A college friend of yours who has been postponing taking any science courses hears you talking about the generation of nuclear energy in the Sun and makes the following observation: "The whole idea of the atomic nucleus is pretty ridiculous. If an oxygen nucleus consists of eight protons and eight neutrons, the charge on that nucleus is positive. Since even I learned in high school that like charges repel, such a nucleus would find all its positive protons repelling and quickly fall apart." How would you answer his argument?

Answer:

300 nm

Explanation:

R = Gas constant = 8.314 J/molK

r = Atomic radii = [tex]1\times 10^{-10}\ m[/tex]

d = Atomic diameter = [tex]2r=2\times 10^{-10}\ m[/tex]

At STP

T = Temperature = 273.15 K

P = Pressure = 100 kPa

[tex]N_A[/tex] = Avogadro's number = [tex]6.022\times 10^{23}[/tex]

The mean free path is given by

[tex]\lambda=\frac{RT}{\sqrt2d^2N_AP}\\\Rightarrow \lambda=\frac{8.314\times 273.15}{\sqrt2 \pi \times (2\times 10^{-10})^2\times 6.022\times 10^{23}\times 100000}\\\Rightarrow \lambda=2.12165\times 10^{-7}\ m=212.165\times 10^{-9}\ m=212.165\ nm[/tex]

The answer that best represents the mean free path for gas molecules is 300 nm

Answer:

option A

Explanation:

given,

velocity to hit home plate = 110 Km/h

a) When the baseball is hit straight back

     Assuming the momentum before hitting be P and after hitting the bal is also equal to P.

change in momentum = P - (-P) = 2P

b) When the catcher catches the ball the change in momentum is equal to zero.

c) when the ball is popped up the change in momentum

   p_x = P and P_y = P

  resultant momentum = [tex]\sqrt{P^2+ P^2} = \sqrt{2}\ P[/tex]

the maximum change in momentum will be in case  A

The correct answer is option A

Final answer:

The scenario in which the baseball has the largest change in momentum is when it is hit straight back to the pitcher at a speed of 110 km/h.

Explanation:

The scenario in which the baseball has the largest change in momentum is when it is hit straight back to the pitcher at a speed of 110 km/h. This is because when the baseball is hit straight back, it completely reverses its direction of motion, resulting in a large change in momentum. The other scenarios, such as the baseball being caught by the catcher or popped straight up, do not involve a complete reversal of direction, so the change in momentum is smaller.

Answer:

The mass of Neptunium is 237.054 u.

Explanation:

Given that,

Mass of Americium = 241.05682 u

Mass of alpha particle = 4.00260 u

The equation is,

[tex]_{95}^{241}Am\rightarrow _{93}^{237}NP+\alpha\ particle+5.5 MeV[/tex]

Let the mass of Neptunium is m.

Since the mass remain same.

We need to calculate the mass of Neptunium

Using formula of mass

Mass of Neptunium =  Mass of Americium -Mass of alpha particle

Put the value into the formula

[tex]m=241.05682 -4.00260[/tex]

[tex]m=237.054\ u[/tex]

Hence, The mass of Neptunium is 237.054 u.

The mass of the Neptunium nucleus is estimated to be 237.05422 u.

The Americium nucleus has a given mass of 241.05682 u and decays by emitting an alpha particle with a mass of 4.00260 u and kinetic energy of 5.5 MeV. To estimate the mass of the Neptunium nucleus, ignoring its recoil, we initially consider the conservation of mass-energy principle.

Mass of Neptunium = Mass of Americium -Mass of alpha particle

The mass of the Neptunium nucleus can be calculated simply by subtracting the mass of the alpha particle from the original Americium nucleus, which results in -:

m = 241.05682 u - 4.00260 u = 237.05422 u

To solve this problem it is necessary to apply the concepts related to constructive interference for multiple split.

The precaution is given by,

[tex]dsin\theta = m\lambda[/tex]

Where,

d = Distance between the slits

[tex]\theta =[/tex] Angle between the path and a line from the slits to the screen

m = Any integer, representing the number of repetition of the spectrum.

[tex]\lambda =[/tex]Wavelength

For first order equation we have that m = 1 then

[tex]d sin\theta = \lambda[/tex]

As the maximum number of lines corresponds to the smallest d values, we have that [tex]\theta = 90[/tex]

[tex]d sin90=\lambda[/tex]

[tex]d = 760nm[/tex]

Therefore the maximum numbers of lines per centimeter would be

[tex]N = \frac{10^{-2}m}{d}[/tex]

[tex]N = \frac{10^{-2}m}{760*10^{-9}m}[/tex]

[tex]N = 13157.89[/tex]

The maximum numbers of lines per centimeter is 13158

Final answer:

The loss in available energy units across the valve is equal to the pressure drop, which is 344,738 N/m².

Explanation:

To determine the loss in available energy units across the valve, we need to calculate the change in kinetic energy and potential energy before and after the valve. The available energy is given by the equation:

Available energy = pressure drop + change in kinetic energy + change in potential energy

Given that the pressure drop across the valve is 50 psi, we can convert it to SI units (N/m²) by multiplying by 6894.76:

Pressure drop = 50 psi × 6894.76 N/m²/psi = 344,738 N/m²

The change in kinetic energy can be calculated using the following formulas:

Change in kinetic energy = (1/2) × mass flow rate × (outlet velocity² - inlet velocity²)

Change in potential energy = mass flow rate × g × (outlet height - inlet height)

Since the flow occurs in a horizontal plane, there is no change in potential energy. Therefore, the loss in available energy units across the valve is equal to the pressure drop:

Loss in available energy units = 344,738 N/m²

Answer:

x = 1.26 sin 3.16 t

Explanation:

Assume that the general equation of the displacement given as

x = A sinω t

A=Amplitude ,t=time ,ω=natural frequency

We know that speed V

[tex]V=\dfrac{dx}{dt}[/tex]

V= A ω cosωt

Maximum velocity

V(max)= Aω

Given that F= 32 N

F = K Δ

K=Spring constant

Δ = 0.4 m

32 =0.4 K

K = 80 N/m

We know that  ω²m = K

8 ω² = 80

ω = 3.16 s⁻¹

Given that V(max)= Aω = 4 m/s

3.16 A = 4

A= 1.26 m

Therefore the general equation of displacement

x = 1.26 sin 3.16 t

Answer:

