Answer:
a
[tex]v_r =8.65 \ ft/s [/tex]
b
[tex] W_{1-2} = 3.24 \ ft \cdot lb[/tex]
Explanation:
From the question we are told that
The mass of the ball is [tex]m = 4 \ lb[/tex]
The radius is [tex]r= 3 \ ft[/tex]
The speed is [tex]v_B_1 = 4.8 \ ft /s[/tex]
The speed of the attached cord is [tex]v_c =2.2 \ ft[/tex]
The position that is been considered is [tex]r_1 = 2 \ ft[/tex]h
Generally according to the law of angular momentum conservation
[tex]L_a = L_b[/tex]
Here [tex]L_a[/tex] is the initial momentum of the ball which is mathematically represented as
[tex]L_a = m* v_B_1 * r[/tex]
while
[tex]L_b[/tex] is the momentum of the ball at r = 2 ft which is mathematically represented as
[tex]L_a = m* v_B_2 * r_1[/tex]
So
[tex]m* v_B_1 * r = m* v_B_2 * r_1[/tex]
=> [tex] 4.8 * 3 = v_B_2 * 2[/tex]
=> [tex] v_B_2 = 7.2 \ ft/s [/tex]
Generally the resultant velocity of the ball is
[tex]v_r = \sqrt{v_B_2^2 + v_B_1^2 }[/tex]
=> [tex]v_r = \sqrt{7.2^2 + 4.8^2 }[/tex]
=> [tex]v_r =8.65 \ ft/s [/tex]
Generally according to equation for principle of work and energy we have that
[tex]K_1 + \sum W_{1-2} = K_2[/tex]
Here [tex]K_1[/tex] is the initial kinetic energy of the ball which is mathematically represented as
[tex]K_1 = \frac{1}{2} * m* v_B_1^2[/tex]
While [tex]\sum W_{1-2}[/tex] is the sum of the total workdone by the ball
and [tex]K_2[/tex] is the final kinetic energy of the ball which is mathematically represented as [tex]K_2 = \frac{1}{2} * m* v_r^2[/tex]
So
[tex]\sum W_{1-2} = \frac{1}{2} * m (v_r^2 - v_B_1^2)[/tex]
Here m is the mass which is mathematically represented as
[tex]m = \frac{W}{g}[/tex] here W is the weight in lb and g is the acceleration due to gravity which is [tex]g = 32 \ ft/s^2[/tex]
So
[tex]\sum W_{1-2} = \frac{1}{2} * \frac{4}{32} * (8.65^2 - 4.8^2)[/tex]
=> [tex] W_{1-2} = 3.24 \ ft \cdot lb[/tex]
A toy car has a mass of 3 Kg is pushed horizontally with a force of 3 N. Friction is so small it can be ignored.
What is the acceleration?
How fast is it going after 8 s if it starts from rest?
How far did it go?
Explanation:
Mass = 3kgForce=3NAcceleration= Force/mass=3/3=1 m/s²v=u+atv=0+1(8) (At rest)v=8m/sDisplacement = velocity× timeDisplacement=8×8 mDisplacement=64mDoes The stretched string of an archer’s bow has kinetic energy in it.
