Enterprising students set an enormous slip-n-slide (a plastic sheet covered in water to reduce friction) on flat ground. If the slip-n-slide is 250 meters long.

a. How small does the average acceleration have to be for a student starting at 5 meters/second to slide to the end?
b. If the acceleration is 0.34 m/s^2, what is the minimum initial speed a student would have to run to make it to the end? Is this speed possible?

(a) The magnitude and direction of the net force on the crate while it is on the rough surface is 34.7 N in opposite direction.

(b) The net work done on the crate while it is on the rough surface is -22.6 J.

(c) The speed of the crate when it reaches the end is 0.5 m/s.

The magnitude and direction of the net force on the crate while it is on the rough surface is calculated as follows;

F (net) = F - Ff

where;

F (net) = F - μmg

where;

F (net) = 289 N  -   (0.359 x 92 X 9.8)

F (net) = -34.7 N

The negative sign indicates opposite direction to the applied force.

The net work done on the crate while it is on the rough surface is calculated as follows;

W = F(net) x L

where;

W = -34.7 x 0.65

W = -22.6 J

The speed of the crate when it reaches the end is calculated as follows;

acceleration of the crate = F(net) / m

a = -34.7 N / 92 kg

a = -0.377 m/s²

v² = u² + 2aL

v² = ( 0.855)²  +  ( 2 x -0.377 x 0.65)

v² = ( 0.855)²   - ( 2 x 0.377 x 0.65)

v² = 0.24

v = √ 0.24

v = 0.5 m/s

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The skater's final angular velocity is approximately 9.86 rad/s.

The skater's final angular velocity can be calculated using the principle of conservation of angular momentum. The equation for angular momentum is given by:

L = Iω

where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Initially, the skater has an angular momentum of:

L_initial = I_initial * ω_initial

Substituting the given values:

L_initial = 2.12 kg m² * 3.25 rad/s

The skater's final angular momentum remains the same, as angular momentum is conserved:

L_final = L_initial

The final moment of inertia is given as 0.699 kg m². Therefore, the final angular velocity can be calculated as:

L_final = I_final * ω_final

0.699 kg m² * ω_final = 2.12 kg m² * 3.25 rad/s

Solving for ω_final:

ω_final = (2.12 kg m² * 3.25 rad/s) / 0.699 kg m²

Hence, the skater's final angular velocity is approximately 9.86 rad/s.

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Car A's time to halt, tA = 0.33T. This is the right response to the question that was asked. The answer has two significant figures.

Provide an illustration of two vehicles with similar speeds but differing velocities. Example: Two 40 km/hr automobiles, one heading east and the other north.

Equal accelerations do not necessarily imply identical velocities, and vice versa. For instance, even if both of your cars have the same acceleration, if one of them accelerates first, it will certainly go more quickly than the other.

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Answer:

Unbalanced

Explanation:

Answer:

Explanation:

f = \(\sqrt{T/(m/L)} / 2L\)

T = 120 N

L = 3.00 m

(m/L) = 120 g/cm(100 cm/m / 1000 g/kg) = 12 kg/m

                                                  (wow that's massive for a "rope")

f = \(\sqrt{120/12} /(2(3))\))

f = \(\sqrt{10\\}\)/6 = 0.527 Hz

This is a completely silly exercise unless this "rope" is in space somewhere as the weight of the rope (353 N on earth) far exceeds the tension applied.

A much more reasonable linear density would be 120 g/m resulting in a frequency of √1000/6 = 5.27 Hz on a rope that weighs only 3.5 N

The gravitational force between two students is 5.336*10^-8 N.

According to universal gravitational law, the force acting on two bodies is given by the formula = F = (G *m1*m2)/r^2

Here mass of one student =m1=75kg,another student m2=54kg

Distance of separation =0.45m, r =0.45/2=0.225m

Force = 6.67*10^-11 (75*54)/(0.225)^2

F=5.336*10^-8 N

The force between two students is 5.336*10^-8 N.

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Answer:

Blood is moved away from the brain

Explanation:

Took the test

The physics behind the movement of electric charges between parallel plates is based on the principles of electrostatics. Electric charges are either positive or negative, and they are affected by electric fields.