Explanation:

mass, m = 8 kg

extension, y = 0.4 m

force, F = 32 N

maximum velocity, v = 4 m/s

maximum velocity , v = ω A

where, ω be the angular velocity and A be the amplitude

4 = ω x 0.4

ω = 10 rad/s

position

x = A Sin ωt

x = 0.4 Sin 10 t

Answer:

16771720740.20324 J

Explanation:

[tex]\rh0[/tex] = Density = 0.97 kg/m³

V = Volume = [tex]\pi r^2h[/tex]

d = Diameter of cylinder = 230 m

r = Radius = [tex]\frac{d}{2}=\frac{230}{2}=115\ m[/tex]

h = Height of the cylinder = 640 m

v = Velocity of cylinder = 51 m/s

Mass of object is given by

[tex]m=\rho V\\\Rightarrow m=0.97\times \pi 115^2\times 640\\\Rightarrow m=25792727.013\ kg[/tex]

Moment of inertia of a cylinder

[tex]I=\frac{1}{2}mr^2\\\Rightarrow I=\frac{1}{2}\times 25792727.013\times 115^2\\\Rightarrow I=170554407373.4625\ kgm^2[/tex]

Angular speed

[tex]\omega=\frac{v}{r}\\\Rightarrow \omega=\frac{51}{115}\ rad/s[/tex]

Kinetic energy is given by

[tex]K=\frac{1}{2}I\omega^2\\\Rightarrow K=\frac{1}{2}\times 170554407373.4625\times \left(\frac{51}{115}\right)^2\\\Rightarrow K=16771720740.20324\ J[/tex]

The kinetic energy contained by the tornado is 16771720740.20324 J

To solve this process it is necessary to consider the concepts related to the relations between pressure and temperature in an adiabatic process.

By definition the relationship between pressure and temperature is given by

[tex](\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}[/tex]

Here

P = Pressure

T = Temperature

[tex]\gamma =[/tex]The ratio of specific heats. For air normally is 1.4.

Our values are given as,

[tex]P_1 = 15lb/in^2\\T_1= 80\°F = 299.817K\\T_2 =400\°F = 408.15K[/tex]

Therefore replacing we have,

[tex](\frac{P_2}{P_1})=(\frac{T_2}{T_1})^{(\frac{\gamma}{\gamma-1})}[/tex]

[tex](\frac{P_2}{15})=(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}[/tex]

Solving for [tex]P_2,[/tex]

[tex]P_2 = 15*(\frac{408.15}{299.817})^{(\frac{1.4}{1.4-1})}[/tex]

[tex]P_2 = 44.15Lbf/in^2[/tex]

Therefore the maximum theoretical pressure at the exit is [tex]44.15Lbf/in^2[/tex]

To determine the maximum theoretical pressure at the exit of the compressor, we can use the ideal gas law. Given the initial pressure and temperature, as well as the final temperature, we can calculate the maximum theoretical pressure using the equation p2,max = (p1 * T2) / T1.

To determine the maximum theoretical pressure at the exit of the compressor, we can use the ideal gas law. The ideal gas law is given by the equation pV = nRT, where p is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. We can rearrange this equation to solve for the pressure:

p = (nRT) / V

Since the volume and the number of moles are constant, we can rewrite the equation as:

p1 / p2 = T1 / T2

where p1 and T1 are the initial pressure and temperature, and p2 and T2 are the final pressure and temperature.

Substituting the given values, we have:

p2,max = (p1 * T2) / T1

Plugging in the values from the problem statement, we have:

p2,max = (15 lbf/in.2 * 275°F) / 80°F

p2,max = 51.56 lbf/in.2

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Answer:

a) Vf = 27.13 m/s

b) It would have been the same

Explanation:

On the y-axis:

[tex]Y=-Vo*sin\theta*t-1/2*g*t^2[/tex]

[tex]-8=-24*sin(21)*t-1/2*10*t^2[/tex]

Solving for t:

t1 = 0.67s     t2= -2.4s

Discarding the negative value and using the positive one to calculate the velocity:

[tex]Vf_y = -Vo*sin\theta-g*t[/tex]

[tex]Vf_y = -15.3m/s[/tex]

So, the module of the velocity will be:

[tex]Vf=\sqrt{(-15.3)^2+(24*cos(21))^2}[/tex]

[tex]Vf=27.13m/s[/tex]

If you throw it above horizontal, it would go up first, and when it reached the initial height, the velocity would be the same at the throwing instant. And starting then, the movement will be the same.

To determine the speed at which the stone strikes the ground, we can break the initial velocity into its horizontal and vertical components and calculate the time and horizontal distance traveled by the stone. By using the equations for vertical and horizontal motion and the initial conditions provided, we can calculate the speed at which the stone strikes the ground. If the stone had been thrown from a greater height above the horizontal, the impact velocity would be different as the horizontal component of the velocity would be greater.

To determine the speed at which the stone strikes the ground, we need to analyze the motion of the stone both horizontally and vertically. Since the stone is thrown with an initial velocity of 24 m/s at an angle of 21° below the horizontal, we can break the initial velocity into its horizontal and vertical components. The horizontal component, Vx, can be calculated using Vx = v0 * cos(θ), where v0 is the initial velocity and θ is the angle. The vertical component, Vy, can be calculated using Vy = v0 * sin(θ).

In this case, Vx = 24 m/s * cos(21°) = 22.4 m/s and Vy = 24 m/s * sin(21°) = 8.7 m/s. The time it takes for the stone to hit the ground can be calculated using the equation y = y0 + Vy0t + 0.5gt^2, where y0 is the initial vertical position, Vy0 is the initial vertical velocity, g is the acceleration due to gravity (approximately 9.8 m/s^2), and t is the time. In this case, y0 = 8.0 m, Vy0 = 8.7 m/s, and g = 9.8 m/s^2. We can solve for t using the quadratic equation.

Once we have the time, we can calculate the horizontal distance traveled by the stone using x = Vx0t. Since Vx0 is constant and equal to Vx, we can substitute Vx into the equation. In this case, x = 22.4 m/s * t.

Therefore, the speed vf at which the stone strikes the ground can be calculated using vf = √(Vx^2 + Vy^2). We can substitute the values we calculated for Vx and Vy to find the answer.

For part (b) of the question, if the stone had been thrown from the clifftop with the same initial speed and the same angle but above the horizontal, the impact velocity will be different. The vertical component of the velocity will be the same, as it only depends on the initial speed and angle of projection. However, the horizontal component will be different. The horizontal component of the velocity depends on the height of the cliff. When thrown from a higher height, the horizontal component will be greater, resulting in a higher impact velocity.

