In a ocean full of storms
A new wave was born
Deep into that darkness flooding
Suddenly, I heard some pummeling
By the grave I saw the winds
And the sun just shined
Answer:
a friendly face that comes with waves,
the waves of all the memorial days,
and with these days we smile with pride,
as for the waves we used to ride,
given up the day has passed,
how it went away like an hour glass,
as if we knew the world was right,
just like the waves, oh so bright,
the time has come the days have passed,
the waves ashore the waves alast,
as if the friendly face was right,
the waves that rode, oh goodnight.
Which change will always result in an increase in the gravitational force between two objects?
O increasing the masses of the objects and increasing the distance between the objects
O decreasing the masses of the objects and decreasing the distance between the objects
O increasing the masses of the objects and decreasing the distance between the objects
• decreasing the masses of the objects and increasing the distance between the objects
Answer:
increasing the masses of the objects and decreasing the distance between the objects
Explanation:
find the rms speed of a sample of oxygen at 30° C and having a molar mass of 16 g/mol.
At 30°C, the rms speed of a sample of oxygen with a molar mass of 16 g/mol is approximately 482.34 m/s.
The root mean square (rms) speed of a gas molecule is a measure of the average speed of the gas particles in a sample. It can be calculated using the formula:
vrms = √(3kT/m)
Where:
vrms is the rms speed
k is the Boltzmann constant (1.38 x 10^-23 J/K)
T is the temperature in Kelvin
m is the molar mass of the gas in kilograms
To calculate the rms speed of oxygen at 30°C (303 Kelvin) with a molar mass of 16 g/mol, we need to convert the molar mass to kilograms by dividing it by 1000:
m = 16 g/mol = 0.016 kg/mol
Substituting the values into the formula, we have:
vrms = √((3 * 1.38 x 10^-23 J/K * 303 K) / (0.016 kg/mol))
Calculating this expression yields the rms speed of the oxygen sample:
vrms ≈ 482.34 m/s
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a glass bottle full of mercury has mass of 500g on being heated through 35 degree celsuis,2.43g of mercury are expelled .calculate the mass of the mercury remaining in the bottle
Answer:
The mass of the mercury remaining in the bottle is 497.57 grams.
Explanation:
The mass of the mercury remaining in the bottle is found by subtracting the mass expeled due to heating from initial mass inside the bottle. That is:
[tex]m_{f} = m_{o}-\Delta m[/tex] (1)
Where:
[tex]m_{o}[/tex] - Initial mass, in grams.
[tex]\Delta m[/tex] - Mass expelled due to heating, in grams.
[tex]m_{f}[/tex] - Final mass, in grams.
If we know that [tex]m_{o} = 500\,g[/tex] and [tex]\Delta m = 2.43\,g[/tex], then the mass of the mercury remaining in the bottle is:
[tex]m_{f} = m_{o}-\Delta m[/tex]
[tex]m_{f} = 497.57\,g[/tex]
The mass of the mercury remaining in the bottle is 497.57 grams.
A 3.5 kg object gains 76 J of potential energy as it is lifted vertically. Find the new height of the object?
Answer:
1.72 m
Explanation:
Potential energy = mgh, where m is mass, g is acceleration due to gravity (9.8), and h is height
76 = (3.5)(9.8)h
76=44.1h
h=1.72335600907 ≈1.72 m
Answer:
:r
Explanation:r
Tobnbv346468this Ishmael
If a reflected ray is 55 degrees from the normal line, they what is the angle of the
incident ray from normal?
Answer:
xplanation:
Angle of reflection is measured between the incident ray and the angle which it makes with the normal at the point where incident ray strikes the mirror surface.
Further on reflection, it makes the same angle i.e. angle of reflection is equal to angle of reflection.
Hence, as angle of incidence is 55∘ angle of reflection too is 55∘ and the angle between the incident ray and the reflected ray is 55∘+55∘=110∘
b. Calculate the kinetic energy of the car for group A.
Answer: Kinectic Energy (KE) is equal to half of an object's mass (1/2*m) multiplied by the velocity.
Explanation: If an object with a mass of 10 kg (m=10 kg) is moving at a velocity of 5 meters per second (v=5 m/s), the kinetic energy is equal to 125 Joules, or (1/2* 10 kg) * 5 m/s^2.
a lens with f = 50.0 cm is held 55.0 cm from an object. what is the image distance? (unit = cm)
Answer: 550 cm
Explanation:
Original equation: 1/f= 1/do + 1/di.
