suppose a 1150-kg elevator car is lifted a distance 42 m by its cable at a constant speed, assuming the frictional force on the elevator has a constant value of 1675 N.

r

\(- \frac{mR^2 }{I } \ v\)Answer:

a)      w = - \(\frac{m r }{I} v\)  ,  b)   W = - ½ m_woman R² (1 + m_woman R / I²) v²

Explanation:

a) To solve this exercise, let's use the conservation of angular momentum.

We define a system formed by the table and the woman, therefore the torques are internal and the moment is conserved

initial instant. Before starting to move the woman

         L₀ = 0

final instant. After starting to move

         L_f = I w + m v r

the moment is preserved

        L₀ = L_f

         0 = Iw + m v r

         w = - \(\frac{m r }{I} v\)                    (1)

the direction of the angular velocity is opposite to the direction of the linear velocity, that is, counterclockwise

b) for this part we use the relationship between work and kinetic energy

        W = ΔK

in this case the initial speed is zero and the final speed of the table, using the relationship between linear and angular variables

         v = w r

we substitute

          W = 0 - ½ I_total w²

          I_total = I + m_{woman} R²

          W = - ½ (I + m_woman R²)  ( \(\frac{m_{woman} R}{I} \ v\)) ²

          W = - ½ (m_woman² R² + m_woman³ R³ / I²) v²

          W = - ½ m_woman R² (1 + m_woman R / I²) v²

Answer:

Extrusive rocks form on the Earth's surface from lava, which is magma that has risen from underground. Intrusive rocks are formed when magma cools and solidifies within the planet's crust.

That's the difference

The magnitude of the traveler's displacement is approximately 39.5 km.

To find the magnitude of the traveler's displacement, we can combine the individual displacements in a vector sum.

The first leg of the journey is a displacement of 21.0 km east. We can represent this as a vector in the positive x-direction: (21.0 km, 0 km).

The second leg of the journey is a displacement of 28.8 km southeast. We can break down this displacement into its x and y components. The southeast direction is a combination of east and south. The angle between the southeast direction and the positive x-axis is 45 degrees. Using trigonometry, we can find the x and y components:

x-component = 28.8 km * cos(45°) = 20.4 km

y-component = 28.8 km * sin(45°) = 20.4 km

So, the second displacement can be represented as a vector: (20.4 km, -20.4 km).

The third and final leg of the journey is a displacement of 10 km south. We can represent this as a vector in the negative y-direction: (0 km, -10 km).

To find the total displacement, we add the individual displacements:

(21.0 km, 0 km) + (20.4 km, -20.4 km) + (0 km, -10 km) = (41.4 km, -30.4 km)

The magnitude of the displacement is given by the Pythagorean theorem:

Magnitude = √((41.4 km)² + (-30.4 km)²) ≈ 39.5 km

Therefore, the magnitude of the traveler's displacement is approximately 39.5 km.

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Answer:

an electron with element with large number of valence electron and large atomic radius

Answer:

an element with a small number of valence electrons and a large atomic radius

Explanation:

a) The unit vector in the direction of a is (0.993, 0.117).

b) The Cartesian vector representing the force is (24.825, 2.925).

c) The mechanical work done is 275 Joules.

The unit vector in the direction of vector a can be found by dividing vector a by its magnitude. In this case, vector a = [6, 1], so the magnitude of a is √(6² + 1²) = √37. Dividing vector a by its magnitude yields (6/√37, 1/√37), which simplifies to approximately (0.993, 0.117).

To find the Cartesian vector representation of the force, we multiply the unit vector from part a) by the magnitude of the force, F = 25 N. The result is (0.993 * 25, 0.117 * 25), which simplifies to (24.825, 2.925).

To determine the mechanical work done, we use the formula W = F * d * cos(θ), where W is the work done, F is the force, d is the displacement, and θ is the angle between the force and displacement vectors.

In this case, the force is F = 25 N, and the displacement is the difference between the x-coordinates of the two points, which is 15 - 4 = 11 meters. Since the force and displacement are in the same direction (θ = 0°), cos(θ) = 1.

