A hanging weight, with a mass of m1 = 0.365 kg, is attached by a string to a block with mass m2 = 0.825 kg as shown in the figure below. The string goes over a pulley with a mass of M = 0.350 kg. The pulley can be modeled as a hollow cylinder with an inner radius of R1 = 0.0200 m, and an outer radius of R2 = 0.0300 m; the mass of the spokes is negligible. As the weight falls, the block slides on the table, and the coefficient of kinetic friction between the block and the table is k = 0.250. At the instant shown, the block is moving with a velocity of vi = 0.820 m/s toward the pulley. Assume that the pulley is free to spin without friction, that the string does not stretch and does not slip on the pulley, and that the mass of the string is negligible. Using energy methods, find the speed of the block (in m/s) after it has moved a distance of 0.700 m away from the initial position shown.

Answers

Answer 1

The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Angular Speed of the pulley

The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;

K.E = P.E

[tex]\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\[/tex]

[tex]\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\[/tex]

Substitute the given parameters and solve for the angular speed;

[tex]\omega = \sqrt{\frac{ 4(9.8)(0.7)(0.365 \ - \ 0.25\times 0.825) - 2(0.825)(0.82)^2}{2(0.03)^2(0.365 \ + \ 0.825)\ \ +\ \ 0.35(0.02^2\ + \ 0.03^2)}} \\\\\omega = \sqrt{\frac{3.25}{0.00214\ + \ 0.000455 } } \\\\\omega = 35.39 \ rad/s[/tex]

Linear speed of the block

The linear speed of the block after travelling 0.7 m;

v = ωR₂

v = 35.39 x 0.03

v = 1.1 m/s

Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.

Learn more about conservation of energy here: https://brainly.com/question/24772394

Related Questions

Two particles each with a charge of 1μC are separated a distance of 0.4 m. A test charge is placed just above them but directly between them. (b) What is the direction of the resultant electric field on the test charge?

Answers

Answer:

Forces on the test (test is positive) charge

Σ Fx = 0 net result of x-compoment charges is zero

Σ Fy = 2 F sin θ where F is the force due to the distance of the test charge from one of the charges and sin θ is the component of that force that is perpendicular to the line separating the charges

The direction of the force is perpendicular to the line separating the two given charges

The sound intensity at a distance of 19 m from a noisy generator is measured to be 0.24 W/m2. What is the sound intensity at a distance of 27 m from the generator

Answers

Answer:

0.1111 W/m²

Explanation:

If all other parameters are constant, sound intensity is inversely proportional to the square of the distance of the sound. That is,

I ∝ (1/r²)

I = k/r²

Since k can be the constant of proportionality. k = Ir²

We can write this relation as

I₁ × r₁² = I₂ × r₂²

I₁ = 0.25 W/m²

r₁ = 16 m

I₂ = ?

r₂ = 24 m

0.25 × 16² = I₂ × 24²

I₂ = (0.25 × 16²)/24²

I₂ = 0.1111 W/m²

If electromagnetic radiation acted like particles in the double-slit experiment, what would be observed? (1 point)

O A series of light and dark bands would appear on the screen.

O Two bright bands would appear on the screen in line with the slits.

O The screen would remain dark because no radiation would reach the screen.

O One bright band would appear in the center of the screen.

Answers

If electromagnetic radiation acted like particles in the double-slit experiment, we would observe one bright band would appear in the center of the screen.

Bahavior of particles in double-slit experiment

In a double-slit experiment, single particles, such as photons, pass one at a time through a screen containing two slits.

The photons behave like wave and the constructive interfernce of the waves of these photons will generate a high amplitude wave seen as a bright band in the center of the screen.

Thus, if electromagnetic radiation acted like particles in the double-slit experiment, we would observe one bright band would appear in the center of the screen.

Learn more about double slit experiment here: https://brainly.com/question/4449144

A DJ is setting up their equipment and measures the sound level near a speaker to be 55.65dB. They decide to add two more speakers next to the first. What would they measure the new sound level to be?

