You push a 45 kg wooden box across a wooden floor at a constant speed of 1.0 m/s. The coefficient of kinetic friction is 0.25. Now you double the force on the box. How long would it take for the velocity of the crate to double to 2.0 m/s

Answer:

[tex]a. C=8X\\\\b. A=\frac{16x^2}{\pi}[/tex]

Explanation:

Given length is [tex]8x[/tex]

Circumference of a circle is given as:

[tex]C=2\pi r[/tex]

a.But the circumference is equal to length of the wire. Therefore circumference as a function of [tex]x[/tex] is:

[tex]C=8x[/tex]

b.Area as function of [tex]x[/tex]

Area is calculated as[tex]A=\pi r^2[/tex]

Rewrite the circumference equation to make r the subject of the formula.

[tex]2\pi r=8x\\r=\frac{4x}{\pi}[/tex]

To express area as a function of [tex]x[/tex]:-

[tex]A=\pi r^2\\=\pi (\frac{4x}{\pi})^2\\=\frac{16x^2}{\pi}[/tex]

Therefore area of circle as a function of [tex]x[/tex] is [tex]\frac{16x^2}{\pi}[/tex]

The circumference of the circle, represented as function of x, is C(x) = 8x. The area of the circle, represented as function of x, is A(x) = (16x²) / π.

The given length of the wire bent into a circle is 8x. This is equivalent to the circumference of this circle because a circle's circumference is the distance around it.

So, the function of the circumference, C, in terms of x is C(x) = 8x.

For the area of the circle, we must understand that the formula for the area of a circle is πr² where r is the radius of the circle. The radius is half the diameter, and since the circumference of a circle is πD where D is the diameter, we can find that r = circumference / (2π), or r = 8x / (2π).  

So, substituting r into the formula for the area, we get A(x) = π(8x / (2π))² = π(4x / π)² = π((4x)² / π²) = (16x²) / π.

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Answer:

No change in velocity of photons...

Explanation:

The reason speed of light is constant is that is doesn't depend upon intensity of light as produced by the source. This is the main reason why the wave function (psi)* in Einstein's Photoelectric experiment remains unaffected by intensity. The monochromatic beam of light will glow lower and lower as the power decreases and eventually vanishes from visible range as the emission from the power source has been stopped...

The angular acceleration of the pencil is 17 rad/sec^2

Explanation:

When the object is balanced at its center of mass then it is said to be motionless because the net force acts through the center of mass. If we shift the location of the net force then which results in the rotation of the object about the center of mass. The angular acceleration possessed by the object is proportional to the torque generated by the net force about the center of mass.

τ [tex]= I \alpha[/tex]

[tex]I[/tex] is the moment of inertia of a body

α  is the angular acceleration

Length of the pencil   L  =  15  c m  =  0.15  m

Mass of the pencil  = 10 g = 10 [tex]\times 10^{-3}[/tex] kg

Let  α be the angular acceleration of pencil

θ  be the inclination pencil from vertical

θ  =  10 °

Assume the pencil as thin rod and moment of inertia of the thin rod with the axis of rotation at one end is

[tex]I = \frac{mL^{2} }{3}[/tex]

When the pencil is balanced then the net torque acting on the pencil is zero and when we release the pencil and begin to fall then net torque acts on the pencil due to the weight of the pencil.

τ [tex]= F \times d[/tex]

τ [tex]= mgsin\theta \times \frac{L}{2}[/tex]

τ [tex]= 10 \times 10^{-3} \times 9.81 \times sin10 \times \frac{0.15}{2}[/tex]

τ [tex]= 1.277 \times 10^{-3} Nm[/tex]

We know that relation between torque and angular acceleration

τ [tex]= I\alpha[/tex]

[tex]\alpha = \frac{\tau}{I}[/tex]

[tex]\alpha = \frac{\tau}{\frac{mL^{2} }{3} }[/tex]

[tex]\alpha = \frac{1.277 \times 10^{-3} \times 3 }{10 \times 10^{-3} \times 0.15^{2} }[/tex]

[tex]\alpha = 17.02[/tex]

[tex]\alpha \approx 17 rad/sec^{2}[/tex]

The angular acceleration of the pencil is 17 rad/sec^2

The student should accelerate and travel through the intersection.