Plugging in the values, we get W = 25 * 11 * 1 = 275 Joules.

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Answer:

1.67s

Explanation:

Given parameters:

Initial velocity  =  20m/s

Acceleration  = 6m/s²

Final velocity  = 30m/s

Unknown:

Time taken  = ?

Solution:

Acceleration is the rate of change of velocity with time. It is mathematically expressed as;

 Acceleration = \(\frac{v - u }{t}\)  

v is the final velocity

u is the initial velocity

 t is the time taken

            6  =\(\frac{30 - 20}{t}\)  

            6  = \(\frac{10}{t}\)  

              10  = 6t

              t  = 1.67s

The factors that determine how much gravity a body in the universe possesses are:

A. Mass: The amount of matter in a body determines its gravitational force. The greater the mass, the stronger the gravitational force.

C. Distance: The distance between two bodies also affects the gravitational force between them. The farther apart two objects are, the weaker their gravitational attraction.

D. Orbital motion: The speed and direction of an object's orbital motion also affect the gravitational force between two bodies. The faster an object is moving, the greater the gravitational force between it and the other body.

B. Weight: Weight is the measure of the gravitational force that acts on a body. It is directly proportional to mass, but it is not a factor that determines how much gravity a body possesses.

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The question given is a 300 N/C uniform electric field points perpendicularly toward the left face of a large neutral conducting sheet.

The formula of the electric field. \(E=\frac{F}{q}\)

The formula of area charge density.\($$\ \sigma = \frac {q} {A} $$\)

where E is the electric field.

F is the force of the electric charge.

q is the charge.

σ is the area charge density.

A is the area.

The electric field is given as E=300 N/C.

As the area is neutral and conductive, thus, there is no net charge and so σ = 0. A neutral conductor sheet doesn't have a charge on its face. Therefore the area charge density on the left and right faces is zero.

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Answer:

5 no

Explanation:

actually the 4kg lying on table has no influence

it slides towards 4kg weight hung

as it has excess 2kg force

force=miu × m ×g

The magnitude of acceleration of the entire system is  \(2.1315 \;\rm m/s^{2}\). Hence, option (5) is correct.

Given data:

The masses  suspended from the table are, m = 2 kg and m' = 4 kg.

And the mass kept on the table is, m'' = 4 kg.

The frictional force will oppose the motion of mass 4 kg, kept on the table. Then the required value of frictional force is,

\(f = \mu \times m'' \times g\\\\f = 0.11 \times 4 \times 9.8\\\\f = 4.312 \;\rm N\)

Since, the mass kept on the right side is more. Then at equilibrium,

-f + m''g + (m' +m)a = m'g

Here,

a is the magnitude of acceleration of the system.

Solving as,

-4.312+ 4(9.8) + (4 -2)a = 4(9.8)

2a = 4.312

\(a \approx 2.1315 \;\rm m/s^{2}\)

Thus, we can conclude that the magnitude of acceleration of the entire system is  \(2.1315 \;\rm m/s^{2}\). Hence, option (5) is correct.

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Answer:

40.90 mph if I did the math right

It takes approximately 5.5 hours for light to travel from the Sun to Pluto

The distance from the Sun to Pluto varies depending on the positions of the two objects in their orbits, as both the Sun and Pluto are constantly moving. However, on average, the distance from the Sun to Pluto is about 5.9 billion kilometers or 3.7 billion miles.

The speed of light is approximately 299,792 kilometers per second or 186,282 miles per second.

Using these values, we can calculate that it takes approximately 5.5 hours for light to travel from the Sun to Pluto

5.9 billion km ÷ 299,792 km/s

=> 19710 seconds

=> 5.5 hours.

