a light truck can go around a flat curve of a 140 m radius with a speed of 26 m/s. what is the acceleration of the truck (in m/s2)

Sunspots are cooler regions on the surface of the Sun that appear as dark spots.

They are cooler because they have a lower temperature than the surrounding photosphere. The photosphere is the visible surface of the Sun and is the layer where the energy generated by nuclear fusion is radiated out into space.

The temperature of the photosphere is around 6000 K, whereas the temperature of sunspots can be as low as 3000 K.  The amount of energy emitted by a surface depends on its temperature and surface area.

The energy emitted by a surface per unit area and per unit time is given by the Stefan-Boltzmann law. According to this law, the energy radiated per unit area per unit time is proportional to the fourth power of the temperature.

This means that if the temperature of a surface is halved, the energy radiated per unit area per unit time decreases by a factor of 16.

Therefore, the ratio of the amount of energy emitted by the sunspot to the amount of energy emitted by the photosphere per unit area and per unit time can be calculated using the Stefan-Boltzmann law.

The ratio of the energy emitted by the sunspot to the energy emitted by the photosphere is given by: (E_s / E_p) = (T_s / T_p)^4,

Where E_s is the energy emitted by the sunspot per unit area and per unit time, E_p is the energy emitted by the photosphere per unit area and per unit time, T_s is the temperature of the sunspot, and T_p is the temperature of the photosphere.

Using the temperatures given in the question, we can calculate the ratio of the energy emitted by the sunspot to the energy emitted by the photosphere: (E_s / E_p) = (3000 K / 6000 K)^4, (E_s / E_p) = (0.5)^4, (E_s / E_p) = 0.0625

Therefore, the ratio of the energy emitted by the sunspot to the energy emitted by the photosphere per unit area and per unit time is 0.0625. This means that the sunspot is emitting much less energy per unit area and per unit time than the photosphere.

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Answer:

B

Explanation:

Chlorine institute

16+1=17

CI has 17 in the bottom.

Answer:C

Explanation:

Finally, we can solve for the specific enthalpy of the feedwater at 100°C and conclude that the feedwater at 100°C is the desired outcome.

An adiabatic open feedwater heater in an electric power plant is a device that mixes steam and feedwater to produce feedwater at a specific temperature. In this case, 0.2 kg/s of steam at 100 kPa and 160°C is mixed with 10 kg/s of feedwater at 100 kPa and 50°C to produce feedwater at 100°C.
To solve this problem, we can use the principle of energy conservation. The energy gained by the feedwater is equal to the energy lost by the steam.
First, we need to determine the specific enthalpy of the steam and feedwater at their respective temperatures and pressures. We can use steam tables to find these values.

For the steam, at 100 kPa and 160°C, the specific enthalpy is h1.

For the feedwater, at 100 kPa and 50°C, the specific enthalpy is h2.

Next, we can use the mass flow rate of the steam and feedwater to calculate the energy gained by the feedwater.

Energy gained by feedwater = mass flow rate of feedwater * (specific enthalpy of feedwater at 100°C - specific enthalpy of feedwater at 50°C)

Now, we can equate the energy gained by the feedwater to the energy lost by the steam.

Energy lost by steam = mass flow rate of steam * (specific enthalpy of steam at 100°C - specific enthalpy of steam at 160°C)

Since the process is adiabatic (no heat transfer), the energy lost by the steam is equal to the energy gained by the feedwater.

Setting the two equations equal to each other, we can solve for the specific enthalpy of the feedwater at 100°C.

mass flow rate of feedwater * (specific enthalpy of feedwater at 100°C - specific enthalpy of feedwater at 50°C) = mass flow rate of steam * (specific enthalpy of steam at 100°C - specific enthalpy of steam at 160°C)

Finally, we can solve for the specific enthalpy of the feedwater at 100°C and conclude that the feedwater at 100°C is the desired outcome.

Please note that I cannot provide an exact numerical answer without the specific values of the specific enthalpy at different temperatures. However, by following the steps outlined above, you should be able to determine the specific enthalpy of the feedwater at 100°C.

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Answer:

2838

Explanation:

Answer:

ummmmmwm whattt

Explanation:

wuhsnnsnsnsnms

Answer:

medicines

band-aids

aqua liquid

paracetamol

Explanation:

Extra items:

alcohol

cottons

bitadine

cotton buds

mask

The change in gravitational potential energy of the couch with 120 kg mass is 9653 J.

