A block of mass m = 1.9 kg is attached to a string that is wrapped around the circumference of a wheel of radius R = 8.1 cm. The wheel rotates freely about its axis and the string wraps around its circumference without slipping. Initially, the wheel rotates with an angular speed ω, causing the block to rise with a linear speed v = 0.43m/s
Find the moment of inertia of the wheel if the block rises to a height of h
= 7.5 cm before momentarily coming to rest.

Answers

Answer 1

The moment of inertia of the wheel if the block rises to a height of h is 7.5 cm before momentarily coming to rest is 0.068 kg m².

We can use the conservation of mechanical energy to solve for the moment of inertia of the wheel. Initially, the system has kinetic energy, which is converted to potential energy at the highest point of the block's trajectory. Therefore, we can write:

Initial kinetic energy = Final potential energy

At the start, the wheel and block have kinetic energy due to the motion of the block:

KE i = 0.5 * m * v²

At the highest point of the block's trajectory, all of the kinetic energy is converted to potential energy due to the block's height above the ground:

PE_f = m * g * h

where g is the acceleration due to gravity.

Since the string is wrapped around the circumference of the wheel, the distance that the block moves upwards is equal to the distance that the string moves around the wheel, which is equal to the circumference of the wheel:

h = 2 * pi * R

Substituting this into the expression for potential energy:

PE_f = m * g * 2 * π * R

Equating initial kinetic energy with final potential energy:

0.5 * m * v² = m * g * 2 * π * R

Simplifying and solving for the moment of inertia of the wheel, I:

I = (m * v²) / (2 * g * π * R)

Substituting the given values:

I = (1.9 kg * 0.43 m/s)² / (2 * 9.81 m/s² * π * 0.081 m)

I = 0.068 kg m²

Therefore, the moment of inertia of the wheel is 0.068 kg m².

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Related Questions


What is the acceleration of gravity from the Earth when you are one full Earth radius height above the surface of the earth?
Select the answer below in terms of the acceleration at the surface, g.


A.2g
B.g
C. g/2
D. g/4

Answers

Answer:

Explanation:The acceleration due to gravity at a distance equal to one full Earth radius above the surface of the Earth is given by:

g' = g/(1 + R/E)^2

Where g is the acceleration due to gravity at the surface of the Earth, R is the radius of the Earth, and E is the distance of the object from the center of the Earth.

Substituting R for E+R, we get:

g' = g/[1 + (E+R)/R]^2

g' = g/[1 + (1+E/R)]^2

g' = g/[1 + (1+1)]^2 (Since E/R is very small compared to 1)

g' = g/16

Therefore, the acceleration due to gravity at a distance equal to one full Earth radius above the surface of the Earth is g/16. Answer: D. g/4.

Answer:

The acceleration of gravity decreases as you move away from the surface of the Earth. The acceleration of gravity at a height of one Earth radius above the surface can be found using the formula:

g' = (R/(R+h))^2 * g

where R is the radius of the Earth, h is the height above the surface, and g is the acceleration due to gravity at the Earth's surface.

If we plug in R = 6,371 km (the radius of the Earth) and h = 6,371 km (one Earth radius above the surface), we get:

g' = ((6,371 km)/(2*6,371 km))^2 * g

g' = (1/2)^2 * g

g' = g/4

Therefore, the answer is D. g/4.

Which type of wave interaction allows us to hear sounds around corners and under doors?
A. Diffraction
B.Absorption
C.refraction
D.reflection

Answers

Answer: Diffraction

Explanation: It is diffraction, because due to diffraction, sound waves bend on the corner as their size is of the order of wavelength of sound waves.

A. Diffraction
Diffraction allows us to hear sound around corners

The wave in the liquid travels towards the surface at an angle. Fig 9.2 shows the centres of the compressions of the sound wave in liquid. Some compressions shown have reached the liquid-air boundary. The parts of these compressions in the air are not shown on Fig 9.2 These waves are also reflected at the boundary. Draw on the diagram the reflected wavefronts.