Answer:
The ball has potential energy just like the Longbow has when it is stretched. The other aspect of energy that the bow portrays is kinetic energy. Kinetic energy is the energy of motion. When the string is released, the potential energy is converted to kinetic energy in movement of the arrow
Explanation:
Answer:
It has potential energy not kinetic
Explanation:
A photovoltaic panel of dimension 2 m × 4 m is installed on the roof of a home. The panel is irradiated with a solar flux of GS = 700 W/m2 , oriented normal to the top panel surface. The absorptivity of the panel to the solar irradiation is αS = 0.83, and the efficiency of conversion of the absorbed flux to electrical power is η = P/αSGSA = 0.553 − 0.001 K−1 Tp, where Tp is the panel temperature expressed in kelvins and A is the solar panel area. Determine the electrical power generated for (a) a still summer day, in which Tsur = T[infinity] = 35°C, h = 10 W/m2 ⋅K, and (b) a breezy winter day, for which Tsur = T[infinity] = −15°C, h = 30 W/m2 ⋅K. The panel emissivity is ε = 0.90
Answer:
(a) the electrical power generated for still summer day is 1013.032 W
(b)the electrical power generated for a breezy winter day is 1270.763 W
Explanation:
Given;
Area of panel = 2 m × 4 m, = 8m²
solar flux GS = 700 W/m²
absorptivity of the panel, αS = 0.83
efficiency of conversion, η = P/αSGSA = 0.553 − 0.001 K⁻¹ Tp
panel emissivity , ε = 0.90
Apply energy balance equation to determine he electrical power generated;
transferred energy + generated energy = 0
(radiation + convection) + generated energy = 0
[tex][\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - \eta \alpha_s G_s = 0[/tex]
[tex][\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - (0.553-0.001T_p)\alpha_s G_s[/tex]
(a) the electrical power generated for still summer day
[tex]T_s = T_{\infty} = 35 ^oC = 308 \ k[/tex]
[tex][0.83*700-0.9*5.67*10^{-8}(T_p_1^4-308^4)]-10(T_p_1-308) - (0.553-0.001T_p_1)0.83*700 = 0\\\\3798.94-5.103*10^{-8}T_p_1^4 - 9.419T_p_1 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_1\\\\T_p_1 = 335.05 \ k[/tex]
[tex]P = \eta \alpha_s G_s A = (0.553-0.001 T_p_1)\alpha_s G_s A \\\\P = (0.553-0.001 *335.05)0.83*700*8 \\\\P = 1013.032 \ W[/tex]
(b)the electrical power generated for a breezy winter day
[tex]T_s = T_{\infty} = -15 ^oC = 258 \ k[/tex]
[tex][0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k[/tex]
[tex]P = \eta \alpha_s G_s A = (0.553-0.001 T_p_2)\alpha_s G_s A \\\\P = (0.553-0.001 *279.6)0.83*700*8 \\\\P = 1270.763 \ W[/tex]
An 80 N bowling ball has an apparent weight of 60 N when completely submerged in water. What is the average density of the bowling ball
Answer:
V * ρB = WB volume of ball * density of ball = weight of ball
V * ρW = Ww volume of ball * density of water = buoyant force
ρB / ρW = WB / Ww = 80 / 20 = 4 water provides 20 N of buoyant force
ρB = 4 ρW = 4 gm/cm^3
ρW = 1000 kg / m^3 = 1 gm / cm^3
(1 gm/cm^3) = .001 kg / .000001 m^3 = 1000 kg/^3
ρW = 1 gm / cm^3
multiply by 9.8 or 980 to get weight densities
In this case
4 * 1000 kg/m^3 * 9.8 m/s^2 = 39200 N /m^3 weight density of ball
Please help me!!! I'm so lost!!
A gold wire carries a current of 76.2 mA.
(a) Find the number of electrons that flow past a given point in the wire in 10.4 minutes.
[Answer] electrons
b)In what direction do the electrons travel with respect to the current?
o opposite direction
o same direction
o The magnitude is zero.
#a
Time=10.4min=10.4(60s)=624sCurrent=76.2mA=76.2×10^{-3}A=INow
Charge=Q[tex]\\ \tt\Rrightarrow Q=\dfrac{I}{t}[/tex]
[tex]\\ \tt\Rrightarrow Q=\dfrac{76.2\times 10^{-3}}{624}[/tex]
[tex]\\ \tt\Rrightarrow Q=0.122\times 10^{-3}[/tex]
[tex]\\ \tt\Rrightarrow Q=122\times 10^{-6}C[/tex]
[tex]\\ \tt\Rrightarrow Q=122\mu C[/tex]
#b
Same directionWhat happens when one side of the tug of war rope has a large net force than the other
Answer:
When one side of the tug-of-war rope has a larger net force than the other side then the basket in the middle will be pulled to whichever side has a larger net force.
Explanation:
When one side of the tug-of-war rope has a larger net force than the other side then the basket in the middle will be pulled to whichever side has a larger net force.
Will give brainlist and many points.
What happens when a plane passes the speed of sound?
How does it sound?
Is the sound super high since the wavelength is super low or maby even non existent?
How does it look and why does it look that way?
How will it affect people on ground?
Answer as many of the questions as possible. You can also send links beside your answers for me to read.
Please make your answers understandable with reasoning backing it.
