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A uniform 190 g rod with length 43 cm rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small 38 g beads are mounted on the rod such that they are able to slide without friction along its length. Initially the beads are held by catches at positions 10 cm on each sides of the center, at which time the system rotates at an angular speed of 12 rad/s. Suddenly, the catches are released and the small beads slide outward along the rod. Find the angular speed of the system at the instant the beads reach the ends of the rod. Answer in units of rad/s.

Answers

Answer 1

Answer:

Answer:

The angular speed of the system at the instant the beads reach the ends of the rod is 14.87 rad/s

Explanation:

Moment of inertia is given as;

I = ¹/₁₂×ML² + 2mr²

where;

I is the moment of inertia

M is the mass of the rod = 0.19 kg

L is the length of the rod = 0.43 m

m is the mass of the bead = 0.038 kg

r is the distance of one bead

Initial moment of inertial is given as;

$I_i = (1)/(12)ML^2 +2mr_1^2$

Final moment of inertia is also given as

$I_f= (1)/(12)ML^2 +2mr_2^2$

Angular momentum is the product of angular speed and moment of inertia;

= Iω

From the principle of conservation of angular momentum;

$I_i \omega_i = I__f } \omega_f$

$((1)/(12)ML^2 +2mr_1^2) \omega_i = ((1)/(12)ML^2 +2mr_2^2) \omega_f$

Given;

ωi = 12 rad/s

r₁ = 10.0 cm = 0.1 m

r₂ = 10.0cm/4 = 2.5 cm = 0.025 m

Substitute these values in the above equation, we will have;

$((1)/(12)*0.19*(0.43)^2 +2*0.038(0.1)^2) 12 = ((1)/(12)*0.19*(0.43)^2 +2*0.038*(0.025)^2) \omega_f\n\n0.04425 =0.002975\ \omega_f\n\n\omega_f = (0.04425)/(0.002975) = 14.87\ rad/s$

Therefore, the angular speed of the system at the instant the beads reach the ends of the rod is 14.87 rad/s

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Consider two aluminum rods of length 1 m, one twice as thick as the other. If a compressive force F is applied to both rods, their lengths are reduced by LaTeX: \Delta L_{thick}ΔLthickand LaTeX: \Delta L_{thin}ΔLthin, respectively.

Answers

Complete Image

Consider two aluminum rods of length 1 m, one twice as thick as the other. If a compressive force F is applied to both rods, their lengths are reduced by $\Delta L_(thick)$ and $\Delta L_(thin)$ , respectively.The ratio ΔLthick rod/ΔLthin rod is:

a.) =1

b.) <1

c.) >1

Answer:

The ratio is less than 1 i.e $(1)/(2) <1$ option B is correct

Explanation:

The Young Modulus of a material is generally calculated with this formula

$E = (\sigma)/(\epsilon)$

Where $\sigma$ is the stress = $(Force)/(Area)$

$\epsilon$ is the strain = $(\Delta L)/(L)$

Making Strain the subject

$\epsilon = (\sigma)/(E)$

now in this question we are that the same tension was applied to both wires so

$(\sigma)/(E)$ would be constant

Hence

$(\Delta L)/(L) = constant$

for the two wire we have that

$(\Delta L_1)/(L_1) = (\Delta L_2)/(L_2)$

Looking at young modulus formula

$E = ((F)/(A) )/((\Delta L)/(L) )$

$E * (\Delta L )/(L) = (F)/(A)$

$A * (\Delta L)/(L) = (F)/(E)$

Now we are told that a comprehensive force is applied to the wire so for this question

$(F)/(E)$ is constant

And given that the length are the same

$A_1 (\Delta L_(thin))/(L_(thin)) = A_2 (\Delta L_(thick))/(L_(thick))$

Now we are told that one is that one rod is twice as thick as the other

So it implies that one would have an area that would be two times of the other

Assuming that

$A_2 = 2 A_1$

$A_1 (\Delta L_(thin))/(L_(thin)) = 2 A_1 (\Delta L_(thick))/(L_(thick))$

$(\Delta L_(thin))/(L_(thin)) = 2 (\Delta L_(thick))/(L_(thick))$

From the question the length are equal

$\Delta L_(thin) =2 \Delta L_(thick)$

$(\Delta L_(thick))/(\Delta L_(thin) ) = (1)/(2)$

Hence the ratio is less than 1

1. Draw a quantitative motion map for the following description: A bicyclist speeds along a road at 10 m/s for 6 seconds. Then she stops for three seconds to make a 180˚ turn and then travels at 5 m/s for 3 seconds.