Explanation:

As per the given problem, the velocity with which the car of the student is moving is 14.7 m/s and he is 20 m away from intersection. Also the width of intersection is 10 m. It is also stated that traffic lights will change from yellow to red in 3 s. So totally , the student has to cover a distance of 30 m in 3 s to avoid getting stopped in intersection.

The velocity or speed required by the car to cover 30 m in 3 s is 30/3 = 15 m/s.

And the car is moving with initial velocity of 14.7 m/s. The student has two options either to decelerate at 4m/s² rate or to accelerate at the rate of 2 m/s².

So the velocity or speed of the car in 3 s with acceleration of 2 m/s² will be

Velocity = Acceleration × Time = 2 × 3 =6 m/s.

Thus on accelerating the car by 2 m/s² in 3 s, it will have the speed of 6 m/s and it can cover a distance of 6 × 3 = 18 m.

And the time taken by the car to reach from 20 m to intersection point with speed of 14.7 m/s is 20/14.7 = 1.36 s.

So, the car will take 1.36 s to reach the entrance of intersection with the speed of 14.7 m/s. But if it gets accelerated to 2 m/s², then it will have an increase in speed which will make the car to cover the complete intersection without stopping. So the student should accelerate the car.

Answer:

The correct answer is

1. the boy 2.

Explanation:

Since Tangential acceleration = v²/r  then the

v = ωr and v² =  ω²r² then the tangential acceleration = ω²r²/r = ω²r

This shows that  since ω is the same for both of them, the boy with greater r has the most acceleration.

The tangential acceleration [tex]a_t =[/tex] α·r

Answer:

neurotransmitters

Explanation:

Neurotransmitters are endogenous chemicals that enable neurotransmission. It is a type of chemical messenger which transmits signals across a chemical synapse, such as a neuromuscular junction, from one neuron (nerve cell) to another "target" neuron, muscle cell, or gland cell.

Answer:

This is because, the flywheel has a very large moment of Inertia and hence sudden piston torques have negligible effect on the flywheel, but every piston combined has a significant torque. This smoothens out the vibrations.

Explanation:

Final answer:

A flywheel reduces engine vibrations by utilizing its angular momentum to smooth out the intermittent power delivery from piston firings, thus ensuring steady rotation of the crankshaft and maintaining engine speed during periods when power is not supplied.

Explanation:

The primary reason why a flywheel reduces engine vibrations in conventional piston engines is due to its ability to store rotational energy. When the pistons fire, they impart energy to the crankshaft, causing the engine to turn. This energy, however, is delivered in pulses rather than smoothly, which can lead to uneven rotation and vibrations. The flywheel, being a heavy rotating disk, has significant angular momentum. This momentum ensures that the flywheel continues to spin between these pulses of power, effectively smoothing out the rotation of the engine's crankshaft. This results in a reduction of the vibrations that occur due to the intermittent nature of power delivery from individual piston firings.

Flywheels also help maintain engine speed during the periods when no power is supplied, such as between combustion events in a four-stroke engine. This steady rotation is crucial for the engine's performance and longevity. Additionally, flywheels play a role in energy conservation, as they can store energy and release it when needed, which is particularly important in stopping and starting scenarios.

Answer:

[tex]\mu_s \geq 0.27[/tex]

Explanation:

The centripetal force acting on the car must be equal to mv²/R, where m is the mass of the car, v its speed and R the radius of the curve. Since the only force acting on the car that is in the direction of the center of the circle is the frictional force, we have by the Newton's Second Law:

[tex]f_s=\frac{mv^{2}}{R}[/tex]

But we know that:

[tex]f_s\leq \mu_s N[/tex]

And the normal force is given by the sum of the forces in the vertical direction:

[tex]N-mg=0 \implies N=mg[/tex]

Finally, we have:

[tex]f_s=\frac{mv^{2}}{R} \leq \mu_s mg\\\\\implies \mu_s\geq \frac{v^{2}}{gR} \\\\\mu_s\geq \frac{(6\frac{m}{s}) ^{2}}{(9.8\frac{m}{s^{2}})(13.5m) }\\\\\mu_s\geq0.27[/tex]

So, the minimum value for the coefficient of friction is 0.27.

