We know that: 1 L = 100 cL. Or 1 cL = 0.01 L. Then we will make the conversion: 34.9 cL = 34.9 / 100 L = 0.349 L. Also: 1 hL = 100 L. 0.349 L = 0.349 / 100 hL = 0.00349 hL. This can be also written as: 3.49 * 10^(-3) hL ( in the scientific notation ). Answer: 3.49 cL = 0.00349 <span>hL </span>
Answer:
therefore critical angle c= 69.79°
Explanation:
Canola oil is less dense than water, so it floats over water.
Given 
which is higher than that of water
refractive index of water 
to calculate critical angle of light going from the oil into water
we know that

now putting values we get

c= 
c=69.79°
therefore critical angle c= 69.79°
Ok, assuming "mj" in the question is Megajoules MJ) you need a total amount of rotational kinetic energy in the fly wheel at the beginning of the trip that equals
(2.4e6 J/km)x(300 km)=7.2e8 J
The expression for rotational kinetic energy is
E = (1/2)Iω²
where I is the moment of inertia of the fly wheel and ω is the angular velocity.
So this comes down to finding the value of I that gives the required energy. We know the mass is 101kg. The formula for a solid cylinder's moment of inertia is
I = (1/2)mR²
We want (1/2)Iω² = 7.2e8 J and we know ω is limited to 470 revs/sec. However, ω must be in radians per second so multiply it by 2π to get
ω = 2953.1 rad/s
Now let's use this to solve the energy equation, E = (1/2)Iω², for I:
I = 2(7.2e8 J)/(2953.1 rad/s)² = 165.12 kg·m²
Now find the radius R,
165.12 kg·m² = (1/2)(101)R²,
√(2·165/101) = 1.807m
R = 1.807m
The distance you free-fall from rest is D = (1/2) (g) (T²) <== memorize this
Height of the platform = (1/2) (9.8 m/s²) (2.4 sec)²
Height = (4.9 m/s²) (5.76 s²)
Height = (4.9/5.76) meters
Height = 28.2 meters (a VERY high platform ... about 93 ft off the water !)
Without air-resistance, your horizontal speed doesn't change. It's constant. Traveling 3.1 m/s for 2.4 sec, you cover (3.1 m/s x 2/4 s) = 7.4 m horizontally.