Imagine two circular plates; one is solid and the other has a hole cut out of the center. Both plates have the same radius, same thickness, and same mass. The same force F is applied tangential to the edge of each plate in such a way that the plates rotate about an axis passing through the center and perpendicular to the surface of the plates.

Which of the following statements helps to explain the question asked above? (Select all that apply.)

Both plates will be subjected to the same torque.

Angular acceleration is inversely proportional to the moment of inertia.

Angular acceleration is directly proportional to the moment of inertia.

The plate with the hole has its mass distributed further out from the axis of rotation, which will increase its moment of inertia.

Because both plates have the same mass, they will have the same moment of inertia.

Answer:

tipping over dominoes

Explanation:

?

One game that represents the concept of a wave is the puzzle game "Wave Wave."

In "Wave Wave," players control a small triangle as it navigates through a series of challenging and rhythmic levels filled with obstacles. The gameplay involves moving the triangle in sync with the pulsating waves that define the level's layout.

The game's visuals and mechanics are designed to mimic the appearance and behavior of a wave, with the obstacles and patterns resembling the peaks and troughs of waves. The player's goal is to maintain control and timing, moving the triangle smoothly through the undulating paths while avoiding collisions.

"Wave Wave" is a creative interpretation of the wave concept in a gaming context, with its gameplay mechanics directly tied to the idea of navigating through waves. Keep in mind that there might be other games that also incorporate wave-like mechanics, so this is just one example.

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Answer:

The minimum speed must the car must be 13.13 m/s.

Explanation:

The radius of the loop is 17.6 m. We need to find the minimum speed must the car traverse the loop so that the rider does not fall out while upside down at the top.

We know that, mg be the weight of car and rider, which is equal to the centripetal force.

[tex]mg = \dfrac{mv^2}{r}\\v = \sqrt{rg}\\v = \sqrt{17.6\times 9.8}\\v = 13.13\ m/s\\[/tex]

So, the minimum speed must the car must be 13.13 m/s.

Answer:

a) 0.144J

b) 0.12J

Explanation:

A 0.200 kg air-track glider moving at 1.20 m/s bumps into a 0.600 kg glider at rest. a.) Find the total kinetic energy after collision if the collision is elastic. b.) Find the total kinetic energy after collision if the collision is completely inelastic.

Given that

M1 = 0.2kg

M2 = 0.6kg

U1 = 1.2 m/s

Since both momentum and energy are conserved in elastic collisions, the total kinetic energy after collision will be

1/2M1U^2 + 1/2M2U^2

1/2 × 0.2 × 1.2^2 + 0

K.E = 0.144J

B) In elastic collision, only momentum is conserved

M1U1 + M2U2 = (M1 + M2)V

U2 = 0 since the object is at rest

0.2×1.2 + 0 = (0.2 + 0.6)V

0.24 = 0.8V

V = 0.24/0.8

V = 0.3 m/s

K.E = 1/2(M1+M2)V

K.E = 1/2 (0.2 + 0.6) × 0.3

K.E = 0.4 × 0.3

K.E = 0.12J

Answer:

The other person is almost right, but I see where the numbers got fudged

Explanation:

Answer:a) 0.144Jb) 0.036J

Explanation: A 0.200 kg air-track glider moving at 1.20 m/s bumps into a 0.600 kg glider at rest. a.) Find the total kinetic energy after collision if the collision is elastic. b.) Find the total kinetic energy after collision if the collision is completely inelastic. Given that M1 = 0.2kgM2 = 0.6kgU1 = 1.2 m/s Since both momentum and energy are conserved in elastic collisions, the total kinetic energy after collision will be1/2M1U^2 + 1/2M2U^21/2 × 0.2 × 1.2^2 + 0K.E = 0.144J

B) This is where the other person's wrong:

m1u1+m2u2=(m1+m2)V

V=(0.200kg*1.20m/s)/(0.200kg+0.600kg)=0.3m/s

KE=1/2m*v^2

KE=1/2(0.200kg+0.600kg)*(0.3m/s)^2=1/2(0.8x0.3)^2=0.036 J

Final answer:

Sound energy is a form of kinetic energy, not potential energy, because it arises from the motion and vibration of particles within a medium as the sound wave propagates.