Yo sup??

a double bond has

1 Sigma bond

1 pi bond

and in total 2 bonds

Hope this helps

Final answer:

In a double bond, there is one sigma bond, which allows for free rotation, and one pi bond, which restricts rotation due to side-to-side overlap of orbitals.

Explanation:

In a double bond, there is always one sigma bond (sigma bond) and one pi bond (pi bond). The sigma bond is formed by the head-on overlap of orbitals, such as those that occur between s-s orbitals, s-p orbitals, or p-p orbitals. This type of bond allows for the free rotation of atoms around the bond axis. On the other hand, a pi bond arises from the side-to-side overlap of p-orbitals, creating electron density above and below the plane of the nuclei of the bonding atoms. This pi bond restricts the rotation of atoms around the bond axis due to the electron cloud above and below the bond plane.

Answer:

A) V = 7.5 V

B) E = 75,000 V/m

C) Q = 16.6 pC

D) V = 7.5 V

E) E = 24,000 V/m

F) Q = 52 pC

Explanation:

Given:

- The Area of plate A = ( 5 x 5 ) mm^2

- The distance between plates d = 0.10 mm

- The thickness of Mylar added t = 0.10 mm

- Voltage supplied by battery V = 7.5 V

Solution:

A) What is the capacitor's potential difference before the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

B) What is the capacitor's electric field before the Mylar is inserted?

- The Electric Field E between the capacitor plates is given by:

                                E = V / k*d

k = 1 (air)                  E = 7.5 / 0.10*10^-3

                                E = 75,000 V/m

C) What is the capacitor's charge Q before the Mylar is inserted?

                                C = k*A*ε / d

k = 1 (air)                   C = ( 0.005^2 * 8.85*10^-12 ) / 0.0001

                                C = 2.213 pF

                                Q = C*V

                                Q = 7.5*(2.213)

                                Q = 16.6 pC

D) What is the capacitor's potential difference after the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

E) What is the capacitor's electric field after the Mylar is inserted?    

- The Electric Field E between the capacitor plates is given by:

                                E = V / k*d

k = 3.13                     E = 7.5 / (3.13)0.10*10^-3

                                E = 24,000 V/m              

F) What is the capacitor's charge after the Mylar is inserted?      

                                C = k*A*ε / d

k = 3.13                    C = 3.13*( 0.005^2 * 8.85*10^-12 ) / 0.0001

                                C = 6.927 pF

                                Q = C*V

                                Q = 7.5*(6.927)

                                Q = 52 pC                                      

Here, we are required to evaluate parameters relating to the capacitor.

Data:

A) The potential difference across the two plates is equal to the voltage provided by the battery prior to the insertion of the Mylar, V = 7.5 V which remains constant throughout.

The potential difference across the two plates is equal to the voltage provided by the battery prior to the insertion of the Mylar, V = 7.5 V which remains constant throughout.B) The capacitor's Electric Field E between the capacitor plates is given as:

C) The capacitor's charge Q before the Mylar is inserted can be evaluated as follows;

ε /d

ε /d C = (1×2.5×10^(-5)×8.85*10^-12)/0.0001

D) The potential difference across the two plates, after the mylar has been inserted, V = 7.5 V which remains constant throughout.

E) The Electric Field, E between the capacitor plates is given by:

where the dielectric constant of the mylar = 3.13

F) C = k×A×ε / d

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Answer:

For further investigation see also the most recent World Population Data ... and Latin America and the Caribbean, and the regions of Melanesia, ... While Germany's death rate exceeds its birth rate, its population ... Population growth accelerated.