Answers

For a DJ is setting up their equipment and measures the sound level near a speaker to be 55.65dB, the new sound level is mathematically given as

L=50.470dB

What would they measure the new sound level to be?

Generally, the equation for the sound level is mathematically given as

l=10logl/io

Therefore

logl/io=5.57

Hence

I=3.715*10^−8

In conclusion

L'=10log(3*3.715*10^−8)/10^{-12}

L=50.470dB

Answers

The speed of the wave with the given frequency and wavelength is 1.2m/s.

Given the data in the question;

Wavelength of the wave; [tex]\lambda = 0.4m[/tex]Frequency; [tex]f = 3Hz = 3s^{-1}[/tex]Speed of the wave; [tex]v = \ ?[/tex]

Wavelength

Wavelength is the distance over which the shapes of waves are repeated. It is the spatial period of a periodic wave.

It is expressed as;

[tex]\lambda = \frac{v}{f}[/tex]

Where v is velocity/speed and f is frequency.

Now, we can easily get the speed of the wave by substituting our given values into the expression above.

[tex]\lambda = \frac{v}{f} \\\\0.4m = \frac{v}{3s^{-1}}\\ \\v = 0.4m * 3s^{-1}\\\\v = 1.2ms^{-1}\\\\v = 1.2m/s[/tex]

Therefore, the speed of the wave with the given frequency and wavelength is 1.2m/s.

Learn more about Speed, Frequency and Wavelength here: brainly.com/question/27120701

the dimensions of a cuboidal block are 2.5 m * 2m * 1.2m its weight 900 n what is the minimum presseure exerted by the block on the surface of its support.?

Answers

Answer:

your answer is here 180

Explanation:

900/2.5 * 2 = 180

Answer:

The answer is 180 Pa.

Explanation:

Given;

Here, weight = thrust = 900 N,

dimensions are 2.5 m × 2 m × 1.2 m

To Find;Pressure exerted by the block on the surface of its support.Formula;Pressure = Thrust ÷ Area

We know that, Pressure is inversely proportional to area. For the minimum pressure the area must be more.

So, if the block is resting with 2.5 m × 2 m surface on the support, the pressure will be minimum.

Pressure = 900 N ÷ 2.5 × 2.0 m² = 180 Pa

Thus, The pressure exerted by the block on the support is 180 Pa.

Which point on the graph would have the most kinetic energy?

-Which point on the graph would have the least kinetic energy?

-What point on the graph would have just a little less potential energy than point E?

-What point on the graph would have just a little more potential energy than point F?

-What point on the graph would have just a little less potential energy than point G?

Answers

The total mechanical energy is kinetic energy + potential energy, so we can look at the object's potential energy to determine the place where it has maximum and minimum kinetic energy.

Potential energy is determined by the equation PE = mgh, where the height is the only changing variable here. Therefore, we look at the height to answer these questions.

Which point on the graph would have the most kinetic energy?

The point where the object's potential energy is the lowest, aka where its height is the lowest. Point D would have the most kinetic energy.

Which point on the graph would have the least kinetic energy?

The point where the object's potential energy is the highest, aka where its height is the highest. Point A would have the least kinetic energy.

What point on the graph would have just a little less potential energy than point E?

Look at the point that is just below point E. Point D has just a little less potential energy than point E.

What point on the graph would have just a little more potential energy than point F?

Look at the point that is just above point F. Point B has just a little more potential energy than point F.

What point on the graph would have just a little less potential energy than point G?

Look at the point that is just below point G. Point B has just a little less potential energy than point E.

The surface of a lake has an area of 15.5 km2. What is the area of the lake in m2?

Answers

Answer:

1.55 x 10⁷ m²

Explanation:

Unit conversion

km² → m² 1 km² = 10⁶ m²

Solving

15.5 km² = 10⁶ x 15.5 m²155 x 10⁵ m²1.55 x 10⁷ m²

Conceptual Example 14 provides useful background for this problem. A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carousel itself (without riders) has a moment of inertia of 105 kg·m2. When one person is standing at a distance of 1.57 m from the center, the carousel has an angular velocity of 0.512 rad/s. However, as this person moves inward to a point located 0.507 m from the center, the angular velocity increases to 0.795 rad/s. What is the person's mass?