F=50.0 cm, and do=55.0.
Since we don't have di, we'll have to subtract do to the other side, making the equation: 1/f - 1/do= 1/di.
Doing the math, 1/f - 1/do is 0.0018181818
Then to get di by itself, you multiply both sides by di. Then you divide by 0.0018181818 to get di by itself. You then get: di= 1/0.0018181818
At that point, you just divide 1 by 0.0018181818, which will give you 550 cm
There could be simpler way, but that is just what I did to get the answer. Answer was right on Acellus
The Solar System formed...
from the collision between two galaxies.
from the merging of two smaller stars.
directly from the leftover gas of the big bang.
from the gravitational collapse of an interstellar cloud of gas.
Answer:The answer is because of the gravity and the mass formed in the sun, the magnetic field reacts to it and leaves a fault on earth. wind goes by the earth and procides to be ok
Explanation:that is it
Great Sand Dunes National Park in Colorado is famous for its giant sand dunes. Sand dunes are landforms that are found in deserts and on beaches. Visitors to the park can surf down the dunes on sleds or boards.
An image of sand dunes in front of a mountain and behind a body of water and grass.
Which process causes the shape of these giant dunes?
A. deposition
B. erosion
C. weathering
D. waves
Answer:
Wind deposits sand into a small mound. So the answer is Deposition
Which energy source can be found on the electromagnetic spectrum? A) sound energy B) chemical energy UV light energy D mechanical energy
Please respond to this for 15 points. Please don’t put in a link.
Answer:
e. Combustion
Explanation:
In Combustion reaction, a substance reacts with oxygen from the air and resultant product is that it releases carbon dioxide and water.
Here,
2C2H6 is the substance that reacted with 7O2 (Oxygen) to release 4CO2 (Carbon Dioxide) and 6H2O (Water).
Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500
Answer:
8 kV
Explanation:
Here is the complete question
Assume a device is designed to obtain a large potential difference by first charging a bank of capacitors connected in parallel and then activating a switch arrangement that in effect disconnects the capacitors from the charging source and from each other and reconnects them all in a series arrangement. The group of charged capacitors is then discharged in series. What is the maximum potential difference that can be obtained in this manner by using ten 500 μF capacitors and an 800−V charging source?
Solution
Since the capacitors are initially connected in parallel, the same voltage of 800 V is applied to each capacitor. The charge on each capacitor Q = CV where C = capacitance = 500 μF and V = voltage = 800 V
So, Q = CV
= 500 × 10⁻⁶ F × 800 V
= 400000 × 10⁻⁶ C
= 0.4 C
Now, when the capacitors are connected in series and the voltage disconnected, the voltage across is capacitor is gotten from Q = CV
V = Q/C
= 0.4 C/500 × 10⁻⁶ F
= 0.0008 × 10⁶ V
= 800 V
The total voltage obtained across the ten capacitors is thus V' = 10V (the voltages are summed up since the capacitors are in series)
= 10 × 800 V
= 8000 V
= 8 kV
A mass of 3 kg stretches a spring 9m. The mass is acted on by an external force of 2 AND. The Mass moves in a medium that imparts a viscous force of 1 N when the speed of the mass is 4m/sec The mass is pulled down 8 cm below its equilibrium position, and then set in motion inthe upward direction with a velocity of 5 m/sec. State the initial value problem describing the motion of the mass. DO NOT SOLVE.
Answer:
k y -b [tex]\frac{dy}{dt}[/tex]dy / dt = m [tex]\frac{d^2y}{dt^2}[/tex]
give us some initial conditions
1) friction force fr = 1N when v = 4m / s
2) an initial displacement of x = 0.08 m for t=0 s
Explanation:
In this exercise, you are asked to state the problem you are posing. We are going to find the equation of motion for this exercise. Let's start with Newton's second law
Let's set a reference system with the y-axis in a vertical and positive direction upwards.