The forces doing non-zero work as the block slides down a frictionless ramp are gravity and any applied forces. The work done can be calculated using the mass, gravitational acceleration, and height. For the pushed block, the initial kinetic energy is combined with gravitational work to predict the final speed.

The forces that do non-zero work on a block as it slides down a frictionless ramp include gravity and any applied forces (like a push or a pull), but not the normal force or friction since the ramp is frictionless. Gravity does positive work as it pulls the block down the plane, increasing its kinetic energy. To calculate the total work done on the 1.85 kg wooden block after sliding down a distance of 1.85 m on the ramp at a 59.3° angle, you can use the formula:

To predict the speed of the block after sliding down, you can use the conservation of energy principle or kinematic equations that relate distance, acceleration, and velocity. For the scenario where the block is given an initial push, the final speed can be predicted using the same principles by accounting for the initial kinetic energy provided by the push.

Here's an example of calculating work done:

Explanation:

Below is an attachment containing the solution.

Answer:

The speed of each ball would be same at the instant each hits the ground.

Explanation:

From the principle of conservation of energy we know that

PE₁ + KE₁ = PE₂ + PE₂

where PE = mgh

The initial and final potential energies would be same for each ball since all the balls are thrown from the same height.

Now what about KE ?

We know that each ball is thrown with the same initial speed so their initial kinetic energies would be same and in turn their final kinetic energies must be same in order to satisfy th e conservation of energy principle. Therefore, we can conclude that the speed of each ball would be same at the instant they hit the ground. The initial angle of ball doesn't have any impact on the speed of the ball.

Answer:

True

Explanation:

The number density of an ideal gas at stp is called the loschmidt number, being named after the Austrian physicist Johann Josef Loschmidt who first estimated this quantity in 1865

n = [tex]\frac{P}{K_{B}T }[/tex]

where [tex]K_{B}[/tex] is the Boltzman constant

T is the temperature

P is the pressure

Loschmidt Number is the number of molecules of gas present in one cubic centimetre of it at STP conditions.Loschmidt constant is also used to define the amagat, which is a practical unit of number density for gases and other substances:

   1 amagat = n = 2.6867811×1025 m−3,

Answer:

True.

Explanation:

The Loschmidt number, n is defined as the number of particles (atoms or molecules) of an ideal gas in a given volume (the number density).

It is also the number of molecules in one cubic centimeter of an ideal gas at standard temperature and pressure which is equal to 2.687 × 10^19.

The ranking from least to greatest kinetic energy is: Set 3 < Set 1 < Set 5 < Set 2, Set 4.

To rank the sets of oranges and cantaloupes based on their kinetic energy, from the formula for kinetic energy:

Kinetic Energy (KE) = 0.5 × mass × speed²

Calculate the kinetic energy for each set and then rank them from least to greatest kinetic energy:

Set 1: KE = 0.5 × m × v²

Set 2: KE = 0.5 ×  4m ×  v² = 2 ×  0.5 ×  m ×  v² (Same as Set 1)

Set 3: KE = 0.5 ×  2m ×  (1/4v)² = 0.125 ×  ×  v² (Less than Set 1)

Set 4: KE = 0.5 ×  4m ×  v² = 2 ×  0.5 ×  m ×  v² (Same as Set 1 and 2)

Set 5: KE = 0.5 ×  4m ×  (1/2v)² = 1 ×  0.5 × m ×  v² (Same as Set 1, 2, and 4)

Ranking from least kinetic energy to greatest kinetic energy:

Set 3 (total mass: 2m, speed: 1/4v)

Set 1 (mass: m, speed: v)

Set 5 (total mass: 4m, speed: 1/2v)

Sets 2 and 4 (mass: 4m, speed: v)

Hence, the ranking from least to greatest kinetic energy is: Set 3 < Set 1 < Set 5 < Set 2, Set 4.

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The kinetic energy of an object is dependent on both its mass and speed. The ranking of the sets of oranges and cantaloupes, based on their kinetic energy (from least to greatest), is: KE3, KE5, KE1 = KE4, KE2. These rankings are calculated by substituting the values of mass and speed provided into the formula for kinetic energy.