Answer:
They create enormous amounts of sound energy, much like explosion. When an aircraft passes through air, it creates a series of pressure waves just like the waves created by the boat. As the speed of aircraft increases these waves are forced to compress.
Answer:
What happens when a plane passes the speed of sound?
Answer - You hear a loud BOOM then you cant hear it
How does it sound?
Answer - there is no sound
Is the sound super high since the wavelength is super low or maybe even non-existent?
Answer - When its about to break the sound barrier it makes a high pitched sound
How does it look and why does it look that way?
Answer - Its air you cant see it ;-;
How will it affect people on ground?
you will hear the sound and it might hurt but nothing else
Explanation:
A transformer has 10,000 windings on the primary side and 5,000 windings on the
secondary side. 100 volts is applied to the primary side. What is the anticipated voltage
on the secondary side? Is this a step-up or step down transformer?
Answer:
50 V; step-down transformer
Explanation:
10,000 : 5,000 = 100 : V
V = 50 V
From 100 V change to 50 V, so it is a step-down transformer
Take another look at line 1. Suppose that you use distance and time between any pair of neighboring dots to calculate speed:
speed = distance ÷ time
Will this speed be the same or different from the average speed you calculated in part F? Why?
Answer:
The time of travel and the distance between neighboring dots is always the same. So, the speed between neighboring dots is constant. This speed will be the same as the average speed in part F. hope it helps!
Why does Venus appear so bright to our eyes?
it is much larger than either Mercury or Mars
it gets closer to us than does any other planet
All of the items listed helps us understand why Venus is so bright.
its sulfuric acid cloud cover reflects almost 60% of the sunlight
it lies closer to the Sun than we do, so sunlight is more intense there
Answer:
GIVE ME BRAINLIEST!!!!!!!!!!
Why does Venus appear so bright to our eyes?
Venus is so bright because its thick clouds reflect most of the sunlight that reaches it (about 70%) back into space, and because it is the closest planet to Earth. Venus can often be seen within a few hours after sunset or before sunrise as the brightest object in the sky (other than the moon).
it is much larger than either Mercury or Mars
Mars, the fourth planet from the sun, is the second smallest planet in the solar system; only Mercury is smaller.
it gets closer to us than does any other planet
In other words, Mercury is closer to Earth, on average, than Venus is because it orbits the Sun more closely. Further, Mercury is the closest neighbor, on average, to each of the other seven planets in the solar system.
All of the items listed helps us understand why Venus is so bright.
Venus is so bright because its thick clouds reflect most of the sunlight that reaches it (about 70%) back into space, and because it is the closest planet to Earth. Venus can often be seen within a few hours after sunset or before sunrise as the brightest object in the sky (other than the moon).
its sulfuric acid cloud cover reflects almost 60% of the sunlight
Clouds. Venusian clouds are thick and are composed mainly (75–96%) of sulfuric acid droplets. These clouds obscure the surface of Venus from optical imaging, and reflect about 75% of the sunlight that falls on them. The geometric albedo, a common measure of reflectivity, is the highest of any planet in the Solar System
it lies closer to the Sun than we do, so sunlight is more intense there
When it is winter in the northern half of Earth, the southern hemisphere, tilted toward our Sun, has summer. During fall and spring, some locations on Earth experience similar, milder, condition.
frances drops a .75kg ball off a balcony sandra standing on the ground below the balcony, catches the ball by exerting a 15.5N force upwards on the ball. determine the acceleration of the ball while aisha is catching it
show diagrams and sketch and explain why
Answer:
[tex]a=-10.86\ m/s^2[/tex]
Explanation:
Given that,
Mass of a ball, m = 0.75 kg
It is dropped off a balcony Sandra standing on the ground below the balcony, catches the ball by exerting a 15.5N force upwards on the ball.
We need to find the acceleration of the ball while Aisha is catching it.
The net force acting on the ball is given by :
F = ma
In this case, when the ball is accelerating downward,
mg-F=ma
a is the acceleration of the ball and g is the acceleration due to gravity
So,
[tex]a=\dfrac{mg-F}{m}\\\\a=\dfrac{0.75\times 9.81-15.5}{0.75}\\\\a=-10.86\ m/s^2[/tex]
A negative sign shows that the ball is accelerating in a downward direction.