Answers

For one person standing at a distance of 1.57 m from the center, the carousel has an angular velocity of 0.512 rad/s, the person's mass is mathematically given as

m = 24.2 kg

What is the person's mass?

Generally, the equation for the conservation angular momentum is mathematically given as

(I + m * r0^2) * w0 = (I + m * r^2) * w^2

Therefore

(105 + m * 1.78^2) * 0.517 = (105 + m * 0.524^2) * 0.841

m = 24.2 kg

In conclusion, the mass

m = 24.2 kg

Answers

Answer:

Current(I)= charge(q)/time

I=15/22

I=0.680.68A

Explanation:

Current is flow of charge per unit time. Always make sure the time is in seconds. istthe

How does the scintillation detector work?

Answers

Answer:

In scintillation detectors the material of the detector is excited to luminescence (emission of visible or near-visible light photons) by the absorbed photons or particles. The number of photons produced is proportional to the energy of the absorbed primary photon. The light pulses are collected by a photo- cathode.

Explanation:

Hope this helps :-)

Enjoy your day or night!

Have fun with your studies and assignments

ps. please don't give full credit to me I used references from other sites

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Find the Product or expression of:
(6x+4)2

Answers

Answer:

20 or 12

Explanation:

if it is 20 u will have to multiple 6 by 4 and the answer u will multiple it by2 and when it is 12 u will have to add 6 +4 and the answer u will add it to 2

6. When can you say that an object is accelerating?
A. when the speed remains the same
B. When the speed is changing
C. when the instantaneous velocity is the same
D. When there is a uniform motion

give explanation

Answers

Answer:

When the speed is changing is the correct answer

Explanation:

Because acceleration is rate of change of velocity

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You ride your bike north for 100 m at a constant speed of 5 m/Your acceleration is…

Answers

Answer:

our acceleration is 0 because there is a constant speed..

a = final speed - initial speed / time

final speed = initial speed ( since speed is constant)

therefore acceleration is 0

What is the effect of the Least Count on the precision/sensitivity of Vernier Callipers?

Answers

Answer:

Least count of vernier caliper=.1/50=.002 CM=.02 MM Least count and precision The least count is directly linked with precision. The smaller is the least count number, the more precise the instrument is.

Explanation:

Is a vernier scale accurate or precise? Every instrument has its accuracy defined as the least count of the instrument.As the least count of vernier callipers is 0.01mm,hence it is not accurate beyond the size measurement of 0.01mm.For which it may be called for vernier scale has presize measurement up to 0.01mm.Ans.. What is a vernier caliper?

Which property of matter is conserved in chemical reactions and shown by balanced equations?
O mass
O volume

O density
O shape

Answers

Answer:

Mass.

Explanation:

I took the quiz and got the answer right

Your answer would be density

list three things that spectra can tell us about objects in space

Answers

Answer:

The temperature, the density, and the element of stars.

Explanation:

The spectral line also can tell us about any magnetic field of the star. The width of the line can tell us how fast the material is moving. We can learn about winds in stars from this.

A potter's wheel is spinning with an initial angular velocity of 11 rad/s . It rotates through an angle of 80.0 rad in the process of coming to rest.
What was the angular acceleration of the wheel?
How long does it take for it to come to rest?

Answers

The angular acceleration of the wheel approximately -0.76 rad/s² or proportionally as deceleration approximately 0.76 rad/s.It need approximately 14.474 s to come to rest.