Explanation:

Sound energy is not considered a type of potential energy because it is inherently associated with the motion of particles; therefore, it is a form of kinetic energy. Sound is a mechanical wave that results from the vibration of particles within a medium. As these particles oscillate, they transfer energy through the medium, resulting in what we perceive as sound. While a sound wave indeed combines kinetic and potential energy due to the motion of particles and the elasticity of the medium, the propagation of the wave itself and the sound energy we detect are manifestations of kinetic energy. In the case of potential energy, this form of energy is related to an object's position or state; for instance, a rock atop a hill has gravitational potential energy due to its elevated position in a gravitational field. Similarly, the chemical energy in a battery is potential energy based on the chemical composition and potential for a chemical reaction.

To help visualize the concept, consider water behind a dam. The water has potential energy because of its elevated position. As it moves and flows through turbines, it converts this potential energy into kinetic energy, which then can be harnessed to produce electricity. In contrast, when a guitar string is plucked, the energy initially stored in the deformed string is potential energy, but as the string returns to its normal state and vibrates, it generates sound—which is an expression of kinetic energy. Thus, when sound is produced, it is the motion and transfer of energy through a medium, reflecting kinetic, not potential, energy characteristics.

The complete question is;

To develop muscle tone, a woman lifts a 2.00-kg weight held in her hand. She uses her biceps muscle to flex the lower arm through an angle of 60.0º . What is the angular acceleration if the weight is 24.0 cm from the elbow joint, her forearm has a moment of inertia of 0.240 kg.m², and the net force she exerts is 759 N at an effective perpendicular lever arm of 2.00 cm?

Answer:

α = 42.76 rad/s²

Explanation:

We are given;

Mass; m = 2 kg

Distance; r = 24cm = 0.24m

Moment of inertia of the forearm; I = 0.24 kg.m²

Muscle Force; F = 759N

Perpendicular distance to lever arm; r' = 2cm = 0.02m

Angle at which she flexes arm; θ =60°

Let's assume that the arm starts extended vertically downwards. and we take the elbow joint as the point about which we calculate the torque.

Thus, we know that torque is given by the formula ;

τ = Force x Perpendicular distance

Thus, the toque exerted by force in the muscle is;

τ_muscle = 759 x 0.02 = 15.18 N.m

Also, torque exerted by the lifted weight is given as 0 because perpendicular distance is zero.

Thus, the net torque on the lower arm is;

τ_net = τ_muscle - τ_weight

τ_net = 15.18 - 0 = 15.18 N.m

Now, let's calculate moment of inertia of lifted weight.

The moment of inertia is given by;

I = mr²

Thus, moment of Inertia of lifted weight is; I_weight = 2 x 0.24² = 0.115 kg.m²

Thus,total moment of inertia is;

I_total = I_arm + I_weight

Thus, I_total = 0.24 + 0.115 = 0.355 kg.m²

Now, we can calculate the angular acceleration.

It's gotten from the equation;

τ_net = I_total•α

Where α is angular acceleration.

Thus making α the subject, we have ; α = τ_net/I_total

α = 15.18/0.355 = 42.76 rad/s²

Answer:

bet

Explanation:

Answer:

Angle θ = 30.82°

Explanation:

From Malus’s law, since the intensity of a wave is proportional to its amplitude squared, the intensity I of the transmitted wave is related to the incident wave by; I = I_o cos²θ

where;

I_o is the intensity of the polarized wave before passing through the filter.

In this question,

I is 0.708 W/m²

While I_o is 0.960 W/m²

Thus, plugging in these values into the equation, we have;

0.708 W/m² = 0.960 W/m² •cos²θ

Thus, cos²θ = 0.708 W/m²/0.960 W/m²

cos²θ = 0.7375

Cos θ = √0.7375

Cos θ = 0.8588

θ = Cos^(-1)0.8588

θ = 30.82°

Explanation:

this is because if your going to touch something dangerouse your hand normally feels it first and its the front not the back so when you pick things up you can tell if its dangerouse hope that helped

The palm of the hand has more touch receptors than the back, due to its glabrous skin type which is thicker and more sensitive. These receptors allow for detection of fine touch and pressure, critical for manual tasks.

Considering the density of touch receptors in different parts of our body, you could expect to have more touch receptors on the palm of your hand than on the back of your hand. This is due to the specific type of skin found on the palm, referred to as glabrous skin. This type of skin is typically thicker and more sensitive as compared to hairy skin, hence it contains more touch receptors.