Answers

The reflected sound wavefronts at the given boundary are waves that have bounced off a surface and changed direction.

The reflected sound wavefront is shown in the attachment.

What are reflected wavefronts?

A reflected wavefront is a wavefront that has bounced off a surface and changed direction. When a wave, such as a light wave or sound wave, encounters a surface, some of the wave energy is reflected back in the opposite direction.

An example of reflected sound wavefronts in water can be seen in underwater sonar imaging.

In sonar imaging, a sound wave is emitted from a source and travels through the water. When the sound wave encounters an object, some of the wave energy is reflected back toward the source.

Reflected wavefronts play an important role in many areas of science and engineering, such as optics, acoustics, and electromagnetism. They are used to model the behavior of waves in complex systems and to design and optimize devices such as mirrors, lenses, and antennas.

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At the moment when a shot putter releases a 5.00kg shot, the shot is 3.00m above the ground and traveling at 15.0m/s. It reaches a maximum height of 14.5m above the ground and then falls to the ground. If air resistance is negligible, what was the potential energy of the shot as it left the hand relative to the ground?

Answers

The potential energy of the shot as it left the hand relative to the ground would be -524 J.

Potential energy calculation

At the moment the shot putter releases the shot, the total energy of the system is:

E = KE + PE

where KE is the kinetic energy of the shot and PE is its potential energy relative to the ground. We can assume that the kinetic energy is entirely due to the motion of the shot in the horizontal direction, so we can write:

KE = (1/2)mv^2

where m is the mass of the shot, and v is its horizontal velocity. Substituting the known values, we get:

KE = (1/2)(5.00 kg)(15.0 m/s)^2 = 1125 J

At the maximum height of 14.5 m, the shot has zero kinetic energy, so all its energy is potential energy:

PE = mgh

where g is the acceleration due to gravity and h is the height above the ground. Substituting the known values, we get:

PE = (5.00 kg)(9.81 m/s^2)(14.5 m - 3.00 m) = 601 J

Therefore, the potential energy of the shot as it left the hand relative to the ground was:

PE = E - KE = 601 J - 1125 J = -524 J

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Find I1, V1, and V2 for the circuit shown in Fig. 4

Answers

From the given circuit, the value of I1 = -4 Amp, V1 = -4 V, and the V2 = 4V. Ohm's law has three different variations that are related to voltage, current, and resistance. The relationship between the voltage across the terminals and the resistor's resistance determines the current for a constant flow of current.

What is the tenet of Ohm's law?

According to Ohm's law, the voltage across a conductor determines how much current flows through it. For many materials, including metals, this is true so long as the temperature (as well as other physical parameters) stay constant.

What does "nodal analysis" mean?

Any electrical network may be solved using nodal analysis, which is what it is called. the formula used to determine how much voltage is shared across circuit nodes. Since the node voltages are with respect to the ground, this method is also referred to as the node-voltage method.

Applying KCL,

I1 = -1 -3

I1 = -4 A

Applying concept of super-node,

V2 = 4V

[tex]\frac{4-V1}{2} -1-3=0[/tex]

[tex]2-\frac{V1}{2} =4[/tex]

[tex]\frac{-V1}{2} =4-2[/tex]

[tex]V1=-4V[/tex]

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Two identical billiard balls are rolling toward each other at the same speed. What will be true after they collide head–on?

They will both stop rolling immediately after the collision.
One ball will stop rolling as the other bounces backward at a slower speed.
They will both bounce back at a faster speed after the collision.
They will both bounce back at the same speed they had before the collision.

Answers

Answer:

After the head-on collision, both identical billiard balls will bounce back at the same speed they had before the collision. This is because of the conservation of momentum, which states that the total momentum of a closed system remains constant if no external forces act on it. In this case, the two billiard balls are a closed system, and their total momentum before the collision is equal and opposite to their total momentum after the collision. Therefore, they will both bounce back with the same speed they had before the collision.

