man stands on a platform that is rotating (without friction) with an angular speed of 1.2 rev/s; his arms are outstretched and he holds a brick in each hand.The rotational inertia of the system consisting of the man, bricks, and platform about the central vertical axis of the platform is 6.0 k g times m squared. If by moving the bricks the man decreases the rotational inertia of the system to 2.0 k g times m squared, what is the resulting angular speed of the platform in rad/s? Express to 3 sig figs.

Explanation:

If the kinetic energy of an object is given and we need to find its velocity of motion, then we can find it by using the formula of kinetic energy as :

[tex]K=\dfrac{1}{2}mv^2[/tex]

m is mass of the object

We can rearrange the above equation such that,

[tex]v=\sqrt{\dfrac{2K}{m}}[/tex]

Hence, this is the velocity at the end of a rollercoaster ride.

Explanation:

It is given that,

Total mass is 70 kg

The truck exerts a constant force of 20 N.

Then the net force is given by :

F = ma

a is acceleration of rider

[tex]a=\dfrac{F}{m}\\\\a=\dfrac{20}{70}\\\\a=\dfrac{2}{7}\ m/s^2[/tex]

Initial velocity of rider is 0. So, using equation of kinematics to find the final velocity as :

[tex]v=u+at\\\\v=at\\\\v=\dfrac{2}{7}\times 15\\\\v=4.28\ m/s[/tex]

Since, 1 m/s = 2.23 mph

4.28 m/s = 9.57 mph

So, the speed of the rider is 4.28 m/s or 9.57 mph.  

Answer:

a. 78 degree

Explanation:

According to Snell's Law, we have:

(ni)(Sin θi) = (nr)(Sin θr)

where,

ni = Refractive index of medium on which light is incident

ni = Refractive index of ethyl alcohol = 1.361

nr = Refractive index of medium from which light is refracted

nr = Refractive index of ethyl alcohol = 1.333

θi = Angle of Incidence

θr = Angle of refraction

So, the Angle of Incidence is know as the Critical Angle (θc), when the refracted angle becomes 90°. This is the case of total internal reflection. That is:

θi = θc

when, θr = 90°

Therefore, Snell's Law becomes:

(1.361)(Sin θc) = (1.333)(Sin 90°)

Sin θc = 1.333/1.361

θc = Sin⁻¹ (0.9794)

θc = 78.35° = 78° (Approximately)

Therefore, correct answer will be:

a. 78 degree

The angle relative to the normal interface of the two liquids at which the light is totally reflected is 78 degrees.

From the information given;

According to Snell's Law of refraction;

[tex]\mathbf{n_1 sin \theta _c = n_2 sin \theta_r}[/tex]

[tex]\mathbf{1.361 \times sin \theta _c = 1.333 \times sin 90}[/tex]

[tex]\mathbf{ sin \theta _c =\dfrac{ 1.333 \times sin 90}{1.361}}[/tex]

[tex]\mathbf{ sin \theta _c =\dfrac{ 1.333 \times 1}{1.361}}[/tex]

[tex]\mathbf{ \theta _c = sin^{-1} (0.9794)}[/tex]

[tex]\mathbf{ \theta _c =78.35^0}[/tex]

[tex]\mathbf{ \theta _c \simeq78^0}[/tex]

Therefore, we can conclude that the angle relative to the normal interface of the two liquids at which the light is totally reflected is 78 degrees.

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Answer:

E) She can perform no measurement to determine this quantity.

Explanation:

A spacecraft is a machine used to fly in outer space.

According to Isaac Newton's third law of motion, every action produces an equal and opposite reaction. When fuel is shoot out of one end of the rocket, the rocket moves forward for which no air is required.

As Leah is moving in a spaceship at a constant velocity away from a group of stars, she cannot measure to determine this quantity.

Answer:

i) A dimension is the physical nature of a quantity. The two limitations of dimensional analysis is as following:

ii) Given:

Velocity = v

Time = t

Force = F

Force = mass x acceleration

         = mass x velocity/time

So, mass= (force x time) / velocity

[mass] = Ftv^-1

Hence, dimesnion of mass is Ftv^-1.

Answer:

Energy Stored = 36000 J = 36 KJ

Explanation:

The power of a battery is given by the formula:

P = IV

where,

P = Power delivered by the battery

I = Current Supplied to the battery

V = Potential Difference between terminals of battery = 12 volt

Now, we multiply both sides by the time period (t):

Pt = VIt

where,

Pt = (Power)(Time) = Energy Stored = E = ?