Introduction

Hi ! I will help you to discuss about Proportionally Changes in Circular Motion. The analogy of proportionally changes in circular motion is same as the analogy of proportionally changes in direct motion. Here you will hear again the terms acceleration and change in speed, only expressed in the form of a certain angle coverage. Before that, in circular motion, it is necessary to know the following conditions:

1 rotation = 2π rad1 rps = 2π rad/s1 rpm = [tex] \sf{\frac{1}{60} \: rps}[/tex] = [tex] \sf{\frac{1}{30}\pi \: rad/s}[/tex]

Formula Used

The following equations apply to proportionally changes circular motion:

Relationship between Angular Acceleration and Change of Angular Velocity

[tex] \boxed{\sf{\bold{\omega_t = \omega_0 + \alpha \times t}}}[/tex]

With the following conditions:

[tex]\sf{\omega_t}[/tex] = final angular velocity (rad/s)[tex]\sf{\omega_0}[/tex] = initial angular velocity (rad/s)[tex]\sf{\alpha}[/tex] = angular acceleration (rad/s²)t = interval of the time (s)Relationship between Angular Acceleration and Change of [tex]\sf{\theta}[/tex] (Angle of Rotation)

[tex] \boxed{\sf{\bold{\theta = \omega_0 \times t + \frac{1}{2} \times \alpha \times t^2}}}[/tex]

[tex] \boxed{\sf{\bold{(\omega_t)^2= (\omega_0)^2 + 2 \times \alpha \times \theta}}}[/tex]

With the following condition :

[tex]\sf{\theta}[/tex] = change of the sudut (rad)[tex]\sf{\alpha}[/tex] = angular acceleration (rad/s²)t = interval of the time (s)[tex]\sf{\omega_t}[/tex] = final angular velocity (rad/s)[tex]\sf{\omega_0}[/tex] = initial angular velocity (rad/s)

Problem Solving

We know that :

[tex]\sf{\omega_t}[/tex] = final angular velocity = 0 rad/s >> see in the sentence "in the process of coming to rest."[tex]\sf{\omega_0}[/tex] = initial angular velocity = 11 rad/s[tex]\sf{\theta}[/tex] = change of the sudut = 80.0 rad

What was asked :

[tex]\sf{\alpha}[/tex] = angular acceleration = ... rad/s²t = interval of the time = ... s

Step by step :

[tex]\sf{\alpha}[/tex] = ... rad/s²

[tex] \sf{(\omega_t)^2= (\omega_0)^2 + 2 \times \alpha \times \theta} [/tex]

[tex] \sf{0^2= (11)^2 + 2 \times \alpha \times 80} [/tex]

[tex] \sf{0 = 121 + 160 \alpha} [/tex]

[tex] \sf{-160 \alpha = 121} [/tex]

[tex] \sf{\alpha = \frac{121}{-160}}[/tex]

[tex] \sf{\alpha = -0.75625 \: rad/s^2 \approx \boxed{-0.76 \: rad/s^2}}[/tex]

t = ... s

[tex] \sf{\alpha = \frac{\omega_0 - \omega_t}{t}}[/tex]

[tex] \sf{-0.76 = \frac{0 - 11}{t}}[/tex]

[tex] \sf{-0.76t = -11}[/tex]

[tex] \sf{t = \frac{- 11}{-0.76}}[/tex]

[tex] \boxed{\sf{t \approx 14.474 \: s}}[/tex]

Conclusion

So :

The angular acceleration of the wheel approximately -0.76 rad/s² or proportionally as deceleration approximately 0.76 rad/s.It need approximately 14.474 s to come to rest.

Which of the following objects would have the greatest: A. momentum if they were all moving with the same velocity? B. Inertia?

A compact car B school bus
C Pick-up truck D road bicycle

Answers

B. because a school bus has the most mass and that Carries over to momentum where it has the most energy over all the other answers

What is the kinetic energy of a 1375 kg car that travels at 8.5 m/s

Answers

Answer:

49671.875

Explanation:

KE = [tex]\frac{1}{2}m[/tex][tex]v^{2}[/tex]

KE = 0.5(1375)(8.5)

= 49671.875 J

Which of the following examples of the streets Newton second law of motion

Answers

Explanation:

Newton’s second law is a quantitative description of the changes that a force can produce on the motion of a body. It states that the time rate of change of the momentum of a body is equal in both magnitude and direction to the force imposed on it.