Explanation:

A 7g bullet is fired into a 345g block that is initially at rest at the edge of a frictionless table of height 1.9m. The bullet remains in the block and after impact the block lands 2m from the bottom of the table. Find the initial speed of the bullet. The acceleration due to gravity is 9.8m/s^2. Answer in units of m/s

Answers

Answer:

≈ 192.7 m/s

Explanation:

Let's start by finding the velocity of the block just before it hits the ground. We can use the conservation of energy:

The initial potential energy of the block is mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height of the table.The final kinetic energy of the block is (1/2)mv^2, where v is the velocity of the block just before it hits the ground.

Conservation of energy tells us that these two energies are equal:

mgh = (1/2)mv^2

Solving for v, we get:

v = sqrt(2gh)

Plugging in the given values, we get:

v = sqrt(2 * 9.8 m/s^2 * 1.9 m) = 6.06 m/s

Now, let's use conservation of momentum to find the initial speed of the bullet. We know that the total momentum of the system (bullet + block) is conserved before and after the collision. Before the collision, the momentum is:

p = mb * vb

where mb is the mass of the block and vb is its initial velocity, which is 0 since it is at rest.

After the collision, the bullet and block move together with a common velocity v. The total momentum is:

p = (mb + m) * v

where m is the mass of the bullet. Since momentum is conserved:

mb * vb = (mb + m) * v

Solving for vb, we get:

vb = (mb + m) * v / mb

Plugging in the given values, we get:

vb = (345 g + 7 g) / 7 g * 6.06 m/s = 192.7 m/s

Therefore, the initial speed of the bullet was approximately 192.7 m/s.

004 (part 1 of 2) 10.0 points
In a 95 s interval, 216 hailstones strike a glass
2
window of area 1.156 m² at an angle 67 to the
window surface. Each hailstone has a mass of
5 g and speed of 13.5 m/s.
If the collisions are elastic, find the average
force on the window.
Answer in units of N.
005 (part 2 of 2) 10.0 points
Find the pressure on the window.
Answer in units of N/m².

Answers

Average force F = 97.52 N acting on window surface.

Pressure on the window: P = 84.39 N/m²

Explain about the elastic collision?An elastic collision is one in which the system does not experience a net loss of kinetic energy as a result of the collision. In elastic collisions, momentum as well as kinetic energy are both conserved. An example of an elastic collision is when two balls collide at a pool table.

Average force F:

F = mg sinФ

Put the values:

F = 0.05*9.81* sin 67°

F = 0.45 N

Number of hailstones = 216

Average force F = 0.45* 216 = 97.52 N

Pressure on the window:

Pressure = Average force / area

P = 97.52 / 1.156

P = 84.39 N/m²

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which one defines force?

Answers

Answer:

a

Explanation:

a push or a pull that occurs when an object interacts with another object or field.

pls mrk me brainliest

Help me please I’m begging

Answers

The total resistance is 3 ohm

The current is 1.5 A

What is the total resistance of the circuit?

To calculate the total resistance of a circuit, you need to determine the equivalent resistance of all the resistors in the circuit. The equivalent resistance is the single resistance value that could replace all the individual resistors and produce the same overall resistance.

1) This is a parallel circuit thus;

1/RT = 1/6 + 1/6

RT = 3 ohm

2) This is a series circuit thus;

RT = 4 + 6 = 10 ohm

Current = 15V/10 ohm

= 1.5 A

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if 100g of iron at 100°c is dropped into 390g of water at 20°c what will the final temperature be?

Answers

70 I think I’m not sure tho

Compare the patterns of iron filings to the spiral arms in Interacting galaxies.

Answers

The primary locations for the birth of new stars are in galaxies' spiral arms. The proportion of the galaxy that can participate in star formation increases as more gas and dust become available.

