Please help me! Show work! Margaret, whose mass is 52 kg, experienced a net force of 1750 N at the bottom of a roller coaster loop during her school's physics field trip to Six Flags. What is her acceleration at the bottom of the loop?

Answer Choices
A. 4
B. 14
C. 24
D. 34

Answer:

[tex]t_2=\sqrt{2}(t_1)[/tex]

Explanation:

The kinetic energy of a body is that energy it possesses due to its motion. In classical mechanics, this energy depends only on its mass and speed, as follows:

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

The speed in an uniformly accelerated motion is given by:

[tex]v=at[/tex]

Replacing this expression in the formula for the kinetic energy, we have:

[tex]K=\frac{ma^2t^2}{2}\\[/tex]

So, if we have [tex]K_2=2K_1[/tex]:

[tex]K_1=\frac{ma^2t_1^2}{2}(1)\\K_2=\frac{ma^2t_2^2}{2}\\2K_1=\frac{ma^2t_2^2}{2}\\K_1=\frac{ma^2t_2^2}{4}(2)\\[/tex]

Equaling (1) and (2) and solving for [tex]t_2[/tex]:

[tex]\frac{ma^2t_1^2}{2}=\frac{ma^2t_2^2}{4}\\t_2=\frac{4t_1^2}{2}\\t_2=\sqrt{2t_1^2}\\t_2=\sqrt{2}(t_1)[/tex]

Answer:0.906 N/m

Explanation:

Given

time period [tex]T=0.66 s[/tex]

mass [tex]m=0.01 kg[/tex]

System can be considered as spring mass system

Time Period of spring mass system is given by

[tex]T=2\pi \sqrt{\frac{m}{k}}[/tex]

squaring

[tex]k=\frac{4\pi ^2m}{T^2}[/tex]

[tex]k=\frac{4\times \pi ^2\times 0.01}{0.66^2}[/tex]

[tex]k=0.906 N/m[/tex]

Answer:

both technicians are correct

Explanation:

both technicians are correct because technician A says the first external component to be installed is the intake manifold and technician B says the exhaust manifold should be installed after the intake manifold has been installed, this implies the same thing because installing the exhaust manifold after installing the intake manifold means you are installing the intake manifold first.

Answer:

f = 63.8 N

Explanation:

initial angle to the vertical = 5 degrees

initial holding force = 40 N

final angle to the vertical = 8 degrees

final holding force = ?

find the final holding force

        m = mass  and  g = acceleration due to gravity

        40 = mgSin5

        mg = [tex]\frac{40}{sin5}[/tex] ....equation 1

        f = mgSin8  ......equation 2

   substituting the value of mg from equation 1 (where mg = [tex]\frac{40}{sin5}[/tex] ) into equation 2

    f = mgSin8 = [tex]\frac{40}{sin5}[/tex] x Sin8

   f = 63.8 N

Answer:

64 N

Explanation:

Given that the angles are very small, the following approximation can be made:

F ≈ m*g*α

where F is the force needed to hold the ball, m is the ball mass, g is the acceleration of gravity and α is the angle between the cord and the vertical.

Let's call F1 the force needed to hold the ball at 5° (α1) and F2 the force needed to hold the ball at 8° (α2).

F1 ≈ m*g*α1

F2 ≈ m*g*α2

Dividing the equations:

F1/F2 = α1/α2

F2 = (α2/α1)*F1

F2 = (8/5)*40

F2 = 64 N

Answer:

The wavelength is 539 nm.

Explanation:

Given that,

Distance of screen L= 10 m

Width s= 5.4 cm

Grating separation d= 0.1 mm

Suppose the pattern is displayed on a screen a distance L from the grating and the spots are separated by s. we need to find the wavelength.

The angle between the central and first maximum is given as

[tex]\tan\theta=\dfrac{s}{L}[/tex]

[tex]\theta=\tan^{-1}\dfrac{s}{L}[/tex]

[tex]\theta=\tan^{-1}\dfrac{5.4\times10^{-2}}{10}[/tex]

[tex]\theta=0.309[/tex]

We need to calculate the wavelength

The condition for maximum of a diffraction grating is

[tex]d\sin\theta=m\lambda[/tex]

[tex]\lambda=\dfrac{d\sin\theta}{m}[/tex]

Put the value into the formula

[tex]\lambda=\dfrac{0.1\times10^{-3}\times\sin0.309}{1}[/tex]

[tex]\lambda=539\times10^{-9}\ m[/tex]

[tex]\lambda=539\ nm[/tex]

Hence, The wavelength is 539 nm.

The wavelength of a laser can be determined using the diffraction grating equation. Calculate the θ value with θ = tan⁻¹(y/L), then substitute the known values into the equation λ = (d sin θ) / m. Convert the resultant wavelength to nanometers.