We have four forces: an external downward force, negative in sign, the but that goes down and is negative, the Hook force that goes up and is positive and the friction force that opposes the movement, in this case it goes down being negative
let's write Newton's second law
F_e -F -fr - W = m a
where
F_e = -kDy = - k y
fr = - b v = -b dy / dt
W = mg
we substitute for the specific case, that is, using the signs
k y -b [tex]\frac{dy}{dt}[/tex] - m g - F = m [tex]\frac{d^2y}{dt^2}[/tex]
In the initial condition of the problem, before starting the movement, the friction force is zero and the acceleration is also zero
k y - m g - F = 0
from this equation you can find the spring constant, y= 9m and F=2 N
It is not clear if when the movement starts this external force becomes zero, but since it balances the weight we can eliminate the two forces that have the same magnitude and opposite direction, so the equation remains
k y - b [tex]\frac{dy}{dt}[/tex]dy / dt = m [tex]\frac{d^2y}{dt^2}[/tex]
give us some initial conditions
1) friction force fr = 1N when v = 4m / s
2) an initial displacement of x = 0.08 m for t=0 s
therefore, to initiate the movement, a small external force F 'is applied that moves the system to a new equilibrium position and this small force F' is made zero, thus initiating an oscillatory movement, described by the equation.
k y -b [tex]\frac{dy}{dt}[/tex]dy / dt = m [tex]\frac{d^2y}{dt^2}[/tex]
This is a differential equation of the second degree, therefore it needs two initial conditions for its complete solution
The initial amount of displacement corresponds to the amplitude of movement A = 0.08 m
A dog runs 51 m west to fetch a ball and brings it back only 27 m before stopping.
The total displacement of the dog is:
how much energy is possessed by 1 mole of nitrogen atoms moving at 35.0 m/s ?
1 mole of nitrogen atoms moving at 35.0 m/s possesses approximately 27.8 joules of energy.
To calculate the energy possessed by 1 mole of nitrogen atoms moving at 35.0 m/s, we need to consider both the kinetic energy and the molecular mass of nitrogen.
The kinetic energy (KE) of an object is given by the equation KE = 1/2 * m * v^2, where m is the mass and v is the velocity.
The molar mass of nitrogen (N₂) is approximately 28.0134 g/mol, which can be converted to kilograms by dividing by Avogadro's number (6.022 × 10^23). This gives us a mass of approximately 4.65 × 10^(-26) kg for one nitrogen atom.
Plugging in the values, we have KE = 1/2 * (4.65 × 10^(-26) kg) * (35.0 m/s)^2.
Evaluating the equation, we find that the kinetic energy possessed by one nitrogen atom is approximately 4.62 × 10^(-23) joules.
Since we are considering 1 mole of nitrogen atoms, we need to multiply this value by Avogadro's number to get the energy possessed by 1 mole. Avogadro's number is 6.022 × 10^23, so the total energy is approximately 2.78 × 10^1 joules, or 27.8 J.
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here is a clever kitchen gadget for drying lettuce leaves after you wash them. It consists of a cylindrical container mounted so that it can be rotated about its axis by turning a hand crank. The outer wall of the cylinder is perforated with small holes. You put the wet leaves in the container and turn the crank to spin off the water. The radius of the container is 10.7 cm. When the cylinder is rotating at 1.65 revolutions per second, what is the magnitude of the centripetal acceleration at the outer wall
Answer:
11.5 m/s²
Explanation:
The centripetal acceleration, a = rω² where r = radius of cylinder = 10.7 cm = 0.107 m and ω = angular speed = 2πN where N = number of revolutions per second = 1.65 rev/s
So, a = rω²
a = r(2πN)²
a = 4π²rN²
substituting the values of the variables into the equation, we have
a = 4π²rN²
a = 4π²(0.107 m)(1.65 rev/s)²
a = 4π²(0.107 m)(2.7225 rev²/s)²
a = 4π² × 0.2913075 mrev²/s)²
a = 11.5 m/s²
1. What is the heat energy when 114.32g of water ( c = 4.18 J/g °C) at 14.85°C is raised to
18.00°C?
At which point is there the most potential energy? At which point is there the most kinetic energy?
A. Potential energy A; Kinetic energy B
B. Potential energy B; Kinetic energy D
C. Potential energy A; Kinetic energy D
D. Potential energy C; Kinetic energy D
Answer:
The cart mark (a) has the most potential energy and the cart marked (b) has the most kinetic energy
what is an example of vaporization?
Answer:
just search it up you'll get ur answer
What forces are used to jump over a wall?
