A propeller blade at rest starts to rotate from t = 0 s to t = 5.0 s with a tangential acceleration of the tip of the blade at 3
1 answer:
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As we know that range of the projectile motion is given by
here we know that range will be same for two different angles
so here we can say the two angle must be complementary angles
so the two angles must be
so it is given that one of the projection angle is 75 degree
so other angle for same range must be 90 - 75 = 15 degree
so other projection angle must be 15 degree
Answer:
0.000003782 m
0.000001891 m
0.000001197125 m
Explanation:
= Wavelength = 248 nm
D = Diameter of beam = 1 cm
f = Focal length = 0.625 cm
The angle is given by
The width is given by
The required width is 0.000003782 m
Minimum resolvable line separation is given by
The minimum resolvable line separation between adjacent lines is 0.000001891 m
when
The new minimum resolvable line separation between adjacent lines is
Answer
The answer and procedures of the exercise are attached in the following archives.
Step-by-step explanation:
You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.
Answer:
a)106.48 x 10⁵ kg.m²
b)144.97 x 10⁵ kgm² s⁻¹
Explanation:
a)Given
m = 5500 kg
l = 44 m
Moment of inertia of one blade
= 1/3 x m l²
where m is mass of the blade
l is length of each blade.
Putting all the required values, moment of inertia of one blade will be
= 1/3 x 5500 x 44²
= 35.49 x 10⁵ kg.m²
Moment of inertia of 3 blades
= 3 x 35.49 x 10⁵ kg.m²
= 106.48 x 10⁵ kg.m²
b) Angular momentum 'L' is given by
L = x ω
where,
= moment of inertia of turbine i.e 106.48 x 10⁵ kg.m²
ω=angular velocity =2π f
f is frequency of rotation of blade i.e 13 rpm
f = 13 rpm=>= 13 / 60 revolution per second
ω = 2π f => 2π x 13 / 60 rad / s
L= x ω =>106.48 x 10⁵ x 2π x 13 / 60
= 144.97 x 10⁵ kgm² s⁻¹