Your question relates to the physics of light and diffraction grating. We can find the wavelength of the laser by using the diffraction grating equation. The equation is: d sin θ = mλ

Where:

In your case, we need to calculate θ first: θ = tan⁻¹(y/L) where y is the distance between the spots (5.4cm) and L is the distance from the grating to the screen (10m).

Substitute the known values and solve for λ: λ = (d sin θ) / m

Remember, the resultant wavelength should be converted to nanometers (1m = 1 x 10⁹ nm).

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a typical asteroid in the asteroid belt, a typical oort cloud object, a typical kuiper belt object

Although Oort cloud comets are now located far beyond the Kuiper belt, they are thought to have formed in the region of the jovian planets. They were then "kicked out" to their current orbits by gravitational encounters with the jovian planets.

The objects in the solar system can be ranked from nearest to farthest based on their formation distance from the Sun: Mercury, Venus, Earth, Mars, and Jupiter.

The objects in the solar system can be ranked based on their distance from the Sun as follows:

Therefore, the ranking from nearest to farthest would be Mercury, Venus, Earth, Mars, and Jupiter.

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Answer:

80% (Eighty percent)

Explanation:

The material has a refractive index (n) of 1.25

Speed of light in a vacuum (c) is 2.99792458 x 10⁸  m/s

We can find the speed of light in the material (v) using the relationship

n = c/v, similarly

v = c/n

therefore v = 2.99792458 x 10⁸  m/s ÷ (1.25) = 239 833 966 m/s

v = 239 833 966 m/s

Therefore the percentage of the speed of light in a vacuum that is the speed of light in the material can be calculated as

(v/c) × 100 = (1/n) × 100 = (1/1.25) × 100 = 0.8 × 100 = 80%

Therefore speed of light in the material (v) is eighty percent of the speed of light in the vacuum (c)

Answer:

The percentage of speed of light in vacuum to the speed of light in the said material is 80%

Explanation:

The common values of refractive index are between 1 and 2 since nothing can travel faster than the speed of light, therefore, no material has a refractive index lower than 1.

According to the formula [tex]n=\frac{c}{v}[/tex]

where [tex]n[/tex] is the index of refraction

[tex]c[/tex] is the speed of light in vacuum

and [tex]v[/tex] is the speed of light in the material, it can be seen that n and v are inversely proportional which means greater the refractive index lower is the speed  of light.

Since we know that speed of light in vacuum is 300,000 km/s using the formula we get,

[tex]v=\frac{c}n}[/tex]

[tex]v=\frac{300000}{1.25}=240,000  km/s[/tex]

for finding percentage,

[tex]=\frac{240000}{300000}*100 = 80[/tex] %

As the gases ( O₂ & CO₂ ) are most abundant here and supports life.

Hope it helps!

Happy Learning!!

:D

In this part, we discuss the expected decrease in angular acceleration when the child is on the merry-go-round due to the greater moment of inertia. We calculate the child's moment of inertia by approximating the child as a point mass at a specific distance from the axis.

b. We expect the angular acceleration for the system to be less in this part because the moment of inertia is greater when the child is on the merry-go-round. To find the total moment of inertia I, we first find the child's moment of inertia Ic by approximating the child as a point mass at a distance of 1.25 m from the axis. Then

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Final answer:

The rotational kinetic energy of the merry-go-round plus child at 29.2 rpm is 394.5834 J. The father must push for approximately 0.167 revolutions to reach this velocity from rest. A force of -8.975 N (in the opposite direction of motion) must be exerted to stop the merry-go-round in seven revolutions.

Explanation:

The problem addresses the concepts of angular velocity, rotational kinetic energy, and the application of work-energy principles within a physics context, specifically relating to a merry-go-round system.

(a) To calculate the rotational kinetic energy (KErot) of the merry-go-round plus child, we use the formula KErot = (1/2)Iω^2 where I is the moment of inertia and ω is the angular velocity in radians per second. First, convert the angular velocity from rpm to rad/s using the conversion factor 1 rpm = (2π/60) rad/s.
KErot = (1/2)(84.4 kg · m^2)(29.2 rpm × 2π/60 rad/s)^2
= (1/2)(84.4)(3.058 rad/s)^2
= (1/2)(84.4)(9.351)
= (1/2)(789.1668) J
= 394.5834 J

(b) The number of revolutions needed to achieve this angular velocity from rest can be found using the work-energy principle, where work done (W) by the father's force equals the rotational kinetic energy of the merry-go-round system.
W = τr(2πN)
394.5834 J = 375 N · m × (2πN)
N = 394.5834 J / (375 N · m × 2π)
N = 0.167 revolutions

(c) The force to stop the merry-go-round in seven revolutions can be calculated using the concept that the work done by the father's force (W) to stop the merry-go-round equals the negative change in the rotational kinetic energy. Thus, W = -KErot, and the rotational kinetic energy is initially computed in part (a). This gives:
W = -394.5834 J
Force (F) = W / (2π×7)
F = -394.5834 J / (2π × 7)
F = -8.975 N (negative sign indicates that the force is applied in the opposite direction of motion)

Based on discoveries to date, the conclusion as “Planetary systems are common and planets similar in size to Earth are also common” is justified.

