Using the formula for kinetic energy of a moving particle k=12mv2, find the kinetic energy ka of particle a and the kinetic energy kb of particle
b. remember that both particles rotate about the y axis. express your answers in terms of m, ω, and r separated by a comma.
Answers
Complete Question
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Answer:
The kinetic energy for particle a is [tex]K_a = \frac{9}{2} mr^2 w^2[/tex]
The kinetic energy for particle b is [tex]K_b = m w^2r^2 [/tex]
Explanation:
From the question we are told that
The mass of particle a is [tex]m[/tex]
The mass of particle b is [tex]2 m[/tex]
Generally the kinetic energy for particle a is mathematically represented as
[tex]K_a = \frac{1}{2} m v^2[/tex]
Here [tex]v_a[/tex] is the linear velocity of particle a which is mathematically represented as
[tex]v_b = w * 3r[/tex]
So
[tex]K = \frac{1}{2} m {w * 3r}^2[/tex]
=> [tex]K_a = \frac{9}{2} mr^2 w^2[/tex]
Generally the kinetic energy for particle a is mathematically represented as
[tex]K_b = \frac{1}{2} 2m v_b^2[/tex]
Here [tex]v_b[/tex] is the linear velocity of particle a which is mathematically represented as
[tex]v_b = w * r[/tex]
So
[tex]K_b = m (wr)^2[/tex]
=> [tex]K_b = m w^2r^2[/tex]
The kinetic energy of the particle in terms of angular speed (ω), mass of the object (m) and the radius of the path is [tex]\frac{1}{2} m \omega ^2 r^2[/tex].
The given parameters;
kinetic energy of the particle, K.E = ¹/₂mv²The kinetic energy of the of the particle in terms of the angular velocity is calculated as follows;
[tex]K_b = \frac{1}{2} m (\omega r)^2\\\\K_b = \frac{1}{2} m \omega ^2 r^2[/tex]
where;
m is the mass of the objectω is the angular speed of the objectr is the radius of the circular pathThus, the kinetic energy of the particle in terms of angular speed (ω), mass of the object (m) and the radius of the path is [tex]\frac{1}{2} m \omega ^2 r^2[/tex].
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Answer: Explanation:
She can measure the volume and mass of the marble, the volume and mass of the water, and the volume and mass of the graduated cylinder. She can measure the volume and mass of the marble, the volume and mass of the water, and the mass of the graduated cylinder.
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Answer:
Li⁻
Explanation:
# of protons = # element on the Periodic Table.
# of protons (+) + # of electrons (-) = overall charge
3 protons is the 3rd element on the Periodic Table: Lithium
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What is the magnitude of force required to accelerate a car of mass
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Answers
Answer:
The answer is 8075 NExplanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × accelerationFrom the question
mass = 1.7 × 10³ kg = 1700 kg
acceleration = 4.75 m/s²
We have
force = 1700 × 4.75
We have the final answer as
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You are riding a bicycle. If you apply a forward force of 172 N, and you and the bicycle have a combined mass of 51 kg, what will be the forward acceleration of the bicycle? (Assume there is no friction)
Answers
Answer:
The answer is 3.37 m/s²Explanation:
The acceleration of an object given it's mass and the force acting on it can be found by using the formula
[tex]acceleration = \frac{force}{mass} \\[/tex]
From the question
force = 172 N
mass = 51 kg
We have
[tex]acceleration = \frac{172}{51} \\ = 3.372549019...[/tex]
We have the final answer as
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Why is the unit of momentum called derived unit
Answers
Answer:
the SI unit of momentum is :- kg.ms-1
and we know that,
kinetic energy = 1/2 mv2
E=p2/2m
p=(2Em)1/2
so the derived units are (J.kg)1/2
Explanation:
On an icy day, you worry about parking your car in your driveway, which has an incline of 12º. Your neighbor's driveway has an incline of 9.0º, and the driveway across the street is at 6.0º. The coefficient of static friction between tire rubber and ice is 0.15. Which driveway(s) will be safe to park in?
Answers
Answer:
Driveway across the street ([tex]\theta = 6^{\circ}[/tex]) is the only choice for a safe parking.
Explanation:
Let suppose that car is represented by a particle, then we proceed to construct its free body diagram and corresponding equations of equilibrium:
[tex]\Sigma F_{x'} = f - W\cdot \sin \theta = 0[/tex] (Eq. 1)
[tex]\Sigma F_{y'} = N-W\cdot \cos \theta = 0[/tex] (Eq. 2)
Where:
[tex]f[/tex] - Static friction force, measured in newtons.
[tex]N[/tex] - Normal force on the car from the ground, measured in newtons.
[tex]W[/tex] - Weight of the car, measured in newtons.
[tex]\theta[/tex] - Driveway inclination, measured in sexagesimal degrees.
By applying definitions of maximum static friction force and weight, we expand the system of equations presented above:
[tex]\mu_{s}\cdot N -m\cdot g\cdot \sin \theta = 0[/tex] (Eq. 1b)
[tex]N - m\cdot g \cdot \cos \theta = 0[/tex] (Eq. 2b)
Where:
[tex]\mu_{s}[/tex] - Static coefficient of friction, dimensionless.
[tex]m[/tex] - Mass of the car, measured in kilograms.
[tex]g[/tex] - Gravitational acceleration, measured in meters per square second.
Then, we substitute normal force and simplify the resulting expression:
[tex]\mu_{s}\cdot m\cdot g \cdot \cos \theta -m\cdot g \cdot \sin \theta = 0[/tex]
[tex]\mu_{s} = \tan \theta[/tex]
[tex]\theta = \tan^{-1}\mu_{s}[/tex] (Eq. 3)
Based on such result, we can conclude that if driveway inclination is greater than value reported, then car shall not be safe in case of parking. Our reference angle is: ([tex]\mu_{s} = 0.15[/tex])
[tex]\theta = \tan^{-1} 0.15[/tex]
[tex]\theta \approx 8.531^{\circ}[/tex]
By direct comparison, we find that driveway across the street ([tex]\theta = 6^{\circ}[/tex]) is the only choice for a safe parking.
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Answer:16 J
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Answer:
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Answer:
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Answer:
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Answer:
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Answer:
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Explanation:
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Answer:
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Answer:
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Answer:
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Answer:
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Answer:
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Answer:
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Answer:
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Answers
Answer:
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Explanation:
Horizontal Motion
When an object is thrown horizontally with a speed v from a height h, the range or maximum horizontal distance traveled by the object can be calculated as follows:
[tex]\displaystyle d=v\cdot\sqrt{\frac {2h}{g}}[/tex]
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[tex]\displaystyle v=d\cdot\sqrt{\frac {g}{2h}}[/tex]
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[tex]\displaystyle t=\frac{d}{v}[/tex]
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[tex]\displaystyle v=3\cdot\sqrt{\frac {9.8}{2\cdot 5}}[/tex]
[tex]\displaystyle v=3\cdot\sqrt{\frac {9.8}{10}}[/tex]
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[tex]v=2.97\ m/s[/tex]
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[tex]\displaystyle t=\sqrt{\frac {2h}{g}}=\sqrt{\frac {10}{9.8}}=1.01\ sec[/tex]