Electric fields are created by a difference in electric potential, which is measured in volts. When a voltage is applied to a set of parallel plates, the charges within the plates will be affected by the electric field, and will move in response to it.

When a voltage is applied to a set of parallel plates, the positive charges in the plate connected to the positive voltage will be attracted to the negative voltage, while the negative charges in the plate connected to the negative voltage will be attracted to the positive voltage.

The movement of charges between the plates is also affected by the presence of any obstacles or resistances in the electric field, such as resistance in the wire. This can slow down the movement of charges and result in a decrease in the current flowing through the circuit.

In all, the movement of charges between electric parallel plates is the result of the electric field created by a difference in electric potential, and the movement of charges is called drift velocity. The movement is also affected by the presence of resistance.

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Answer:

Δr = 0.425 m

Explanation:

This is a sound interference exercise, the expression for destructive interference is

          Δr = (2n + 1) λ / 2

in this case the movement is in the same direction as the sound, therefore the movement is one-dimensional

let's use the relationship between the speed of sound and its frequency and wavelength

          v = λ f

          λ = v / f

the first minium occurs for n = 0

           Δr = λ / 2

            Δr = v / 2f

            Δr = \(\frac{340}{2 \ 400}\)

             Δr = 0.425 m

this is the distance from the current position that we assume in the center of the room

the frequency of the wave detected by the bat after the wave reflects off the moth is 54.24 KHz .

Calculation :

given ,

speed of sound(v) = 343 m/s

speed of bat(v bat) = 8.0 m/s

speed of moth (V moth) = 4.6 m/s

real frequncy of moth = 50.4 KHz

the first apparent heard by moth is :

f moth = \(\frac{V+Vmoth}{V-Vbat}*f\)

apparent frequently heard by the bat when the sound is reflected from the moth :

fbat  :  \((\frac{V+Vbat}{V-Vmoth}) (\frac{V+Vmoth}{V-Vbat} )f\)

f bat = \((\frac{343+8}{343-4.6} )(\frac{343+4.6}{343-8} )(50.4)\)

= 54.24

Frequency is the number of occurrences of a repeating event per unit time. [1] For clarity, also called temporal frequency, which is different from angular frequency. Frequency is measured in Hertz (Hz) and corresponds to one event per second. Duration is the reciprocal of frequency because duration is the time interval between events. [2]

For example, if the heart beats 120 times per minute (2 hertz), the period T (the interval between repeated beats) is 0.5 seconds (60 seconds divided by 120 beats). value). Frequency is an important parameter used in science and engineering to describe the speed of vibrations and vibrational phenomena such as mechanical vibrations, audio signals (sound), radio waves, and light.

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Explanation:

Ethics, also called moral philosophy, the discipline concerned with what is morally good and bad and morally right and wrong. The term is also applied to any system or theory of moral values or principles. Its subject consists of the fundamental issues of practical decision making, and its major concerns include the nature of ultimate value and the standards by which human actions can be judged right or wrong.

hope it helps!

Answer:

0.14 J

Explanation:

The maximum velocity is the amplitude times the angular frequency.

vmax = Aω

ω = vmax / A

ω = (3.2 m/s) / (0.06 m)

ω = 53.3 rad/s

For a spring-mass system:

ω = √(k / m)

ω² = k / m

k = ω²m

k = (53.3 rad/s)² (0.050 kg)

k = 142 N/m

The elastic potential energy is:

EE = ½ kx²

EE = ½ (142 N/m) (0.044 m)²

EE = 0.14 J

The portion of the range of electromagnetic waves that is visible to the human eye is much less than 1/10. Therefore, the correct answer is e. much less than 1/10. Visible light is a very small portion of the electromagnetic spectrum, with wavelengths ranging from about 400 to 700 nanometers. This represents only a small fraction of the entire electromagnetic spectrum, which includes radio waves, microwaves, infrared radiation, ultraviolet radiation, X-rays, and gamma rays.

Combined energy of the two sleds will be conserved when they collides.

Energy is nothing but the ability to do work. there are different energies in different form which are thermal energy, mechanical energy, electric energy and sound energy etc.

According to first law of thermodynamic, Energy neither be created nor be destroyed. it can only be transferred from one form into another form. Energy is expressed in joule (J). its dimensions are [M¹ L² T⁻²].