Answer: Option C

Explanation:

Some studies show that on average, each star has at least single planet. This means that most stars, such as the Solar System, possess planets (otherwise exoplanets). It is known that small planets (more or less Earthly or slightly larger) are more common than giant planets. The mediocrity principles state that planet like Earth should be universal in the universe, while the rare earth hypothesis says they are extremely rare.

Size is often considered an important factor, because planets the size of the Earth are probably more terrestrial and can hold the earth's atmosphere. The planetary system is a series of gravitational celestial objects orbiting a star or galaxy. Generally, planetary systems describe systems with one or more planets, although such systems may also consist of bodies such as dwarf planets, asteroids and the like.

The principle of conservation of momentum states that in a closed system, an object's momentum before a collision will equal its momentum after the collision.

The principle of conservation of momentum states that in a closed system, the total momentum of the objects before a collision will be equal to the total momentum of the objects after the collision. This means that in a closed system, an object's momentum before a collision will equal its momentum after the collision.

Answer: A hazard is a condition that has the potential to cause harm.

A natural hazard is a potentially harmful situation, where a person places himself in a naturally unsafe zone.

A natural disaster is a large scale destruction of life and properties by the forces of nature.

A natural hazard can become a natural disaster in some cases where the natural harmful situation a person places himself in, is acted upon by the forces of nature on a large scale.

Answer:

Explanation:

Facts about magnetic flux;

(1). Change in magnetic flux is equal to the current generated.

(2). The magnetic field moves from the north to the south.

(3). Magnetic flux is equal to how much magnetic field goes through a direction.

(4). Magnetlc direction can be found found by using Lenz's law.

Faraday's law= emf= N(BA/ ∆t.

Where N= number of loop

∆B/∆T = rate at which flux changes.

The magnitude φ(initial) of the magnetic flux through one turn of the coil before it is rotated is

φ(initial) = BA while φ(final)= zero(0).

The initial magnetic flux through one turn of the coil, with the coil initially perpendicular to Earth's magnetic field, is 6.10 × 10⁻⁹ T×m².

To calculate the magnitude of the initial magnetic flux (φinitial) through one turn of the coil, we use the formula:

φ = B × A × cos(θ)

where:

Since initially, the plane of the coil is perpendicular to Earth's magnetic field, θ = 0°, and hence cos(θ) = 1. First, we must convert the area from cm² to m²:

A = 12.2 cm² × (1 m² / 10,000 cm²) = 12.2 × 10⁻⁴ m²

Now, we can calculate the magnetic flux:

φinitial = 5.00 × 10⁻⁵ T × 12.2 × 10⁻⁴ m² × cos(0°)

φinitial = 5.00 × 10⁻⁵ T × 12.2 × 10⁻⁴ m²

φinitial = 6.10 × 10⁻⁹ T×m²

This is the magnitude of the initial magnetic flux through one turn of the coil before it is rotated.

Complete Question:

A coil has 230 turns enclosing an area of 12.2 cm³. In a physics laboratory experiment, the coil is rotated during the time interval 3.00 × 10⁻² s from a position in which the plane of each turn is perpendicular to Earth's magnetic field to one in which the plane of each turn is parallel to the field. The magnitude of Earth's magnetic field at the lab location is 5.00 × 10⁻⁵ T.

What is the magnitude φinitial of the magnetic flux through one turn of the coil before it is rotated?

Answer:

The tension in the string connecting block 50 to block 51 is 50 N.

Explanation:

Given that,

Number of block = 100

Force = 100 N

let m be the mass of each block.

We need to calculate the net force acting on the 100th block

Using second law of newton

[tex]F=ma[/tex]

[tex]100=100m\times a[/tex]

[tex]ma=1\ N[/tex]

We need to calculate the tension in the string between blocks 99 and 100

Using formula of force

[tex]F_{100-99}=ma[/tex]

[tex]F_{100-99}=1[/tex]

We need to calculate the total number of masses attached to the string

Using formula for mass

[tex]m'=(100-50)m[/tex]

[tex]m'=50m[/tex]

We need to calculate the tension in the string connecting block 50 to block 51

Using formula of tension

[tex]F_{50}=m'a[/tex]

Put the value into the formula

[tex]F_{50}=50m\times a[/tex]

[tex]F_{50}=50\times1[/tex]

[tex]F_{50}=50\ N[/tex]

Hence, The tension in the string connecting block 50 to block 51 is 50 N.