Answer:

$364.29

Explanation:

given,

Packing of crates per month (u)= 800

annual carrying cost of 35 percent of the purchase price per crate.

Ordering cost(S) = $ 28

D = 800 x 12 = 9600 crates/year

H = 0.35 P

H = 0.35 x $10

H = $3.50/crate per yr.

Present Total cost

= [tex]\dfrac{800}{2}\times 3.50 + \dfrac{9600}{800}\times 28[/tex]

= 1400 + 336

= $ 1,736

[tex]Q_0 = \sqrt{\dfrac{2DS}{H}}[/tex]

[tex]Q_0 = \sqrt{\dfrac{2\times 9600 \times 28}{3.50}}[/tex]

[tex]Q_0 =\$ 391.92[/tex]

Total cost at EOQ

= [tex]\dfrac{391.92}{2}\times 3.50 + \dfrac{9600}{391.92}\times 28[/tex]

= 685.86 + 685.85

= $ 1,371.71

the firm save annually in ordering and carrying costs by using the EOQ

    = $ 1,736 - $ 1,371.71

    = $364.29

The question focuses on Economic Order Quantity (EOQ) in inventory management for a produce distributor's packing crates. EOQ helps determine the optimal number of units a company should add to its inventory to minimize total holding and setup costs.

The subject of this question is

Economic Order Quantity (EOQ)

, which is an operational efficiency measure used in inventory management. It refers to the number of units that a company should add to its inventory at a time to minimize total holding and setup costs.

In this case, the produce distributor uses 800 crates a month, purchased at $10 each. The annual carrying cost is 35% of the purchase price per crate, and the ordering cost is $28. Currently, the distributor orders once a month. Hence, to determine the optimal number of orders, or the EOQ, various factors such as ordering cost, holding cost, and demand rate need to be considered.

This cost minimization issue can be mathematically represented and solved using the EOQ formula: EOQ = √((2DS)/H) where D is the demand rate, S is the setup (or order) cost, and H is the holding (or carrying) cost.

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Answer:

a ) = 381.48 J

b )=  84.25 cm

Explanation:

Kinetic energy of the runner

= 1/2 m v²

= .5 x 66 x 3.4²

= 381.48 J

The final kinetic energy of the runner is zero .

Loss of mechanical energy

= 381.48 J

This  loss in  mechanical energy is due to action of frictional force .

b )

Let s be the distance of slide

deceleration due to frictional force

= μmg/m

.7 x 66 x 9.8 / 66

a = - 6.86 m s⁻¹

v² = u² - 2 a s

0 = 3.4² - 2x6.86 s

s = 3.4² / 2x6.86

= .8425 m

84.25 cm

Answer: 8.6 µm

Explanation:

At a long distance from the source, the components (the electric and magnetic fields) of the electromagnetic waves, behave like plane waves, so the equation for the y component of the electric field obeys an equation like this one:

Ey =Emax cos (kx-ωt)

So, we can write the following equality:

ω= 2.2 1014 rad/sec

The angular frequency and the linear frequency are related as follows:

f = ω/ 2π= 2.2 1014 / 2π (rad/sec) / rad = 0.35 1014 1/sec

In an electromagnetic wave propagating through vacuum, the speed of the wave is just the speed of light, c.

The wavelength, speed and frequency, are related by this equation:

λ = c/f

λ = 3.108 m/s / 0.35. 1014 1/s = 8.6 µm.

The wavelength of the electromagnetic wave is 8.57 micrometers.

The wavelength of an electromagnetic wave can be calculated using the following equation:

λ = c / f

where:

λ is the wavelength

c is the speed of light in vacuum (3.0 × 10^8 m/s)

f is the frequency

The frequency of the electromagnetic wave can be calculated from the angular frequency ω using the following equation:

f = ω / 2π

where:

ω is the angular frequency (2.20 × 10^14 rad/s)

Plugging in the values for ω and c, we get:

f = (2.20 × 10^14 rad/s) / (2π)

f = 3.50 × 10^13 Hz

Now we can calculate the wavelength:

λ = (3.0 × 10^8 m/s) / (3.50 × 10^13 Hz)

λ = 8.57 × 10^-6 m

Therefore, the wavelength of the electromagnetic wave is 8.57 micrometers.

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Answer:

A) 35.5N/m b) 20.1cm

Explanation:

Using Hooke's law;

F = Ke where F is the weight of the object = mass of the object in kg * acceleration due to gravity in m/s^2 and k if the force constant of the spring in N/m and e is the extension of the spring which original length of the spring - new length after extension in meters

For the first body, m*g = K * (0.25- li)

Where li is the initial length of the spring

0.175*9.81 = k(0.25-li)

1.72 = k(0.25-li) as equation 1

For the second body, m *g = K* ( 0.775-li)

2.075*9.81 = k (0.775-li) equation 2

20.36 = k(0.775-li)

Make li subject of the formula;

li = 0.775 - 20.36/k

Substitute for li in equation 1

1.72 = k(0.25- (0.775 - 20.36/k))

1.72 = k ( 0.25 - 0.775 + 20.36/k)

Open the bracket with k

1.72 = 0.25k - 0.775k + 20.36 (since k cancel k)

Collect the like terms:

1.72 - 20.36 = - 0.525k

- 18.64 = -0.525k

Divide both side by -0.525

-18.64/-0.525 = -0.525/-0.525k

K = 35.5N/m

B) substitute for k in using

li = 0.775 - 20.36/k

li = 0.775 - 20.36/35.5

li = 0.775 - 0.574

li = 0.201 in meters

li = 0.201 * 100 centimeters = 20.1cm

(A) The force constant of the spring is obtained as 34.66 N/m

(B) The unloaded length of the spring is 20 cm.

(A) According to Hooke's law;

[tex]F = kx[/tex]

Where 'F' is the force needed to extend or compress the spring, 'k' is the force constant and 'x' is the extended length or compressed length.

Here, the gravitational force of the mass provides the necessary force for the spring to elongate.

Therefore, we can say that in the case of first mass;

[tex]k\,(0.25-l_0) = m_1g = (0.175\,kg \times 9.8\,m/s^2)[/tex]

Where [tex]l_0[/tex] is the unloaded length of the spring.