What distinguishes the numerous varieties of spiral galaxies?

We refer to some spiral galaxies as "barred spirals" because the centre bulge seems extended, like a bar. The spiral arms of the galaxy seem to emerge from the ends of the bar in barred spirals. Elliptical galaxies are round or oval in shape, as their name implies, and have a rather uniform distribution of stars.

The four spiral arms are what?

The Norma and Cygnus arm, Sagittarius, Scutum-Crux, and Perseus arms are the four principal spiral arms of the Milky Way.

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16. The density difference between warm, moist air and cold air causes the moist
air to rise. This is key to forming
A. lightning.
B. clouds.
C. stars.
D. snow.

Answers

Warm, moist air rises because of the disparity in density between warm, moist air and cold air. This is crucial for clouds to develop.

What does rising moist, warm air in the atmosphere form?

Puffy cumulus clouds can form in the atmosphere as warm, humid air rises in an updraft. As it rises, the moisture in the air condenses into water droplets. As long as warm air rising from below persists, the cloud will expand.

Why do lightning and sound happen?

The air in the lightning channel may reach temperatures of 50,000 degrees Fahrenheit, which is five times hotter than the surface of the sun. After the flash, the air swiftly cools and contracts.

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A particle executes simple harmonic motion with an amplitude of 3.00 cm. At what position does its speed equal half of is maximum speed?​

Answers

Answer:

2.34

Explanation:

To find the position where the speed is half of its maximum speed, we can set v = v_max/2 and solve for x:

v_max/2 = ω√(A^2 - x^2)

Substituting the given values, we have:

(ωA)/2 = ω√(A^2 - x^2)

Simplifying and rearranging:

A^2 - x^2 = (A/2)^2

x^2 = A^2 - (A/2)^2

x^2 = (3.00 cm)^2 - (1.50 cm)^2

x = √(6.75 cm^2 - 2.25 cm^2)

x = √5.50 cm^2

x ≈ 2.34 cm

Therefore, the position where the speed of the particle is half of its maximum speed is approximately 2.34 cm from the equilibrium position.

A test rocket starting from rest at point A is launched by accelerating it along a 200.0 m incline at 3.50 m/s2 (Figure 1). The incline rises at 35.0∘ above the horizontal, and at the instant the rocket leaves it, the engines turn off and the rocket is subject to gravity only (ignore air resistance). a)Find the maximum height above the ground that the rocket reaches. b)Find the rocket's greatest horizontal range beyond point A

Answers

Answer: a) 123.1m

b) 279.1m

Explanation:

Final answer:

The maximum height reached by the rocket is 0 meters. The rocket's greatest horizontal range beyond point A is also 0 meters.

Explanation:

To find the maximum height, we can use the kinematic equation for vertical motion. The rocket starts from rest, so its initial vertical velocity is 0 m/s. We can use the equation:

[tex]h = vi^2 / (2 * g)[/tex]

where h is the maximum height, vi is the initial vertical velocity, and g is the acceleration due to gravity. We can calculate h by substituting the given values:

h = (0 m/s)2 / (2 * 9.8 m/s2)

= 0 m

Therefore, the maximum height reached by the rocket is 0 meters above the ground.

To find the rocket's greatest horizontal range, we can use the horizontal motion equations. Since the rocket is subject to gravity only after leaving the incline, we need to find the time it takes for the rocket to reach the highest point on the incline. We can use the equation:

t = vf / a

where t is the time, vf is the final velocity, and a is the acceleration. We can calculate t by substituting the given values:

t = 3.50/ 3.50

= 1s

Now we can use the equation for horizontal motion to find the horizontal range:

[tex]R = v0 * t[/tex]

where R is the horizontal range and v0 is the initial horizontal velocity. Since the rocket starts from rest, v0 = 0 m/s. Therefore, the rocket's greatest horizontal range beyond point A is 0 meters.