Answers

Answer:

Please find the attached file for the figure.

Explanation:

Given that a bicyclist speeds along a road at 10 m/s for 6 seconds.

Its acceleration = 10/6 = 1.667 m/s^2

The distance covered = 1/2 × 10 × 6

Distance covered = 30 m

That is, displacement = 30 m

Then she stops for three seconds to make a 180˚ turn and then travels at 5 m/s for 3 seconds.

The acceleration = 5/3 = 1.667 m/s^2

The displacement = 1/2 × 5 × 3

Displacement = 7.5 m

The resultant acceleration will be equal to zero.

While the resultant displacement will be:

Displacement = 30 - 7.5 = 22.5 m

Please find the attached file for the sketch.

A stream of water emerging from a faucet narrows as fails. The cross-sectional area of the soutis As -6.40 cm. water comes out of the spout at a speed of 33.2 cm/s, and the waterfalls h = 7.05 cm before iting the bottom of sink What is the cross-sectional area of the water stream just before it is the sink? a. 0.162 cm3 b. 1.74 cm3c. 6.21cm3d. 0.943cm3

Answers

Answer:

The area of the water stream will be 1.74 cm^2

Explanation:

initial velocity of water u = 33.2 cm/s

initial area = 6.4 cm^2

height of fall = 7.05 cm

final area before hitting the sink = ?

as the water falls down the height, it accelerates under gravity; causing the speed to increase, and the area to decrease.

first we find the velocity before hitting the sink

using

$v^(2) = u^(2) + 2gh$ -----Newton's equation of motion

where v is the velocity of the water stream at the sink

u is the initial speed of the water at the spout

h is the height of fall

g is acceleration due to gravity, and it is positive downwards.

g = 981 cm/s^2

imputing relevant values, we have

$v^(2) = 33.2^(2) + (2 * 981 * 7.05)$

$v^(2) = 1102.24 + 13832.1 = 14934.34$

$v = √(14934.34)$ = 122.206 cm/s

according to continuity equation,

A1v1 = A2v2

where A1 is the initial area

V1 = initial velocity

A2 = final area

V2 = final velocity

6.4 x 33.2 = 122.206 x A2

212.48 = 122.206 x A2

A2 = 212.48 ÷ 122.206 ≅ 1.74 cm^2

7. If the impact of the golf club on the ball in the previous question occurs over a time of 2 x 10 seconds, whatforce does the ball experience to accelerate from rest to 73 m/s?

Answers

Answer:

3.65 x mass

Explanation:

Given parameters:

Time = 20s

Initial velocity = 0m/s

Final velocity = 73m/s

Unknown:

Force the ball experience = ?

Solution:

To solve this problem, we apply the equation from newton's second law of motion:

F = m $(v - u)/(t)$

m is the mass

v is the final velocity

u is the initial velocity

t is the time taken

So;

F = m ( $(73 - 0)/(20)$ ) = 3.65 x mass

Final answer:

To calculate the force experienced by the ball to accelerate from rest to 73 m/s, use Newton's second law of motion.

Explanation:

To calculate the force experienced by the ball to accelerate from rest to 73 m/s, we can use Newton's second law of motion, which states that force equals mass times acceleration (F = m * a).

Since the ball starts from rest, its initial velocity (vi) is 0 m/s. The final velocity (vf) is 73 m/s. The time (t) taken for the impact is given as 2 x 10 seconds. So, the acceleration (a) can be calculated using the formula a = (vf - vi) / t.

Substituting the given values into the equation, we have a = (73 - 0) / (2 x 10) = 3.65 m/s^2.

Now, we can find the force (F) using the formula F = m * a. If the mass of the ball is known, we can substitute it into the equation to find the force experienced by the ball.

Learn more about Force here:

brainly.com/question/33422584

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A 580-mm long tungsten wire, with a 0.046-mm-diameter circular cross section, is wrapped around in the shape of a coil and used as a filament in an incandescent light bulb. When the light bulb is connected to a battery, a current of 0.526 A is measured through the filament. (Note: tungsten has a resistivity of 4.9 × 10-8 Ω • m.). How many electrons pass through this filament in 5 seconds?

