A light platform is suspended from the ceiling by a spring. A student with a mass of 90 kg climbs onto the platform. When it stops bouncing and reaches its new equilibrium position (x=0), the student notices that the spring has stretched 0.82 m. The student's friend pulls the platform down 0.32 m further and then releases it at t=0. What is the amplitude of the motion of the student on the platform?

Answer:

2.19 m/s

Explanation:

The momentum (p) is the product of the mass (m) and velocity (v) of an object.

p = m × v

The momentum of a 0.0075 kg bullet traveling at a speed of 365 m/s is:

p(bullet) = 0.0075 kg × 365 m/s = 2.74 kg.m/s

We want the 1.25 kg-coffee cup to have the same momentum than the bullet.

p(cup) = p(bullet) = 2.74 kg.m/s

p (cup) = m(cup) × v(cup)

2.74 kg.m/s = 1.25 kg × v(cup)

v(cup) = 2.19 m/s

In terms of physics, the regular dart goes further. Despite both darts being fired with the same energy from the spring-operated dart guns, the heavier weighted dart experiences a greater recoil force, which slows it down more than the regular dart.

The question asks which of two darts fired from identical spring-loaded dart guns would go farther: a regular dart or a weighted dart. This relates to the principles of physics, specifically, the concepts of energy conservation and Newton's third law.

In a spring-operated dart gun, energy is stored in the compressed spring and then transferred to the dart when the spring is released. This total energy put into the system is conserved in the closed system of the dart gun and dart, and it is mostly converted into the kinetic energy of the dart (ignoring the system's internal energy losses).

Newton's third law states that for every action, there's an equal and opposite reaction. In this case, the spring's force propels both darts forward. However, the reaction - the backward push on the dart - will be greater for the heavier weighted dart, slowing it more compared to the regular one. Therefore, the regular dart will go further given that it faces less opposing force from the gun itself.

https://brainly.com/question/35917326

#SPJ11

To find the initial separation between two cars before braking, we use the kinematic equation related to uniform acceleration. By calculating the deceleration of car 1, we can determine the total stopping distance for car 2, which includes the initial separation distance plus the stopping distance of car 1.

The question involves the concept of uniform acceleration and the use of kinematic equations to determine the initial separation between two cars before they start braking. Under the assumption that both cars decelerate with the same constant acceleration and car 2 stops exactly at the position where car 1 initially started to brake, we can use the kinematic equation:

\[ v^2 = u^2 + 2as \]

Where \(v\) is the final velocity, \(u\) is the initial velocity, \(a\) is the acceleration, and \(s\) is the distance. Since car 1 stops in 10 m, we can find the acceleration by setting \(v=0\), and since car 2 stops at the back of car 1 exactly, the initial separation
\(s\) is actually the distance car 2 travelled while car 1 was braking plus the 10 m stopping distance of car 1.

To solve this, we first calculate the acceleration of car 1 as it comes to a stop:

\[ 0 = u^2 + 2a(10) \]

This gives us the value of \(a\), and since car 2 has the same acceleration and stops in the same way, it will travel a distance of 10 m after starting to brake, plus the initial separation distance while car 1 was braking. This total distance equals the stopping distance of car 1, which is 10 m, plus the distance car 2 travels before it starts braking.

Therefore, the initial separation between the two cars before braking started is the distance car 2 travels while car 1 is braking (which can be calculated using the found acceleration and the fact that car 2 starts braking 10 m after car 1 has come to a stop).

https://brainly.com/question/28641931

#SPJ12

Assuming pulley is frictionless. Let the tension be ‘T’. See equation below.

The expression for the acceleration of the red block after it is released from the rest is

[tex]\fbox{\begin{minispace}\\{a = \dfrac{{\left( {{m_g} - \mu {m_r}} \right)g}}{\left( {{m_g} + {m_r}} \right)}}\end{minispace}}[/tex]

Further Explanation:

The friction force acting on the red block placed on the table is acting opposite to the direction of motion of block. The motion of the system, of red block and grey block, is vertically downward direction. It implies that the tension on the string is less than the weight on the grey block.

Given:

The mass of the red block is [tex]{m_r}[/tex].

The mass of the grey block is [tex]{m_g}[/tex].

The coefficient of friction between red block and table is [tex]\mu[/tex].

Concept:

The net force on the red block is acting along the direction of motion of the system and is equal to the difference between the tension in the string and friction force.

The net force on the red block is:

[tex]{m_r}a = T - f[/tex]

Rearrange the above expression.

[tex]T = {m_r}a + f[/tex]

Substitute [tex]\mu {m_r}g[/tex] for [tex]f[/tex] in above equation.

[tex]\fbox{\begin\\T = {m_r}a + \mu {m_r}g\end{minispace}}[/tex]                                                           …… (1)

Here,  [tex]T[/tex] is the tension on the string, [tex]{m_r}[/tex] is the mass of the red block, [tex]a[/tex] is the acceleration of the system and [tex]g[/tex] is the acceleration due to gravity.