[tex]\implies k\,(0.25-l_0)=1.715\,N[/tex]

In the case of the second mass, we can write;

[tex]k\,(0.775\,m-l_0) = m_2g = (2.075\,kg \times 9.8\,m/s^2)[/tex]

[tex]\implies k\,(0.775\,m-l_0) = 20.335\,N[/tex]

From both these equations, we can write;

[tex]k\,(0.775\,m-l_0-0.25\,m+l_0)=20.335\,N- 1.715\,N[/tex]

[tex]\implies0.525\, k=18.2\,N[/tex]

[tex]\therefore k=\frac{18.2\,N}{0.525\,m}=34.66\,N/m[/tex]

(B) Applying the value of 'k' in any of the equations for force of given masses, we get;

[tex]k\,(0.25-l_0)=1.715\,N[/tex]

[tex]\implies (34.66\,N/m)\times\,(0.25-l_0)=1.715\,N[/tex]

[tex]\implies 8.665N-34.66\,l_0=1.715\,N[/tex]

[tex]\implies \,l_0=\frac{-6.95N}{-34.66} =0.2\,m=20\,cm[/tex]

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Answer:

Explanation:

Given

mass of cart m=20 kg

Force applied F=12 N at angle of [tex]\theta =20^{\circ}[/tex] with horizontal

Force can be resolved in two components i.e. [tex]F\cos \theta [/tex]and [tex]F\sin \theta [/tex]

[tex]F\sin \theta [/tex]is will act Perpendicular to the motion of cart and [tex]F\cos \theta [/tex]will help to move the cart .

Since cart is moving with constant velocity therefore net force is zero on the cart as friction is opposing the [tex]F\cos \theta [/tex]

Work done by [tex]F\cos \theta [/tex]is

[tex]W_1=F\cos \theta \cdot L[/tex]

[tex]W_1=12\cos (20)\cdot 18=202.978 J[/tex]

Work done by Friction will be same but in Opposite Direction

[tex]W_f=-202.978 J[/tex]

Work done by [tex]F\sin \theta [/tex]  is

[tex]W_2=0[/tex] because direction of force and motion is Perpendicular to each other.              

Answer:

D. There is no way to know which skier has the greater speed at the finish.

Explanation::

Total energy is conserved because there is no friction

(Et)₀ = (Et)f

(Et)₀ = Initial total energy (J)

(Et)f  = Final total energy (J)

(Et)₀= K₀ + U₀

(Et)f= Kf + Uf

K₀ : Initial kinetic energy

U₀ : Initial potential energy

Kf  : final kinetic energy

Uf :  final potential energy

The formulas to calculate the kinetic energy (K) and potential energy (U) are:

K = ( 1/2)m*v²

U = m* g*h

m : mass (kg)

v: speed ( m/s)

h: hight ( m)

Problem development

Skier A

(Et)₀ = (Et)f

K₀ + U₀ =  Kf + Uf

( 1/2)mA*(v₀A)² + mA*g*h₀ =  ( 1/2)mA*(vfA)² + mA*g*hf ,

We divide by mA on both sides of the equation

( 1/2)*(v₀A)²+ g*h₀ =  ( 1/2)(vfA)² + g*hf

( 1/2)*(v₀A)²+ g*h₀ - g*hf  =  ( 1/2)(vfA)²  

We multiply by 2 both sides of the equation

(v₀A)²+2g(h₀ -hf) = (vfA)²  

(vfA)²  = (v₀A)²+2g(h₀ -hf) Equation (1)

Skier B

(Et)₀ = (Et)f

K₀ + U₀ =  Kf + Uf

(1/2)mB*(v₀B)² + mB*g*h₀ =  ( 1/2)mB*(vfB)² + mB*g*hf

We perform the same procedure above:

(vfB)²  = (v₀B)²+2g(h₀ -hf) Equation (2)

Comparison of equation (1) with equation (2)

The term 2g (h₀ -hf) is the same in both equations because the paths of the two skiers start in the same place and end in the same place.

The final speed (vf) of skiers depends on their initial speed (v₀).

Because the initial speed of the skiers is unknown, it cannot be determined which has the highest final speed

Answer:

6.86 m/s

Explanation:

This problem can be solved by doing the total energy balance, i.e:

initial (KE + PE)  = final (KE + PE). { KE = Kinetic Energy and PE = Potential Energy}

Since the rod comes to a halt at the topmost position, the KE final is 0. Therefore, all the KE initial is changed to PE, i.e, ΔKE = ΔPE.

Now, at the initial position (the rod hanging vertically down), the bottom-most end is given a velocity of v0. The initial angular velocity(ω) of the rod is given by ω = v/r , where v is the velocity of a particle on the rod and r is the distance of this particle from the axis.

Now, taking v = v0 and r = length of the rod(L), we get ω = v0/ 0.8 rad/s

The rotational KE of the rod is given by KE = 0.5Iω², where I is the moment of inertia of the rod about the axis of rotation and this is given by I = 1/3mL², where L is the length of the rod. Therefore, KE = 1/2ω²1/3mL² = 1/6ω²mL². Also, ω = v0/L, hence KE = 1/6m(v0)²

This KE is equal to the change in PE of the rod. Since the rod is uniform, the center of mass of the rod is at its center and is therefore at a distane of L/2 from the axis of rotation in the downward direction and at the final position, it is at a distance of L/2 in the upward direction. Hence ΔPE = mgL/2 + mgL/2 = mgL. (g = 9.8 m/s²)

Now, 1/6m(v0)² = mgL ⇒ v0 = [tex]\sqrt{6gL}[/tex]

Hence, v0 = 6.86 m/s

To find the value of v0 that will cause the rod to come to a momentary halt in a straight-up orientation, we need to consider conservation of energy and angular momentum. The equation (3/2)gh = 4(v0)^2 can then be used to solve for v0. The value of v0 is equal to the square root of (3/8)gh.