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a) y(t) = te⁻³ᵗ is a possible position function.

b) General equation: y(t) = C₁e^(-2t) + C₂e^(-3t) + t(e^(-3t))(At + B).

c) Exact motion equation: y(t) = (23/5)e^(-2t) - (8/5)e^(-3t) + t(e^(-3t))(At + B), with initial conditions y(0) = 3 and v(0) = -7.

a) To show that y(t) = te⁻³ᵗ is a possible position function, we substitute it into the differential equation:

y(t) = te⁻³ᵗ

y'(t) = e⁻³ᵗ - 3te⁻³ᵗ

y''(t) = -6e⁻³ᵗ + 9te⁻³ᵗ

Substituting these expressions into the differential equation, we have:

-6e⁻³ᵗ + 9te⁻³ᵗ + 7(e⁻³ᵗ - 3te⁻³ᵗ) + 6(te⁻³ᵗ) = -6te⁻³ᵗ - e³ᵗ

Simplifying this equation, we find that both sides are equal, thus confirming that y(t) = te⁻³ᵗ is a possible position function.

b) The general equation for all possible position functions can be written as:

y(t) = C₁e^(-2t) + C₂e^(-3t) + t(e^(-3t))(At + B)

c) Given the initial conditions y(0) = 3 and y'(0) = v(0) = -7, we substitute these values into the general equation and solve for the constants:

3 = C₁ + C₂

-7 = -2C₁ - 3C₂

Solving these equations, we find C₁ = 23/5 and C₂ = -8/5.

The exact motion equation for the weight is:

y(t) = (23/5)e⁻²ᵗ - (8/5)e⁻³ᵗ + t(e⁻³ᵗ)(At + B)

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Suppose a weight of 1 kg is attached to an oscillating spring with friction constant b = 7 and stiffness constant k = 6. Suppose the external forces on the weight are: Fₑₓₜ(t) = - 6te⁻³ᵗ - e³ᵗ

y" + 7y' + 6y = - 6te⁻³ᵗ - e³ᵗ

a) Show y(t) = te⁻³ᵗ is a possible position function for this weight.

b) Find a general equation for all possible position functions.

c) Find the exact motion equation for this weight if its initial position is y(0) = 3, and its initial velocity is v(0) = y'(0) = -7.

The initial velocity of a car that reached 32 m/s in 3.4 at a rate of 2.6 m/s  is 23.16 m/s.

Velocity is described as the directional speed of a object in motion as an indication of its rate of change in position as observed from a particular frame of reference and as measured by a particular standard of time.

The given parameters include:

final velocity: 32 m/s

time: 3.4 seconds

acceleration = 2.6 m/s

V = u + at

32 = u + 3.4 * 2.6

32 = u + 8.84

32- 8.84 = u

initial velocity = 23.16.

So the initial velocity of the car that travels 32m/s in 3.4 at a rtae of 2.6 m/s is 23.16 m/s.

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Answer:

Kinetic energy

Explanation:

The ranking of the buoyant forces on the weighted balloon in water, from greatest to least, is as follows:

c. Pushed 2 m beneath the surface (highest buoyant force)

b. Pushed 1 m beneath the surface

a. Barely floating with its top at the surface (lowest buoyant force)

Let's consider the scenarios mentioned and rank the buoyant forces on a weighted balloon in water from greatest to least:

a. Barely floating with its top at the surface:

In this scenario, the balloon is floating at the water's surface, with only a small portion of the balloon submerged. The buoyant force is equal to the weight of the water displaced by the submerged portion of the balloon, which is relatively small. The top part of the balloon is exposed to air, so it doesn't contribute to buoyancy. The buoyant force in this case is relatively low.

b. Pushed 1 m beneath the surface:

When the balloon is pushed 1 meter beneath the surface, more of the balloon becomes submerged. As the depth increases, the volume of water displaced by the balloon also increases. The buoyant force on the balloon becomes greater than in scenario (a), as a larger volume of water is displaced by the balloon. Therefore, the buoyant force in this case is higher than in scenario (a).

c. Pushed 2 m beneath the surface:

When the balloon is pushed 2 meters beneath the surface, even more of the balloon becomes submerged, displacing an even larger volume of water. The buoyant force further increases compared to scenarios (a) and (b) because a greater volume of water is displaced by the balloon. Therefore, the buoyant force in this case is the highest among the three scenarios.

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Therefore, the amount of work done on the lawnmower is: W=Fdcosθ=(37 N)(20 m)cos15∘≈710J if is the amount of work the person.