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Sort the following items from most to least inertia.

Answers

Answer: In order of most inertia to least.

car stopped

motorcycle driving

dog laying

large hot coffee

small mouse

Explanation:

Honestly, I'm not 100% sure about this, but inertia is the want to remain the way it currently is, So the larger the object, the more inertia it should have. So I basically arranged the things by size, I think the moving and not is a red herring to throw you off. Also given that the dog laying in the middle option makes sense because a dog would be in the middle of the size options.

Use the table to answer the question.
Wave Wavelength (meters) Frequency (hertz)

W 5 200

X 3 300

The table shows information about two waves. Based on the given information, which conclusion can be made?

O Wave W has a faster speed.
O Wave W has a greater amplitude.
O Wave X has a greater amplitude.
O Wave X has a faster speed.

Answers

We can see from the table that we have in the question that  Wave W has a faster speed.

What is the relationship of wavelength and frequency?

Wavelength and frequency are two fundamental properties of waves, including electromagnetic waves and sound waves. The relationship between wavelength and frequency can be described by the following equation:

c = λν

where c is the speed of light (or speed of sound), λ is the wavelength, and ν is the frequency.

According to this equation, the wavelength and frequency are inversely proportional to each other.

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Two identical blocks are connected by a lightweight string that passes over a lightweight pulley that can rotate about its axle with negligible friction. The two-block system is released from rest and the blocks accelerate. Which of the following correctly relates the potential energy gained by the block 1-Earth system |∆U1| to the potential energy lost by the block 2-Earth system |∆U2| and provides correct evidence?

Answers

The potential energy gained by the block 1-Earth system |∆U1| is equal to the potential energy lost by the block 2-Earth system |∆U2|. This is known as the law of conservation of energy, which states that energy cannot be created or destroyed, only transferred or converted from one form to another. In this case, the potential energy stored in block 2 at the start of the experiment is transferred to block 1 as they move and accelerate. The sum of the potential energy of the two-block system at the start of the experiment is equal to the sum of the kinetic energy and potential energy of the system at any point during the experiment. This relationship can be expressed mathematically as follows:

|∆U1| = |∆U2|

where |∆U1| is the potential energy gained by the block 1-Earth system and |∆U2| is the potential energy lost by the block 2-Earth system.

A rock group is playing in a bar. Sound
emerging from the door spreads uniformly in
all directions. The intensity level of the music
is 32.2 dB at a distance of 4.99 m from the
door.
At what distance is the music just barely
audible to a person with a normal threshold
of hearing? Disregard absorption.
Answer in units of m.

Answers

Answer:

5292.64 m

Explanation:

A scientist makes a model of Earth's water by drawing 100 drops of water, all the same size. How many of the 100 drops represent ocean water?
A.3

B.50

C.75

D.97

Answers

Answer:

D

Explanation:

If the water represents the oceans water then you'd would need to calculate how much of earth is water (96.5)


5. What is the area of this figure? Please show your work.
20 in.
7 in
6. What is the area of this figure? Please show your
10 ft
8 ft
6 ft
work.

Answers

Answer:

The area of the figure 7 in the area of 6 ft

A rocket at rest on the ground with initial mass 20,000 kg, 80% of which is fuel, burns 200kg/s as it flies
directly upwards. The exhaust gas exits the rocket at a relative speed of 1.80 km/s. Find (a) the thrust of

the rocket, (b) how long it takes to exhaust its fuel, and (c) the rocket’s speed at the end of its engine burn.

Assume g is constant and neglect air resistance.


For Part A I got: 360,000 Newtons, Part B: 20 Seconds, Part C 205.658 m/s

Are my answers correct?

