A 0.5 kg block of aluminum (caluminum=900J/kg⋅∘C) is heated to 200∘C. The block is then quickly placed in an insulated tub of cold water at 0∘C (cwater=4186J/kg⋅∘C) and sealed. At equilibrium, the temperature of the water and block are measured to be 20∘C.If the original experiment is repeated with a 1.0 kg aluminum block, what is the final temperature of the water and block?If the original experiment is repeated with a 1.0 kg copper (ccopper=387 J/kg⋅∘C) block, what is the final temperature of the water and block?If the original experiment is repeated but 100 g of the 0∘C water is replaced with 100 g of 0∘C ice, what is the final temperature of the water and block? If the original experiment is repeated but 100 g of the 0∘C water is replaced with only 25 g of 0∘C ice, what is the final temperature of the water and block?

Answer:

a) -1.97 rad/sec² b) -8.09*10⁻³ N.m

Explanation:

a) Assuming a constant angular acceleration, we can apply the definition of angular acceleration, as follows:

γ = (ωf -ω₀) / t

We know that the final state of the grindstone is at rest, so ωf =0

In order to be consistent in terms of units, we can convert ω₀ from rev/min to rad/sec:

ω₀ = 730 rev/min* (1 min/60 sec)* (2*π rad / 1 rev) = 73/3*π

⇒ γ = (0-73/3*π) / 38.8 sec = - 1.97 rad/sec²

b) In order to get the value of the frictional torque exerted on the grindstone, that caused it to stop, we can apply the rotational equivalent of the Newton's 2nd law, as follows:

τ = I * γ (1)

As the grindstone can be approximated by a solid disk, the rotational inertia I can be expressed as follows:

I = m*r² / 2, where m=1.5 kg and r = 0.074 m.

Replacing in (1) , m. r and γ (the one we calculated in a)), we get:

τ = (1.5 kg* (0.074)² m² / 2) * -1.97 rad/sec = -8.09*10⁻³ N.m

(The negative sign implies that the frictional torque opposes to the rotation of the grindstone).

To solve this problem we will use the mathematical definition of the light years in metric terms, from there, through the kinematic equations of motion we will find the distance traveled as a function of the speed in proportion to the elapsed time. Therefore we have to

[tex]1Ly =9.4605284*10^{15}m \rightarrow 'Ly'[/tex]means  Light Year

Then

[tex]14.4Ly = 1.36231609*10^{17} m[/tex]

If we have that

[tex]v= \frac{x}{t} \rightarrow t = \frac{x}{t}[/tex]

Where,

v = Velocity

x = Displacement

t = Time

We have that

[tex]t = \frac{1.36231609*10^{17}}{0.96c} \rightarrow c[/tex]= Speed of light

[tex]t = \frac{1.36231609*10^{17}}{0.96(3*10^8)}[/tex]

[tex]t= 454105363 s (\frac{1hour}{3600s})[/tex]

[tex]t= 126140 hours(\frac{1day}{24hours})[/tex]

[tex]t= 5255.85 days(\frac{1 year}{365days})[/tex]

[tex]t = 14.399 years[/tex]

Therefore will take 14.399 years

It takes Rob approximately 15.0 years to complete the trip relative to a frame of reference fixed with respect to Earth.

  To solve this problem, we can use the concept of time dilation from the theory of special relativity. According to this concept, time measured in a frame of reference moving at a high velocity relative to a stationary observer (in this case, Earth) will be dilated, or ""slowed down,"" compared to the time measured by the stationary observer.

 The time dilation formula is given by:

 [tex]\[ t' = \frac{t}{\sqrt{1 - \frac{v^2}{c^2}}} \][/tex]

 where:

- [tex]\( t' \)[/tex]is the time interval measured by the moving observer (Rob),

- [tex]\( t \)[/tex]is the time interval measured by the stationary observer (Earth),

-[tex]\( v \)[/tex] is the relative velocity between the two frames of reference, and

- [tex]\( c \)[/tex] is the speed of light in a vacuum.

 Given that Rob's speed is [tex]\( 0.960c \)[/tex]and the distance to the star system is [tex]\( 14.4 \)[/tex] light-years, we can calculate the time it takes for Rob to complete the trip according to Earth's frame of reference.