The kinetic energy of an object can be defined as the energy which it possesses due to its motion. It is dependent on both its mass and speed, as outlined by the formula KE = 1/2mv². Where KE is the kinetic energy, m is the mass of the object, and v is the velocity (or speed).

Considering the information provided on the sets of oranges and cantaloupes:

Ranking them according to their kinetic energy, from least to greatest, gives us: KE3, KE5, KE1 = KE4, KE2.

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Answer: A. Study conceptually the nature of the electric charge and force

Explanation: Since the electrons are the mobile charge carriers, they will therefore be the particles that are transferred. For this case, since the electrophorus is positively- charged, the electrons from the sphere will be attracted to the electrophorus and thus will be transfer there.

Answer

given,

R₁= 4 Ω

R₂ = 3 Ω

When two resistors are connected in series

R = R₁ + R₂

R = 4 + 3

R = 7 Ω

When two resistors are connected in series then their effective resistance is equal to 7 Ω .

When two resistors are connected in parallel.

[tex]\dfrac{1}{R}=\dfrac{1}{R_1}+\dfrac{1}{R_2}[/tex]

[tex]\dfrac{1}{R}=\dfrac{1}{3}+\dfrac{1}{4}[/tex]

[tex]\dfrac{1}{R}=\dfrac{7}{12}[/tex]

[tex]R = 1.714 \Omega[/tex]

Hence, the equivanet resistance in parallel is equal to [tex]R = 1.714 \Omega[/tex]

Answer:

5.0 Ω

Explanation:

[tex]R_{1}[/tex] = 4.0 Ω,     [tex]R_{2}[/tex] = 4.0 Ω,     [tex]R_{3}[/tex] = 3.0 Ω

[tex]\frac{1}{R_{parallel}} = \frac{1}{R_{1}} + \frac{1}{R_{2}} \\\frac{1}{R_{parallel}} = \frac{1}{4.0} + \frac{1}{4.0}\\\frac{1}{R_{parallel}} = \frac{2}{4.0}\\\frac{1}{R_{parallel}} = 0.5\\R_{parallel} = (0.5)^{-1} \\R_{parallel} = 2.0[/tex]

[tex]R_{T} = R_{parallel} + R_{3} \\R_{T} = 2.0 + 3.0\\R_{T} = 5.0[/tex]

Therefore, the effective resistance of this combination is 5.0 Ω.

Answer:

Option A is correct.

Inheritance, superclass, subclass

A new class of objects can be created conveniently by **inheritance**; the new class (called the **superclass**) starts with the characteristics of an existing class (called the **subclass**), possibly customizing them and adding unique characteristics of its own.

Explanation:

A class is a term in object oriented programming that is extensible program code used to create, name and implement what you want the objects to do.

Inheritance is used to describe the making of new classes from already existing classes. The general properties of the old classes remain in the new classes created from them.

Hence, the new classes (because they're usually an improvement on the old classes as they hold on to the wanted properties of the old classes and additional great ones are added or modified) are called superclasses and the old, already existing ones are called subclasses.

The missing words from the given paragraph are inheritance, superclass and subclass respectively.

The given problem is based on the concept of Object Oriented Programming (OOPs). A class is a term in object oriented programming that is extensible program code used to create, name and implement what you want the objects to do.

Inheritance is used to describe the making of new classes from already existing classes. The general properties of the old classes remain in the new classes created from them.

Hence, the new classes (because they're usually an improvement on the old classes as they hold on to the wanted properties of the old classes and additional great ones are added or modified) are called super classes and the old, already existing ones are called subclasses.

So, the complete paragraph will be,

A new class of objects can be created conveniently by inheritance; the new class (called the superclass) starts with the characteristics of an existing class (called the subclass), possibly customizing them and adding unique characteristics of its own.

Thus, we can conclude that the missing words from the given paragraph are inheritance, superclass and subclass respectively.