It = Battery Current Rating = 50 A.min

Converting this to A.sec;

It = Battery Current Rating = (50 A.min)(60 sec/min) = 3000 A.sec

Therefore,

E = (12 volt)(3000 A.sec)

E = 36000 J = 36 KJ

Question: The circular handle of a faucet is attached to a rod that opens and closes a valve when the handle is turned. If the rod has a diameter of 1cm and the IMA of the machine is 6, what is the radius of the handle?

Answer:

Radius of the handle = 3 cm = 0.03 m

Explanation:

Mechanical Accuracy, MA = (Radius of the handle)/(Radius of the rod).......(1)

Diameter of the rod = 1 cm

Radius of the rod = Diameter/2

Radius of the rod = 1/2

Radius of the rod = 0.5 cm

Mechanical Accuracy of the machine, MA = 6

Substitute the values into equation (1)

6 =  (Radius of the handle)/0.5

Radius of the handle = 6 * 0.5

Radius of the handle = 3 cm

The Radius of the handle is = 3 cm = 0.03 m

Since

Mechanical Accuracy, MA = (Radius of the handle)/(Radius of the rod)

Here,

Diameter of the rod = 1 cm

We know that

The radius of the rod = Diameter/2

So,

Radius of the rod = 1/2

So,

Radius of the rod = 0.5 cm

Now

Mechanical Accuracy of the machine, MA = 6

Now

6 =  (Radius of the handle)/0.5

Radius of the handle = 6 * 0.5

Radius of the handle = 3 cm

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Answer:

Force, [tex]F=4.86\times 10^{-4}\ N[/tex]

Explanation:

We have,

Masses of two twins are 54 kg each

They are placed at a distance of 0.02 m

It is required to find the force of gravity between them. The formula used to find the gravitational force between masses is given by :

[tex]F=G\dfrac{m_1m_2}{r^2}[/tex]

plugging all the known values:

[tex]F=6.67\times 10^{-11}\times \dfrac{54^2}{(0.02)^2}\\\\F=4.86\times 10^{-4}\ N[/tex]

So, the force of gravity between them is [tex]4.86\times 10^{-4}\ N[/tex].  

Answer:

Explanation:

According to newton's secomd law, ∑F = ma

∑F is the summation of the force acting on the body

m is the mass of the body

a is the acceleration

Given the normal force when the elevator starts N1 = 592N

Normal force after the elevator stopped N2 = 400N

When the elevator starts, its moves upward, the sum of force ∑F = Normal (N)force on the elevator - weight of the person( Fg)

When moving up;

N1 - Fg = ma

N1 = ma + Fg ...(1)

Stopping motion of the elevator occurs after the elevator has accelerates down. The sum of forces in this case will give;

N2 - Fg = -ma

N2 = -ma+Fg ...(2)

Adding equation 1 and 2 we will have;

N1+N2 = 2Fg

592N + 400N = 2Fg

992N 2Fg

Fg = 992/2

Fg = 496N

The weight of the person is 496N

\b) To get the person mass, we will use the relationship Fg = mg

g = 9.81m/s

496 = 9.81m

mass m = 496/9.81

mass = 50.56kg

c) To get the magnitude of acceleration of the elevator, we will subtract equation 1 from 2 to have;

N1-N2 = 2ma

592-400 = 2(50.56)a

192 = 101.12a

a = 192/101.12

a = 1.90m/s²

Answer:

v = 46.99 m/s

Explanation:

The velocity of the ball just before it touches the ground, is given by the following formula:

[tex]v=\sqrt{v_x^2+v_y^2}[/tex]           (1)

vx: horizontal component of the velocity

vy: vertical component of the velocity

The vertical component vy is calculated by using the following formula:

[tex]v_y^2=v_{oy}^2+2gh[/tex]   (2)

vy: final velocity

voy: initial vertilal velocity = 0m/s  (because it is a semi parabolic motion)

g: gravitational acceleration = 9.8 m/s^2

h: height = 1.60m

You replace the values of the parameters in the equation (2):

[tex]v_y=2(9.8m/s^2)(1.60m)=31.36\frac{m}{s}[/tex]

vx is calculated by using the information about the horizontal range of the ball:

[tex]R=v_o\sqrt{\frac{2h}{g}}[/tex]    (3)

R: horizontal range of the ball = 20.0 m

You solve the previous equation for vo, the initial horizontal velocity:

[tex]v_o=R\sqrt{\frac{g}{2h}}=(20.0m)\sqrt{\frac{9.8m/s^2}{2(1.60m)}}\\\\v_o=35\frac{m}{s}[/tex]

The horizontal component of the velocity is constant in the complete trajectory, hence, you have that

vx = vo = 35 m/s

Finally, you replace the values of vx and vy in the equation (1):

[tex]v=\sqrt{(35m/s)^2+(31.36m/s)^2}=46.99\frac{m}{s}[/tex]

The velocity of the ball just before it touches the ground is 46.99 m/s

Answer:

Explanation:

  P = VI

  I = P/V = P/120

  R = V/I = V/(P/V) = V^2/P = 14400/P

Toaster: I = 1800/120 = 15 . . . amps

  R = 14400/1800 = 8 . . . ohms

Fry pan: I = 1300/120 = 10.833 . . . amps

  R = 14400/1300 = 11.08 . . . ohms

Lamp: I = 100/120 = 0.833 . . . amps

  R = 14400/100 = 144 . . . ohms

The total current exceeds 20 A, so will blow the fuse.

Answer:

0.00000009512

Explanation:

Scientific notation is a very useful and abbreviated way of writing quantities that are very large or small. It consists of placing the number with an integer and multiplying by an exponent to arrive at the same number.

let's pass the number 9,512 10⁻⁸ to decimal notation

       9,512 / 10⁸ = 9,512 / 100000000

        0.00000009512

As we see writing this number, it is very easy to make mistakes

Answer:

[tex]\Delta p=1.3475\ kg-m/s[/tex]

Explanation:

The computation of magnitude of the change in the ball's momentum in kg · m/s is shown below:-

We represent

The ball mass =  m = 275 g = 0.275 kg

Thus it goes to the floor and resurfaces upward.

The ball hits the ground at 3.30 m/s speed that is

u = -3.30 m/s which represents the Negative since the ball hits the ground)

It rebounds at a speed of 1.60 m / s i.e. v = 1.60 m/s (positive as the ball rebounds upstream)

[tex]\Delta p=p_f-p_i[/tex]

[tex]\Delta p=m(v-u)[/tex]

[tex]\Delta p=0.275\ kg(1.60\ m/s-(-3.30\ m/s))[/tex]

[tex]\Delta p=1.3475\ kg-m/s[/tex]

Answer:

2000 m

Explanation:

since the boat is 1000 m from the river bank, the beam must be focused at this point. This indicates that the focal length is 1000 m

for a spherical mirror, the focal length is given by

f = R/2

where R is the radius of curvature

1000 = R/2

R = 2000 m

R = 2000 m

this means that the radius of curvature must be 2000 m

Answer:

B. energy

Explanation:

A vector has direction.

Energy does not have a direction.

Answer:

= 0.0532 cm^3

Explanation:

The computation of volume of the sphere is shown below:-

[tex]Density = \frac{Mass}{Volume}[/tex]

Where,

Density = 13.6 g/cm^3

Mass of sphere = 0.723 g

now we will put the values into the above formula to reach volume of the sphere which is here below:-

[tex]Volume = \frac{0.723}{13.6}[/tex]

= 0.0532 cm^3

Therefore for computing the volume of the sphere we simply applied the above formula.

Answer:

d = 41.91 m

Explanation:

In order to calculate the distance traveled by the train while superman passes it, you write the equations of motion for both superman and train:

For train, you have a motion with constant speed. You write the equation of motion of the position of the front of the train:

[tex]x=x_o+v_1t[/tex]    (1)

xo: initial position of the front of the train = 500m

v1: speed of the train = 50km/h

For superman, you take into account that the motion is an accelerated motion (you assume superman is at the origin of coordinates):

[tex]x'=v_2t+\frac{1}{2}at^2[/tex]   (1)

v2: initial speed of superman = 100km/h

a: acceleration = 10m/s^2

When superman passes the train, both positions x and x' will be equal. Hence, you equal the equations (1) and (2) and you calculate the time t. But before you convert the units of the velocities v1 and v2 to m/s:

[tex]v_1=50\frac{km}{h}*\frac{1000m}{1km}*\frac{1h}{3600s}=13.88\frac{m}{s}\\\\v_2=100\frac{km}{h}=\frac{1000m}{1km}*\frac{1h}{3600s}=27.77\frac{m}{s}[/tex]