To find the value of v0 that will cause the rod to come to a momentary halt in a straight-up orientation, we need to consider conservation of energy and angular momentum. When the rod is released, its potential energy is converted to kinetic energy and rotational kinetic energy. At the moment the rod comes to a halt, all of its initial kinetic energy will be converted back into potential energy. Since the rod is initially hanging vertically downward, we can equate the initial potential energy to the potential energy at the moment of momentary halt:

mgh = (1/2)Iω2

Where m is the mass of the rod, g is the acceleration due to gravity, h is the height of the rod, I is the moment of inertia of the rod about the pivot point, and ω is the angular velocity of the rod at the moment of momentary halt. The moment of inertia of a rod rotating about one end is given by I = (1/3)mL2, where L is the length of the rod. Therefore, we can rewrite the equation as:

mgh = (1/2)(1/3)mL2ω2

Simplifying the equation, we have:

(3/2)gh = ω2L2

To find ω, we need to relate it to the linear speed v0. Since the rod is rotating about a pivot point, the linear speed of a point on the rod is related to the angular velocity by the equation v = ωr, where r is the distance of the point from the pivot. In this case, the distance r is equal to half the length of the rod, so r = L/2. Substituting this into the equation, we have:

v0 = ω(L/2)

From this equation, we can solve for ω:

ω = (2v0)/L

Substituting this into the previous equation, we get:

(3/2)gh = ((2v0)/L)2L2

Simplifying further, we have:

(3/2)gh = 4(v0)2

Finally, solving for v0, we get:

v0 = √((3/8)gh)

The momentum conservation allows to find the results for the questions about the collision of the balls are:

  A) The velocities after the collision are: vₐ = 3.7 i m/s and [tex]v_b[/tex] = 2 j  m/s

  B) The direction of ball A is:  x-axis

  C) The total moment is: [tex]p_{total}[/tex] = 4.2m  and the direction is θ = 28.4º

      The total kinetic energy is: [tex]K_{total}[/tex]  = 8.84m J

Given parameters

To find

A) The speeds after the collision.

B) direction of the ball A.

C) The moment and energy after the crash.

Momentum is defined by the product of mass and velocity. When a system is isolated, there is no external force, the momentum is conserved. This is an important conservation law of physics.

          p = m v

Where the bold letters indicate vectors, m is the mass and v the velocity.

A)   The system is formed by the two balls, therefore it is isolated and the momentum is conserved. Since the momentum is a vector quantity we write the components on each axis, see attached for a schematic.

Initial instant. Before the crash.

They indicate that ball B moves in the axis and after the collision.

x-axis

       p₀ₓ = m [tex]v_{obx}[/tex]vobx

y-axis  

       [tex]p_{oy} = m v_{oay}[/tex]

Final moment. After the crash.

x-axis

       [tex]p_{fx} = m v_{fax}[/tex]  

y- axis  

      [tex]p_{fy} = m v_{fay} + m v_{fby}[/tex]  

The momentum is preserved

x- axis

       p₀ₓ = [tex]p_{fx}[/tex]

       m [tex]v_{obx}[/tex] = m [tex]v_{fax}[/tex]  

       [tex]v_{fax} = v _{obx}\\v_{fax} = 3.7 m/s[/tex]

y- axis  

       [tex]p_{oy} = p_[fy} \\m v_{oay} = m v_{fay} + m v_[fby}\\v_{oay} = v_{fay} + v_{fby}[/tex]

        2 = [tex]v_{fay} + v_fby}[/tex]

Ball A moves in the x-axis therefore it has no final velocity in the y-axis.

       [tex]v_{fay}=0[/tex]

       [tex]v_{fby}[/tex] = 2 m / s

B) Ball A moves on the x-axis.

C) the total momentum is the sum of the momentum of each particle

       [tex]p_{total} = p_a+p_b[/tex]  

To find the module let's use the Pythagorean theorem.          

          [tex]p_{total} = m \sqrt{2^2 + 3.7^2}[/tex]  

           p = 4.2m

Let's use trigonometry for the direction.

         tan θ = [tex]\frac{p_{fb}}{p_{fa}}[/tex]  

         θ = tan⁻¹  [tex]\frac{p_{fb}x}{p_{fa}}[/tex]

         θ = tam⁻¹ [tex]\frac{2}{3.7}[/tex]  

         θ = 28.4º

The total kinetic energy is

         [tex]K_{total}= K_1+K_2[/tex]  

         [tex]K_{total}[/tex] = ½ m [tex]v_a^2[/tex] + ½ m [tex]v_b^2[/tex]

         [tex]K_{total}[/tex] = [tex]\frac{m}{2}[/tex]  (3.7² + 2²)

         [tex]K_{total}[/tex] = 8.845 m

In conclusion using momentum conservation we can find the results for the questions about the collision of the balls are:

  A) The velocities after the collision [tex]v_a[/tex] = 3.7 i m/s and [tex]v_b[/tex] = 2 j  m/s

  B) The direction of ball A is: x-axis

  C) The total momentum is: [tex]p_{total}[/tex] = 4.2m and the direction is θ = 28.4º

      The total kinetic energy is: K_ {total} = 8.84m J

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Answer:

[tex]3.81972\times 10^{-7}\ A[/tex]

Explanation:

B = Magnetic field = [tex]3\times 10^{-8}\ G[/tex]

d = Diameter of loop = 16 cm

r = Radius = [tex]\frac{d}{2}=\frac{16}{2}=8\ cm[/tex]

i = Current

[tex]\mu_0[/tex] = Vacuum permeability = [tex]4\pi \times 10^{-7}\ H/m[/tex]

The magnetic field of a loop is given by

[tex]B=\frac{\mu_0i}{2r}\\\Rightarrow i=\frac{B2r}{\mu_0}\\\Rightarrow i=\frac{3\times 10^{-8}\times 10^{-4}\times 2\times 0.08}{4\pi\times 10^{-7}}\\\Rightarrow i=3.81972\times 10^{-7}\ A[/tex]

The current needed to produce such a field at the center of the loop is [tex]3.81972\times 10^{-7}\ A[/tex]

Answer:

13.54 N/m

0.6 m

4.37 m/s

32.496 m/s²

Explanation:

m = Mass of block = 0.25 kg

k = Spring constant

A = Amplitude

x = Compression of spring = 0.24 m

a = Acceleration = -13 m/s²

v = Velocity = 4 m/s

The weight of the block and force on spring is equal

[tex]ma=-kx\\\Rightarrow k=-\frac{ma}{x}\\\Rightarrow k=-\frac{0.25\times -13}{0.24}\\\Rightarrow k=13.54\ N/m[/tex]

The spring's force constant is 13.54 N/m

Total energy of the system is given by

[tex]E=\frac{1}{2}(mv^2+kx^2)\\\Rightarrow E=\frac{1}{2}(0.25\times 4^2+13.54\times 0.24^2)\\\Rightarrow E=2.39\ J[/tex]

At maximum displacement v = 0

[tex]E=\frac{1}{2}(mv^2+kA^2)\\\Rightarrow E=\frac{1}{2}(0+kA^2)[/tex]