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Human eye could perceive this visible light and used it to look at the objects.

Explanation:

There are 3 forms of radiation emitted by the sun. They are Infrared radiation, Ultra Violet radiation as well as visible radiation.

Visible light is the type of light can perceive by the human eye and utilizes to see the objects.  

Since the sun is hotter (5800 K) than the earth, it emits mostly the radiation energy in the form of this visible light.

Answer:

a) [tex]U_{e} = 3 \times 10^{10}\,J[/tex], b) [tex]v \approx 7745.967\,\frac{m}{s}[/tex]

Explanation:

a) The potential energy is:

[tex]U_{e} = Q \cdot \Delta V[/tex]

[tex]U_{e} = (30\,C)\cdot (1.0\times 10^{9}\,V)[/tex]

[tex]U_{e} = 3 \times 10^{10}\,J[/tex]

b) Maximum final speed:

[tex]U_{e} = \frac{1}{2}\cdot m \cdot v^{2}\\v = \sqrt{\frac{2\cdot U_{e}}{m} }[/tex]

The final speed is:

[tex]v=\sqrt{\frac{2\cdot (3 \times 10^{10}\,J)}{1000\,kg} }[/tex]

[tex]v \approx 7745.967\,\frac{m}{s}[/tex]

The change in the energy transferred or potential energy is 3 x 10¹⁰ J.

The final speed of the charge released is 7745.97 m/s.

The change in the energy transferred or potential energy is calculated as follows;

U = QV

U = 30 x 1 x 10⁹

U = 3 x 10¹⁰ J

The speed of the charge released is calculated by applying law of conservation of energy.

[tex]K.E = U\\\\\frac{1}{2} mv^2 = U\\\\v= \sqrt{\frac{2U}{m} } \\\\v = \sqrt{\frac{2\times 3\times 10^{10} }{1000} }\\\\v = 7745.97 \ m/s[/tex]

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Answer:

The final temperature when the lead and water reach thermal equilibrium is [tex]23.14^{o} C[/tex]

Explanation:

Using the law of conservation of energy the heat lost by the lead ball is gained by the water.

[tex]-q_{L} =+q_{w}[/tex] .............................1

where

[tex]q_{L}[/tex] is the heat energy of the lead ball;

[tex]q_{w}[/tex] is the heat energy of water;

but q = msΔ

T...............................2

substituting expression in equation 2 into the equation 1;

Δ

T =Δ

T .............................3

where [tex]m_{L}[/tex] is the mass of the lead ball = 5.2 g

[tex]s_{L}[/tex] is the specific heat capacity of the lead ball = 0.128 Joules/g-deg

[tex]m_{w}[/tex] is the mass of water = volume x density = 34.5 ml x 1 g/ml = 34.5 g

[tex]s_{w}[/tex] is the specific heat capacity of water = 4.184 J/g-deg

Δ

T is the temperature changes encountered by the lead ball and water   respectively

inputting the values of the parameters into equation 3

5.2 g x 0.128 Joules/g-deg x (183-22.4) = 34.5 g x 4.184 J/g-deg x ([tex]T_{2}[/tex] -22.4)

[tex]T_{2}[/tex] -22.4 = [tex]\frac{106.9}{144.3}[/tex]

[tex]T_{2}[/tex] -22.4 = 0.7408

[tex]T_{2}[/tex] = [tex]23.14^{o} C[/tex]

Therefore the final temperature  when the lead and water reach thermal equilibrium is [tex]23.14^{o} C[/tex]

Answer:

The ratio of moment of inertia of larger sphere to that of smaller sphere = 4

Explanation:

The moment of inertia of solid sphere is given by I = 2/5MR² where M = mass of sphere and R = radius of sphere.

Radius of smaller sphere = D/2

Radius of larger sphere = 2D/2 = D.