Thus, you equal x=x'

[tex]x=x'\\\\x_o+v_1t=v_2t+\frac{1}{2}at^2\\\\500m+(13.88m/s)t=(27.77m/s)t+\frac{1}{2}(10m/s^2)t^2\\\\(50\frac{m}{s^2})t^2+(13.89\frac{m}{s})t-500m=0[/tex]

You solve the last equation for t by using the quadratic formula:

[tex]t_{1,2}=\frac{-13.89\pm \sqrt{(13.89)^2-4(50)(-500)}}{2(50)}\\\\t_{1,2}=\frac{-13.89\pm 316.53}{100}\\\\t_1=3.02s\\\\t_2=-3.30s[/tex]

You only use t1 = 3.02s because negative times do not have physical meaning.

Next, you replace this value of t in the equation (1) to calculate the position of the train (for when superman just passed it):

[tex]x=500m+(13.88m/s)(3.02s)=541.91m[/tex]

x is the position of the front of the train, then, the dstance traveled by the train is:

d = 541.91m - 500m = 41.91 m

Answer:

the correct answer is B

Explanation:

We analyze this exercise a little, the block goes into the air and is under the acceleration of gravity. The ball is fired by the hand and is describing a parabolic movement, subjected to the acceleration of gravity.

For the ball to hit the block we must have the distance the ball goes up equal to the distance the block moves, therefore we must shoot the ball at the block at its highest point.

Let's write the kinematic equation for the two bodies

The block. At the highest point of the path

      y = - ½ g t2

The ball, in its vertical movement

     y = vo t - ½ g t2

therefore the correct answer is B

Answer:

r = 20 m

Explanation:

The formula for the angular momentum of a rotating body is given as:

L = mvr

where,

L = Angular Momentum = 10000 kgm²/s

m = mass

v = speed = 2 m/s

r = radius of merry-go-round

Therefore,

10000 kg.m²/s = mr(2 m/s)

m r = (10000 kg.m²/s)/(2 m/s)

m r = 5000 kg.m   ------------- equation 1

Now, the moment of inertia of a solid uniform disc about its axis through its center is given as:

I = (1/2) m r²

where,

I = moment of inertia = 50000 kg.m²

Therefore,

50000 kg.m² = (1/2)(m r)(r)

using equation 1, we get:

50000 kg.m² = (1/2)(5000 kg.m)(r)

(50000 kg.m²)/(2500 kg.m) = r

r = 20 m

Answer:

a) 5.143 m/s^2

b) T = 15.43 N

c) Fr = 10.29 N

d) 5.143 m/s^2 , T = 15.43 N

Explanation:

Given:-

- The mass of left most block, m1 = 1.0 kg

- The mass of center block, m2 = 2.0 kg

- The mass of right most block, m3 = 4.0 kg

- A force that acts on the right most block, F = 36 N

Solution:-

a)

- For the first part we will consider the three blocks with masses ( m1 , m2 , and m3 ) as one system on which a force of F = 36 N is acted upon. The masses m1 and m3 are connected with a string with tension ( T ) and the m1 and m2 are in contact.

- We apply the Newton's second law of motion to the system with acceleration ( a ) and the combined mass ( M ) of the three blocks as follows:

                       [tex]F = M*a\\\\36 = ( 1 + 2 + 4 )*a\\\\a = \frac{36}{7}\\\\a = 5.143 \frac{m}{s^2}[/tex]

Answer: The system moves in the direction of external force ( F ) i.e to the right with an acceleration of 5.143 m/s^2

b)

- The blocks with mass ( m1 and m3 ) are connected with a string with tension ( T ) with a combined acceleration of ( a ).

- We will isolate the massive block ( m3 ) and notice that two opposing forces ( F and T ) act on the block.

- We will again apply the Newton's 2nd law of motion for the block m3 as follows:

                       [tex]F_n_e_t = m_3 * a\\\\F - T = m_3 * a\\\\36 - T = 4*5.143\\\\T = 36 - 20.5714\\\\T = 15.43 N[/tex]

Answer:- A tension of T = 15.43 Newtons acts on both blocks ( m1 and m3 )

c)

- We will now isolate the left most block ( m1 ) and draw a free body diagram. This block experiences two forces that is due to tension ( T ) and a reaction force ( Fr ) exerted by block ( m2 ) onto ( m3 ).