An atom has 5 protons, 4 neutrons, and 6 electrons. Which one of the following could be its isotope
Answer:
9 :)
Explanation:
4. Brian stands at the edge of a cliff and throws a stone horizontally over the edge with a
speed of 18 m/s. The cliff is 50 m above a flat, horizontal beach: (2 pts each)
a. How long does it take for the stone to reach the bottom of the cliff?
b. What is the range of the stone?
5. A student shot a basketball at an initial velocity of 25 m/s at an angle of 30 degrees with
respect to the horizontal: (2 pts eac)
a. What is the vertical component of the initial velocity?
b. What is the maximum height reached by the arrow?
6. A fireman stands 50 m away from a burning building and directs a stream of water from
ground level at an angle of 30° above the horizontal. If the speed of the stream as it
leaves the hose is 40 m/s: (2 pts each)
a. How long will it take for the stream to reach the building?
b. At what height will the water strike the building?
Answer:A).3.16 s B).56.92
A).12.5m/s
B).7.81m
Explanation:
Question 1: What is the before image showing?
Question 2: What is the after image showing?
Answer:
the after image is showing precipitation. i dont know what the before image is though.
Explanation:
A block-spring system oscillates on a frictionless horizontal surface. The time needed for the block to complete one cycle is 0.1 sec. Determine the time needed for the block to travel from -A/2 to A/2, where A is the amplitude of motion.
The time required for the block to travel or complete it's half-cycle is 0.05s
Data;
time = 0.1sdistance = -A/2 to A/2Amplitude of a Simple Harmonic MotionThis is the maximum displacement of a body from it's mean position.
Assuming a system start from its mean position i.e mean initial position and go to A/2 towards right and then back to his mean position after going through -A/2 then it is said to have completed one cycle. If the block oscillate only between -A/2 to A/2 then it will complete half cycle then it's time period will be 0.05 sec.
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Find the magnitude of acceleration (ft/s^2) a person experiences when he or she is texting and driving 58mph, hits a wall, and comes to a complete stop .24 seconds after impact.
Answer:
350 ft/s²
Explanation:
First, convert mph to ft/s.
58 mi/hr × (5280 ft/mi) × (1 hr / 3600 s) = 85.1 ft/s
Given:
v₀ = 85.1 ft/s
v = 0 ft/s
t = 0.24 s
Find: a
v = at + v₀
a = (v − v₀) / t
a = (0 ft/s − 85.1 ft/s) / 0.24 s
a = -354 ft/s²
Rounded to two significant figures, the magnitude of the acceleration is 350 ft/s².
A car is uniformly accelerated at a rate of 2 m/sec2 for 12 sec. If the original speed of the car is 36 m/sec, what is its final speed?
Answer:
60m/s
Explanation:
v=u+at
v=36+(2×12)
v=36+24
v=60m/s
Why are force fields necessary to describe gravity?
A. Gravity is a type of air resistance.
B. Gravity acts between any two objects with charges.
C. Gravity is a noncontact force.
D. Gravity can only push objects apart.
Answer:
C
Explanation:
Gravity pulls objects towards earth's core without needing any contact with objects
Gravity is a force that acts upon a ball after it is thrown
true
false
Answer:
true.
Explanation:
The forces on a thrown ball after it leaves the thrower's hand are the force of gravity .
A car slows from a velocity of 26m/s to 18m/s in a distance of 52 meters. What was the cars acceleration
Recall that
v² - u² = 2 a ∆x
where u and v denote initial and final velocities, respectively; a is acceleration; and ∆x is the distance traveled.
Then we get
(18 m/s)² - (26 m/s)² = 2 a (52 m)
a = ((18 m/s)² - (26 m/s)²) / (104 m)
a ≈ -3.4 m/s²
a ball is rolled at a velocity of 12 miles per second. after 36 seconds, it comes to a stop. what is the acceleration of the ball?
Question :-
A Ball is Rolled at a Velocity of 12 m/s. After 36 sec , it comes to a stop. What is the Acceleration of the ball ?Answer :-
Acceleration is -0.33 m/s² .Explanation :-
As per the provided information in the given question, we have been given that the Velocity of the ball is 12 m/s . Time is given as 36 sec . And, we have been asked to calculate the Acceleration .
