Urgent help with Questions regarding Gen. Physics

[Badal]

As a 2.0-kg block travels around a 0.50-m radius circle it has an angular speed of 12 rad/s.
The circle is parallel to the xy plane and is centered on the z axis, 0.75m from the origin. The
magnitude of its angular momentum around the origin is:
A. 6.0kg · m2/s
B. 9.0kg · m2/s
C. 11 kg · m2/s
D. 14 kg · m2/s
E. 20 kg · m2/s
ans: C

Okay i know that
L = Iw and L = r x p
I can use equation of motion to find W.and I=wrp2
but how will that get me to the answer? i have tried everything and the answer i get everytime is 12 or 18.. Both of them are not one of the choices. I have a test tonight and have to go to work right before the test so if anyone of you can please help me.
I would really appreciate it.

Another question i needed help with is

Venus has a mass of about 0.0558 times the mass of Earth and a diameter of about 0.381 times
the diameter of Earth. The acceleration of a body falling near the surface of Venus is about:
A. 0.21m/s2
B. 1.4m/s2
C. 2.8m/s2
D. 3.8m/s2
E. 25m/s2
ans: D

So what i thaught on this question was to take the Earth's mass and diameter to be 1
so the mass of venus would be
Mass(venus)=0.0558xMass(earth)
Mass(venus)/Mass(earth)=0.0558
is this the correct aproach? i got the right answer at the end and just want to know if i was doing it right?

 

[Astronuc]

L = r x p = mvr = m$\omega$r², because p = mv acting at r.

This is a useful reference - http://hyperphysics.phy-astr.gsu.edu/hbase/mi.html

http://hyperphysics.phy-astr.gsu.edu/hbase/amom.html


For the second problem g = GM/r², where g = acceleration of gravity (9.81 m/s², M = Earth's mass, and r = Earth's radius.

Substituting the in the appropriate mass and radius of venus would give the acceleration of gravity at the surface of Venus.

 

[wysard]

Are you sure that's VENUS? The numbers look suspiciously like the relative mass and diameter of MERCURY to me. Could be just a typical textbook typo. Just an observation, has no bearing on the info Astronuc already provided to solve the problem at hand.