[tex]\\\Rightarrow E=\frac{1}{2}kA^2\\\Rightarrow A=\sqrt{\frac{E2}{k}}\\\Rightarrow A=\sqrt{\frac{2\times 2.39}{13.54}}\\\Rightarrow A=0.6\ m[/tex]

The amplitude of the motion is 0.6 m

Speed of the block

[tex]E=\frac{1}{2}mv_m^2+0\\\Rightarrow v_m=\sqrt{\frac{2E}{m}}\\\Rightarrow v_m=\sqrt{\frac{2\times 2.39}{0.25}}\\\Rightarrow v_m=4.37\ m/s[/tex]

The maximum speed of the block during its motion is 4.37 m/s

Forces in the spring

[tex]ma_m=kA\\\Rightarrow a_m=\frac{kA}{m}\\\Rightarrow a_m=\frac{13.54\times 0.6}{0.25}\\\Rightarrow a_m=32.496\ m/s^2[/tex]

Maximum magnitude of the block's acceleration during its motion is 32.496 m/s²

The force constant of the spring is 13.54 N/m. The amplitude of the oscillatory motion is 0.240 m. The maximum velocity and acceleration are 4.72 m/s and 13 m/s² respectively.

First, we can calculate the spring's force constant k using Newton's second law (F = ma) and Hooke's Law (F = kx), by equating the forces F (Note that the negative acceleration is due to the force acting opposite to displacement). Hence, having the force due to spring F = m * a = 0.250 kg * -13.0 m/s² = -3.25 N and applying force due to spring as per Hooke's law as -k * x. Equating these forces gives k = 13.54 N/m.

Secondly, the amplitude A of the motion can be identified as |x| = 0.240 m, since for a simple harmonic oscillator, the amplitude is the maximum displacement from equilibrium (x = 0).

For the third part, the maximum speed Vmax during its motion can be calculated using the energy conservation principle, as given by the formula, Vmax = sqrt(k/m * A²), which gives Vmax = 4.72 m/s.

Finally, the maximum acceleration of the block Amax happens at the points of maximum displacement, i.e., at the amplitude points, which is Amax = k/m * A = 13m/s².

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Answer:

The resistance is [tex]7.47\times10^{6}\ \Omega[/tex].

Explanation:

Given that,

Power = 259 kV

Current = 429 A

Resistance [tex]R=0.71\times10^{9}\ Omega[/tex]

We need to calculate the current in each insulator

Using formula of current

[tex]I=\dfrac{P}{R}[/tex]

Put the value into the formula

[tex]I=\dfrac{259\times10^{3}}{0.71\times10^{9}}[/tex]

[tex]I=3.64\times10^{-4}\ A[/tex]

So all 95 insulators are in parallel

We need to calculate the resistance

Using formula of resistance

[tex]\dfrac{1}{R}=\sum_{i=1}^{95}\dfrac{1}{R_{i}}[/tex]

Put the value into the formula

[tex]\dfrac{1}{R}=\dfrac{95}{0.71\times10^{9}}[/tex]

[tex]\dfrac{1}{R}=1.338\times10^{-7}[/tex]

[tex]R=\dfrac{1}{1.338\times10^{-7}}[/tex]

[tex]R=7473841.5=7.47\times10^{6}\ \Omega[/tex]

Hence, The resistance is [tex]7.47\times10^{6}\ \Omega[/tex].

Answer:

Explanation:

Given

Radius of Track [tex]r_1=75 m[/tex]

coefficient of Static Friction [tex]\mu _s=0.78[/tex]

Here centripetal Force is Balanced by Friction Force      

thus

[tex]\frac{mv^2}{r}=\mu _sg[/tex]

[tex]\frac{v^2}{r}=\mu _sg[/tex]

[tex]v=\sqrt{\mu _srg}[/tex]

[tex]v=\sqrt{0.78\times 75\times 9.8}[/tex]

[tex]v=23.94 m/s[/tex]

(b)For [tex]r_2=25 m[/tex]

[tex]\mu _s=0.12[/tex]

[tex]v=\sqrt{\mu _sr_2g}[/tex]

[tex]v=\sqrt{0.12\times 25\times 9.8}[/tex]

[tex]v=5.42 m/s[/tex]

Final answer:

The maximum speed that a car can maneuver over a circular turn without sliding depends on the coefficient of static friction and the radius of the turn. This can be calculated using the formula v = sqrt(μs * g * r). For the given scenarios, the maximum speeds are 30.5 m/s and 7.36 m/s respectively.

Explanation:

The maximum speed that a car can maneuver over a circular turn without sliding depends on the coefficient of static friction and the radius of the turn. The maximum speed can be calculated using the formula:

v = sqrt(μs * g * r)

where v is the maximum speed, μs is the coefficient of static friction, g is the acceleration due to gravity (9.8 m/s²), and r is the radius of the turn.

For the first scenario with r = 75.0 m and μs = 0.780, the maximum speed is:

v = sqrt(0.780 * 9.8 * 75.0) = 30.5 m/s

For the second scenario with r = 25.0 m and μs = 0.120, the maximum speed is:

v = sqrt(0.120 * 9.8 * 25.0) = 7.36 m/s

To solve this problem it is necessary to apply the concepts of thermal expansion. Thermal expansion can be expressed in longitudinal terms such as

[tex]\Delta L = L_0 \alpha \Delta T[/tex]

Where,

[tex]\alpha =[/tex] Thermal Expanssion Coefficient

[tex]L_0 =[/tex] Initial Length

[tex]\Delta T =[/tex] Change in Temperature

Our values are given as

[tex]\alpha = 11*10^{-6}/\°C \rightarrow[/tex] from Steel

[tex]L_0 = 15m[/tex]

[tex]T_1 = -23\°C[/tex]

[tex]T_2 = 32\°C[/tex]

Replacing we have that,

[tex]\Delta L = L_0 \alpha \Delta T[/tex]

[tex]\Delta L = (17)(11*10^{-6})(32-(-21))[/tex]

[tex]\Delta L = 0.009911m[/tex]

[tex]\Delta L = 9.911mm[/tex]

Therefore the difference in length of the beam between these two temperature extremes is 9.911mm

Answer:

0.00384 kg/m

Explanation:

The fundamental frequency of string waves is given by

[tex]f=\frac{1}{2L}\sqrt{\frac{F}{\mu}}[/tex]

For some tension (F) and length (L)

[tex]f\propto\frac{1}{\mu}[/tex]