Answer:
Potential and kinetic
Explanation:
A mass weighing 24 pounds, attached to the end of a spring, stretches it 4 inches. Initially, the mass is released from rest from a point 4 inches above the equilibrium position. Find the equation of motion. (Use g
Answer:
The equation of motion is [tex]x(t)=-[/tex][tex]\frac{1}{3} cos4\sqrt{6t}[/tex]
Explanation:
Lets calculate
The weight attached to the spring is 24 pounds
Acceleration due to gravity is [tex]32ft/s^2[/tex]
Assume x , is spring stretched length is ,4 inches
Converting the length inches into feet [tex]x=\frac{4}{12} =\frac{1}{3}feet[/tex]
The weight (W=mg) is balanced by restoring force ks at equilibrium position
mg=kx
[tex]W=kx[/tex] ⇒ [tex]k=\frac{W}{x}[/tex]
The spring constant , [tex]k=\frac{24}{1/3}[/tex]
= 72
If the mass is displaced from its equilibrium position by an amount x, then the differential equation is
[tex]m\frac{d^2x}{dt} +kx=0[/tex]
[tex]\frac{3}{4} \frac{d^2x}{dt} +72x=0[/tex]
[tex]\frac{d^2x}{dt} +96x=0[/tex]
Auxiliary equation is, [tex]m^2+96=0[/tex]
[tex]m=\sqrt{-96}[/tex]
=[tex]\frac{+}{} i4\sqrt{6}[/tex]
Thus , the solution is [tex]x(t)=c_1cos4\sqrt{6t}+c_2sin4\sqrt{6t}[/tex]
[tex]x'(t)=-4\sqrt{6c_1} sin4\sqrt{6t}+c_2[/tex] [tex]4\sqrt{6}[/tex] [tex]cos4\sqrt{6t}[/tex]
The mass is released from the rest x'(0) = 0
[tex]=-4\sqrt{6c_1} sin4\sqrt{6(0)}+c_2[/tex] [tex]4\sqrt{6}[/tex] [tex]cos4\sqrt{6(0)}[/tex] =0
[tex]c_2[/tex] [tex]4\sqrt{6} =0[/tex]
[tex]c_2=0[/tex]
Therefore , [tex]x(t)=c_1[/tex] [tex]cos 4\sqrt{6t}[/tex]
Since , the mass is released from the rest from 4 inches
[tex]x(0)= -4[/tex] inches
[tex]c_1 cos 4\sqrt{6(0)} =-\frac{4}{12}[/tex] feet
[tex]c_1=-\frac{1}{3}[/tex] feet
Therefore , the equation of motion is [tex]-\frac{1}{3} cos4\sqrt{6t}[/tex]
A wire is oriented along the x-axis. It is connected to two batteries, and a conventional current of 2.6 A runs through the wire, in the x direction. Along 0.17 m of the length of the wire there is a magnetic field of 0.52 tesla in the y direction, due to a large magnet nearby. At other locations in the circuit, the magnetic field due to external sources is negligible. What is the magnitude of the magnetic force on the wire
Answer:
the magnitude of the magnetic force on the wire is 0.2298 N
Explanation:
Given the data in the question;
we know that, the magnitude of magnetic force is given as;
|F[tex]_{mg}^>[/tex] | = I([tex]B^>[/tex] × [tex]L^>[/tex] )
given that
I = 2.6 A
[tex]B^>[/tex] = 0.17
[tex]L^>[/tex] = 0.52
so we substitute
|F[tex]_{mg}^>[/tex] | = 2.6( 0.17i" × 0.52j" )
|F[tex]_{mg}^>[/tex] | = 0.2298 N
Therefore, the magnitude of the magnetic force on the wire is 0.2298 N
A dog runs 51 m west to fetch a ball and brings it back only 27 m before stopping.
The total displacement of the dog is:
Daryl ties a rope to a brick and lifts the brick straight up. The free-body
diagram below shows the brick when it is suspended above the ground.
Force 1
Force 2
What is force 1 in this diagram?
O A. Friction
OB. Tension
O C. Normal force
O D. Weight
The force 1 is tension force.
To find the correct statement among all the options, we need to know more about friction, tension, normal force and weight.
What is friction?Friction force is found between two surfaces when one is kept or moved on another surface.It is directed opposite to the direction of motion.What is tension force?When any object is hanged by an thread or rope, that object exerts a force on that rope. This force is called as tensional force.It's directed from along the rope towards the point of hanging.What is normal force?When an object is kept on a surface, the surface exerts a force on the object to oppose the weight of the object which is the normal force.It's perpendicular to the surface that an object contacts.What is weight?Weight is the gravitational force exerted by earth on that object. It's always directed towards the center of the earth.Thus, we can conclude that the correct option is (B).