 First, we calculate the Lorentz factor [tex]\( \gamma \),[/tex] which is the factor by which time is dilated:

[tex]\[ \gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}} \][/tex]

 Substituting \( v = 0.960c \):

 [tex]\[ \gamma = \frac{1}{\sqrt{1 - (0.960)^2}} \][/tex]

[tex]\[ \gamma = \frac{1}{\sqrt{1 - 0.9216}} \][/tex]

[tex]\[ \gamma = \frac{1}{\sqrt{0.0784}} \][/tex]

[tex]\[ \gamma = \frac{1}{0.280} \][/tex]

[tex]\[ \gamma \approx 3.5714 \][/tex]

 Now, we calculate the time [tex]\( t \)[/tex] it takes for Rob to travel 14.4 light-years at a speed of [tex]\( 0.960c \):[/tex]

[tex]\[ t = \frac{d}{v} \][/tex]

[tex]\[ t = \frac{14.4 \text{ light-years}}{0.960c} \][/tex]

[tex]\[ t = \frac{14.4}{0.960} \text{ years} \][/tex]

[tex]\[ t = 15 \text{ years} \][/tex]

 Since the Lorentz factor [tex]\( \gamma \)[/tex] is approximately 3.5714, the time measured on Earth for Rob's journey, taking into account time dilation, would be:

[tex]\[ t' = \gamma \cdot t \][/tex]

[tex]\[ t' \approx 3.5714 \cdot 15 \text{ years} \][/tex]

[tex]\[ t' \approx 53.571 \text{ years} \][/tex]

 However, since the question asks for the time according to the frame of reference fixed with respect to Earth, we do not need to multiply by the Lorentz factor. The time [tex]\( t \)[/tex] is already the time measured by the stationary observer on Earth.

 Therefore, the time it takes Rob to complete the trip relative to Earth is approximately 15.0 years.

Answer:

Explanation:

Fission means breaking or splitting i.e. Nuclear fission involves breaking up of unstable nucleus into smaller pieces and simultaneously producing energy.  

Fusion means joining of different elements to form a unified whole. In Nuclear fusion smaller nuclei combined to form a new heavy nucleus associated with a large amount of energy release.

Thus nuclear Fusion and fission can release Energy.                                        

The weight of a 15kg dog on Earth is 147 Newtons. Weight is calculated by multiplying the object's mass by the acceleration due to gravity on Earth, which is approximately 9.8 m/s². The result is rounded to the tenths place if necessary, though not required in this calculation.

The weight of an object is the force due to gravity acting on that object's mass. On Earth, the weight of an object can be calculated by multiplying its mass by the acceleration due to gravity, which is approximately 9.8 m/s2. Therefore, the weight of a 15kg dog on Earth is calculated as 15 kg × 9.8 m/s2, which equals 147 Newtons (N).

When working with measurements and answers in physics, it's important to consider precision. Since the least precise measurement provided in the reference information is 13.7 kg, expressed to the 0.1 decimal place, any calculated weight should also be expressed to the tenths place. However, since our calculation already gives a whole number without a decimal component, further rounding is not required for this particular scenario.

The time taken for the stone to hit the ground can be calculated using vertical motion equations, and the stone's speed at impact can be determined using the vertical velocity equation.

Time taken for the stone to hit the ground can be calculated using the vertical motion equation: t = sqrt(2h/g). Substituting the values, t = √(2×45/9.81) ≈ 3.0 seconds.

Stone's speed at impact can be calculated using the vertical velocity equation: vf = u + gt, where vf is the final velocity (0 m/s at impact), u is the initial vertical speed, and g is the acceleration due to gravity.

With the given information, the stone's speed at impact is approximately 29.4 m/s.

Answer:

74 N

Explanation:

T = Tension in the string = 74 N

[tex]\rho[/tex] = Density of the steel = 7800 kg/m³

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

r = Radius = [tex]8.5\times 10^{-4}\ m[/tex]

Linear density is given by

[tex]\mu=\rho A\\\Rightarrow \mu=7800\times \pi (8.5\times 10^{-4})^2\\\Rightarrow \mu=0.0177\ kg/m[/tex]

The linear density is 0.0177 kg/m

Velocity is given by

[tex]v=\sqrt{\dfrac{T}{\mu}}\\\Rightarrow v=\sqrt{\dfrac{74}{0.0177}}\\\Rightarrow v=64.65903\ m/s[/tex]

The velocity of the wave on the guitar string is 64.65903 m/s

Final answer:

The net force F120 on the pressure cooker's lid when the inside air is heated to 120°C is the difference between the force exerted by the heated air inside and the atmospheric force outside on the lid's area A. Calculate pressure using the ideal gas law, then calculate the forces exerted at 120°C and at atmospheric conditions, and find their difference.

Explanation:

The student asks about the magnitude of the net force F120 on the lid of a pressure cooker when the air inside is heated to 120°C, assuming room temperature outside the cooker is 20°C, the lid's area is A, and atmospheric pressure is Pa. To find the net force on the lid, we must compare the force exerted by the heated air inside the cooker with the force exerted by the outside atmosphere.

First, using the ideal gas law (PV = NkBT), we determine the pressure inside the cooker at 120°C. Then, we calculate the force exerted by this pressure over area A. Similarly, the outside atmospheric pressure Pa also exerts a force over area A. The net force on the lid, F120, is the difference between these two forces.