The gravitational potential energy is defined as the energy possessed by an object due to its position above the ground. If an object is located near the surface of the earth then the change in gravitational potential energy is given by

ΔU - mgΔh

Where

m = the mass of the object

g = acceleration of gravity

g = 9.8 m / \(s^{2}\)

Δh =  change in height of the object

For the couch in this problem, we have m = 120 kg which is its mass.

Δh = 8.2 cm

Now solving the equation, we get:

ΔU = 120 x 9.8 x 8.2 = 9653 J

The energy gained by the couch is 9653 J

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Answer:The answer is C

Explanation:

To calculate the net force on particle q₁, we need to calculate the electric forces between particle q₁ and particles q₂ and q₃, and then add them together taking into account their direction. The electric force between two point charges is given by Coulomb's law:

F = k * (q₁*q₂) / r²

where k is Coulomb's constant (k = 9.0 x 10⁹ N*m²/C²), q₁ and q₂ are the charges of the particles, and r is the distance between them.

Let's first calculate the force between particles q₁ and q₂:

F₁₂ = k * (q₁*q₂) / r₁₂²

F₁₂ = 9.0 x 10⁹ N*m²/C² * (8.0 x 10⁻⁶ C) * (3.5 x 10⁻⁶ C) / (0.10 m)²

F₁₂ = 2.52 x 10⁻² N

The force between particles q₁ and q₂ is positive, since both particles have the same sign (+8.0 μC and +3.5 μC). Therefore, the force points to the right.

Now let's calculate the force between particles q₂ and q₃:

F₂₃ = k * (q₂*q₃) / r₂₃²

F₂₃ = 9.0 x 10⁹ N*m²/C² * (3.5 x 10⁻⁶ C) * (-2.5 x 10⁻⁶ C) / (0.15 m)²

F₂₃ = -8.40 x 10⁻³ N

The force between particles q₂ and q₃ is negative, since they have opposite signs (+3.5 μC and -2.5 μC). Therefore, the force points to the left.

Now we can calculate the net force on particle q₁:

F_net = F₁₂ + F₂₃

F_net = 2.52 x 10⁻² N + (-8.40 x 10⁻³ N)

F_net = 1.68 x 10⁻² N

The net force on particle q₁ is positive, since the force between particles q₁ and q₂ is greater than the force between particles q₂ and q₃, and both forces have the same direction.

Therefore, the net force points to the right. The magnitude of the net force is 1.68 x 10⁻² N.

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Given that the radius of the circular path is r = 142 m and the coefficient of the friction is

The condition for the car to travel safely is

Frictional force = centrifugal force

Here, m is the mass of the car and the acceleration due to gravity is g = 9.8 m/s^2.

v is the maximum speed of the car.

Answer:

3

Explanation:

i took the quiz

The speed, relative to the frame of the observer making the measurement, at which the length of a meter stick appears to be 0.60 m is approximately 0.92 times the speed of light (c).

According to the theory of special relativity, objects in motion experience length contraction along their direction of motion as observed from a different inertial reference frame. This phenomenon is described by the Lorentz transformation equations. The equation for length contraction is given by L' = L₀ / γ, where L' is the contracted length, L₀ is the rest length, and γ is the Lorentz factor.

In this case, the rest length of the meter stick is 1 meter (L₀ = 1 m), and the observed length is 0.60 m (L' = 0.60 m). We can rearrange the length contraction equation to solve for the Lorentz factor: γ = L₀ / L'.

Substituting the given values, we have γ = 1 m / 0.60 m = 1.67.

The Lorentz factor is related to the velocity (v) of the meter stick relative to the observer by the equation γ = 1 / √(1 - v²/c²), where c is the speed of light. Rearranging this equation, we can solve for v: v = c √(1 - 1/γ²).

Substituting γ = 1.67, we find v ≈ 0.92c, where c is the speed of light.

Therefore, the speed at which the meter stick is moving relative to the observer's frame of reference, such that its length appears to be 0.60 m, is approximately 0.92 times the speed of light.

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Answer:

See below

Explanation:

X component = 120 * cos 35° =  98.298 m/s

Y component = 120 * sin 35° = 68.829 m/s

Any object launched into space with only gravity acting on it is referred to as a projectile. Gravity is the main force affecting a projectile. The x and y components of the velocity vector are 98.298 m/s and 68.829 m/s.

