Answer:
vcyl / vsph = 1.05
Explanation:
The kinetic energy of a rolling object can be expressed as the sum of a translational kinetic energy plus a rotational kinetic energy.The traslational part can be written as follows:[tex]K_{trans} = \frac{1}{2}* M* v_{cm} ^{2} (1)[/tex]
The rotational part can be expressed as follows:[tex]K_{rot} = \frac{1}{2}* I* \omega ^{2} (2)[/tex]
where I = moment of Inertia regarding the axis of rotation.ω = angular speed of the rotating object.If the object has a radius R, and it rolls without slipping, there is a fixed relationship between the linear and angular speed, as follows:[tex]v = \omega * R (3)[/tex]
For a solid cylinder, I = M*R²/2 (4)Replacing (3) and (4) in (2), we get:[tex]K_{rot} = \frac{1}{2}* \frac{1}{2} M*R^{2} * \frac{v_{cmc} ^{2}}{R^{2}} = \frac{1}{4}* M* v_{cmc}^{2} (5)[/tex]
Adding (5) and (1), we get the total kinetic energy for the solid cylinder, as follows:[tex]K_{cyl} = \frac{1}{2}* M* v_{cmc} ^{2} +\frac{1}{4}* M* v_{cmc}^{2} = \frac{3}{4}* M* v_{cmc} ^{2} (6)[/tex]
Repeating the same steps for the spherical shell:[tex]I_{sph} = \frac{2}{3} * M* R^{2} (7)[/tex]
[tex]K_{rot} = \frac{1}{2}* \frac{2}{3} M*R^{2} * \frac{v_{cms} ^{2}}{R^{2}} = \frac{1}{3}* M* v_{cms}^{2} (8)[/tex]
[tex]K_{sph} = \frac{1}{2}* M* v_{cms} ^{2} +\frac{1}{3}* M* v_{cms}^{2} = \frac{5}{6}* M* v_{cms} ^{2} (9)[/tex]
Since we know that both masses are equal each other, we can simplify (6) and (9), cancelling both masses out.And since we also know that both objects have the same kinetic energy, this means that (6) are (9) are equal each other.Rearranging, and taking square roots on both sides, we get:[tex]\frac{v_{cmc}}{v_{cms}} =\sqrt{\frac{10}{9} } = 1.05 (10)[/tex]
This means that the solid cylinder is 5% faster than the spherical shell, which is due to the larger moment of inertia for the shell.the higher the objects " ? ", the more kinetic energy
a 4.5 Hz wave has a wavelength of 0.8m. what is the speed
0.18 m/s
5.6m/s
5.3m/s
3.6m/s
Answer:
Explanation
The liquid emerges into a vertical jet as it drains from the container, with the velocity profile in the jet remaining uniform. The outlet of the container is located 2.0 m above ground, and the radius of the emerging liquid jet changes with vertical distance from the bottom of the container as it accelerates under the action of gravity. Neglecting viscous losses and surface tension effects in the liquid jet, what is the velocity of the water jet as it strikes the ground when the container begins to drain
Answer:
6.26 m/s
Explanation:
Since we are neglecting viscous losses and surface tension effects in the liquid jet, by conservation of energy, the potential energy loss of the jet = kinetic energy gain of the jet
So, mgh = 1/2mv² where m = mass of water in jet, g = acceleration due to gravity = 9.8 m/s², h = height of outlet = 2.0 mand v = velocity of liquid jet
So, mgh = 1/2mv²
gh = 1/2v²
v² = 2gh
v = √(2gh)
v = √(2 × 9.8 m/s² × 2.0 m)
v = √(39.2 m²/s²)
v = 6.26 m/s
One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. The first force has a magnitude of 2.00 N and is applied perpendicular to the length of the stick at the free end. The second force has a magnitude of 6.00 N and acts at a 42.9o angle with respect to the length of the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the end of the stick that is pinned.
Answer:
x = 0.455 L
Explanation:
For this exercise we must use the rotational equilibrium condition
Σ τ = 0
it has two forces, the first is perpendicular to the rod, so its stub is
τ₁ = F₁ L
the second force is applied with an angle, so we can use trigonometry to find its components
sin θ = F_parallel / F₂
cos θ = F_perpendicular / F₂
F_parallel = F₂ sin θ
F _perpendicular = F₂ cos θ
torque is
τ₂ = F_perpendicular x + F_parallel 0
the parallel force is on the rod therefore its distance is zero
we apply the equilibrium equation
τ₁ - τ₂ = 0
F₁ L = F₂ cos θ x
x = [tex]\frac{L}{cos \theta} \ \frac{F_1}{F_2}[/tex]
let's calculate
x = [tex]\frac{L}{cos \ 42.9} \ \frac{2.00}{6.00}[/tex]
x = 0.455 L
What is the speed of a ball that is attached to a string and swings in a horizontal circle of radius 2.0 m with the central acceleration of 15 m/s^2?
Answer:
5.48 m/s.
Explanation:
Use the formula a=v^2/r.
The following statements address the science behind the pulley system illustrated:
A. The pulleys increase the entropy of the system.
B. The force applied to the rope is less than the force needed to lift the object.
C. The pulleys help generate as much energy as possible.
D. The pulleys multiply energy input, resulting in more energy output.
E. The pulleys generate no thermal energy.
Which of these statements is/are true?
i. Statements A and B
ii. Statements D and E
iii. Only statement C
iv. All of the statements
Answer:
i. Statements A and B
Explanation:
Sana nakatulong
9.2 True/False Questions
1) Unsatisfying relationships can interfere with your well-being.
Answer:
2) One way to tell if a relationship is unhealthy is that you are unhappy.
Answer:
3) Managers at large companies are more involved in the day-to-day operations of their businesses than entrepreneurs.
Answer:
4) Being an effective member of a team depends entirely on the project's outcome.
Answer:
5) Teams are influenced by the personal qualities of the team members.
Answer:
why is potassium and sodium considered as reactive metals?
Answer:
because they are found freely in nature uncombined so they are highly reactive with other elements
If an object went from 0 m/s to 6 m/s in 1.7 seconds after a 10 N force was applied to it; what is the object's mass? No links pls
The force acting on the object is constant, so the acceleration of the object is also constant. By definition of average acceleration, this acceleration was
a = ∆v / ∆t = (6 m/s - 0) / (1.7 s) ≈ 3.52941 m/s²
By Newton's second law, the magnitude of the force F is proportional to the acceleration a according to
F = m a
where m is the object's mass. Solving for m gives
m = F / a = (10 N) / (3.52941 m/s²) ≈ 2.8 kg