The importance of the high density of touch receptors in areas such as palms and fingertips can be seen in the daily actions, such as the ability to sense fine touch and detailed information, which is necessary for fine motor tasks.

Furthermore, mechanoreceptors such as Merkel's disks, Meissner's corpuscles, Pacinian corpuscles, and Ruffini endings play a vital role in interpreting the touch-related sensory information of these regions. For instance, Merkel's disks and Meissner's corpuscles that are found deeper in the skin and in the upper parts of the skin respectively, detect fine touch. On the other hand, Pacinian corpuscles and Ruffini endings can sense deeper touch like pressure.

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Answer:

258 Hz or 266 Hz

Explanation:

When two sounds of similar frequency occur at the same time, a phenomenon called "beat" occurs. The beat is an interference patter between the two sounds; the frequency of this beats (called beat frequency) is given by the difference between the frequencies of the two sounds:

[tex]f_B = |f_1 -f_2|[/tex]

where

[tex]f_B[/tex] is the beat frequency

[tex]f_1[/tex] is the frequency of the 1st sound

[tex]f_2[/tex] is the frequency of the 2nd sound

In this problem,

[tex]f_B=4 Hz[/tex] is the beat frequency

[tex]f_1=262 Hz[/tex] is the frequency of the middle C tuning fork

[tex]f_2[/tex] is the frequency of the middle C piano string

Solving for f2, we find two solutions:

[tex]f_2=f_1-f_B = 262-4 = 258 Hz\\f_2 = f_1+f_B = 262+4=266 Hz[/tex]

Answer:

Resistance

Explanation:

Resistance "resists" the flow of electricity and makes it more difficult to travel. The higher the resistance the less current there is in a circuit. Example being, open circuit (infinity ohms) means there is no current flowing with ohms law.

Answer:

0.00156875 m

Explanation:

y = Distance from the 20th fringe = 9.6 mm

l = Distance to screen = 1.5 m

[tex]\lambda[/tex] = Wavelength = [tex]502\ nm[/tex]

m = Number of fringe = 20

We have the relation

[tex]\dfrac{y}{l}=\dfrac{m\lambda}{d}\\\Rightarrow d=\dfrac{m\lambda l}{y}\\\Rightarrow d=\dfrac{20\times 502\times 10^{-9}\times 1.5}{9.6\times 10^{-3}}\\\Rightarrow d=0.00156875\ m[/tex]

The seperation of the two slits is 0.00156875 m

Answer:

from 7.69×10¹⁵ Hz

Explanation:

Frequency: This can be defined as the number of complete cycle a wave complete in one seconds, the S.I unit of frequency is Hertz (Hz).

From the question,

v = λf............. Equation 1

Where v = speed of visible light in air, λ = wave length of visible light, f = frequency of visible light.

make f the subject of the equation

f = v/λ.................... Equation 2

Note: all electromagnetic wave have the same speed = 3×10⁸ m/s

Given: v = 3×10⁸ m/s, λ = from 3.9×10⁻⁷ m

Substitute into equation 2

f = 3×10⁸/ 3.9×10⁻⁷

f = 7.69×10¹⁵ Hz

Hence the range of the frequencies for visible light = from 7.69×10¹⁵ Hz

Final answer:

The range of frequencies for visible light, corresponding to wavelengths from about 400 nm to 700 nm, extends from approximately 4.28 × 10¹⁴ Hz to 7.50 × 10¹⁴ Hz.

Explanation:

The range of wavelengths for visible light is commonly accepted to be from about 400 nm to 700 nm. To find the corresponding range of frequencies for visible light, we use the formula c = λf, where c is the speed of light (≈ 3.00 × 10¸ m/s), λ is the wavelength in meters, and f is the frequency in hertz (Hz). For the visible light range, converting nanometers to meters (× 10⁻⁹), we calculate the frequencies at both ends of the spectrum: for 400 nm (4.00 × 10⁻⁷ m), the frequency is approximately 7.50 × 10¹⁴ Hz, and for 700 nm (7.00 × 10⁻⁷ m), it's approximately 4.28 × 10¹⁴ Hz. Therefore, the range of frequencies for visible light extends approximately from 4.28 × 10¹⁴ Hz to 7.50 × 10¹⁴ Hz.