Moment of inertia of smaller sphere I₁ = 2/5M × D²/4 = MD²/10

Moment of inertia of larger sphere I₂ = 2/5M × D² = 2MD²/5

The ratio of moment of inertia of larger sphere to that of smaller sphere = I₂/I₁ = 2MD²/5 ÷ MD²/10 = 10 × 2/5 = 4

Final answer:

To calculate the final temperature after placing ice cubes into water, principles of thermodynamics are used, involving calculations for warming the ice, melting it, and adjusting the water temperature. The steps highlight the application of heat transfer and phase change concepts within a closed system.

Explanation:

The student's question involves the calculation of the final temperature of the water after two 20.0 g ice cubes at -20.0 °C are placed into 285 g of water at 25.0 °C. The principles of thermodynamics, specifically the concept of heat transfer and the conservation of energy, are essential to solving this problem. However, with the given information, an exact numerical solution cannot be provided without knowing the specific heats of ice and water, as well as the heat of fusion of ice. Generally, the solution involves several steps:

Calculate the amount of energy required to warm the ice from -20.0 °C to 0 °C using the specific heat of ice.

Calculate the energy needed to melt the ice into water at 0 °C using the enthalpy of fusion.

Calculate the energy released by the water as it cools down to the new equilibrium temperature.

Set the energy gained by the ice equal to the energy lost by the water to find the final temperature of the mixture.

Without the numerical values, the step-by-step method demonstrates how principles of heat transfer and phase change are applied to predict changes in temperature and state within a closed system.

The student's question involves calculating the maximum speed a car can take a banked curve without sliding, using physic principles of circular motion and the coefficient of static friction.

The student's question revolves around determining the maximum speed at which a 1400 kg car can navigate a banked highway curve without sliding, given a static coefficient of friction between rubber and concrete. To solve this, we need to use principles of circular motion and friction.

To find the maximum speed (v), we can use the following equation that balances gravitational, frictional, and centripetal forces:

F_{friction} = μ_s × N

N = m × g × cos(θ)

F_{centripetal} = m × v^2 / r

Where μ_s is the static coefficient of friction (1.0), N is the normal force, m is the mass of the car (1400 kg), g is the acceleration due to gravity (9.8 m/s^2), θ is the banking angle (19.0 degrees), v is the maximum speed, and r is the radius of the curve (60.0 m). The forces due to friction and centripetal need to equal for the car to navigate the curve without sliding.

After applying trigonometry and mechanics principles, we can solve for v which gives us the maximum speed without sliding.

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Answer:

If [tex]\large{P_{H}}[/tex] and [tex]\large{P_{S}}[/tex] be the power consumed by the headlight and starter respectively, then the total power consumed will be [tex]\large{P = \dfrac{P_{H}P_{S}}{P_{H} + P_{S}}}[/tex]

Explanation:

We know that electric power ([tex]P[/tex]) is given by

[tex]\large{P = \dfrac{V^{2}}{R}}[/tex]

where '[tex]V[/tex]' is the applied voltage and '[tex]R[/tex]' is the resistance.

If we consider that '[tex]\large{P_{H}}[/tex]' and '[tex]\large{P_{S}}[/tex]' be the power consumed by the headlight and starter respectively, then the resistance ([tex]\large{R_{H}}[/tex]) of the headlight and the resistance ([tex]\large{R_{S}}[/tex]) of the starter can be written as

[tex]&&\large{R_{H} = \dfrac{V^{2}}{P_{H}} = \dfrac{12^{2}}{P_{H}}}\\&and,& \large{R_{S} = \dfrac{V^{2}}{P_{S}} = \dfrac{12^{2}}{P_{S}}}[/tex]

If the headlight and the starter in connected in series, then equivalent resistance ([tex]\large{R_{eq}}[/tex]) will be

[tex]\large{R_{eq} = R_{H} + R_{S} = 12^{2}(\dfrac{1}{P_{H} + P_{S}})}[/tex]

So the total power ([tex]P[/tex]) consumed will be given by

[tex]\large{P = \dfrac{12^{2}}{R_{eq}} = \dfrac{1}{(\dfrac{1}{P_{H}} + \dfrac{1}{P_{S}})} = \dfrac{P_{H}P_{S}}{P_{H} + P_{S}}}[/tex]

When the 30.0-W headlight and the 2.40-kW starter are connected in series to a 12.0-V battery, they consume a total power of 2430 W.