Answers

Answer: 1.64 *10^19 electrons

Explanation: In order to the explain this problem we have to consider the following:

The current= charge/time; so

as the electrons move in the tungsten wire we have:

0.526 C/s= N electrons per second* charge of electron=

N electrons/s= 0.526/1.6*10^-19= 3.28 *10^18 electrons/s

Then, during 5 seconds will pass:

3.28 *10^18 electrons/s*5 5s= 1.64 *10^19 electrons

Answer:

1.64 x 10^19 electrons

Explanation:

The current is defined as I=ΔQ/Δt where ∆Q is the amount of charge flowing past a point in the filament. This charge is comprised of electrons that each carry charge of e = 1.602 × 10^-19 C. So ΔQ=Ne=IΔt and the number of electrons flowing through the filament in 5 s is N=IΔte=(0.526 A)(5 s)1.602×10^−19 C=1.64×10^19 electrons.

A small ball of mass m is held directly above a large ball of mass M with a small space between them, and the two balls are dropped simultaneously from height H. (The height is much larger than the radius of each ball, so you may neglect the radius.) The large ball bounces elastically off the floor and the small ball bounces elastically off the large ball. a) For which value of the mass m, in terms of M, does the large ball stop when it collides with the small ball? b) What final height, in terms of H, does the small ball reach?

Answers

a) The large sphere has 3 times the mass of the small sphere

b) The final height at which small ball reach y = 4H

What will be the mass of the sphere and height covered by the small ball?

We must start this problem by calculating the speed with which the spheres reach the floor

$V_f^2=V_o^2-2gy$

As the spheres are released v₀ = 0

$V_f^2=2gH$

$V_f=√(2gH)$

The two spheres arrive at the same speed to the floor.

The largest sphere clashes elastically so that with the floor it has a much higher mass, the sphere bounces with the same speed with which it arrived, the exit speed of the spheres

$V_(10)=\sqrt2gH$

The big sphere goes up and the small one down, the two collide, let's form a system that contains the two spheres, let's use moment conservation

Let's call

$V_h=\sqrt2gH$

Small sphere m₂ and $V_(20)=-\sqrt2gH=-V_h$

Large sphere m₁ and $V_(10)=√(2gH)=V_h$

Before crash

$P_o=m_1V_(10)+m_2V_(20)$

After the crash

$P_f=m_1V_(1f)+m_2V_(2f)$

$P_o=P_f$

$m_1V_(10)+m_2V_(20)=m_1V_(1f)+m_2V_(2f)$

The conservation of kinetic energy

$K_o=(1)/(2) m_1V_(10)^2+(1)/(2) m_2V_(20)^2$

$K_f=(1)/(2) m_1V_(1f^2)+(1)/(2) m_2V_(2f)^2$

$K_o=K_f$

$K_o=(1)/(2) m_1V_(10)^2+(1)/(2) m_2V_(20)^2$ $=$ $K_f=(1)/(2) m_1V_(1f^2)+(1)/(2) m_2V_(2f)^2$

Let's write the values

$-m_1V_h+m_2V_h=m_1V_(1f)+m_2V_(2f)$

$m_1V_h^2+m_2V_h^2=m_1V_(1f)^2+m_2V_(2f)^2$

The solution to this system of equations is

$m_t=m_1+m_2$

$V_(1f)=((m_1-m_2))/(m_tV_(10)) +(2m_2)/(m_tV_2)$

$V_(2f)=(2m_1)/(m_tV_(10)) +(m_2-m_1)/(m_tV_2)$

The large sphere is labeled 1, we are asked for the mass so that $V_(1f)$ = 0, let's clear the equation

$V_(1f)=(m_1-m_2)/(m_tV_(10)) +(2m_2)/(m_tV_(20))$

$0=(m_1-m_2)/(m_tV_h) +(2m_2)/(m_t(-V_h))$

$(m_1-m_2)/(m_tV_h) =(2m_2)/(m_tV_h)$

$(m_1-m_2)=2m_2$

$m_1=3m_2$

The large sphere has to complete 3 times the mass of the sphere1 because it stops after the crash.