We have that for the Question "What is the tension in the string connecting block 100 to block 99? What is the tension in the string connecting block 50 to block 51?"

it can be said that

From the question we are told

Each of 100 identical blocks sitting on a frictionless surface is connected to the next block by a massless string. The first block is pulled with a force of 100 N.

Assuming mass of each block is 1 kg

The equation for the force is given as

[tex]F = ma\\\\a = \frac{F}{m}\\\\ = \frac{100}{100*1}\\\\ = 1 m/s^2[/tex]

Now, between block 100 and 99,

[tex]F = ma\\\\F = 1*1 \\\\= 1 N[/tex]

Now between block 50 and 51. There are 50 blocks behind 51 st block,  so,

[tex]m = 50 Kg\\\\a = 1 m/s2 (assuming all blocks accelerate at same rate)\\\\F = 50 * 1\\\\F= 50 N[/tex]

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Answer:

7.65 m

Explanation:

[tex]P_1[/tex] = Initial pressure = 0.03 atm

[tex]P_2[/tex] = Final pressure = 1 atm

[tex]r_1[/tex] = Inital radius = 21 m

[tex]V_1[/tex] = Intial volume of gas = [tex]\frac{4}{3}\pi r_1^3[/tex]

[tex]V_2[/tex] = Final volume of gas = [tex]\frac{4}{3}\pi r_2^3[/tex]

[tex]T_1[/tex] = Initial temperature = 200 K

[tex]T_2[/tex] = Final temperature = 323 K

From ideal gas law we have

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\Rightarrow \frac{P_1\frac{4}{3}\pi r_1^3}{T_1}=\frac{P_2\frac{4}{3}\pi r_2^3}{T_2}\\\Rightarrow \frac{P_1 r_1^3}{T_1}=\frac{P_2r_2^3}{T_2}\\\Rightarrow r_2=\frac{P_1r_1^3T_2}{T_1P_2}\\\Rightarrow r_2=\left(\frac{0.03\times 21^3\times 323}{200\times 1}\right)^{\frac{1}{3}}\\\Rightarrow r_2=7.65\ m[/tex]

The radius at liftoff is 7.65 m

The initial radius of the weather balloon at liftoff can be calculated using the ideal gas law by comparing the conditions at the ground and at the balloon's working altitude.

The subject of this question deals with the physics principle of the behavior of gases, namely the ideal gas law. The ideal gas law links the pressure, temperature and volume of a gas, represented by the equation PV=nRT. Here, 'P' denotes pressure, 'V' is volume, 'n' is the number of moles, 'R' is the universal gas constant, and 'T' is temperature.

In the case of the weather balloon, when it is filled at atmospheric pressure and 323 K on the ground, and then expands to a maximum radius of 21m at a working altitude where the air pressure is 0.030 atm and the temperature is 200K, the principle of the ideal gas law can be applied. Assuming the same quantity of gas (n) we have P1V1/T1=P2V2/T2 where indices 1 and 2 refer to conditions at lift off and at altitude respectively.

We know that the volume of the balloon (V) is related to its radius (r) by the equation for volume of a sphere V=(4/3)*pi*r^3. So we can substitute the volumes in above equation with their expressions in terms of radius and solve for r1 (radius at liftoff).

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Answer:

Earth attract the less massive moon with a force that is the same as the force with which the moon attracts the Earth.

Explanation:

This can be explained by Newton's third law:

[tex]F_{12}=-F_{21}[/tex]

The force exerted by body 1 on body 2 is the same as that exerted by body 2 on body 1, only with the opposite sign.

In this case that force is the gravitational force, but the law still applies.

So the moon and the earth are attracted with the same magnitude of force.

Answer:

[tex]c_q=0.7409\ J.g^{-1}.^{\circ}C^{-1}[/tex]

Explanation:

Given:

We have:

Assuming that heat loss is neither from any of the components before mixing nor from the mixture just after mixing. Also the container does not absorbs any heat.

Therefore,

heat gained by the water = heat lost by the quartz

[tex]Q_w=Q_q[/tex]

[tex]m_w.c_w. (T_f-T_{wi})=m_q.c_q.(T_{qi}-T_f)[/tex]

    where: [tex]c_q=[/tex] specific heat capacity of quartz

[tex]250\times 4.186\times (27.9-25)=58.1\times c_q\times (98.4-27.9)[/tex]

[tex]c_q=0.7409\ J.g^{-1}.^{\circ}C^{-1}[/tex]

To calculate the specific heat capacity of quartz, use the equation q = mcΔT and set it equal to the equation for the heat transferred to the water. Solve for c to find the specific heat capacity of quartz.