The net force on the grey block is along the direction of motion of the system and it is equal to the difference between the weight on grey block and tension on the string.

The net force on the grey block is:

[tex]{m_g}a = {m_g}g - T[/tex]                                                                        …… (2)

Here, [tex]{m_g}[/tex] is the mass of the grey block.

Substitute [tex]{m_r}a + \mu {m_r}g[/tex] for [tex]T[/tex] in equation (2).

[tex]{m_g}a = {m_g}g - \left( {{m_r}a + \mu {m_r}g} \right)[/tex]

Rearrange the above expression.

[tex]{m_g}a + {m_r}a = {m_g}g - \mu {m_r}g[/tex]

Simplify the above expression.

[tex]\fbox{\begin{minispace}\\{a = \dfrac{{\left( {{m_g} - \mu {m_r}} \right)g}}{\left( {{m_g} + {m_r}} \right)}}\end{minispace}}[/tex]

Learn more:

1. Change in momentum  https://brainly.com/question/9484203

2.  The motion of a body under friction brainly.com/question/4033012

3. A ball falling under the acceleration due to gravity brainly.com/question/10934170

Answer Details:

Grade: College

Subject: Physics

Chapter: Kinematics

Keywords:

Acceleration, friction, tension, system of two blocks, pulley mass system, motion, net force, string, block placed on the table, block hanging through a pulley, block is released from rest.

Answer:a) a force acting on an object in motion causes it to stop moving

Explanation: gradpoint

Contact forces occur when two objects are in contact while Non-contact forces occur without any physical contact.

Force is the action of push or pull which cause an object to move or stop.

Contact forces are the forces that come into existence when two objects come in contact to apply a force, For example: Frictional Force.

Non-contact forces takes place without any physical contact. For example: Gravitational force, Electric force, Magnetic force.

Thus, Contact forces occur when two objects are in contact while Non-contact forces occur without any physical contact.

Learn more about force.

https://brainly.com/question/13191643

#SPJ2

There are various examples of material culture like Tool, Monuments, Machines, etc.

The physical components of a society—the things that humans have created or altered—are referred to as material culture. These things are identified by their characteristics, whether chemical, physical, or biological, and they surround a population and its activities.

Shape, size, surface roughness, chemical composition and organization, molecular or cellular structure, color, and weight are examples of intrinsic properties of an entity.

Some more examples of material culture are :

Get more details about Material Culture  :

https://brainly.com/question/14306069

#SPJ2

A common finding with couples who can’t connect or resolve their problems is that they reason out or feel that it’s the other individual who can’t connect, not them. It’s the other individual that has the problem and they remain to think "I’m right and you’re wrong." The answer is letter a. In 40 percent of new weddings, one or both partners has been wedded no less than once before. Men are more probable to re-wed than women who've been wedded before.

What was the correct answer?

Answer:

Work done, [tex]W=1.03\times 10^6\ J[/tex]

Explanation:

It is given that,

Mass of the car, m = 950 kg

Distance up to the incline, d = 710 m

Angle of inclination of the car is 9 degrees.

When the car is push in the inclined path, the vertical component of the force will act on it such that the work done is given by :

[tex]W=mg\ sin\theta\times d[/tex]

[tex]W=950\times 9.8\times \ sin(9) \times 710[/tex]  

[tex]W=1.03\times 10^6\ J[/tex]

So, the minimum work needed to push a car is [tex]1.03\times 10^6\ J[/tex]. Hence, this is the required solution.

The minimum work needed to push a car up an incline can be found by calculating the work done against gravity. This is done by multiplying the mass of the car by the gravitational acceleration and the height, which can be deduced from the sin of the incline angle and the distance.

The question is asking for the minimum work done to push a car up an inclined plane. To solve this, you use the concept of work done against gravity. The formula for work done against gravity is Work = m * g * h, where m is the mass, g is the gravitational acceleration and h is the height. Since the distance and incline angle are provided, we can calculate the height with the trigonometric function sine, h = sin(angle) * distance. Substituting these values, we get Work = 950kg * 9.81m/s² * sin(9.0°) * 710m. The required calculation will then provide the answer to the work done in pushing the car up the incline.

https://brainly.com/question/31538865

#SPJ3

Answer: The entire process is mentioned in  step-wise below-

The motion of an unanchored rowboat when a water wave passes, the boat moves along with the waves. The boat doesn't move forward as waves could not take the boat along with it.

The water waves are waves that have a combination of both longitudinal and transverse waves. Longitudinal waves are the waves that cause the medium to move parallel to the direction of the waves.

Transverse waves are the waves that cause the medium to move perpendicular to the direction of the waves. When the waves travel through the water, the particles move in a clockwise direction. Most of the waves are formed by the winds.

The motion of the unanchored rowboat, when water pass is a boat, is moved up and down along with the wave. But the boat does not move forwards along with the wave. Because the wave does not carry the boat with it.

To learn more about the motion of a boat:

https://brainly.com/question/2599064

#SPJ2

The velocity of a 2kg mass with a kinetic energy of 16 J is 4 m/s. This result is obtained by rearranging and plugging the given values into the kinetic energy formula.

