the sheet, therefore they are parallel to the normal of the area, which reduces the scalar product to the algebraic product
E A = q_{int} /ε₀
area let's use the concept of density
σ = q_{int}/ A
q_{int} = σ A
E = σ /ε₀
as the leaf emits bonnet towards both sides, for only one side the field must be
E = σ / 2ε₀
we substitute in equation 1 and integrate
V = - σ x / 2ε₀
V = - x ( σ / 2ε₀)
if the area of the sheeta is 100 cm² = 10⁻² m²
V = - x (10⁻²/(2 8.85 10⁻¹²) = - x ( 5.6 10⁻¹⁰)
x = 1 cm V = -1 V
x = 2cm V = -2 V
This value is relative to the loaded sheet if we combine our reference system the values are inverted
V ’= V (inf) - V
x = 1 V = 5
x = 2 V = 4
x = 3 V = 3
These surfaces are perpendicular to the electric field lines, so they are parallel to the sheet.
In the attachment we can see a schematic representation of the equipotential surfaces
b) From the equation we can see that the equipotential surfaces are parallel to the sheet and equally spaced
c) parallel to the flat sheet of paper
