Answer:
Work done by the machine (W) = 500 × 1.5 = 750 J
Work supplied to the machine (W) = 100 × 10 = 1000 J
Here, work supplied to the machine is input work = 1000 J
Answer:
Explanation:
a )
Each blade is in the form of rod with axis near one end of the rod
Moment of inertia of one blade
= 1/3 x m l²
where m is mass of the blade
l is length of each blade.
Total moment of moment of 3 blades
= 3 x
x m l²
ml²
2 )
Given
m = 5500 kg
l = 45 m
Putting these values we get
moment of inertia of one blade
= 1/3 x 5500 x 45 x 45
= 37.125 x 10⁵ kg.m²
Moment of inertia of 3 blades
= 3 x 37.125 x 10⁵ kg.m²
= 111 .375 x 10⁵ kg.m²
c )
Angular momentum
= I x ω
I is moment of inertia of turbine
ω is angular velocity
ω = 2π f
f is frequency of rotation of blade
d )
I = 111 .375 x 10⁵ kg.m² ( Calculated )
f = 11 rpm ( revolution per minute )
= 11 / 60 revolution per second
ω = 2π f
= 2π x 11 / 60 rad / s
Angular momentum
= I x ω
111 .375 x 10⁵ kg.m² x 2π x 11 / 60 rad / s
= 128.23 x 10⁵ kgm² s⁻¹ .
Answer:
995.12 N/C
Explanation:
R = 9 cm = 0.09 m
σ = 9 nC/m^2 = 9 x 10^-9 C/m^2
r = 9.1 cm = 0.091 m
q = σ x 4π R² = 9 x 10^-9 x 4 x 3.14 x 0.09 x 0.09 = 9.156 x 10^-10 C
E = kq / r^2
E = ( 9 x 10^9 x 9.156 x 10^-10) / (0.091 x 0.091)
E = 995.12 N/C
Answer:
3360 N
Explanation:
In a first-class lever, the effort force and load force are on opposite sides of the fulcrum.
The lever is 5 m long. The load force is 1.50 m from the fulcrum, so the effort force must be 3.50 m from the fulcrum.
The torques are equal:
Fr = Fr
(1440 N) (3.5 m) = F (1.5 m)
F = 3360 N
Answer:
5.09 x 10⁵ Nm²/C
Explanation:
The electric flux φ through a planar area is defined as the electric field Ε times the component of the area Α perpendicular to the field. i.e
φ = E A
From the question;
E = (8.0j + 2.0k) ✕ 10³ N/C
r = radius of the circular area = 9.0m
A = area of a circle = π r² [Take π = 3.142]
A = 3.142 x 9² = 254.502m²
Now, since the area lies in the x-y plane, only the z-component of the electric field is responsible for the electric flux through the circular area.
Therefore;
φ = (2.0) x 10³ x 254.502
φ = 5.09 x 10⁵ Nm²/C
The electric flux is 5.09 x 10⁵ Nm²/C
