Recent content by tem_osu

Thank you for your explanation. I understood it now. Solve for T: Mug 0.350*9.8* 0.280= 0.9604 Nm Plate 1.00*9.8*0.140= 1.372 Nm Tray 0.220*9.8*0.100= 0.2156 Nm Total Clockwise torque = 2.548 Nm 2.548 = 0.04T T = 63.7 N Solve for F: F is the upward force so it must be equal to the...

Using distance measure from the same spot: Mug: 0.350 x 9.8 x 0.380 = 1.3034 Plate: 1.00 x 9.8 x 0.240 = 2.352 Tray: 0.220 x 9.8 x 0.400 = 0.8624 Total = 4.5178 Nm 4.5178Nm / 0.06 m = 75.297 N for T?

Do I use the distance between the thumb & the mug? = 0.320 m

[SOLVED] Rigid Objects in Equilibrium & Center of Gravity Homework Statement http://img149.imageshack.us/img149/8568/0968lr6sc4.gif A lunch tray is being held in one hand, as the drawing illustrates. The mass of the tray itself is 0.220 kg, and its center of gravity is located at...

ω = ω[SIZE="1"]0 + αt ω - αt = ω[SIZE="1"]0 (1.70 rad/s) - (-5.05 rad/s^2)(10.0 s) = ω[SIZE="1"]0 ω[SIZE="1"]0 = 52.5 rad/s ω^2 = ω0^2 + 2αθ (1.70^2) = (52.2^2) + 2(-5.05)θ θ = [(1.70^2) - (52.2^2)]/(2 x -5.05) = 269.5 rad GOT IT! Thank you so much for helping me. ^-^

(a) 13.5 x 2Pi = 9.0t t = 9.42 s (b) (0-18.0 rad/s)/9.42s = -1.91 rad/s^2 Thank you so much. I'm still confuse about the second problem though. I tried several ways, but I've gotten some weird answers.

equation 1: θ = ω[SIZE="1"]0t + 1/2αt^2 equation 2: ω = ω0 + αt (ω - ω[SIZE="1"]0)/t = α solve for α from equation 2 & plug into equation 1: θ = ω[SIZE="1"]0t + 1/2[(ω - ω[SIZE="1"]0)/t]t^2 θ = ω[SIZE="1"]0t + 1/2[(ω - ω[SIZE="1"]0)]t 13.5 = 18.0t + 1/2(0 - 18.0)t 13.5 = 18.0t - 9t...

[SOLVED] Rotational Kinematics Homework Statement A person is riding a bicycle, and its wheels have an angular velocity of +18.0 rad/s. Then, the brakes are applied and the bike is brought to a uniform stop. During braking, the angular displacement of each wheel is +13.5 revolutions. (a)...

So, the circumference is (2)(pi)(2100) = 13,194.69 m the car traveled for 410 s, 13194.69/410 = 32.18 m/s? EDIT: I got it! Thank you so much. You are a lifesaver!

Thank you for your help. I got the 2nd problem. The answer came out to be 28.35 m/s. For the 1st problem: a = (v^2)/r r = 2100/(2*3.14) = 334.39 velocity = 334.39m/410s = 0.8156 m/s? then the acceleration would be a= 0.8156/410 = 0.001989 m/s^2?

Thank you so much for replying. For the 1st problem: 2.1 km = 2,100 m velocity = 2,100 m / 410 s = 5.12 m/s (did anything went wrong in this step?) a = (v^2)/r a = (5.12^2)/2100 = 0.01248 --- For the 2nd problem: F - mg = m(v^2)/r Force is 5 times the weight, so F=5m 5m -...

[SOLVED] Vertical Circular Motion & Centripetal Acceleration Homework Statement 1) A car travels at a constant speed around a circular track whose radius is 2.1 km. The car goes once around the track in 410 s. What is the magnitude of the centripetal acceleration of the car? Homework...