b) Let us calculate with the other equation the speed with which the sphere comes out2 (small)

$V_(2f)=(2m_1)/(m_tV_(10)) +(m_2-m_1)/(m_tV_(20))$

$V_(2f)=(2m_1)/(m_tV_h) +(m_2-m_1)/(m_t(-V_h))$

In addition, we know that m₁ = 3 m₂

$m_t=3m_2+m_2$ mt = 3m2 + m2

$m_t=4m_2$

$V_(2f)=(2* 3m_2)/(4m_2V_h-(m_2-3m_2)4m_2V_h)$

$V_(2f)=(3)/(2) V_h+(1)/(2) V_h$

$V_(2f)=2V_h$

$V_(2f)=2\sqrt{2gh$

This is the rate of rising of sphere 2 (small. At the highest point, it zeroes velocity $V_f$ = 0

$V^2=V_(2f)^2-2gy$

$0=(2√(2gh))^2-2gy$

$y=4H$

Thus

a) The large sphere has 3 times the mass of the small sphere

b) The final height at which small ball reach y = 4H

To know more about the Laws of collisions follow

brainly.com/question/7538238

Answer:

a) the large sphere has 3 times the mass of the small sphere

b) y = 4H

Explanation:

We must start this problem by calculating the speed with which the spheres reach the floor

vf² = vo² - 2g y

As the spheres are released v₀ = 0

vf² = - 2g H

vf = √ (2g H)

The two spheres arrive with the same speed to the floor.

The largest sphere clashes elastically so that with the floor it has a much higher mass, the sphere bounces with the same speed with which it arrived, the exit speed of the spheres

V₁₀ = √2gH

The big sphere goes up and the small one down, the two collide, let's form a system that contains the two spheres, let's use moment conservation

Let's call vh = √2gH

Small sphere m₂ and v₂o = - √2gH = -vh

Large sphere m₁ and v₁o = √ 2gh = vh

Before crash

p₀ = m₁ v₁₀ + m₂ v₂₀

After the crash

pf = m₁ v₁f + m₂ v₂f

po = pf

m₁ v₁₀ + m₂ v₂₀ = m₁ v₁f + m₂ v₂f

The conservation of kinetic energy

Ko = ½ m₁ v₁₀² + ½ m₂ v₂₀²

Kf = ½ m₁ v₁f² + ½ m₂ v₂f²

Ko = KF

½ m₁ v1₁₀² + ½ m₂ v₂₀² = ½ m₁ v₁f² + ½ m₂ v₂f²

Let's write the values

-m₁ vh + m₂ vh = m₁ v₁f + m₂ v₂f

m₁ vh² + m₂ vh² = m₁ v₁f² + m₂ v₂f²

The solution to this system of equations is

mt = m₁ + m₂

v1f = (m₁-m₂) / mt v₁₀ + 2m₂ / mt v₂

v₂f = 2m₁ /mt v₁₀ + (m₂-m₁) / mt v₂

The large sphere is labeled 1, we are asked for the mass so that v1f = 0, let's clear the equation

v₁f = (m₁-m₂) / mt v₁₀ + 2m₂ / mt v₂₀

0 = (m₁-m₂) / mt vh + 2 m₂ / mt (-vh)

(m₁-m₂) / mt vh = 2 m₂ / mt vh

(m₁-m₂) = 2m₂

m₁ = 3 m₂

The large sphere has to complete 3 times the mass of the sphere1 because it stops after the crash.

b) Let us calculate with the other equation the speed with which the sphere comes out2 (small)

v₂f = 2m₁ / mt v₁₀ + (m₂-m₁) / mt v₂₀

v₂f = 2 m₁ / mt vh + (m₂-m₁) mt (-vh)

In addition, we know that m₁ = 3 m₂

mt = 3m2 + m2

mt= 4m2

v₂f = 2 3m₂ / 4m₂ vh - (m₂-3m₂) 4m₂ vh

v₂f = 3/2 vh +1/2 vh

v₂f = 2 vh

v₂f = 2 √ 2gh

This is the rate of rise of sphere 2 (small. At the highest point its zero velocity vf = 0

V² = v₂f² - 2 g Y

0 = (2√2gh)² - 2gy

2gy = 4 (2gH)

y = 4H

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