An electric turntable 0.750 m in diameter is rotating about a fixed axis with an initial angular velocity of 0.250 rev/s and a constant angular acceleration of 0.900 rev/s 2 , (a) Compute the angular velocity of the turntable after 0.200 s. (b) Through how many revolutions has the turntable spun in this time interval? (c) What is the tangential speed of a point on the rim of the turn-table at t = 0.200 s? (d) What is the magnitude of the resultant acceleration of a point on the rim at t = 0.200 s?
An electric turntable 0.750 m in diameter is rotating about a fixed axis with an initial angular velocity of 0.250 rev/s and a constant angular acceleration of 0.900 rev/s 2 , (a) Compute the angular velocity of the turntable after 0.200 s. (b) Through how many revolutions has the turntable spun in this time interval? (c) What is the tangential speed of a point on the rim of the turn-table at t = 0.200 s? (d) What is the magnitude of the resultant acceleration of a point on the rim at t = 0.200 s?
An electric turntable 0.750 m in diameter is rotating about a fixed axis with an initial angular velocity of 0.250 rev/s and a constant angular acceleration of 0.900 rev/s2, (a) Compute the angular velocity of the turntable after 0.200 s. (b) Through how many revolutions has the turntable spun in this time interval? (c) What is the tangential speed of a point on the rim of the turn-table at t = 0.200 s? (d) What is the magnitude of the resultant acceleration of a point on the rim at t = 0.200 s?
Definition Definition Rate of change of angular velocity. Angular acceleration indicates how fast the angular velocity changes over time. It is a vector quantity and has both magnitude and direction. Magnitude is represented by the length of the vector and direction is represented by the right-hand thumb rule. An angular acceleration vector will be always perpendicular to the plane of rotation. Angular acceleration is generally denoted by the Greek letter α and its SI unit is rad/s 2 .
Please help me on the following question (Please refer to image below)An Olympic lifter (m = 103kg) is holding a lift with a mass of 350 kg. The barexerts a purely vertical force that is equally distributed between both hands. Each arm has amass of 9 kg, are 0.8m long and form a 40° angle with the horizontal. The CoM for each armis 0.5 m from hand. Assuming the lifter is facing us in the diagram below, his right deltoidinserts 14cm from the shoulder at an angle of 13° counter-clockwise from the humerus.A) You are interested in calculating the force in the right deltoid. Draw a free body diagramof the right arm including the external forces, joint reaction forces, a coordinate system andstate your assumptions.B) Find the force exerted by the right deltoidC) Find the shoulder joint contact force. Report your answer using the magnitude and directionof the shoulder force vector.
I need help with part B. I cant seem to get the correct answer. Please walk me through what youre doing to get to the answer and what that could be
Question 6:
Chlorine is widely used to purify municipal water supplies and to treat swimming pool
waters. Suppose that the volume of a particular sample of Cl₂ gas is 8.70 L at 895 torr
and 24°C.
(a) How many grams of Cl₂ are in the sample?
⚫ Atomic mass of CI = 35.453 g/mol
• Molar mass of Cl₂ = 2 x 35.453 = 70.906 g/mol
Solution:
Use the Ideal Gas Law:
Step 1: Convert Given Values
• Pressure: P = 895 torr → atm
PV=
= nRT
1
P = 895 ×
= 1.1789 atm
760
•
Temperature: Convert to Kelvin:
T24273.15 = 297.15 K
• Gas constant: R = 0.0821 L atm/mol. K
Volume: V = 8.70 L
Step 2: Solve for n
.
PV
n =
RT
n =
(1.1789)(8.70)
(0.0821)(297.15)
10.25
n =
= 0.420 mol
24.405
Step 3: Calculate Mass of Cl₂
Final Answer: 29.78 g of Cl₂.
mass nx M
mass=
(0.420)(70.906)
mass=
29.78 g
Chapter 9 Solutions
University Physics with Modern Physics (14th Edition)
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