Calculate the power developed by a 68.0 kg athlete who, in 20.0 seconds. runs up stairs that have a vertical height of 8.00 m

Answers

Mass=m=68kgHeight=h=8mg=10m/s^2(Acceleration due to gravity)

Potential energy:-

mgh68(8)(10)680(8)5440J

Now

time=t=20s

Power:-

PE/t5440/20272W

The diagram shows a heating curve for water.
Which statement best describes what is happening at
segment W-X?
Heating Curve for Water
O Water is absorbing heat.
O Water is releasing heat.
O Aliquid is changing to a gas.
O A gas is changing to a liquid.
z
110
100
90
x
х
Y
80
Temperature (Cº)
70
60
50
40
w
Heat (J)

Answers

Answer:

a water is heating up

Explanation:

the water's temperature is inceasing

In the image, not to scale, which phase of the moon would you observe from earth?

Answers

Answer:

The "new" moon phase is being shown because the moon shown is directly between the sun and earth and is not visible from the earth because of the brightess of the sun.

[tex]Question: $x_{1}=\frac{5}{2}[\sin (2 \pi t)+\cos (2 \pi t)], x_{2}=5\left[\sin \left(2 \pi t+\frac{\pi}{4}\right)\right]$[/tex]
Find the ratio of the amplitude of the given motion?
Options:
[tex](a) $\sqrt{2}: 1$\\(b) $2: 1$\\(c) $1: \sqrt{2}$\\(d) $1: 2$[/tex]

Answers

Condense x₁ to get

sin(2πt) + cos(2πt) = √2 • (sin(2πt) + cos(2πt))/√2

… = √2 (cos(2πt) cos(π/4) + sin(2πt) sin(π/4))

… = √2 cos(2πt - π/4)

So the amplitude of x₁ is (5/2) • √2 = 5/√2, while the amplitude of x₂ is 5. The ratio between them is then 5/√2 : 5, or equivalently 1/√2 : 1 or 1 : √2.

The force of a hammer drives a nail into wood. This is an example of. A. An unbalanced force. B. Gravitational force. C. Friction. D. Balanced forces

Answers

B gravitational force sorry if I’m wrong!

Answer:

What i got was A. An unbalanced force.

Hope This Helps

If the intensity of radiation from an x-ray tube at 2 feet is 100 mR/min, the intensity at a distance of 5 feet will
be?

Answers

Answer:

100 mR

Explanation:

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A 0.14 kg baseball is dropped from rest. It has a momentum of 0.78 kg.m/s just before it lands on the ground. For what amount of time was the ball in the air?

Answers

Answer:

0.568

Explanation:

velocity = -9.8 (t) + Vo

Vo= 0 so v= 9.8(t)

Momentum = mass x velocity

v * 0.14 kg = 0.78 kg m/s

v = 5.571 m/sec

v= 9.8(t) , so substituting v we get , 5.571 = 9.8 * t

t = 5.571/9.8 = 0.568 seconds

what’s is the vertical colum of the elements in the periodic table called

Answers

it is called groups or families.

Answer:

groups or families

Explanation:

The vertical column of the periodic table are called groups or famileis because of their similar chemical behavior.

A system of four capacitors is connected across a 90-V voltage source as shown in the figure.
(a) What is the potential difference across the plates of the 6.0-µF capacitor? (b) What is the charge on the 2.0-µF capacitor?

Answers

For a system of four capacitors is connected across a 90-V voltage source, the potential difference across the plates of the 6.0-µF and 2uF capacitors is mathematically given as

a) vd=30v

b) Q=119.7uc

What is the potential difference across the plates of the 6.0-µF and 2uF capacitors?

Generally, the equation for the charge across 6uf is mathematically given as

Q=C6Vd

Where

1/c36=1/3+1/6

c36=2

Q3=2*90

Q3=180

Therefore

vd=180/6

vd=30v

In conclusion

1/c24=1/2+1/4

C24=1.33uf

Q2=1.33*90

Q=119.7uc