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Grade 10 My smart Physics people help me with this review question please
Answer:
sorry I am not confident you the answer
A ball weighs 5.7 N on Earth. What is its mass?
Answer:
55.897905
Explanation:
1 Newton in Earth gravity is the equivalent weight of 1/9.80665 kg on Earth
9.80665 times 5.7=55.897905
Brainliest?
2.- a person weighing 70 kg travels at 2m / s. What is the value of his kinetic energy?
Answer:18 watts
Explanation:i just got this question trust me
\
Someone help me like please thank you
Chris used a non plane mirror to check out an box resting on a shelf. He wanted to find
the focal length of the mirror. The image of the box was located 15 cm behind the mirror
and the box was placed 19 cm from the mirror.
Chris used a non-plane mirror to check out a box resting on a shelf, the focal length of the mirror is mathematically given as
f=8.38cm
What is the focal length of the mirror?Question Parameter(s):
The image of the box was located 15 cm behind the mirror
and the box was placed 19 cm from the mirror.
Generally, the equation for the focal length is mathematically given as
1/f=1/u+1/v
Therefore
1/f=1/15+1/19
f=8.3823529cm
In conclusion, the focal length of the mirror
f=8.3823529cm
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A wooden cylinder (in the form of a thin disk) of uniform density and a steel hoop are set side by side, released from rest at the same moment, and roll down an inclined plane towards a wall at the bottom. The cylinder has a larger radius than the hoop, but the hoop weighs more than the cylinder.
Required:
Who reaches the bottom first and why?
Answer:
a. The wooden cylinder b. the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.
Explanation:
a. Who reaches the bottom first
The kinetic energy of the objects is given by
K = 1/2mv² + 1/2Iω² where m = mass of object, v = velocity of object, I = moment of inertia and ω = angular velocity = v/r where r = radius of object
For the wooden cylinder, I = mr²/2 where m = mass of wooden cylinder and r = radius of wooden cylinder and v = velocity of wooden cylinder
So, its kinetic energy, K = 1/2mv² + 1/2(mr²/2)(v/r)²
K = 1/2mv² + 1/4mv²
K = 3mv²/4
For the steel hoop, I' = mr'² where m' = mass of steel hoop and r' = radius of steel hoop and v' = velocity of steel hoop
So, its kinetic energy, K' = 1/2m'v'² + 1/2(m'r'²)(v'/r')²
K' = 1/2m'v'² + 1/2m'v'²
K' = m'v'²
Since both kinetic energies are the same, since the drop from the same height,
K = K'
3mv²/4 = m'v'²
v²/v'² = 4m/3m'
v²/v'² = 4/3(m/m')
v/v' = √[4/3(m/m')]
Since the hoop weighs more than the cylinder m/m' < 1 and 4/3(m/m') < 4/3 ⇒ √ [4/3(m/m')] < √4/3 ⇒ v/v' < 1.16 ⇒ v'/v > 1/1.16 ⇒ v'/v > 0.866. Since 0.866 < 1, it implies v' < v.
Since v' = speed of steel hoop < v = speed of wooden cylinder, the wooden cylinder reaches the bottom first.
b. Why
Since the kinetic energy, K = translational + rotational
We find the translational kinetic energy of each object.
For the wooden cylinder,
K = K₀ + 1/2Iω² where K₀ = translational kinetic energy of wooden cylinder
K - 1/2Iω² = K₀
3/4mv² - 1/2(mr²/2)(v/r)² = K₀
3/4mv² - 1/4mv² = K₀
K₀ = 1/2mv²
For the steel hoop,
K' = K₁ + 1/2I'ω'² where K₁ = translational kinetic energy of steel hoop
K' - 1/2I'ω'² = K₁
m'v'² - 1/2(m'r'²)(v'/r')² = K₁
m'v'² - 1/2m'v'² = K₁
K₁ = 1/2m'v'²
So, K₀/K₁ = 1/2mv²÷1/2m'v'² = mv²/m'v'² = (m/m')(v²/v'²) = (m/m')4/3(m/m') = 4/3(m/m')².