For calculating the Acceleration , we will use the Formula :-
[tex] \bigstar \: \: \boxed{ \sf{ \: Acceleration \: = \: \dfrac{v \: - \: u}{t} \: }} [/tex]
Where ,
V denotes to the Final VelocityU denotes to the Initial VelocityT denotes to the Time TakenTherefore , by Substituting the given values in the above Formula :-
[tex] \dag \: \: \: \sf { Acceleration \: = \: \dfrac{Final \: Velocity \: - \: Initial \: Velocity}{Time} } [/tex]
[tex] \longmapsto \: \: \sf { Acceleration \: = \: \dfrac{0 \: - \: 12}{36}} [/tex]
[tex]\longmapsto \: \: \sf {Acceleration \: = \: \dfrac{ \: 12 \: }{36}}[/tex]
[tex]\longmapsto \: \: \sf {Acceleration \: = \: \dfrac{ \: 1 \: }{3}}[/tex]
[tex] \longmapsto \: \bf {Acceleration \: = \: 0.33 \: m/s^{2}} [/tex]
Hence :-
Acceleration of Ball is -0.33 m/s² .[tex] \underline {\rule {212pt} {4pt}} [/tex]
6.When light bends it is called
a - broken
b- benzene
C - reflection
d - refraction
Answer:
D refraction
Explanation:
I had a question like this so and I got it right so I believe this is the correct answer
Robert throws a 3 kg rock at 20 m/s. What is the rock's momentum?
a
O 0.15 kgm/s
O 6.67 kgm/s
O 20 m/s
O 60 kg . m/s
A spring has a relaxed length of 7 cm and a stiffness of 200 N/m. How much work must you do to change its length from 10 cm to 15 cm
The work W needed to stretch/compress a spring from rest by a distance x is
W = 1/2 kx²
where k is the spring constant.
This means the work needed to change the length of this spring by 10 cm = 0.01 m is
W = 1/2 (200 N/m) (0.01 m)² = 0.01 J
and by 15 cm = 0.015 m is
W' = 1/2 (200 N/m) (0.015 m)² = 0.0225 J
Then the total work performed on the spring by stretching from 10 cm to 15 cm is
∆W = W' - W = 0.0225 J - 0.01 J = 0.0125 J
if an astronaut weighs 981 N on Earth and only 160 N on the Moon, then what is his mass on the Moon?
Answer:
if an astronaut weighs 981 N on Earth and only 160 N on the Moon, then what is his mass on the Moon?
ANSWER 256n
The mass of the astronaut on the moon as compared to the earth will be [tex]M_m=97.85\ lg[/tex]
What will be the mass?The mass of any substance or body is defined as how much quantity of matter is present.
Now it is given in the question:
Weight of the astronaut on earth [tex]W_E=981\ N[/tex]
Weight of the astronaut on earth [tex]W_M=160\ N[/tex]
The mass of the astronaut on the moon will be calculated as:
Weight on the moon will be given as:
[tex]W_M=M_M\times g_m[/tex]
Weight on the earth will be given as:
[tex]W_E=M_E\times g_e[/tex]
The ratio of the gravity of the earth to the moon is given as
[tex]\dfrac{g_e}{g_m} =\dfrac{9.81}{1.62} =6[/tex]
The mass of the earth will be calculated as
[tex]W_E=M_E\times ge[/tex]
[tex]M_E=\dfrac{981}{9.81} =100\ kg[/tex]
Now taking the ratio of the weight of the earth to the moon :
[tex]\dfrac{W_E}{W_M} =\dfrac{M_E\times g_e}{M_M\times g_m}[/tex]
[tex]M_M= \dfrac{W_M\times M_E}{W_E} \times \dfrac{g_e}{g_m}[/tex]
Now by putting the value in the formula we get;
[tex]M_M=\dfrac{160\times 100}{981} \times 6=97.85\ kg[/tex]
Thus the mass of the astronaut on the moon as compared to the earth will be [tex]M_m=97.85\ lg[/tex]
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fluid meaning in Nepali
Answer:
Tarala
Explanation:
It's Nepali
Answer:
Fluid meaning in Nepali is Tarala (तरल )
Which of the following statements are true about the motion of an object?
a. To change the magnitude of the momentum you need to apply a force with a component parallel to the momentum (either in the same direction as the momentum or the opposite direction).
b. A force perpendicular to the momentum changes the direction of the momentum but not its magnitude.
c. When an object moves at constant speed along a curving path, the net force on the object must act straight outward from the center of the kissing circle.
d. To make an object turn to the left, something has to exert a force to the right on the object.