Fundamental frequency of G string

[tex]f_G=196\ Hz[/tex]

Fundamental frequency of E string

[tex]f_E=659.3\ Hz[/tex]

Linear mass density of E string is

[tex]\mu_E=3.4\times 10^{-4}\ kg/m[/tex]

So,

[tex]\frac{F_G}{F_E}=\sqrt{\frac{\mu_E}{\mu_G}}\\\Rightarrow \frac{F_G^2}{F_E^2}=\frac{\mu_E}{\mu_G}\\\Rightarrow \mu_G=3.4\times 10^{-4}\times \frac{659.3^2}{196^2}\\\Rightarrow \mu_G=0.00384\ kg/m[/tex]

The linear density of the G string is 0.00384 kg/m

Answer:

E=930.84 N/C

Explanation:

Given that

I = 1150 W/m²

μ = 4Π x 10⁻⁷

C = 2.999 x 10⁸ m/s

E= C B

C=speed of light

B=Magnetic filed  ,E=Electric filed

Power  P = I A

A=Area=4πr²  ,I=Intensity

[tex]I=\dfrac{CB^2}{2\mu_0}[/tex]

[tex]I=\dfrac{CE^2}{2\mu_0 C^2}[/tex]

[tex]E=\sqrt{{2I\mu_0 C}}[/tex]

[tex]E=\sqrt{{2\times 1150\times 4\pi \times 10^{-7}(2.99792\times 10^8)}}[/tex]

E=930.84 N/C

Therefore answer is 930.84 N/C

To find the magnitude Em of the electromagnetic waves at the top of the earth's atmosphere, we use the intensity of electromagnetic wave and solving the equation Em = sqrt(2Icμo), we can find the magnitude of Em in units of N/C.

To find the magnitude Em of the electromagnetic waves at the top of the Earth's atmosphere, we use the fact that the power received per unit area is the intensity I of the electromagnetic wave. According to the given information, this intensity is 1150 W/m2. The relationship between the intensity and electromagnetic fields is given by the equation I = 0.5 * E²/c * μo. Solving for Em, we get Em = sqrt(2Icμo), where μo = 4π × 10-7 T N/A² is the permeability of free space and c = 2.99792 × 10⁸ m/s is the speed of light.

Subbing in the given values, we can compute Em as:

Em = sqrt[2 * 1150 W/m² * 2.99792 × 10⁸ m/s * 4π × 10-7 T N/A²]

This computation will give the strength of the electric field at the top of the earth’s atmosphere in units of N/C.

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To develop this problem it is necessary to use the continuity equations and Bernoullie's theorem.

It is known from Bernoullie's theorem that

[tex]P_1 + \rho gh_1+\frac{1}{2} \rho v_2^2 =P_2 + \rho gh_2+\frac{1}{2} \rho v_2^2[/tex]

Where

P = Pressure

g = Gravity

h= Height

v = Velocity

[tex]\rho[/tex] = Density

On the other hand we have that the continuity equation is given by

[tex]A_1v_1 = A_2 v_2[/tex]

Where A is the Cross-sectional area and v the velocity.

For our values we know that

[tex]A_1v_1 = A_2 v_2[/tex]

[tex](\frac{\pi d_1^2}{4})v_1 =(\frac{\pi d_2^2}{4})v_2[/tex]

[tex]d_1^2v_1=d_2^2v_2[/tex]

[tex](11.5cm)^2(9.39)=(15.3)^2v_2[/tex]

[tex]v_2 = 5.305m/s[/tex]

Using Bernoulli's expression we can now find the pressure difference,

[tex]P_1 + \rho gh_1+\frac{1}{2} \rho v_2^2 =P_2 + \rho gh_2+\frac{1}{2} \rho v_2^2[/tex]

[tex]P_1-P_2=-\rho gh_1-\frac{1}{2}\rho v_2^2 +\rho gh_2+\frac{1}{2} \rho v_2^2[/tex]

[tex]P_1-P_2 = \rho g (h_1-h_2)+\frac{1}{2}\rho(v_1^2-v_2^2)[/tex]

[tex]P_1-P_2 = (1.3*10^3)(9.8)(9.59)+\frac{1}{2}(1.3*10^3)((9.39)^2-(5.305)^2)[/tex]

[tex]P_1-P_2 = 1.612*10^5Pa[/tex]

Answer:

439.587 kgm²

Explanation:

[tex]m_p[/tex] = Mass of person = 71 kg

R = Radius of platform = 1.62 m

[tex]m_d[/tex] = Mass of disc = 193 kg

The moment of inertia of the system is given by

[tex]I=m_pR^2+\frac{1}{2}m_dR^2\\\Rightarrow I=71\times 1.62^2+\frac{1}{2}\times 193\times 1.62^2\\\Rightarrow I=439.587\ kgm^2[/tex]

The total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk is 439.587 kgm²

Answer:

The magnitude of the linear acceleration of block A is 1.21 m/s² in downward.

Explanation:

Given that,

Mass of block A = 6.00 kg

Mass of block B = 3.00 kg

Moment of inertia = 0.220 kg.m²

Radius = 0.120 m

Suppose we need to find the the magnitude of the linear acceleration of block A

Let a is the acceleration of the blocks.

Let [tex]T_{a}[/tex] and [tex]T_{b}[/tex] are the tension in the A and B cord.

According to figure,

We need to calculate the magnitude of the linear acceleration of block A

Net force acting on block A,

[tex]F_{A}=m_{A}g-T_{A}[/tex]

[tex]m_{A} a=m_{A}g-T_{A}[/tex]

[tex]T_{A}=m_{A}g-m_{A}a[/tex]...(I)

Net force acting on block B,

[tex]F_{B}=T_{B}-m_{B}g[/tex]

[tex]m_{B}a=T_{B}-m_{B}g[/tex]

[tex]T_{B}=m_{B}a+m_{B}g[/tex]...(II)

Net torque acting on pulley

[tex]T_{net}=I\times\alpha[/tex]

[tex]T_{A}r-T_{B}r=I\times \dfrac{a}{r}[/tex]

[tex]T_{A}-T_{B}=I\times\dfrac{a}{r^2}[/tex]

[tex]m_{A}g-m_{A}a-(m_{B}g+m_{B}a)=I\times\dfrac{a}{r^2}[/tex]

[tex]g(m_{A}-m_{B})-a(m_{A}+m_{B})=I\times\dfrac{a}{r^2}[/tex]

[tex]g(m_{A}-m_{B})=I\times\dfrac{a}{r^2}+a(m_{A}+m_{B})[/tex]

[tex]g(m_{A}-m_{B})=a(\dfrac{I}{r^2}+(m_{A}+m_{B}))[/tex]

[tex]a=\dfrac{g(m_{A}-m_{B})}{(\dfrac{I}{r^2}+(m_{A}+m_{B}))}[/tex]

Put the value into the formula

[tex]a=\dfrac{9.8\times(6.00-3.00)}{\dfrac{0.220}{(0.120)^2}+(6.00+3.00)}[/tex]

[tex]a=1.21\ m/s^2[/tex]

Hence, The magnitude of the linear acceleration of block A is 1.21 m/s² in downward.

