Answer:
Assume two identical cans filled with two types of soup having same mass are rolling down on an inclined plane in same conditions. In terms of inertia different types of soup will indicate different viscosity. The higher viscosity fillings indicates more part of the soup mass is rotating together with the can’s body. This means that for the can with lower viscosity soup has a lower moment of inertia and the can with higher viscosity has higher moment of inertia while the same gravity makes them to roll.
incline angle = θ ; can's mass = m ; Radius of the can's = R , Angular acceleration for Can 1 = α1 ; Angular acceleration for Can 2 = α2
T1 = Inertia of Can with high viscosity soup
T2 = Inertia of Can with low viscosity soup
M1 rolling moment of Can 1
M2 rolling moment of Can 2
equation is given by
T1*α1 = M1 - (a)
T2*α2 = M2 - (b)
M1 = M2 = m*g*R*sin(θ). (c)
as assumed T1 > T2
from the three equation (a), (b) & (c)
the α2 > α1
Angular acceleration of Can 2 is higher than Can 1. Already stated that Can 1 has more viscous soup as compared to Can 2.
The answer of this question is D. All of the above
Acceleration happen when an obeject change its velocity. It has nothing to do with speed.
The huge misconception about acceleration is when we thought it only aply if we increase our speed ( in a sport match, sportcaster often describe acceleration as an increase in players speed)
slower, faster, right , left, it does not matter, as long as that object change its velocity, it accelerates
Answer:
CB = 4.45 x 10⁻⁹ F = 4.45 nF
Explanation:
The capacitance of a parallel plate capacitor is given by the following formula:
C = ε₀A/d
where,
C = Capacitance
ε₀ = Permeability of free space
A = Area of plates
d = Distance between plates
FOR CAPACITOR A:
C = CA = 17.8 nF = 17.8 x 10⁻⁹ F
A = A₁
d = d₁
Therefore,
CA = ε₀A₁/d₁ = 17.8 x 10⁻⁹ F ----------------- equation 1
FOR CAPACITOR B:
C = CB = ?
A = A₁/2
d = 2 d₁
Therefore,
CB = ε₀(A₁/2)/2d₁
CB = (1/4)(ε₀A₁/d₁)
using equation 1:
CB = (1/4)(17.8 X 10⁻⁹ F)
<u>CB = 4.45 x 10⁻⁹ F = 4.45 nF</u>
Given:
Gasoline pumping rate, R = 5.64 x 10⁻² kg/s
Density of gasoline, D = 735 kg/m³
Radius of fuel line, r = 3.43 x 10⁻³ m
Calculate the cross sectional area of the fuel line.
A = πr² = π(3.43 x 10⁻³ m)² = 3.6961 x 10⁻⁵ m²
Let v = speed of pumping the gasoline, m/s
Then the mass flow rate is
M = AvD = (3.6961 x 10⁻⁵ m²)*(v m/s)*(735 kg/m³) = 0.027166v kg/s
The gasoline pumping rate is given as 5.64 x 10⁻² kg/s, therefore
0.027166v = 0.0564
v = 2.076 m/s
Answer: 2.076 m/s
The gasoline moves through the fuel line at 2.076 m/s.