The specific heat capacity of quartz can be calculated using the equation:

q = mcΔT

where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

The heat transferred to the water can be calculated using the equation:

q = mcΔT

where q is the heat transferred, m is the mass, c is the specific heat capacity of water (4.184 J/g °C), and ΔT is the change in temperature.

By setting the two equations equal to each other and solving for c, the specific heat capacity of quartz can be determined.

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Answer:

11m

Explanation:

Given:

Resistivity ρ = 5.6e-8 Ωm

Radius r = 0.045 mm [tex]=\frac{0.045}{1000}[/tex] = 4.5 x 10⁻⁵ m

Voltage V = 120V

Current I = 1.24A

From Ohm's law, [tex]R = \frac{V}{I}[/tex]

                            [tex]R = \frac{120}{1.24}[/tex]

                            R = 96.77 Ω

Resistivity = (Resistance × Area)/ length

      ρ = (RA)/L

Therefore, the length of a wire is given by;

       L = (RA)/ρ

Calculating the area A of the wire;

       A = πr²

       A = π × (4.5 x 10⁻⁵)²

       A = 6.36 x 10⁻⁹ m²

Substituting area of the wire A =  6.36 x 10⁻⁹ m² into the equation of the length of wire

       L = (96.77 × 6.36×10⁻⁹ ) / 5.6×10⁻⁸

       L = 10.9977m

       L = 11m (approximately)

Answer:

2.35 x 10^{5} J

Explanation:

force (f) = 1080 N

distance (d) = 218 m

find the work done

work is said to be done when force applied to an object moves the object, therefore work done = force x distance

work done = 1080 x 218 = 235,440 J = 2.35 x 10^{5} J

Answer:

4400 K approximately

Explanation:

Stefan-Boltzmann law establish that the power radiated from a black body in terms of its temperature which is proportional to [tex]T^4[/tex]

[tex]i = aT^4[/tex]  with a Stefan-Boltzmann constant

Also we know that [tex]i_{sunspot}[/tex] is three times [tex]i_{sun}[/tex]

[tex]i_{sun}=3i_{sunspot}[/tex] or [tex]\frac{i_{sun}}{i_{sunspot}} = 3[/tex]

Using the Stefan-Boltzmann we can write

[tex]\frac{i_{sun}}{i_{sunspot}} = 3 = \frac{aT_{sun}^4}{aT_{sunspot}^4}[/tex]

solving for [tex]T_{sunspot}[/tex]

[tex]T_{sunspot} = \left(\frac{T_{sun}^4}{3}\right)^{1/4}[/tex]

Replacing the value of [tex]T_{sun}[/tex] (5800 K) it is obtained that [tex]T_{sunspot}[/tex] is 4407.05

Answer:

20 metric tons

Explanation:

Using the Principle of Conservation of Linear Momentum which states that the total momentum of a system is constant in any direction in which no external force acts.

momentum before collision= momentum after collision

mathematically expressed as

m1u1 + m2u2 = m1v1+m2v2 =0

where:

m1= mass of rolling freight car = 10 metric tons = 10Mg

1 ton = 1000kg= 1Mg

u1 =initial velocity of freight car = 108km/h,

m2 = mass of stationary freight car = ?, this the unknown we are finding

u2 = initial velocity of

m2 = 0, because it is at rest,

v1= final velocity of m1,

v2= final velocity of m2

since they move together in the same direction they share a common final velocity as such

v1=v2= 36km/h

substituting valves in expression give

m1u1 + m2u2 = m1v1 + m2v2= 0

10Mgx 108km/hr + m2 x 0 = 10Mg x 36km/h + m2x36km/h

1080( Mg x km/hr) + 0 = 360 (Mg x Km/h) + 36m2 (km/h)

1080 (Mgx km/hr) - 360(Mg x km/h) = 36m2 (km/h)

720(Mg x km/h) = 36m2 (km/h)

divide both side by  36 (km/h) gives

m2 = 20Mg =20 metric tons

Answer: B). 20 metric tons

Explanation: If a freight car with a mass of 10 metric tons is going at 108 km/h collides with another freight car, that is parked. If the speed of the cars, after they crash together, is 36 km/h, the mass of the second car is 20 metric tons.