Answer: B
Explanation: A force perpendicular to the momentum changes the direction of the momentum but not its magnitude.
Answer:
a. To change the magnitude of the momentum you need to apply a force with a component parallel to the momentum (either in the same direction as the momentum or the opposite direction).
d. To make an object turn to the left, something has to exert a force to the right on the object.
Explanation:
For motion along a circle or curve, the net force on the particle must be parallel with the change in momentum.
This change in momentum of the system is given by;
[tex]\delta P_{sys} = F_{net} \delta t[/tex]
Thus, option A is correct.
a. To change the magnitude of the momentum you need to apply a force with a component parallel to the momentum (either in the same direction as the momentum or the opposite direction).
Also, applying the principle of torque, to make an object turn to the left, something has to exert a force to the right on the object.
Option D is also correct.
Kiting during a storm. The legend that Benjamin Franklin flew a kite as a storm approached is only a legend — he was neither stupid nor suicidal. Suppose a kite string of radius 2.02 mm extends directly upward by 0.823 km and is coated with a 0.506 mm layer of water having resistivity 159 Ω·m. If the potential difference between the two ends of the string is 186 MV, what is the current through the water layer? The danger is not this current but the chance that the string draws a lightning strike, which can have a current as large as 500 000 A (way beyond just being lethal).
Answer:
The current is [tex]I = 1.1434*10^{-5}}\ A[/tex]
Explanation:
From the question we are told that
The radius of the kite string is [tex]R = 2.02 mm = 0.00202 \ m[/tex]
The distance it extended upward is [tex]D = 0.823 km = 823 \ m[/tex]
The thickness of the water layer is [tex]d = 0.506 mm = 0.000506 \ m[/tex]
The resistivity is [tex]\rho = 159\ \Omega \cdot m[/tex]
The potential difference is [tex]V = 186 MV = 186 *10^{6} \ V[/tex]
Generally the cross sectional area of the water layer is mathematically represented as
[tex]A = \pi r^2[/tex]
Here r is mathematically represented as
[tex]r = [(R + d ) - R][/tex]
=> [tex]r = [(0.00202 + 0.000506 ) - 0.00202][/tex]
=> [tex]r = 0.000506[/tex]
=> [tex]A = 3.142 * [0.000506]^2 [/tex]
=> [tex]A = 8.0447*10^{-7}\ m^2 [/tex]
Generally the resistance of the water is mathematically represented as
[tex]R = \frac{\rho * D }{A}[/tex]
=> [tex]R = \frac{159 *823 }{8.0447*10^{-7}}[/tex]
=> [tex]R = 1.62662 * 10^{11} \ \Omega [/tex]
Generally the current is mathematically represented as
[tex]I = \frac{V}{R}[/tex]
=> [tex]I = \frac{186 *10^{6} }{1.62662 * 10^{11}}[/tex]
=> [tex]I = 1.1434*10^{-5}}\ A[/tex]
You want to double the radius of a rotating solid sphere while keeping its kinetic energy constant. (The mass does not change.) To do this, the final angular velocity of the sphere must be Group of answer choices
The Final angular velocity of the sphere is : Half of its initial value ( W/2 )
Kinetic energy of a rotating body
Given that The kinetic energy of a rotating body ( K ) = 1/2 IW² --- ( 1 )
where : I = moment of inertia, w = angular velocity
I = [tex]\frac{2}{5} MR^{2}[/tex]
Therefore equation ( 1 ) becomes
K = [tex]\frac{1}{5}MR^2W^2[/tex] ----- ( 2 )
After doubling the radius
R' = 2R
K' = K
Mass = unchanged
therefore :
[tex]\frac{1}{5}MR'^2W'^2 = \frac{1}{5}MR^2W^2[/tex]
= ( 2R )² W'² = R²W²
Resolving equation above
Hence : W' ( new angular velocity ) = W / 2
In conclusion The Final angular velocity of the sphere is : Half of its initial value ( W / 2).
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