The total power consumed by the headlight and starter when connected in series to a 12.0-V battery can be calculated by adding the individual powers.

Answer:

Current = 8696 A

Fraction of power lost = [tex]\dfrac{80}{529}[/tex] = 0.151

Explanation:

Electric power is given by

[tex]P=IV[/tex]

where I is the current and V is the voltage.

[tex]I=\dfrac{P}{V}[/tex]

Using values from the question,

[tex]I=\dfrac{1000\times10^6 \text{ W}}{115\times10^3\text{ V}} = 8696 \text{ A}[/tex]

The power loss is given by

[tex]P_\text{loss} = I^2R[/tex]

where R is the resistance of the wire. From the question, the wire has a resistance of [tex]0.050\Omega[/tex] per km. Since resistance is proportional to length, the resistance of the wire is

[tex]R = 0.050\times40 = 2\Omega[/tex]

Hence,

[tex]P_\text{loss} = \left(\dfrac{200000}{23}\right)^2\times2[/tex]

The fraction lost = [tex]\dfrac{P_\text{loss}}{P}=\left(\dfrac{200000}{23}\right)^2\times2\div (1000\times10^6)=\dfrac{80}{529}=0.151[/tex]

Answer:

Force(F) = -80,955.01 N

Explanation:

We need to first determine the impulse that the truck driver received from the car during the collision

So; m₁v₁ - m₂v₂ = (m₁m₂)v₀

where;

m₁ = mass of the truck = 4280 kg

v₁ = v₂ =  speed of the each vehicle = 7.69 m/s

m₂ = mass of the car = 810 kg

Substituting our data; we have:

(4280×7.69) - (810×7.69) = (4280+810)v₀

32913.2 - 6228.9 = (5090)v₀

26684.1 =  (5090)v₀

v₀ = [tex]\frac{26684.1}{5090}[/tex]

v₀ = 5.25 m/s

NOW, Impulse on the truck = m (v₀ - v)

= 4280 × (5.25 - 7.69)

= 4280 ×  (-2.44)

= -10,443.2 kg. m/s

Force that the seat belt exert on the truck driver can be calculated as:

Impulse = Force × Time

-10,443.2 kg. m/s = F (0.129)

F = [tex]\frac{-10,443.2}{0.129}[/tex]

Force(F) = -80,955.01 N

Thus, the Force that the seat belt exert on the truck driver = -80,955.01 N

Answer:

The scientist will be looking for the velocity of the wave in air which is equivalent to 10^7m/s

Explanation:

If an object in space is giving off a frequency of 10^13Hz and wavelength of 10^-6m then the scientist will be looking for the velocity of the object in air.

The relationship between the frequency (f) of a wave, the wavelength (¶) and the velocity of the wave in air(v) is expressed as;

v = f¶

Given f = 10^13Hz and ¶ = 10^-6m,

v = 10¹³ × 10^-6

v = 10^7 m/s

The value of the velocity of the object in space that the scientist will be looking for is 10^7m/s

Answer: Rolling friction

Explanation:

Rolling friction is said to act when an object rolls over a surface. Machines often have moving parts. These moving parts roll over each other as the machine works. Ball bearings are commonly used to reduce friction between the two surfaces in contact as the machine parts roll over each other. This is a typical example of rolling friction as applied to machine parts.

Answer:

Rolling friction.

Explanation:

Rolling friction is the force resisting the motion when a body ( such as a ball, tires, wheel) rolls on a surface. It is applicable for bodies whose point of contact keeps changing. It is the force that opposes the motion of a body which is rolling over the surface of another.

Rolling resistance, sometimes called rolling friction or rolling drag, is the force resisting the motion when a body rolls on a surface. It is mainly caused by non-elastic effects; that is, not all the energy needed for deformation of the wheel, roadbed, etc. is recovered when the pressure is removed.