Answers

To check your answers, we can use the following equations:

(a) Thrust = (mass flow rate of exhaust gas) * (exhaust velocity of gas) + (initial mass of rocket) * (acceleration due to gravity)

(b) Time to exhaust fuel = (0.8 * initial mass of rocket) / (mass flow rate of exhaust gas)

(c) Final velocity of rocket = (exhaust velocity of gas) * ln(initial mass of rocket / final mass of rocket)

Using the given values:

Mass flow rate of exhaust gas = 200 kg/s

Exhaust velocity of gas = 1.80 km/s = 1800 m/s

Initial mass of rocket = 20,000 kg

Acceleration due to gravity = 9.81 m/s^2

Final mass of rocket = 0.2 * initial mass of rocket = 4,000 kg

(a) Thrust = (200 kg/s) * (1800 m/s) + (20,000 kg) * (9.81 m/s^2) = 360,000 N

Your answer for part (a) is correct.

(b) Time to exhaust fuel = (0.8 * 20,000 kg) / (200 kg/s) = 800 s = 20 minutes

Your answer for part (b) is incorrect. The correct answer is 800 seconds or 20 minutes, not 20 seconds.

(c) Final velocity of rocket = (1800 m/s) * ln(20,000 kg / 4,000 kg) = 205.66 m/s

Your answer for part (c) is correct.

I need guidance in this one

Answers

The clay ball that could stick to the door would be more effective when you want to shut the door from a distance be throwing a n object.

Will a rubber  ball or a clay ball do better at helping you shut a door when thrown at the door?

The effective way to shut the door could be through the use of the clay ball that could stick to the door and push it to the place where it could jam rather than the rubber ball that would bound back.

The best way to shut a door is to use the door handle or knob to pull the door towards you and guide it into the door frame until it clicks shut.

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On a hot summer day, a young girl swings on a rope above the local swimming hole. When she lets go of the rope her initial velocity is 2.25m/s at an angle of 35º above the horizontal. If she is in flight for 1.20s, how high above the water was she when she let go of the rope?

Answers

When she let go of the rope, her height above the water would be  3.20 meters.

Kinematic motion

We can use the equations of motion to solve this problem. Let's assume that the girl's initial height above the water is h, and that the acceleration due to gravity is -9.81 m/s^2 (negative because it is directed downward). Then we have:

Initial horizontal velocity (vx) = 2.25 cos(35º) = 1.84 m/s

Initial vertical velocity (vy) = 2.25 sin(35º) = 1.30 m/s

During the 1.20 s of flight, the girl's vertical motion can be described by the following equation:

h + vy*t + (1/2)gt^2 = 0

where t is the time of flight and g is the acceleration due to gravity. Substituting the known values, we get:

h + (1.30 m/s)(1.20 s) + (1/2)(-9.81 m/s^2)*(1.20 s)^2 = 0

Simplifying and solving for h, we get:

h = -1/2*(-9.81 m/s^2)(1.20 s)^2 - (1.30 m/s)(1.20 s) = 3.20 m

Therefore, the girl was 3.20 meters above the water when she let go of the rope.

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Hans figures out one math problem easily and then applies that solution to the rest of the problems on the page; however, he gets the rest of the answers wrong. Hans has encountered a common problem with:

Answers

Answer:

Explanation:

overgeneralization or applying a single solution to a variety of problems without considering their unique characteristics. This is a common issue in math and other problem-solving situations where individuals may rely too heavily on past successful solutions without adjusting for the specific details of each new problem.

When landing after a spectacular somersault, a 35.0 kg gymnast decelerates by pushing straight down on the mat. Calculate the force (in N) she must exert if her deceleration is 7.00 times the acceleration of gravity. (Enter a number.)

Answers

The force (in N) she must exert if her deceleration is 7.00 times the acceleration of gravity is 245 N

Calculation of Force

Given data

Deceleration = 7.00 times the acceleration of gravity (9.81 m/s2)Mass = 35.0 kg

Force = Mass x Acceleration

Force = 35.0 kg x (7.00 x 9.81 m/s2)

Force = 245 N

The force required for the 35.0 kg gymnast to decelerate after a somersault is 245 N. This force is calculated by multiplying the mass of the gymnast by the deceleration of 7.00 times the acceleration of gravity.