- Again we will apply the the Newton's 2nd law of motion for the block m3 as follows:

                         [tex]F_n_e_t = m_1*a\\\\T - F_r = m_1*a\\\\15.43 - F_r = 1*5.143\\\\F_r = 15.43 - 5.143\\\\F_r = 10.29 N[/tex]

- The reaction force ( Fr ) is contact between masses ( m1 and m2 ) exists as a pair of equal magnitude and opposite direction acting on both the masses. ( Newton's Third Law of motion )

Answer: The block m2 experiences a contact force of ( Fr = 10.29 N ) to the right.

d)

- If we were to stack the block ( m2 ) on-top of block ( m1 ) such that block ( m2 ) does not slip we the initial system would remain the same and move with the same acceleration calculated in part a) i.e 5.143 m/s^2

- We will check to see if the tension ( T ) differs or not as the two block ( m1 and m2 ) both experience the same Tension force ( T ) as a sub-system. with a combined mass of ( m1 + m2 ).

- We apply the Newton's 2nd law of motion for the block m3 as follows:

                            [tex]T = ( m_1 + m_2 ) *a\\\\T = ( 1 + 2 ) * 5.143\\\\T = 15.43 N[/tex]

Answer: The acceleration of the whole system remains the same at a = 5.143 m/s^2 and the tension T = 15.43 N also remains the same.

Answer:

Explanation:

Sensorimotor Infants "think" by acting on the world with their eyes, ears, hands, and mouth.

Preoperational. Development of language and make-believe play takes place.

Concrete Operational children think in a logical, organized fashion only when dealing with concrete information they can perceive directly.

Formal Operational.  Adolescences  can also evaluate the logic of verbal statements without referring to real-world circumstances.

Sensorimotor, preoperational, concrete operational, and formal operational are Piaget's four phases of cognitive development.

The way youngsters think, investigate, and figure things out is referred to as cognitive development.

Piaget defined four stages of cognitive development:

1. Sensorimotor. From birth through the age of 18-24 months.

2. Preoperational.From infancy (18-24 months) until toddlerhood (age 7)

3. Operational concrete. 7 to 11 years old

4. Formal operational. From adolescence to adulthood

Hence, sensorimotor, preoperational, concrete operational, and formal operational are Piaget's four phases of cognitive development.

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Answer:

Maximum weight that can be lifted = 18,000 N

Explanation:

Given:

Cross-sectional area of input (A1) = 0.004 m²

Cross-sectional area of the output (A2) = 1.2 m ²

Force (F) = 60 N

Computation:

Pressure on input piston (P1) = F / A1

Assume,

Maximum weight lifted by piston = W

Pressure on output piston (P2) = W / A2

We, know that

P1 = P2

[F / A1]  = [W / A2]

[60 / 0.004] = [W / 1.2]

150,00 = W / 1.2

Weight = 18,000 N

Maximum weight that can be lifted = 18,000 N

Answer:

a. global warming

Explanation:

that's the definitain of global warming

Answer:

A climate

Explanation:

Answer:

  More than 48%

Explanation:

If the interest is computed monthly on the outstanding balance, it has an effective annual rate of ...

  (1 +4%)^12 -1 = 60.1% . . . .  more than 48%

The effective annual or yearly interest rate would be=30.56% which is Between 4% and 48%

The formula used to calculate annual Interest rate =

[tex](1+ \frac{i}{n} ) {}^{n} - 1[/tex]

where i= nominal interest rate = 4%

n= number of periods= 12 months

Annual Interest rate=

[tex](1 + \frac{4\%}{12} ) {}^{12} - 1[/tex]

= (1+0.333)^12 -1

= (1.333)^12-1

= 31.56 - 1

= 30.56%

Therefore, the effective annual or yearly interest rate would be= 30.56%

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Answer:

[tex]52.25\times10^4W\\699.1 hp[/tex]

Explanation:

According to the energy conversation:

ΔK=[tex]-f_kd+W[/tex]

ΔK=[tex]K_f-K_i ; K=1/2 mv^2[/tex]

where,

[tex]k_i, k_f[/tex] are initial and final kinetic energy of the system.