Answer:

Acceleration, a = [tex]2.8\ m/s^2[/tex]

Explanation:

Given that,

Initial speed of the skier, u = 8 m/s

Final speed of the skier, v = 6 m/s

Distance, d = 5 m

We need to find the magnitude of her acceleration. It is equal to the rate of change of speed. It is given by :

[tex]v^2-u^2=2as\\\\a=\dfrac{v^2-u^2}{2s}\\\\a=\dfrac{6^2-8^2}{2\times 5}\\\\a=-2.8\ m/s^2[/tex]

So, the magnitude of her acceleration is [tex]2.8\ m/s^2[/tex].

The magnitude of the skier's acceleration as she slows down is 0.4 m/s^2.

To find the magnitude of the skier's acceleration, we can use the formula:

Acceleration = (Final Velocity - Initial Velocity) / Time

In this case, the initial velocity is 8.0 m/s, the final velocity is 6.0 m/s, and the time is 5.0 seconds.

Plugging these values into the formula, we get:

Acceleration = (6.0 m/s - 8.0 m/s) / 5.0 s = -0.4 m/s^2

So the magnitude of the skier's acceleration as she slows down is 0.4 m/s^2.

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Final answer:

The 65 kg baseball player slides a distance of 5.2 meters before coming to a stop, calculated using kinetic energy and the work done by friction.

Explanation:

To calculate the distance the 65 kg baseball player slides using energy considerations, we start by noting that the work done by friction is equal to the loss in kinetic energy as the player comes to a stop. The kinetic energy (KE) of the player can be calculated using the formula KE = (1/2)mv², where m is mass and v is velocity. Since the force of friction (f) is constant, the work (W) done by friction is W = fd, with d being the distance over which the friction acts.

Setting the initial kinetic energy equal to the work done by friction, we get:

(1/2)mv² = fd

Inserting the given values, we have:

(1/2)(65 kg)(6 m/s)² = (450 N)d

Upon calculating, we find:

d = (1/2)(65 kg)(6 m/s)² / (450 N)

d = 5.2 meters

Therefore, the baseball player slides a distance of 5.2 meters before coming to a stop.

Answer:

they have a cooling effect! :)

Explanation:

Answer:

A cooling effect

Explanation:

As strange as it seems, volcanoes cool during eruption. All of the magma pent up in the core gets released into lava, which drastically changes the internal temperatures. This is kind of like how for some people, yelling/beating on inanimate objects helps them to "cool off".

Answer:

Evaporation, condensation, precipitation

Explanation:

Answer:

Evaporation, condensation, precipitation & collection

Explanation:

Answer:

4.157 m/s

Explanation:

Average velocity: This can be defined as the ratio change in position to time interval. The S.I unit of average velocity is m/s

The expression for average velocity is given as,

V = Δx/t.............. Equation 1

Where V = average velocity, Δx = change in position on the x- axis, t = time.

But,

Δx = x₂-x₁........... Equation 2

Substitute equation 2 into equation 1

V = (x₂-x₁)/t................ Equation 3

Given: x₂ = 32.4 m, x₁ = -5 m, t = 8.9 s

Substitute into equation 3

V = [32.4-(-5)]/8.9

V = (32.4+5)/8.9

V = 37.4/8.9

V = 4.157 m/s

Hence the average velocity = 4.157 m/s

Answer:

v = 4.20m/s

Average velocity is 4.20m/s

Explanation:

Average velocity is the change in displacement per unit time;

v = ∆x/t

Given;

∆x = 32.4 - (-5) = 37.4 m

t = 8.9 seconds

Substituting the values;

v = 37.4m/8.9s

v = 4.20m/s

Average velocity is 4.20m/s

Explanation:

The direction of magnetic field in a straight wire is given by Maxwell's right hand rule. It states that if the thumb of right hand points towards the direction of electric current, then the curled finger will give the direction of magnetic field.

In this case, a vertical wire carries a current straight down. It means that current will flow in the clockwise direction.

The magnetic field to the east of a vertical wire carrying a current straight down is directed into the page, which is determined by the right-hand rule.