Energy is conserved throughout the motion,

according to conservation law of energy, initial energy is equal to final energy.

When two sleds coming in opposite direction, it is having mass as well as velocity. hence it has kinetic energy, when they get collide with each other some of total energy gets converted into sound energy as collision cause sound(boom). some of the energy will use to break the material which made the sled and remaining energy will convert in kinetic energy of broken sleds it can be moved in opposite direction or in same direction depending initial energies.

in this collision combined kinetic energy can be converted into sound energy, mechanical energy and again kinetic energy but totally energy is conserved.

Hence combined energy will be conserved.

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meters per second

What is Velocity?

The pace at which an object's position changes in relation to a frame of reference and time is what is meant by velocity. Although it may appear sophisticated, velocity is just the act of moving quickly in one direction. Since it is a vector quantity, the definition of velocity requires both magnitude (speed) and direction. Its SI equivalent is the meter per second (ms-1). A body is considered to be accelerating if the magnitude or direction of its velocity changes.

Therefore, it will measured in meters per second.

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(a) The work done by the force applied by the tractor is 79,968.47 J.

(b) The work done by the frictional force on the tractor is 55,977.93 J.

(c) The total work done by  all the forces is 23,990.54 J.

The work done by the force applied by the tractor is calculated as follows;

W = Fd cosθ

W = (5000 x 20) x cos(36.9)

W = 79,968.47 J

W = Ffd cosθ

W = (3500 x 20) x cos(36.9)

W = 55,977.93 J

W(net) = work done by applied force  -  work done by friction force

W(net) = 79,968.47 J -  55,977.93 J

W(net) = 23,990.54 J

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 a. The function that describes the oscillations of the object can be described using a sinusoidal function.

b. Yes, the function is shifted in the x-direction.

c. No, the function is not squeezed or stretched along the y-axis.

d. No, the function is not squeezed or stretched along the x-direction. The equation describing the position of the object in time can be expressed as :x(t) = A * cos(ωt + φ)

e. x(0) = A * cos(ω * 0 + φ) = 2 is the equation describing position of the object in time.

a. The function that describes the oscillations of the object can be described using a sinusoidal function. A common function used to represent oscillatory motion is the cosine function or the sine function. In this case, since the object is in the equilibrium position at t=0 and t=2s, we can use the cosine function to describe its oscillations.

b. Yes, the function is shifted in the x-direction. The equilibrium position of the object is at x=2m, which means the function is shifted 2 units to the right along the x-axis.

c. No, the function is not squeezed or stretched along the y-axis. The amplitude of the oscillations remains constant at 0.2 m.

d. No, the function is not squeezed or stretched along the x-direction. The period of the oscillations remains constant, indicating that the function is not compressed or stretched along the x-axis.

The equation describing the position of the object in time can be expressed as follows:

x(t) = A * cos(ωt + φ)

Where:

- x(t) represents the position of the object at time t.

- A is the amplitude of the oscillations (0.2 m in this case).

- ω is the angular frequency, given by ω = 2π / T, where T is the period of oscillation.

- φ is the phase constant, representing the initial phase of the oscillation.

e. To determine the equation describing the position of the object in time, we need to find the values of ω and φ. Since the object is in the equilibrium position at t=0 and t=2s, we can use this information to determine the values.

At t=0, the object is in the equilibrium position, so we have:

x(0) = A * cos(0 + φ) = 2

At t=2s, the object is in the equilibrium position again, so we have:

x(2) = A * cos(2ω + φ) = 2

Now,

At t=0, we have:

x(0) = A * cos(ω * 0 + φ) = 2

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Answer:

A. They'll bounce off each other

Explanation:

Nothing sticks together in a totally elastic collision.