Answer:

[tex]T_H = 1335\ J[/tex]

Explanation:

Given,

Work done by engine = 19900 J

rejected energy of heat =  5300 J

temperature = 285 K

efficiency of the engine

[tex]\eta = \dfrac{W}{Q_H}[/tex]

where

[tex]W = Q_H - Q_C[/tex]

[tex]19900 = Q_H - 5300[/tex]

[tex]Q_H = 25200\ J[/tex]

efficiency in terms of temperature

[tex]\eta = \dfrac{T_H-T_C}{T_H}[/tex]

from the above two equation

[tex]\dfrac{Q_H-Q_C}{Q_H}= \dfrac{T_H-T_C}{T_H}[/tex]

[tex]1-\dfrac{Q_C}{Q_H}= 1- \dfrac{T_C}{T_H}[/tex]

[tex]\dfrac{Q_C}{Q_H}= \dfrac{T_C}{T_H}[/tex]

[tex]T_H=T_C\ \dfrac{Q_H}{Q_C}[/tex]

[tex]T_H=285\times \dfrac{25200}{5300}[/tex]

[tex]T_H = 1335\ J[/tex]

The smallest possible temperature of the hot reservoir, given the work done by the engine, heat rejected into the cold reservoir, and temperature of the cold reservoir, is approximately 1357.14 Kelvin.

The problem is based on the Carnot heat engine, a theoretical model used for understanding the thermodynamics of heat conversion into work. The efficiency of Carnot's engine depends on the temperatures of the hot and cold reservoirs (Th and Tc respectively). The efficiency, represented as e, is calculated as e = (Th-Tc) / Th.

Before we do the calculation, it's important to note that temperatures must be expressed in Kelvin for this formula to be accurate. Tc, the temperature of the cold reservoir, is given as 285 K. The work done by the engine, represented as W, is 19900 J, and the heat rejected by the engine into the cold reservoir, represented as Qc, is 5300J. The heat absorbed by the engine from the hot reservoir, represented as Qh, can be calculated using the first law of thermodynamics, which states Qh = W + Qc. Therefore, Qh = 19900 J + 5300 J = 25200 J.

Now, to find Th, we rearrange the efficiency formula to Th = Tc / (1 - e), where e is now W / Qh = 19900 J / 25200 J ≈ 0.79. Substituting Tc = 285 K and e ≈ 0.79 into the formula, Th = 285 K / (1 - 0.79) ≈ 1357.14 K. This is the smallest possible temperature of the hot reservoir for the given conditions.

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Answer:

[tex]0.836cm[/tex] in the positive direction.

Explanation:

The equation that describes the motion of this mass-spring system is given by;

[tex]y=Asin\omega t...................(1)[/tex]

Where A is the amplitude, which defined as the maximum displacement from the equilibrium position for a body in simple harmonic motion.

[tex]\omega[/tex] is the angular velocity measured in [tex]rads^{-1[/tex], this is the angle turned through per unit time.

[tex]y[/tex] is the displacement along the axis of the amplitude, and [tex]t[/tex] is any instant of time in the motion.

Given; A = 8.8cm = 0.088m

The angular velocity is given by the following relationship also;

[tex]\omega=2\pi/T[/tex]

Where T is the period, which is defined as the time taken for a body in simple harmonic motion to make one complete oscillation.

Given; T=0.66s

Therefore;

[tex]\omega=2\pi/0.66\\\omega=3.03\pi rads^{-1[/tex]

Substituting all values into equation (1) we obtain the following;

[tex]y=0.088sin3.03\pi t................(2)[/tex]

Equation (2) is the equation describing the motion of the mass on the spring.

At an instant of time t = 2.3s, the displacement is therefore given as follows;

[tex]y=0.088sin[3.03\pi(2.3)]\\y=0.088sin6.97\pi\\[/tex]

By conversion, [tex]6.97\pi rad=6.97*180^o=1254.55^o[/tex]

Therefore

[tex]y=0.088sin1254.54=0.00836m\\[/tex]

[tex]y=0.00836m=0.836cm[/tex]

Answer:

ball D will fall toward the ground at the same time as ball C

Explanation:

both balls experience the same downward (vertical) force of gravity as such they will both fall down at the same time, given that all other factors are equal.

although the ball were through with different forces,

those forces where in the horizontal direction but the force of gravity (downward force) will act on them equally to bring them down at the same time

Answer:

e). [tex]a = 0.066 m/s^2[/tex]

Explanation:

As we know that wheel is turned by 90 degree angle

so the angular speed of the wheel is given as

[tex]\omega_f^2 - \omega_i^2 = 2\alpha \theta[/tex]

now we have

[tex]\omega_f^2 - 0 = 2(0.01)(\frac{\pi}{2})[/tex]

[tex]\omega = 0.177 rad/s[/tex]

now the centripetal acceleration of the point P is given as

[tex]a_c = \omega^2 R[/tex]