Work is force applied over distance. Examples of work include lifting an object against the Earth's gravitation, driving a car up a hill, and pulling down a captive helium balloon. Work is a mechanical manifestation of energy. The standard unit of work is the joule (J), equivalent to a newton - meter.

Work is defined as transferring energy into an object so that there is some displacement. Energy is defined as the ability to do work. Work done is always the same. Energy can be of different types such as kinetic and potential energy.

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A galaxy is an enormous collection of gas, dust, billions of stars, and their solar systems all locked together by gravity.

A nebula is a very large cloud of gas and dust in outer space. When a dying star explodes, such as during a supernova, gas and dust are released into space, creating certain nebulae. Other nebulae are places where brand-new stars are only starting to develop.

The substance that fills the void between stars is, in a nutshell, the interstellar medium. Dust makes up the remaining 1% of the interstellar medium, which is 99% gas (mainly hydrogen).

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Answer:

The wavelength of the wave would increase to 10 meters.

Explanation:

We can use the formula:

velocity = frequency × wavelength

to relate the velocity, frequency, and wavelength of a wave.

Given that the wave has a speed of 20 m/s and a wavelength of 5 meters, we can solve for its frequency as follows:

frequency = velocity ÷ wavelength = 20 m/s ÷ 5 meters = 4 Hz

Now, if the same wave is created in the same medium, but with half the original frequency, its new frequency will be:

new frequency = 4 Hz ÷ 2 = 2 Hz

To find the new wavelength of the wave, we can rearrange the formula above to solve for wavelength:

wavelength = velocity ÷ frequency

Using the new frequency of 2 Hz, we get:

new wavelength = 20 m/s ÷ 2 Hz = 10 meters

Therefore, if the same wave was created in the same medium, with half the original frequency, the wavelength of the wave would increase to 10 meters.

225.0 g of water releases 32.07 kJ of heat when it cools from 85.5 °C to 50.0 °C.

When water cools, it releases heat. To calculate the amount of heat that 225.0 g of water releases as it cools from 85.5 °C to 50.0 °C,

we can use the formula q = mct. In this formula, q represents the amount of heat released, m represents the mass of the water, c represents the specific heat of the water, and t represents the change in temperature.

Plugging in the values given in the question, we get:q = 225.0 g × 4.18 J/g•°C × (85.5 °C − 50.0 °C) = 32,067.75 J or 32.07 kJ

Therefore, 225.0 g of water releases 32.07 kJ of heat when it cools from 85.5 °C to 50.0 °C.

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The specific heat of water is 4.18 J/g•°C. How much heat does 225.0 g of water release when it cools from 85.5°C to 50.0°C?

Use the formula q = mC T.

Answer: B

3.34 x 10 exponent- 4 J

Answer:

the impulse and change in momentum of the football is 18.83 N/s.

Explanation:

Given;

applied force, F = 420 N

time of force action, t = 22.3 s

The impulse and change in momentum of the football is calculated as;

J = ΔP = \(\frac{F}{t}\)

where;

J is the impulse

ΔP is the change in momentum

J = ΔP = \(\frac{F}{t}= \frac{420}{22.3} = 18.83 \ N/s\)

Therefore, the impulse and change in momentum of the football is 18.83 N/s.

Answer:

blue

Explanation:

blue is the only color being reflected, meaning it's the only one that will be visible

Answer:

4.4 seconds

Explanation:

use the equation

v(final)=v(initial)+a(t)

what we know:

v(initial)=28mps

v(final)=0 because it comes to a stop

a= -6.4    it is negative because the 6.4 is resisting forward motion

using these you get

0= 28 + (-6.4)t

6.4(t)=28

t=4.375 sec

and with rounding

t=4.4 sec

Answer:

4.4

Explanation:

i got it correct

Answer:

-compressions

-rarefactions

Explanation: research

The distance between the two charges is 3.12 m.

The distance between the two charges is determined by applying Coulomb's law of electrostatic force.