To calculate the torque required to create an angular acceleration, we can use the formula:

Torque = Moment of Inertia × Angular Acceleration

The moment of inertia of a disk can be calculated using the formula:

Moment of Inertia = (1/2) × Mass × Radius^2

Given:

Mass = 15,000 kg

Radius = 6.14 m

Angular Acceleration = 0.0500 rad/s^2

First, calculate the moment of inertia:

Moment of Inertia = (1/2) × Mass × Radius^2

Moment of Inertia = (1/2) × 15,000 kg × (6.14 m)^2

Next, calculate the torque:

Torque = Moment of Inertia × Angular Acceleration

Torque = Moment of Inertia × 0.0500 rad/s^2

Now, let's plug in the values and calculate:

Moment of Inertia = (1/2) × 15,000 kg × (6.14 m)^2

Moment of Inertia ≈ 283,594.13 kg·m^2

Torque = 283,594.13 kg·m^2 × 0.0500 rad/s^2

Torque ≈ 14,179.71 N·m

Rounding to three significant figures, the torque required to create an angular acceleration of 0.0500 rad/s^2 is approximately 14,180 N·m.

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The right sphere will acquire an equal and opposite net positive charge to balance the negative charge on the left sphere.

Since the left sphere is attracted to the positively charged rod, it means that the left sphere acquires a temporary negative charge due to induction.

The positive charge on the rod repels electrons in the left sphere, causing them to move away from the rod side and accumulate on the opposite side, resulting in a net negative charge on the left sphere.

According to the principle of charge conservation, the net charge on the system must remain zero. Therefore, the right sphere acquires an equal and opposite net positive charge to balance the negative charge on the left sphere.

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Answer:

The definition is defined in the clarification section down, and according to the particular circumstance.

Explanation:

The value of the inductance, in , microhenrys of Tarik's solenoid is approximately 1.6239 microhenrys.

To calculate the inductance of a solenoid, we can use the formula:

L = (μ₀ * N² * A) / l

where L is the inductance, μ₀ is the permeability of free space (4π × 10⁻⁷ T·m/A), N is the number of turns, A is the cross-sectional area of the solenoid, and l is the length of the solenoid.

In this case, we have:

N = 175 turns

Diameter = 8.05 mm = 0.00805 m (converted to meters)

Radius = Diameter / 2 = 0.00805 m / 2 = 0.004025 m

Length (l) = 2.37 cm = 0.0237 m (converted to meters)

First, let's find the cross-sectional area (A) using the formula for the area of a circle:

A = π * r² = π * (0.004025 m)² ≈ 5.08398 × 10⁻⁵ m²

Now we can plug in the values into the formula for inductance:

L = (4π × 10⁻⁷ T·m/A * (175)² * 5.08398 × 10⁻⁵ m²) / 0.0237 m

L ≈ (1.2566 × 10⁻⁶ * 30625 * 5.08398 × 10⁻⁵) / 0.0237

L ≈ (38.5086 × 10⁻⁶) / 0.0237

L ≈ 1.6239 × 10⁻⁶ H

Now, let's convert the inductance from henrys to microhenrys:

L ≈ 1.6239 × 10⁻⁶ H * 10⁶ μH/H ≈ 1.6239 μH

So the inductance of Tarik's solenoid is approximately 1.6239 microhenrys.

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The first harmonic of the standing wave on a string with a loose end represents half a wavelength.

The fraction of a wavelength represented by the first harmonic is 1/2.

The higher harmonics of a standing wave on a string with a loose end are odd-number multiples of the first harmonic.

1. The first harmonic of a standing wave on a string with a loose end occurs when there is a node near the driver end and an antinode at the loose end. To measure the frequency of the first harmonic, we need to determine the fraction of a wavelength represented by this standing wave.

The first harmonic of the standing wave on a string with a loose end represents half a wavelength.

The first harmonic of a standing wave on a string with a loose end consists of a node near the driver end and an antinode at the loose end. This configuration creates the simplest standing wave pattern.

In a standing wave, a node is a point where the amplitude of the wave is always zero, representing a point of minimum displacement. An antinode, on the other hand, is a point of maximum displacement, where the amplitude is at its highest.

Since the loose end does not invert the phase of the wave upon reflection, the reflected wave and the original wave constructively interfere at the loose end, resulting in an antinode.

In the first harmonic, there is exactly half a wavelength between the node near the driver end and the antinode at the loose end.

Therefore, the fraction of a wavelength represented by the first harmonic is 1/2.

2. To predict the frequencies of higher harmonics, we can use the relationship that the frequency of each harmonic is a multiple of the frequency of the first harmonic. The higher harmonics can be calculated as follows:

Second Harmonic: The second harmonic consists of two nodes and one additional antinode compared to the first harmonic. The fraction of a wavelength for the second harmonic is 1/2 * 2 = 1. Thus, the second harmonic has a frequency that is twice that of the first harmonic.