[tex]a_c = (0.177)^2(2)[/tex]

[tex]a_c = 0.063 m/s^2[/tex]

tangential acceleration is given as

[tex]a_t = R\alpha[/tex]

[tex]a_t = 2(0.01)[/tex]

[tex]a_t = 0.02 m/s^2[/tex]

now net acceleration is given as

[tex]a = \sqrt{a_t^2 + a_c^2}[/tex]

[tex]a = \sqrt{0.02^2 + 0.063^2}[/tex]

[tex]a = 0.066 m/s^2[/tex]

The question involves the calculation of linear acceleration of a point on a wheel undergoing angular acceleration. The linear acceleration involves both tangential and radial components, and these are combined to provide the total acceleration. The closest answer is 0.066 m/s^2.

The subject of this question is physics, specifically the concepts of angular acceleration and linear acceleration. Given an angular acceleration, we can find the linear acceleration by using the formula a = rα, where a is linear acceleration, r is the radius, and α is angular acceleration. Substituting the given values, we get a = 2.0 m * 0.01 rad/s2 = 0.02 m/s2. This value is the tangential acceleration of point P.

Since the point P is on the y-axis, the linear acceleration, a, is given by the radial acceleration which is equal to ω2r, where ω is the angular velocity and r is the radius of the circle. But we also have ω2 = 2αr, where ρ is the angular acceleration and r is the radius. So, the radial acceleration is (2αr)2r = 4α2r3. Substituting for α and r we get: 4*(0.01 rad/s2)2*(2.0 m)3 = 0.0016 m/s2.

The total linear acceleration is the vector sum of the radial and tangential accelerations. Since these vectors are perpendicular, we calculate their resultant by Pythagoras' theorem: √(at2 + ar2) = √(0.022 m/s2 + 0.00162 m/s2) = 0.02016 m/s2. Hence, the choice that's closest is (a) 0.066 m/s2.

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Answer:

a= 0.063 m

b = 0.116 m

Explanation:

First of all, we need the spring constant in order to solve this problem. You are not giving that data, but I will tell you how to solve this assuming a value of k, In this case, let's assume the value of k is 1600 N/m. (I solved an exercise like this before, using this value).

Now, we need to use the expressions to calculate the distance of the spring.

The elastic potential energy (Uel) is given with the following formula:

Uel = 1/2 kx²

Solving for x:

x = √2*Uel/k

Replacing the data in the above formula (And using the value of k os 1,600):

x = √2 * 3.2 / 1600

x = 0.063 m

b) For this part, we need to apply the work energy theorem which is:

K1 + Ugrav1 + Uel1 + Uo = K2 + Ugrav2 + Uel2

Since in this part, the exercise states that the book is dropped, we can say that the innitial and the end is 0, therefore, K1 = K2 = 0.

The spring at first is not compressed, so Uel1 = 0, and Uo which is the potential energy of other factors, is also 0, because there are no other force or factor here. Therefore, our theorem is resumed like this:

Ugrav1 = Uel2

The potential energy from gravity is given by:

Ug = mgy

And as the spring is placed vertically, we know the height which the book is dropped, so the distance y is:

y = x + h

And this value of x, is the one we need to solve. Replacing this in the theorem we have:

mg(h+x) = 1/2kx²

g would be 9.8 m/s²

Now, replacing the data:

1.2*9.8(0.8 + x) = 1/2*1600x²

Rearranging and solving for x we have:

1.2*9.8*2(0.8 + x) = 1600x²

18.82 + 23.52x = 1600x²

1600x² - 23.52x - 18.82 = 0

Now we need to solve for x, using the general formula:

x = - (-23.52) ± √(-23.52)² - 4 * 1600 * (-18.82) / 2*1600

x = 23.52 ± √553.19 + 120,448 / 3200

x = 23.52 ± 347.85 / 3200

x1 = 23.52 + 347.85 / 3200 = 0.116 m

x2 = 23.52 - 347.85 / 3200 = -0.101 m

Using the positive value, we have that the distance is 0.116 m.