When a particle is thrown obliquely close to the Earth's surface, it travels along a curving route while experiencing constant acceleration. Always pointing in the direction of the Earth's center is this curving path. Such a particle's motion is referred to as projectile motion, and its trajectory is known as the projectile's trajectory.

The components of the projectile are:

X component = 120 × cos 35° =  98.298 m/s

Y component = 120 × sin 35° = 68.829 m/s

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Answer:

Beta Particles

Explanation:

Answer:

C. oversaturated

Explanation:

Sana nakatulong

13.0 m/s   that's how fast he was going

The platy or sheet-like form that foliated rocks develop serves as a representation of the direction of pressure application. A few examples of rocks with foliated metamorphosed tectonic plates are slate, sandstone, and granite.

Foliated metamorphic rock must grow in a directed force or shear stress environment. When pressure is applied to a parent rock during recrystallization, the platy or extended minerals inside the parent rock become aligned or foliated. The platy and paper structure that foliated rocks take on is a representation of the direction in which pressure is applied. Tectonic plates that were foliated during metamorphism produced the rock types slate, schist, and gneiss. The many different types of foliated metamorphic rocks are classified by grade or quantity of metamorphic rocks, and the type includes slate, phyllite, schist, and gneiss.

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The platy or sheet-like form that foliated rocks develop serves as a representation of the direction of pressure application. A few examples of rocks with foliated metamorphosed tectonic plates are slate, sandstone, and granite.

Foliated metamorphic rock must grow in a directed force or shear stress environment. When pressure is applied to a parent rock during recrystallization, the platy or extended minerals inside the parent rock become aligned or foliated.

The platy and paper structure that foliated rocks take on is a representation of the direction in which pressure is applied. Tectonic plates that were foliated during metamorphism produced the rock types slate, schist, and gneiss. The many different types of foliated metamorphic rocks are classified by grade or quantity of metamorphic rocks, and the type includes slate, phyllite, schist, and gneiss.

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Starting from rest on a circular track with a radius of 300 m and a constant tangential acceleration of 0.75 m/s², the car will travel a distance of approximately 0.2119 meters or 21.19 centimeters in 0.75 seconds.

To determine the distance traveled and the time elapsed by the car starting from rest on a circular track with a radius of 300 m and a constant tangential acceleration of 0.75 m/s², we can use the equations of circular motion.

The tangential acceleration is the rate of change of tangential velocity. Since the car starts from rest, its initial tangential velocity is zero (v₀ = 0).

Using the equation:

v = v₀ + at

where v is the final tangential velocity, v₀ is the initial tangential velocity, a is the tangential acceleration, and t is the time, we can solve for v:

v = 0 + (0.75 m/s²) * t

v = 0.75t m/s

The tangential velocity is related to the angular velocity (ω) and the radius (r) of the circular track:

v = ωr

Substituting the values:

0.75t = ω * 300

Since the car starts from rest, the initial angular velocity (ω₀) is zero. So, we have:

ω = ω₀ + αt

ω = 0 + (0.75 m/s²) * t

ω = 0.75t rad/s

We can now substitute the value of ω into the equation:

0.75t = (0.75t) * 300

Simplifying the equation gives:

0.75t = 225t

t = 0.75 seconds

The time elapsed is 0.75 seconds.

To calculate the distance traveled (s), we can use the equation:

s = v₀t + (1/2)at²

Since the initial velocity (v₀) is zero, the equation becomes:

s = (1/2)at²

s = (1/2)(0.75 m/s²)(0.75 s)²

s = (1/2)(0.75 m/s²)(0.5625 s²)

s = 0.2119 meters or approximately 21.19 centimeters

Therefore, the car travels a distance of approximately 0.2119 meters or 21.19 centimeters.

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Answer:

Given

mass (m) =5kg

velocity (v) =6m/s

momentum (p) =?

Form

p=mv

5kgx6m/s

p=30kg.m/s

momentum =30kg.m/s

Answer:

As one goes "forward" in time, the second law of thermodynamics says, the entropy of an isolated system can increase, but not decrease. ... Thus, entropy measurement is a way of distinguishing the past from the future.

