5 Question from Practice Exam, Please Help Me!

The Practice Exam I was given contains answer, but getting the answer is the hard part.

**1. A rotating wheel requires 3.00 s to rotate 37 revolutions. At the end of 3.00 s the angular velocity will be 98.0 rad/s. What is the constant angular acceleration? (1 rev = 2pi rad)**

Ans:13.7 rad/s^2

Im using the formula delta(theta) = (omega)(time) + 1/2at^2

delta(theta)= (37*2pi)

(omega)(time)= 98.0 rad/s * 3sec

1/2at^2= 1/2a(3sec)^2

I plug everything in, but i get -13.7rad/s^2. Am I just lucky I came close to negative answer or did i so something wrong?

**2. A meter stick can be balanced at the 49.7-cm mark when balanced on a fulcrum (pivot). When a 50.0-gram mass is attached at the 10.0-cm mark, the fulcrum must be moved to the 39.2-cm mark for balance. What is the mass of the meter stick in grams?**

Ans: 139 gm

Yea, I'm clueless

**3. Two spheres of equal volume 100 cm3 contain an inert gas at 0°C and 1.00 atm pressure. They are joined by a small tube of negligible volume, allowing the gas to flow between the spheres. What common pressure will exist in the two spheres if one is raised to 100°C while the other is kept at 0° C? Assume an ideal gas.**

Ans: 1.16 atm

I try every way possible to manipulate Pv=nRT, but I can't get the answer

**4.A 5.00–g bullet moving with an initial speed of 400 m/s is fired horizontally along the positive x-axis, passing completely through a 1.00-kg block, which initially is at rest on a frictionless horizontal surface at x = 0.00 cm. The block is connected to a spring having a force constant of 900 N/m. If the block moves 5.00 cm to the right (x = 5.00 cm) along the positive x-axis after impact, what is the speed of the bullet when it emerges from the block?**

Ans:100 m/s

From the formual (KE+PEg+PEs)i = (KE+PEg+PEs)f

I'm got the formula: 1/2mv^2 = 1/2mv^2 + 1/2Kx^2

I don't know what I am doing wrong.

**5. A thin spherical shell of mass 0.400 kg and diameter 0.200 m is filled with alcohol (rho = 806 kg/m3). It is released from rest at the bottom of a tank of water (rho = 1000 kg/m3). What is the acceleration of the alcohol spherical shell as it starts to rise to the surface?**

Ans:1.06 m/s^2

Yea, another clueless one

Thank you, for whoever helps me on this