Answer:

X rays, ultraviolet light, green light, red light, infrared radio waves

Explanation:

The electromagnetic spectrum can be thought of as being arranged either in decreasing frequency or increasing wavelength. The full spectrum is as follows, from the shortest wavelength to the longest:

Gamma rays, X rays, ultraviolet light, violet light, indigo light, blue light, green light, yellow light, orange light, red light, infrared,  radio waves, radar waves, microwaves, television waves, radio waves.

The electromagnetic radiation types can be arranged from shortest to longest wavelength as follows: X rays, ultraviolet light, green light, red light, infrared, and radio waves.

The electromagnetic radiation types provided can be arranged in order of increasing wavelength as follows: X rays, ultraviolet light, green light, red light, infrared, radio waves. This order represents the electromagnetic spectrum from shortest to longest wavelengths. In simpler terms, X rays have the shortest wavelength, followed by ultraviolet light, green light, red light, infrared, and finally radio waves with the longest wavelength. Wavelength and frequency are inversely related - the shorter the wavelength, the higher the frequency.

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To solve this problem it is necessary to apply the related concepts to the scalar value of displacement current, which can be expressed in terms of electric flux as

[tex]I_d = \epsilon_0 \frac{d\Phi_E}{dt}[/tex]

Where,

[tex]\epsilon_0[/tex] = Permitibitty of free space constant

[tex]\Phi_E[/tex] = Magnetic flux

t = time

We know as well that the Flux can be expressed as

[tex]\Phi = EA[/tex]

Here

A= Cross-sectional area

E = Electric Potential in a Uniform electric field

At the same time the electric potential is expressed in terms of Voltage and distance, that is

[tex]E = \frac{V}{d}[/tex]

Using this equation we have then that

[tex]I_d = \epsilon_0 \frac{d\Phi_E}{dt}[/tex]

[tex]I_d = \epsilon_0 \frac{d(EA)}{dt}[/tex]

[tex]I_d = \epsilon_0*A (\frac{d(E)}{dt})[/tex]

[tex]I_d = \epsilon_0*A (\frac{d(V)}{dt*d})[/tex]

[tex]I_d = \frac{\epsilon_0*A}{d} (\frac{d(V)}{dt})[/tex]

According to our values we have that

[tex]\frac{dV}{dt} = 107V/s[/tex]

[tex]A = 0.174m^2[/tex]

[tex]d = 1.06*10^{-3}m[/tex]

[tex]\epsilon = 8.85418^{-12} m^{-3}kg^{-1}s^4A^2[/tex]

Replacing,

[tex]I_d = \frac{\epsilon_0*A}{d} (\frac{d(V)}{dt})[/tex]

[tex]I_d = \frac{(8.85418^{-12})*(0.174)}{1.06*10^{-3}} (107)[/tex]

[tex]I_d = 7.565*10^{-8}A[/tex]

Therefore the displacement current is [tex]7.565*10^{-5}mA[/tex]

The displacement current (in mA) between these plates is equal to [tex]1.56 \times 10^{-5}\;mA[/tex]

Given the following data:

Scientific data:

Mathematically, the displacement current (in mA) between the plates in an electric field is given by this formula:

[tex]I_d=\frac{\epsilon _o A}{d} (\frac{d(v)}{dt} )\\\\[/tex]

Where:

Substituting the given parameters into the formula, we have;

[tex]I_d=\frac{8.854 \times 10^{-12} \times 0.174 \times 107}{1.06 \times 10^{-2}} \\\\I_d=1.56 \times 10^{-8}\\\\I_d=1.56 \times 10^{-5}\;mA[/tex]

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Answer:Hollow sphere

Explanation:

Given

same mass for solid and hollow sphere

same [tex]v_{cm}[/tex] before they start up incline

Moment of inertia of solid Sphere

[tex]I_1=\frac{2}{5}Mr^2[/tex]

Moment of inertia of hollow sphere

[tex]I_2=\frac{2}{3}Mr^2[/tex]

Conserving Energy at bottom and top point for solid sphere

kinetic energy +Rotational Energy=Potential energy

[tex]\frac{1}{2}Mv_{cm}^2+\frac{1}{2}I\omega ^2=mgh_1[/tex]

for pure rolling [tex]v_{cm}=\omega r[/tex]

[tex]\frac{1}{2}Mv_{cm}^2+\frac{1}{2}\times \frac{2}{5}Mr^2=Mgh_1[/tex]

[tex]\frac{7}{10}Mv_{cm}^2=Mgh_1[/tex]

[tex]h_1=\frac{7v_{cm}^2}{10g}[/tex]

For hollow sphere

[tex]\frac{1}{2}Mv_{cm}^2+\frac{1}{2}\times \frac{2}{3}Mr^2=Mgh_2[/tex]

[tex]h_2=\frac{5v_{cm}^2}{6g}[/tex]

therefore height gained by hollow sphere is more

The hollow sphere reached the greatest height.

The height reached by each object is determined by applying the principle of conservation of energy as show below;

K.E = P.E

¹/₂mv² + ¹/₂Iω² = mgh

where;

I =  ²/₅mr²

¹/₂mv² + ¹/₂Iω² = mgh

¹/₂mv² + ¹/₂(²/₅mr²)(v/r)² = mgh

¹/₂v² + ¹/₅v² = gh

7v² = 10gh

h = 7v²/10g

h = 0.7(v²/g)

I = ²/₃mr²

¹/₂mv² + ¹/₂Iω² = mgh

¹/₂mv² + ¹/₂(²/₃mr²)(v/r)² = mgh

¹/₂v² + ¹/₃v² = gh

5v² = 6gh

h = 5v²/6g

h = 0.83(v²/g)

Thus, the hollow sphere reached the greatest height.

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