Explanation:

Answer:

\(V_y = 16.44\ m/s\)

\(V_y = 7.32\ m/s\)

Explanation:

Given

\(\theta = 24\)

\(Velocity\ (V) = 18m/s\)

Required

Determine the vertical and horizontal components

The vertical (Vy) and horizontal (Vx) components is calculated as thus:

\(V_x = Vcos\theta\)

\(V_y = Vsin\theta\)

Calculating Vertical Components:

\(V_y = Vsin\theta\)

\(V_y = 18 * sin24\)

\(V_y = 18 * 0.40673664307\)

\(V_y = 7.32125957526\)

\(V_y = 7.32\ m/s\) --- Approximated

Calculating Horizontal Components:

\(V_x = Vcos\theta\)

\(V_x = 18 * cos24\)

\(V_x = 18 * 0.91354545764\)

\(V_x = 16.4438182375\)

\(V_x = 16.44\ m/s\) --- Approximated

Answer:

=> Black colour do not reflect the wavelength of light and is a good emitter of heat . This, black is a good absorber of radiation.

Answer:

Black coloured objects absorb all the wavelengths of light and do not reflect any of them. Therefore, it looks black in colour and is considered to be a good absorber and a good emitter of heat. On the other hand, white coloured objects reflect all the light that falls into it and therefore they look white

The value of the expression \(E^2 − (pc)^2\) for this proton is: \(881879 MeV^2.\)

To find the value of the expression \(E^2 − (pc)^2\) for this proton, follow these steps:

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You weigh slightly more at the beach than on top of a mountain, because at the beach, you're closer to the center of the Earth.

Answer:

400 m

Explanation:

The swimmer swims 50 meters to one end of the pool but has to swim back therefore you double 50 which would be 100 meters. Then you have to multiply 100 by 4 since the swimmer did it 4 times.

so, 444 eggs would have been released in 37yrs

If a woman releases one egg every month for 37 years, then she releases 444 eggs.

A year is a unit of time that is commonly used to measure the period of time it takes for the Earth to make one revolution around the Sun.

The time it takes for the Earth to complete one orbit around the Sun is approximately 365.24 days, which is why a year is usually defined as 365 days, with an extra day added every four years (in what's called a leap year) to account for the fractional part of the year. This is known as the Gregorian calendar, which is the most widely used calendar system in the world.

A year is also commonly used to refer to the period of time it takes for an event or cycle to repeat, such as the school year, fiscal year, or a company's financial year.

There are 12 months in a year, and they are:

January

February

March

April

May

June

July

August

September

October

November

December

Each month has either 30 or 31 days, except for February which has 28 days (or 29 days in a leap year). The number of days in a month determines the number of days in a year, which is why a year typically has either 365 or 366 days, depending on whether or not it is a leap year.

Here in the question,

37 years is equivalent to 37 x 12 = 444 months (since there are 12 months in a year).

To calculate this, we simply multiply the number of eggs released each month (which is one) by the total number of months:

Number of eggs released = 1 egg/month x 444 months

Number of eggs released = 444 eggs

Therefore, over 37 years, a woman releases 444 eggs.

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Answer:

Yes, its true

Explanation:

hdheheishdhieijzuwhs

Answer: yes its true

Explanation:

nmo

Answer:

mgi

Explanation:

Answer: x = 7

Explanation: 12 = 14 -2 = 12.

The period of the electromagnetic wave is 5×10⁻¹⁰ seconds

Period is the time taken for a wave to complete one rotation.

To calculate the period of the wave, we use the formula below.

Formula:

Where:

From the question,

Given:

substitute these values equation 1

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Answer:

Explanation:

Line Spectrum

The quantum number associated with the intensity of spectral lines and the spin of the electron is called the spin quantum number or simply the spin. The spin quantum number determines the intrinsic angular momentum of a particle, such as an electron.

The spin quantum number has a value of either +1/2 or -1/2, representing the two possible spin states of an electron. These states are commonly denoted as "spin-up" (+1/2) and "spin-down" (-1/2). The spin of an electron is an intrinsic property and plays a crucial role in determining the electronic structure and behavior of atoms, as well as in various quantum mechanical phenomena.

It is important to note that the spin quantum number is not related to the intensity of spectral lines directly. The intensity of spectral lines is primarily determined by other factors such as the probability of electronic transitions between energy levels and the population of energy states.

In summary, the spin quantum number is associated with both the intensity of spectral lines (indirectly) and the spin of the electron.

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When a ferromagnetic material is placed in an electromagnetic coil and a magnetic field is applied, the magnetic induction (B) is increased.

Therefore, the correct answer is:

(a) There is a large increase in the magnetic induction (B)

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