F = kq₁q₂ / r²

where;

The distance between the two charges is calculated as;

r² = kq₁q₂  / F

r = √ ( kq₁q₂  / F )

r = √ ( 9 x 10⁹ x 40 x 10⁻⁶ x 108 x 10⁻⁶ / 4 )

r = 3.12 m

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Given:

The mass on the table is,

The hanging mass is,

THe coefficient of friction is,

let the acceleration of the whole system is a (for the hanging mass it is downward and for the mass on the table it is rightward). the tension towards The fixed point of the pulley is T.

we can write,

Adding these equations we get,

Substituting the values we get,

Hence the acceleration is 6.66 m/s^2.

Answer:

B. people with OCD know their disorder is irrational

Explanation:

Got it right

Answer:

B

Explanation:

Edge2021

Answer:

C.) The sperm cells swim to the egg cells through their watery surroundings.

a. The translational kinetic energy of this system ( K trans) is 3462.112 J

b. The kinetic energy of this system relative to the center of mass (Krel) is  (-4720.612, -357.112, -2463.112) J

To find the total kinetic energy of the system, we need to calculate the kinetic energy of each particle and add them together.

The kinetic energy of a particle is given by:

K = 1/2 mv^2

where m is the mass of the particle and v is its velocity.

For particle 1:

K1 = 1/2 * 3 kg * (7, -4, 14) m/s^2 = (73.5, -42, 220.5) J

For particle 2:

K2 = 1/2 * 7 kg * (-13, 12, -3) m/s^2 = (-318.5, 252, -31.5) J

For particle 3:

K3 = 1/2 * 5 kg * (-29, 34, 18) m/s^2 = (-1012.5, 2895, 810) J

The total kinetic energy of the system is the sum of the kinetic energies of the particles:

Ktot = K1 + K2 + K3

= (73.5 - 318.5 - 1012.5, -42 + 252 + 2895, 220.5 - 31.5 + 810) J

= (-1257.5, 3105, 999) J

To find the translational kinetic energy of the system, we need to find the velocity of the center of mass and use it to calculate the total kinetic energy of the system as if all the mass were concentrated at the center of mass.

The velocity of the center of mass is given by:

Vcm = (m1V1 + m2V2 + m3V3) / (m1 + m2 + m3)

where V1, V2, and V3 are the velocities of the particles, and m1, m2, and m3 are their masses.

Plugging in the values, we get:

Vcm = (3 kg * <7, -4, 14> m/s + 7 kg * <-13, 12, -3> m/s + 5 kg * <-29, 34, 18> m/s) / (3 kg + 7 kg + 5 kg)

= (-215/15, 242/15, 111/15) m/s

= (-14.33, 16.13, 7.4) m/s

The total kinetic energy of the system as if all the mass were concentrated at the center of mass is given by:

Ktrans = 1/2 (m1 + m2 + m3) |Vcm|^2

Plugging in the values, we get:

Ktrans = 1/2 (3 kg + 7 kg + 5 kg) |(-14.33, 16.13, 7.4) m/s|^2

= 1/2 (15 kg) (461.0149 m^2/s^2)

= 3462.112 J

To find the kinetic energy of the system relative to the center of mass, we need to subtract the translational kinetic energy from the total kinetic energy:

Krel = Ktot - Ktrans

= (-1257.5, 3105, 999) J - 3462.112 J

= (-4720.612, -357.112, -2463.112) J

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As the chest expands (volume increases), the pressure in the chest becomes lower than that of the atmospheric pressure around the person's nose and mouth. This process is called inhalation or inspiration.

Inspiration has an unusual history in that its figurative sense appears to predate its literal one. It comes from the Latin inspiratus (the past participle of inspirare, “to breathe into, inspire”) and in English has had the meaning “the drawing of air into the lungs” since the middle of the 16th century. This breathing sense is still in common use among doctors, as is expiration (“the act or process of releasing air from the lungs”). However, before inspiration was used to refer to breath it had a distinctly theological meaning in English, referring to a divine influence upon a person, from a divine entity; this sense dates back to the early 14th century. So, As the chest expands (volume increases), the pressure in the chest becomes lower than that of the atmospheric pressure around the person's nose and mouth. This process is called inhalation or inspiration.

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