Third Harmonic: The third harmonic consists of three nodes and two additional antinodes compared to the first harmonic. The fraction of a wavelength for the third harmonic is 1/2 * 3 = 1.5. Thus, the third harmonic has a frequency that is three times that of the first harmonic.

Fourth Harmonic: The fourth harmonic consists of four nodes and three additional antinodes compared to the first harmonic. The fraction of a wavelength for the fourth harmonic is 1/2 * 4 = 2. Thus, the fourth harmonic has a frequency that is four times that of the first harmonic.

In general, the higher harmonics of a standing wave on a string with a loose end are odd-number multiples of the first harmonic.

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Red-shift refers to the shift in the frequency or wavelength of light towards the red end of the spectrum. This phenomenon is observed when light is emitted from a source that is moving away from an observer.

In astronomy, red-shift is an important tool used to study the motion of celestial objects. The observed red-shift of light from distant galaxies indicates that they are moving away from us and the universe is expanding. This phenomenon is known as the Doppler effect and is a consequence of the relative motion between the source and the observer.

The amount of red-shift can be used to determine the velocity of the object and its distance from the observer. This information is used to estimate the size and age of the universe, as well as the distribution and motion of galaxies within it. The red-shift is also used to study other cosmic phenomena, such as the properties of supernovae and the presence of dark energy.

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Sound waves are a type of mechanical wave that propagate through a medium, typically air but also other materials such as water or solids.

Frequency: the frequency of a sound wave refers to the number of cycles or vibrations it completes per second and is measured in Hertz (Hz).

Amplitude: the amplitude of a sound wave refers to the maximum displacement or intensity of the wave from its equilibrium position. It represents the loudness or volume of the sound, with larger amplitudes corresponding to louder sounds and smaller amplitudes corresponding to softer sounds.

Wavelength: the wavelength of a sound wave is the distance between two consecutive points in the wave that are in phase, such as from one peak to the next or one trough to the next. It is inversely related to the frequency of the wave.

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He goes 400 m straight then goes 500m to the left side then goes backwards 400m so walking north is cancelled out he covers only 500 m to the west.

If u mean how much distance he travelled from starting point then this is the answer

Answer: He goes 400 m straight then goes 500m to the left side then goes backwards 400m so walking north is cancelled out he covers only 500 m to the west.

If you mean how much distance he travelled from starting point then this is the answer

Answer:

Work= Fcos∆×S

W=60N×cos 30⁰×100

W=60×0.866×100=5196.1J

PLEASE GIVE BRAINLIEST

Answer:

\(W=\frac{773}{4.45}=173.76 l b f\)

Explanation:

\(W=\frac{G \cdot m_{e} \cdot m}{(R+h)^{2}}\)

The law of gravitation

\(G=6.673\left(10^{-11}\right) m^{3} /\left(k g \cdot s^{2}\right)\)

Universal gravitational constant [S.I. units]

\(m_{e}=5.976\left(10^{24}\right) k g\)

Mass of Earth [S.I. units]

\(m=89 kg\)

Mass of a man in a spacecraft [S.I. units]

\(R=6371 \mathrm{~km}\)

Earth radius [km]

Distance between man and the earth's surface

\(h=261 \mathrm{~km} \quad[\mathrm{~km}]\)

ESULT \(W=\frac{6.673\left(10^{-11}\right) \cdot 5.976\left(10^{24}\right) \cdot 89}{\left(6371 \cdot 10^{3}+261 \cdot 10^{3}\right)^{2}}=773.22 \mathrm{~N}\)

\(W=\frac{773}{4.45}=173.76 l b f\)

Therefore, the answer is (a) 1292 cm is stick has a shadow when held vertical.

Additionally, since light moves in a straight path from its source to an object, the shadow of the object moves with the light source.

Let's use h centimetres to represent the tree's height. We have the following percentage in the problem:

height of tree/length of its shadow = height of meter stick/length of its shadow

or

h / 4200 = 100 / 325

We can solve this proportion for h:

h = 4200 * 100 / 325 = 1292.31 cm

Rounding to the nearest centimeter, we get:

h ≈ 1292 cm

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