Answer:
Explanation:
90 rpm = 90 / 60 rps
= 1.5 rps
= 1.5 x 2π rad /s
angular velocity of flywheel
ω = 3π rad /s
Let I be the moment of inertia of flywheel
kinetic energy = (1/2) I ω²
(1/2) I ω² = 10⁷ J
I = 2 x 10⁷ / ω²
=2 x 10⁷ / (3π)²
= 2.2538 x 10⁵ kg m²
Let radius of wheel be R
I = 1/2 M R² , M is mass of flywheel
= 1/2 πR² x t x d x R² , t is thickness , d is density of wheel .
1/2 πR⁴ x t x d = 2.2538 x 10⁵
R⁴ = 2 x 2.2538 x 10⁵ / πt d
= 4.5076 x 10⁵ / 3.14 x .1 x 7800
= 184
R= 3.683 m .
diameter = 7.366 m .
b ) centripetal accn required
= ω² R
= 9π² x 3.683
= 326.816 m /s²
Answer:
45.88297 m
Violet
Explanation:
x = Gap between holes = 5.9 mm
= Wavelength = 527 nm
D = Diameter of eye = 5 mm
L= Distance of observer from holes
From Rayleigh criteria we have the relation

A person could be 45.88297 m from the tile and still resolve the holes
Resolving them better means increasing the distance between the observer and the holes. It can be seen here that the distance is inversely proportional to the wavelength. Violet has a lower wavelength than red so, violet light would resolve the holes better.
We can solve the problem by using Newton's second law of motion:

where
F is the net force applied to the object
m is the object's mass
a is the acceleration of the object
In this problem, the force applied to the car is F=1050 N, while the mass of the car is m=760 kg. Therefore, we can rearrange the equation and put these numbers in, in order to find the acceleration of the car:

The equation also tells us that the acceleration and the force have same directions: therefore, since the force exerted on the car is horizontal, the correct answer is
<span>
B) 1.4 m/s2 horizontally.</span>
<span>If you have only two
data from two recording stations then you will be having a hard time finding
the correct location of the epicenter. This is because triangulation method requires
3 recording station. If you have 2 recording station, the 2 circles will
intersect at 2 points giving you 2 locations that could possibly be the
epicenter.</span>