Since (m/m') < 1 ⇒ (m/m')² < 1 ⇒ 4/3(m/m')² < 4/3 ⇒ K₀/K₁ < 1.33 ⇒ K₀ > K₁
So, the kinetic energy of the wooden cylinder is greater than that of the steel hoop.
So, the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.
a. The wooden cylinder b. the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.
What is Kinetic energy?
The energy of the body due to its movement in a particular direction under the influence of a force like a free-falling body due to gravitaional force is called Kinetic energy.
The kinetic energy of the objects is given by
[tex]K = \dfrac{1}{2}mv^2 + \dfrac{1}{2}Iw^2[/tex]
where
m = mass of object,
v = velocity of object,
I = moment of inertia and
ω = angular velocity = v/r where r = radius of object
For the wooden cylinder, I = mr²/2 where m = mass of wooden cylinder and r = radius of wooden cylinder and v = velocity of wooden cylinder
So, its kinetic energy,
[tex]K = \dfrac{1}{2}mv^2 + \dfrac{1}{2}(\dfrac{mr^2}{2})\dfrac{v}{r}^2[/tex]
[tex]K = \dfrac{3mv^2}{4}[/tex]
For the steel hoop,
I' = mr'²
where
m' = mass of steel hoop and
r' = radius of steel hoop and
v' = velocity of steel hoop
So, its kinetic energy,
[tex]K' = \dfrac{1}{2}m'v'^2 + \dfrac{1}{2}(m'r'^2)\dfrac{v'}{r'}^2[/tex]
[tex]K' = \dfrac{1}{2}m'v'^2 + \dfrac{1}{2}m'v'^2[/tex]
K' = m'v'²
Since both kinetic energies are the same, since the drop from the same height,
K = K'
[tex]\dfrac{3mv^2}{4 }= m'v'^2[/tex]
[tex]\dfrac{v^2}{v'^2} =\dfrac{ 4m}{3m'}[/tex]
[tex]\dfrac{v^2}{v'^2} = \dfrac{4}{3}(\dfrac{m}{m'})[/tex]
[tex]\dfrac{v}{v'} = \sqrt{[\dfrac{4}{3}(\dfrac{m}{m'})][/tex]
Since the hoop weighs more than the cylinder m/m' < 1 and 4/3(m/m') < 4/3 ⇒ √ [4/3(m/m')] < √4/3 ⇒ v/v' < 1.16 ⇒ v'/v > 1/1.16 ⇒ v'/v > 0.866. Since 0.866 < 1, it implies v' < v.
Since v' = speed of steel hoop < v = speed of wooden cylinder, the wooden cylinder reaches the bottom first.
(b) Since the kinetic energy, K = translational + rotational
We find the translational kinetic energy of each object.
For the wooden cylinder,
[tex]K = K_o + \dfrac{1}{2}Iw^2[/tex]
where
K₀ = translational kinetic energy of wooden cylinder
[tex]K - \dfrac{1}{2}Iw^2 = K_o[/tex]
[tex]\dfrac{3}{4}mv^2 - \dfrac{1}{2}(\dfrac{mr^2}{2})(\dfrac{v}{r})^2 = K_a[/tex]
[tex]\dfrac{3}{4}mv^2 - \dfrac{1}{4}mv^2 = K_o[/tex]
[tex]K_o = \dfrac{1}{2}mv^2[/tex]
For the steel hoop,
[tex]K' = K_1 + \dfrac{1}{2}I'w'^2[/tex]
where
K₁ = translational kinetic energy of steel hoop
[tex]K' - \dfrac{1}{2}I'w'^2 = K_1[/tex]
[tex]m'v'^2 - \dfrac{1}{2}(m'r'^2)(\dfrac{v'}{r'})^2 = K_1[/tex]
[tex]m'v'^2 - \dfrac{1}{2}m'v'^2 = K_1[/tex]
[tex]K_1= \dfrac{1}{2}m'v'^2[/tex]
So, K₀/K₁ = 1/2mv²÷1/2m'v'² = mv²/m'v'² = (m/m')(v²/v'²) = (m/m')4/3(m/m') = 4/3(m/m')².
Since (m/m') < 1 ⇒ (m/m')² < 1 ⇒ 4/3(m/m')² < 4/3 ⇒ K₀/K₁ < 1.33 ⇒ K₀ > K₁
So, the kinetic energy of the wooden cylinder is greater than that of the steel hoop.
So, the wooden cylinder reaches the bottom first because its translational kinetic energy is greater.
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