To find the velocity of a 2kg mass when its kinetic energy (KE) is 16 J, we can use the kinetic energy formula:

KE = (1/2)mv²

where:

By rearranging the formula to solve for v, we get:

v = √(2KE/m)

Substituting the given values:

v = √(2 × 16 J / 2 kg)

v = √(16 J / kg) = 4 m/s

So, the velocity of the 2kg mass when its kinetic energy is 16 J is 4 m/s.

The velocity of a 2 kg mass with 16 J of kinetic energy is 4 m/s, calculated using the kinetic energy formula KE = 0.5 * m * v² and solving for v.

To find the velocity of a 2 kg mass when its kinetic energy is 16 J, we can use the kinetic energy formula:

KE = ½mv²

where:

KE is the kinetic energy,

m is the mass,

v is the velocity.

First, we can solve for v:

v = √(2KE/m)

Substitute the given values (KE = 16 J and m = 2 kg) into the formula:

v = √(2*16 J / 2 kg)

v = √(16 J/kg)

v = 4 m/s

Therefore, the velocity of the 2 kg mass is 4 m/s.

Answer:

it will become charged by Q = 5C

Explanation:

When electron is removed from matter then the system will get positively charged

So we will have

[tex]Q = Ne[/tex]

here N = number of electrons that is removed

[tex]N = 31.25 \times 10^{18}[/tex]

also we know that charge on one electron is given as

[tex]e = 1.6 \times 10^{-19}[/tex]

here we have

[tex]Q = Ne[/tex]

[tex]Q = (31.25 \times 10^{18})(1.6 \times 10^{-19})[/tex]

[tex]Q = 5 C[/tex]

so it will become charged by Q = 5C

We have first velocity equals 8 m/s and the kinetic energy is 480 J

The order is the mass of the object

We have this formula K= mv2/2 isolating the mass we have m=k/v2

Now we will have m=480/8×8 > m= 480/64 m= 7.5kg now we deternimed the mass it's 7.5 kg

Final answer:

To determine the maximum volume flow rate of water, we need to calculate the power supplied by the pump. Plugging in the values, the maximum volume flow rate of water is approximately 0.0019 m³/s.

Explanation:

Answer:

To determine the maximum volume flow rate of water, we need to calculate the power supplied by the pump. The formula to calculate power is power = force * velocity. If we convert the horsepower of the pump to watts (

1 hp = 746 watts), we can use the formula power = (7 hp * 746 W/hp) * (0.82 efficiency) to find the power supplied by the pump. Next, we can use the formula power = work/time = force * distance/time = force * velocity to find the force exerted by the pump. Finally, we can use the formula volume flow rate = velocity * area, where the area can be calculated using the equation area = π * (diameter/2)^2, to find the maximum volume flow rate of water. Plugging in the values, the maximum volume flow rate of water is approximately 0.0019 m³/s.

By the Kepler's third law, the time period of this planet in earth years can be calculated to be 8 years.

Kepler's Third Law states that the squares of the orbital periods of the planets are directly proportional to the cubes of the semi-major axes of the planet's orbits. Kepler's Third Law implies that the period for a planet to orbit around the Sun increases rapidly with the radius of its orbit.

In planetary motion, the planet's net external torque is equal to zero. As a result, the angular momentum will not change.

T² is proportional to r³, and the orbital period square is proportional to the distance cubes.

(Tearth/ Tplanet)² = (rearth/ r planet)³

The planet's distance from the sun is 4 times the semi-major axis of the earth's orbit.

(Tearth/ Tplanet)² = (1/4)³

Tearth = Tplanet × √1/64 = Tplanet × (1/8)

Tplanet = 8 × Tearth

Tearth = 1 year

Tplanet = 8 years

Therefore, the time period of this planet is 8 years.

Learn more about Kepler's law here:

https://brainly.com/question/1608361

#SPJ5

Your question is incomplete, most probably the complete question is:

A planet is in a circular orbit around the sun. It distance from the sun is 4 times the semi major axis of earth's orbits. The period of the planets is

earth years

Using Kepler's third law of planetary motion, the period of a planet in circular orbit around the Sun at a distance of nine times the Earth's from the Sun is calculated to be 27 Earth years.

The question asks to determine the orbital period of a planet that is nine times farther from the Sun than the Earth, using Kepler's third law, which relates a planet’s orbital period to its distance from the Sun. The law states that the orbital period squared (P^2) is proportional to the average distance from the Sun cubed (a^3). Given that the Earth's average distance from the Sun is about 150 million km (1 astronomical unit or AU), the planet in question is 9 AU from the Sun.

According to Kepler's third law, (P_earth/P_planet)^2 = (a_earth/a_planet)^3. We know P_earth is 1 year and a_earth is 1 AU, so substituting for a_planet with 9 AU, we get (1 year / P_planet)^2 = (1 AU / 9 AU)^3.
Solving this equation for P_planet gives us an orbital period of 3^3, or 27 Earth years for the planet.