A straightforward method of finding the density of an object is to measure its mass and then measure its volume by submerging it
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b and e are the largest and equal in magnitude.
A and d are next. aR = (3rad/s2)R = 3R
c is zero. wR = v = 0; Angular acceleration is zero.
<h3>What is angular acceleration?</h3>- The temporal rate at which angular velocity changes is known as angular acceleration. The standard unit of measurement is radians per second per second. Therefore, = d d t. Rotational acceleration is another name for angular acceleration.
- Angular velocity divided by acceleration time can be used to define angular acceleration. (t). As an alternative, use pi times the drive speed (n) divided by the acceleration time (t) times 30. Radians per second squared (Rad/sec2) is the standard SI unit for rotational acceleration resulting from this equation.
- To calculate angular velocity, we can use one of three formulas. The definition itself provides the first. Theta = position angle, t = time, and w = angular velocity, where w = angular velocity, theta = position angle, and t = time. Angular velocity is the rate of change of an object's position angle with respect to time.
- The symbol for angular acceleration is, and it is measured in rad/s2, or radians per second square.
If two items are equal, show them as equal in your ranking. If a quantity is equal to zero, show that fact in your ranking:
b and e are the largest and equal in magnitude.
A and d are next. aR = (3rad/s2)R = 3R
c is zero. wR = v = 0; Angular acceleration is zero.
To learn more about angular acceleration, refer to:
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Answer:
(a) L = 0·73 m
(b) 4·39 × J
Explanation:
(a) From the figure, consider the torque about the point where the rod is attached because if we consider another point then there will be hinge forces acting on the rod at the point of attachment
Let L m be the length of the rod and β be the angle between the rod and the vertical
Let α be the angular acceleration of the rod
As the force of gravity acts at the centre so from the figure, the torque about the point of attachment will be 0·53 × g ×(L ÷ 2) ×sinβ
Assuming that the value of amplitude of this oscillation to be small
As torque = moment of inertia × angular acceleration
0·53 × g ×(L ÷ 2) ×sinβ = ((0·53 × L²) ÷ 3) × α (∵ moment of inertia of the rod from the point of attachment)
<h3>For small oscillations, α = ω² × β</h3>After substituting the value of α and solving we get
ω = √((3 × g) ÷ (2 × L))
Time period = (2 × π) ÷ ω = (2 × π) ÷ √((3 × g) ÷ (2 × L))
∴ (2 × π) ÷ √((3 × g) ÷ (2 × L)) = 1·4
Substituting the value of g as 9·8 m/s² and solving we get
L = 0·73 m
(b) At the maximum amplitude condition the velocity will be 0 and potential energy will be maximum and maximum kinetic energy will be attained at the lowest point and hinge forces will not do work as the point of attachment is not moving
∴ Taking the reference for finding the potential energy as the lowest point
<h3>Maximum potential energy = Maximum kinetic energy </h3><h3>As total energy is constant, since there is no dissipative force</h3>Maximum potential energy = (0·53 × g × L ×(1 - cosβ)) ÷ 2 (∵ increment in height is (L × (1 - cosβ)) ÷ 2
∴ Maximum potential energy = (0·53 × g × L ×(1 - cosβ)) ÷ 2 After substituting the value we get
Maximum potential energy = 4·39 × J
∴ Maximum kinetic energy = 4·39 × J
Answer:
525 V
Explanation:
A = Area =
= Rate of change of magnetic field =
(assumed)
Induced electromotive force is given by
The induced electromotive force is 525 V