To determine the direction of the magnetic field around a current-carrying wire, we can use the right-hand rule. For a wire that carries a current downward, you would point the thumb of your right hand in the direction of the current (downward), and then your fingers would naturally curl in the direction of the magnetic field. To the east of the wire, your fingers would point into the page, indicating that the magnetic field is directed into the page.

a) 0.0028 rad/s

b) [tex]23.68 m/s^2[/tex]

c) [tex]0 m/s^2[/tex]

Explanation:

a)

When an object is in circular motion, the angular speed of the object is the rate of change of its angular position. In formula, it is given by

[tex]\omega = \frac{\theta}{t}[/tex]

where

[tex]\theta[/tex] is the angular displacement

t is the time interval

The angular speed of an object in circular motion can also be written as

[tex]\omega = \frac{v}{r}[/tex] (1)

where

v is the linear speed of the object

r is the radius of the orbit

For the spaceship in this problem we have:

[tex]v=29,960 km/h[/tex] is the linear speed, converted into m/s,

[tex]v=8322 m/s[/tex]

[tex]r=2925 km = 2.925\cdot 10^6 m[/tex] is the radius of the orbit

Subsituting into eq(1), we find the angular speed of the spaceship:

[tex]\omega=\frac{8322}{2.925\cdot 10^6}=0.0028 rad/s[/tex]

b)

When an object is in circular motion, its direction is constantly changing, therefore the object is accelerating; in particular, there is a component of the acceleration acting towards the  centre of the orbit: this is called centripetal acceleration, or radial acceleration.

The magnitude of the radial acceleration is given by

[tex]a_r=\omega^2 r[/tex]

where

[tex]\omega[/tex] is the angular speed

[tex]r[/tex] is the radius of the orbit

For the spaceship in the problem, we have

[tex]\omega=0.0028 rad/s[/tex] is the angular speed

[tex]r=2925 km = 2.925\cdot 10^6 m[/tex] is the radius of the orbit

Substittuing into the equation above, we find the radial acceleration:

[tex]a_r=(0.0028)^2(2.925\cdot 10^6)=23.68 m/s^2[/tex]

c)

When an object is in circular motion, it can also have a component of the acceleration in the direction tangential to its motion: this component is called tangential acceleration.

The tangential acceleration is given by

[tex]a_t=\frac{\Delta v}{\Delta t}[/tex]

where

[tex]\Delta v[/tex] is the change in the linear speed

[tex]\Delta t[/tex] is the time interval

In this problem, the spaceship is moving with constant linear speed equal to

[tex]v=8322 m/s[/tex]

Therefore, its linear speed is not changing, so the change in linear speed is zero:

[tex]\Delta v=0[/tex]

And therefore, the tangential acceleration is zero as well:

[tex]a_t=\frac{0}{\Delta t}=0 m/s^2[/tex]

Answer:

magnetic field is the correct answer.

Explanation:

Answer:

The time that will have to pass before one has under 1 oz of caffeine remaining is 24.53 hours

Explanation:

Here, we have the formula for half life as follows;

[tex]N(t) = N_0(\frac{1}{2})^{\frac{t}{t_{1/2}}[/tex]

Where:

N(t) = Remaining quantity of the substance = 1 oz

N₀ = Initial quantity of the substance = 30 oz

t = Time duration

[tex]t_{1/2}[/tex] = Half life of the substance = 5 hours

Therefore, plugging in the values, we have

[tex]1= 30(\frac{1}{2})^{\frac{t}{5}}[/tex]

[tex]\frac{1}{30} = (\frac{1}{2})^{\frac{t}{5}}\\ln(\frac{1}{30}) =\frac{t}{5} ln(\frac{1}{2})\\\frac{t}{5} = \frac{ln(\frac{1}{30}) }{ ln(\frac{1}{2})} = 4.91\\ t = 4.91 \times 5 = 24.53 \ hours[/tex]

The time that will have to pass before one has under 1 oz of caffeine remaining = 24.53 hours.

Answer:200×10^5 meters

Explanation:

Answer:

200×10^5 meters

Explanation:

EDGE

The magnetic field produced by a current loop is not uniform. It is strongest at the center of the loop and decreases as you move away from the center. This change in strength across the space surrounding the loop makes it non-uniform.

No, the magnetic field created by a current loop is not uniform. A current loop means a closed electrical circuit or a current flowing in a loop of conducting wire. When a current flows through the loop, a magnetic field is generated surrounding the loop. This magnetic field is strongest at the center of the loop and diminishes as you move away from the center. This variance in strength implies that the magnetic field is not uniform.

Example:

If you place a compass around a loop with flowing current, you'd notice that the compass needle's direction (which aligns with the magnetic field) changes as you move the compass around. This indicates that the magnetic field strength (and direction) changes across the space surrounding the current loop, which makes it non-uniform.