[tex]v_i[/tex]= initial velocity of the system

[tex]v_f[/tex]=final velocity of the system

W= total work done on the system

[tex]f_k[/tex]= friction force

d= distance traveled

Given: [tex]v_f[/tex]=110m/s

d=400m

[tex]f_k[/tex]=1200N

[tex]v_i[/tex]=0m/s

t=7.3s

ΔK=[tex]-f_kd+W[/tex]

W= ΔK + [tex]f_kd[/tex]

  =[tex]K_f-K_i+f_kd\\[/tex]

  [tex]=1/2 mv_f^2-1/2 mv_i^2+f_kd\\=\frac{1}{2} \times 550\times110^2 - \frac{1}{2} \times 550\times0^2+ (1200\times400)\\=3807500[/tex]

[tex]P=\frac{W}{t} =\frac{3807500}{7.3} \\P=52.15 \times10^4w\\P=\frac{52.15 \times10^4}{746} \\P=699.1 hp[/tex]

Answer:

t = 5.03 s

Explanation:

To find the time interval when Sheila encounter her brother, you first calculate the angular speed of both Sheila and her brother.

You use the following formula:

[tex]\omega = \frac{v}{r}[/tex]

w: angular speed

v: tangential speed

r: radius of the trajectory = 8 m

For  you have:

[tex]\omega=\frac{4m/s}{8m}=0.5\frac{rad}{s}[/tex]

For her brother:

[tex]\omega'=\frac{6m/s}{8m}=0.75\frac{rad}{s}[/tex]

Next, they will encounter to each other when the angular distance of the Brother of sheila equals the angular distance of Sheila in the opposite direction. This can be written as follow:

[tex]\theta=\omega t\\\\\theta'=\omega ' t[/tex]

They encounter for θ = 2π-θ':

[tex]\omega t=2\pi-\omega' t[/tex]

You replace the values of the parameters in the previous equation and solve for t:

[tex]0.5t=2\pi-0.75t\\\\1.25t=2\pi\\\\t=5.026\approx5.03[/tex]

Hence, Sheila encounter her brother in 5.03 s

Explanation:

The relationship between electric force and distance between charged objects is given by the formula as follows :

[tex]F=\dfrac{kq_1q_2}{d^2}[/tex]

k is electrostatic constant and d is distance between charges

The electric force between charges is inversely proportional to the square of distance between them.

Answer:

(a) t = 3.74 s

(b) H = 136.86 m

(c) Vₓ = 41.83 m/s,  Vy = 0 m/s

(d) ax = 0 m/s²,  ay = 9.8 m/s²

Explanation:

(a)

Time to reach maximum height by the projectile is given as:

t = V₀ Sinθ/g

where,

V₀ = Launching Speed = 55.6 m/s

Angle with Horizontal = θ = 41.2°

g = 9.8 m/s²

Therefore,

t = (55.6 m/s)(Sin 41.2°)/(9.8 m/s²)

t = 3.74 s

(b)

Maximum height reached by projectile is:

H = V₀² Sin²θ/g

H = (55.6 m/s)² (Sin²41.2°)/(9.8 m/s²)

H = 136.86 m

(c)

Neglecting the air resistance, the horizontal component of velocity remains constant. This component can be evaluated by the formula:

Vₓ = V₀ₓ = V₀ Cos θ

Vₓ = (55.6 m/s)(Cos 41.2°)

Vₓ = 41.83 m/s

Since, the projectile stops momentarily in vertical direction at the highest point. Therefore, the vertical component of velocity will be zero at the highest point.

Vy = 0 m/s

(d)

Since, the horizontal component of velocity is uniform. Thus there is no acceleration in horizontal direction.

ax = 0 m/s²

The vertical component of acceleration is always equal to the acceleration due to gravity during projectile motion:

ay = 9.8 m/s²

Answer:

0.002kg and 0.01cm

Explanation:

500 leaves has a thickness is 5cm

Means I leaf has a thickness of 5/500= 0.01cm

Similarly the mass of one leaf would be 1/500 =0.002kg

Answer:

Explanation:

angular momentum of the putty about the point of rotation

= mvR   where m is mass , v is velocity of the putty and R is perpendicular distance between line of velocity and point of rotation .

= .045 x 4.23 x 2/3 x .95 cos46

= .0837 units

moment of inertia of rod = ml² / 3 , m is mass of rod and l is length

= 2.95 x .95² / 3

I₁ = .8874 units

moment of inertia of rod + putty

I₁ + mr²

m is mass of putty and r is distance where it sticks

I₂  = .8874 + .045 x (2 x .95 / 3)²

I₂ = .905

Applying conservation of angular momentum

angular momentum of putty = final angular momentum of rod+ putty

.0837 = .905 ω

ω is final angular velocity of rod + putty

ω = .092 rad /s .