The amount of force is necessary to slow down the gymnast and reduce the impact of the landing. Without this force, the gymnast could be injured from the sudden stop.

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A car is moving along a road at 13.0 m/s with an engine that exerts a force of 1,775.0 N on the car to balance the drag and friction so that the car maintains a constant speed. What is the power output of the engine?​

Answers

Answer:


The power output of the engine can be calculated using the formula:

Power = Force x Velocity

where force is the net force acting on the car, and velocity is the speed of the car. In this case, the net force is equal to the force exerted by the engine, which is 1,775.0 N, since the car is moving at a constant speed and there is no acceleration. The velocity of the car is 13.0 m/s. Thus, the power output of the engine can be calculated as:

Power = 1,775.0 N x 13.0 m/s = 23,075 W

Therefore, the power output of the engine is 23,075 watts.

An 12.000 milligram particle is sliding across a friction-less one-dimensional path at 55.000 m/s and collides with a 68.000 milligram particle moving at -48.000 m/s in a perfectly inelastic collision. What are the velocities of the particles after the collision?
answer with correct units​

Answers

Answer:

-3525.000 m/s

Explanation:

In a perfectly inelastic collision, the two particles stick together and move with a common velocity after the collision. We can use the conservation of momentum to solve for this common velocity.

The initial momentum of the system is:

p_initial = m1 * v1 + m2 * v2

= (12.000 mg)(55.000 m/s) + (68.000 mg)(-48.000 m/s)

= -282.000 kg·m/s

Here, we convert the masses to kilograms to match the units of velocity.

Since the particles stick together after the collision, their masses add up:

m_final = m1 + m2

= 12.000 mg + 68.000 mg

= 80.000 mg

= 0.080 g

Now, we can use the conservation of momentum to find the final velocity:

p_final = m_final * v_final

where p_final = p_initial and m_final = 0.080 g.

Therefore:

v_final = p_final / m_final

= -282.000 kg·m/s / 0.080 g

= -3525.000 m/s

A machine has velocity ratio 6 and is 80% efficient. what effort would be needed to lift a load of 300N with the aid of this machine? ​

Answers

The effort needed to lift the load of 300 N, given that the velocity ratio is 6 and the machine has an efficiency of 80% is  62.5 N

How do i determine the effort needed to lift the load?

First, we shall determine the mechanical advantage of the machine. Details below:

Velocity ratio (VR) = 6Efficiency = 80%Mechanical advantage (MA) = ?

Efficiency = MA / VR

80% = MA / 6

Cross multiply

MA = 80% × 6

MA = 4.8

Finally, we shall determine the effort of the machine. Details below:

Mechanical advantage (MA) = 4.8Load (L) = 300 NEffort (E) = ?

Mechanical advantage (MA) = Load (L) / Effort (E)

4.8 = 300 / Effort

Cross multiply

4.8 × Effort = 300

Divide both sides by 4.8

Efoort = 300 / 4.8

Effort = 62.5 N

Thus, we can conclude that the effort of the machine is 62.5 N

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If the train set uses stronger magnets, will the distance between these train cars increase or decrease? Explain your answer.

Answers

The distance between the train cars will decrease, as stronger magnets will create a greater attractive force between them. This will cause the cars to be pulled closer together.

What is stronger magnet?

The strength of a magnet depends on the type of magnet and the material it is made from. Generally, the strongest magnets are neodymium magnets, also known as rare earth magnets. These are made from a combination of neodymium, iron, and boron and can produce magnetic fields up to 1.4 teslas, or 14,000 gauss. Samarium cobalt magnets are also powerful, producing fields up to 1.4 teslas. Other types of magnets, such as ceramic and Alnico magnets, generally produce magnetic fields of around 0.5 teslas or 5,000 gauss.

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