Answer:

111.6 g

Explanation:

Given

m + M = 2.58 / 9.8

= 0.2632 kg

When the can of fruit juice is balance from scale , we get following relation

M x ( 50 - 21.2 ) = m x 21.2 ( balancing the torque due to weight of scale and can about the balancing point )

M x 28.8 = 21.2 m

= 21.2 ( 0.2632 - M )

= 5.58 - 21.2 M

M ( 28.8 + 21.2 ) = 5.58

M = .1116 kg

= 111.6 g

Answer:

E)  E₂= 20000 N/C

Explanation:

The electric field between two parallel conductive plates is thus calculated

E= V/d

Where:

E: Electric field (N/C)

V:  voltage (V)

d : distance between the plates (m)

Problem development

E₁= V₁/d₁  = 2000 N/C

E₂= V₂/d₂

[tex]E_{2} = \frac{2V_{1} }{\frac{d_{1} }{5} }[/tex]

[tex]E_{2} = 10(\frac{V_{1} }{d_{1} })[/tex]

E₂= 10* (2000)

E₂= 20000 N/C

Final answer:

When the voltage between two parallel conducting plates is doubled and the distance between them is reduced to 1/5 of the original distance, the magnitude of the new electric field is 20,000 N/C.

Explanation:

The question relates to the electric field between two parallel conducting plates connected to a voltage source. The original electric field's magnitude is given as 2,000 N/C, and we're asked to find the new electric field magnitude when the voltage is doubled and the distance between the plates is reduced to 1/5 of the original distance. The electric field (E) between two parallel plates is given by the equation E = V/d, where V is the potential difference (voltage) between the plates, and d is the distance between them. When the voltage is doubled, V becomes 2V, and when the distance is reduced to 1/5, d becomes d/5. Thus, the new electric field E' = (2V)/(d/5) = 10(V/d) = 10E. So, the new electric field strength is 10 * 2,000 N/C = 20,000 N/C, which corresponds to option E).

The magnitude of the resultant force on the race car and driver is 357.375 kN.

First, we need to find the centripetal acceleration of the race car traveling around the circular track. The centripetal acceleration can be found using the formula:

a = v² / r

where v is the velocity of the car and r is the radius of the circular track. Plugging in the values, we get:

a = (99 m/s)² / 52 m = 188.25 m/s²

Next, we can use Newton's second law of motion to find the magnitude of the resultant force acting on the car. The formula is:

F = m * a

where F is the force, m is the mass of the car, and a is the centripetal acceleration. Plugging in the values, we get:

F = (1900 kg) * (188.25 m/s²) = 357,375 N

Finally, we can convert the force to kilonewtons (kN) by dividing by 1000:

F = 357,375 N / 1000 = 357.375 kN

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Answer:

Explanation:

Time period of oscillation of spring mass system

T = 2π[tex]\sqrt{\frac{m}{k} }[/tex]

m = [tex]\frac{T^2\times k}{4\pi^2}[/tex]

= [tex]\frac{(2.6)^2\times 13.2}{4\pi^2}[/tex]

= 2.26 kg

Amplitude is the maximum displacement of the mass from the _equilibrium point .

maximum displacement from equilibrium of +14.8 cm at time, t = 0.00 s

x(t) = 14.8cosωt

ω = 2πn = 2π x 9.73 = 61.1 rad /s

x(t) = 14.8cos61.1 t

when t = 1.25

x(t) = 14.8cos61.1 x 1.25

= 14.8cos61.1 x 1.25

= 14.8cos76.38

= 8.22 cm

A ball of mass 1.55 kg is hung from a spring which stretches a distance of 0.3800 m

spring constant

k = mg / x

= 1.55 x 9.8 / .38

= 40 N / m

T = 2π[tex]\sqrt{\frac{m}{k} }[/tex]

so  if mass is hung from a spring with a spring constant of 3k

T decreases by a factor of the square root of 3

For the first question, the unknown mass is approximately 0.71 kg. For the second question, the term 'equilibrium point' completes the statement. For the third question, the displacement at time 1.25 seconds is -14.2 cm. For the fourth question, the spring constant k is approximately 40.0 N/m. For the fifth question, the period of the motion will be decreased by a factor of the square root of 3.

The unknown mass hung from a spring can be calculated using the formula for the period of a mass-spring system, T = 2π√(m / k), where T is the period, m is the mass, and k is the spring constant. Rearranging for m gives us m = (T²k) / (4π²). Given that T = 2.60 seconds and k = 13.2 N/m, we find m = (2.60^2 * 13.2) / (4*π^2), which gives a mass of approximately 0.71 kg.

The amplitude is the maximum displacement of the mass from the equilibrium point.

The displacement of a harmonically oscillating mass can be found using the equation:    x(t) = A*cos(2πft + φ), where A is the amplitude, f is the frequency, and φ is the phase angle. Given the information we have a displacement at 1.25 seconds of -14.2 cm.

The spring constant can be calculated using the equation  k = mg / x, where m is the mass, g is gravity (9.81 m/s²), and x is the distance stretched. Given that the mass is 1.55 kg and the distance stretched is 0.3800 m, we find that the spring constant k is approximately 40.0 N/m.

If the same mass is hung from a spring with a spring constant of 3k, the period of the motion will be decreased by a factor of the square root of 3.

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