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The magnitude of the normal force and acceleration is calculated using Newton's second law and free body diagrams, considering all forces acting on the object.

The question pertains to finding the magnitude of the acceleration and the normal force acting on a chair, block, or similar object using Newton's second law of motion. To determine these quantities, one must analyze the forces acting on the object in a free body diagram and apply the equations of motion. The normal force is the force exerted by a surface to support the weight of an object resting on it, and it acts perpendicular to the surface. To find the normal force (FN), we typically assume static equilibrium if the object is at rest, which implies that the net force in the vertical direction is zero; thus, FN equals the object's weight (mg), where m is the mass and g is the acceleration due to gravity. The magnitude of acceleration (a) can be calculated using the net force acting on the object (resultant force) divided by its mass, as stated by Newton's second law (F = ma).

For example, if a block is placed on a table, the normal force is equal to the gravitational force acting on it (FN = Fg). If the block is accelerating, you would calculate the net force by including all forces acting on the block, such as tension, friction, or applied forces, and then use F = ma to find the acceleration.

Answer:

need more info

Explanation:

Answer:

(A) Strain = 0.0012

(b) Stress [tex]=77.44\times 10^5N/m^2[/tex]

(c) Young's modulus [tex]=6.45\times 10^9N/m^2[/tex]    

Explanation:

Mass of the rock m = 75 kg

So weight of the rock [tex]F=mg=75\times 9.8=735N[/tex]

Length of the rope l = 18 m

Diameter of the rope d = 11 mm

Change in length of rope [tex]\Delta l=2.1cm =0.021m[/tex]

So radius r = 5.5 mm = 0.0055 m

Cross sectional area [tex]A=\pi r^2[/tex]

[tex]A=3.14\times 0.0055^2=9.49\times 10^{-5}m^2[/tex]

(a) Strain is equal to ratio change in length to original length

So strain [tex]=\frac{\Delta l}{l}=\frac{0.021}{18}=0.0012[/tex]

(b) Stress [tex]=\frac{Weight}{area}[/tex]

[tex]=\frac{735}{9.49\times 10^{-5}}=77.44\times 10^5N/m^2[/tex]

(c) Young's modulus is equal to ratio of stress and strain

So young's modulus [tex]=\frac{77.44\times 10^5}{0.0012}=6.45\times 10^9N/m^2[/tex]

Answer:

C

Explanation:

Answer:

In a parallel circuit, each resistor has: its own connection to the voltage source.

Explanation:

Answer:

Centripetal acceleration

Explanation:

An object moving around a xirxular path maintains its route as a result of centripetal force. However, its acceleration is caused by centripetal acceleration. Despite centripetal acceleration not being among the choices, it is the right answer.

Centripetal acceleration helps an object that navigates around a circular path to accelerate while centripetal force enables the movement of an object around a circular path to move inwards. Momentum, given as one of the choices is product of mass and velocity while friction is the force opposing movement of an object around a surface.

Answer:

Explanation:

Given that,

The peak wavelength is

λp = 475nm

Then, we want to find the temperature,

From Wein's displacement law,

When the maximum is evaluated from the Planck radiation formula, the product of the peak wavelength and the temperature is found to be a constant (k = 2.898 ×10^-3 mK)

So, applying this we have

λp•T = 2.898 ×10^-3

T = 2.898 × 10^-3 / λp

T = 2.898 × 10^-3 / 475 × 10^-9

T = 6101.05 K

The surface temperature of the distant star having the given peak wavelength is 6.1 × 10³K.

Given the data in the question;'

Peak wavelength; [tex]\lambda _p = 475nm = 4.75 *10^{-7}m[/tex]

To determine the surface temperature the star, we use the expression for peak wavelength by Wien's Displacement Law:

[tex]\lambda _p = \frac{0.2898 * 10^{-2}m.K}{T}[/tex]

Where T is the surface temperature

We substitute our value into the formula and solve for for "T"

[tex]4.75*10^{-7}m = \frac{0.2898 * 10^{-2}m.K}{T}\\\\T = \frac{0.2898 * 10^{-2}m.K}{4.75*10^{-7}m}\\\\T = 6. 1 * 10^3K[/tex]

Therefore, the surface temperature of the distant star having the given peak wavelength is 6.1 × 10³K.

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