Without numerical values for parameters like the number of moles of air inside the pressure cooker or its volume, a specific numeric answer cannot be provided. The student must apply the ideal gas law to find the pressure at 120°C, then calculate the two forces and find the difference to obtain F120.

The tensile strength and ductility cannot be directly calculated from the provided information, but the proportion of cold work (or percentage deformation) can be computed based on the diameter change.

Despite the detailed context provided, the question lacks sufficient data or formulas to directly compute the tensile strength nor the ductility of a copper rod from its diameter change due to cold working. Usually, the tensile strength and ductility mean the ultimate tensile strength (the maximum stress that a material can withstand while being stretched or pulled before failing or breaking) and the percent elongation after a material specimen has been pulled and rupture occurs. To get these values, experimenting with the material and measuring would be necessary.

However, we can calculate the percentage of cold work using the geometric deformation change. The percentage of cold work (%CW) based on change in diameter can be calculated with the formula: %CW = [(initial area - final area)/initial area] x 100%. Here, the area is that of the rod cross-section, which for a cylinder is pi*(d/2)². Thus you can substitute and calculate for %CW with the given diameters.

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Answer:

5.5 × 10^14 Hz or s^-1

no orange light has less frequency so no photoelectric effect

Explanation:

hf = hf0 + K.E

HERE h is Planck 's constant having value 6.63 × 10 ^-34 J s

f is frequency of incident photon and f0 is threshold frequency

hf0 = hf- k.E

6.63 × 10 ^-34 × f0 = 6.63 × 10 ^-34× 6.66 × 10^14 - 7.74× 10^-20

6.63 × 10 ^-34 × f0 = 3.64158×10^-19

                           f0 = 3.64158×10^-19/ 6.63 × 10 ^-34

                           f0 = 5.4925 × 10^14

                            f0 =5.5 × 10^14 Hz or s^-1

frequency of orange light is 4.82 × 10^14 Hz which is less than threshold frequency hence photo electric effect will not be observed for orange light

Answer:

Energy stored in inductor will be 20.797 J

Explanation:

We have given inductance L = 3.54 H

And resistance R = 7.76 ohm

Battery voltage V = 26.6 VOLT

After very long time means at steady state inductor behaves as short circuit

So  current [tex]i=\frac{V}{R}=\frac{26.6}{7.76}=3.427Amp[/tex]

Now energy stored in inductor [tex]E=\frac{1}{2}Li^2=\frac{1}{2}\times 3.54\times 3.427^2=20.797J[/tex]

So energy stored in inductor will be 20.797 J

To find the energy stored in an inductor in an RL circuit at equilibrium, use Ohm's law to calculate the steady current, and then apply the energy formula E = (1/2)LI^2.

When the switch in an RL series circuit is closed for a long time, the current in the circuit reaches an equilibrium value due to the voltage provided by the battery. The current, I, can be calculated using Ohm's law, I = V/R, where V is the voltage of the battery and R is the resistance in the circuit. In this case, with a 26.6 V battery and a 7.76 Ω resistor, the current would be I = 26.6 V / 7.76 Ω. After calculating the current, the energy stored in the inductor at equilibrium can be found using the formula for energy stored in an inductor, which is E = (1/2)LI^2, where L is the inductance and I is the current.

The corresponding energy stored in the inductor after a long time can be determined using these calculations.

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Entropy is the energy that cannot be used to do useful work.

The correct answer is b. entropy. Entropy is a measure of the disorder or randomness in a system. It is a form of energy that cannot be converted into useful work. For example, when energy is transferred from one form to another, such as from chemical energy to thermal energy, some of the energy is lost as heat, which is a form of entropy.

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Answer:

Both equilibrium price and quantity will increase.

Explanation:

An increase in the income of video game players will surely lead to an upward shift in the supply and demand curve. This shift in the supply and demand curve would affect the equilibrium price and quantity positively leading to a correspondent increase in the equilibrium price and quantity.

If video game disks are a normal good, an increase in the income of video game players would likely lead to an increased demand for video games. This is because, with increased income, players now have more purchasing power ability to buy more video games. Other factors such as popularity, population, and price of substitutes can also influence this demand.

Your question relates to the concept of normal goods within the field of economics. According to economic theory, a normal good is a good for which demand increases as income increases. Therefore, if the income of video game players increases, according to economic principles, you can expect the demand for video games (assuming they are a normal good) to also increase.

This increase in demand in the market for video games originates from the increased purchasing power. That is, players now have the ability to buy more video games than before. Additionally, factors such as taste shift to greater popularity, the population likely to buy rises, and price of substitutes can further influence this demand.

However, it's essential to also consider that economic theory often simplifies real-world conditions. In reality, numerous variables could influence the market demand for video games beyond income changes.

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The distance between your eye and the image of the spider in the mirror is 85 cm. This is found by adding the distance from the spider to the mirror and from your eyes to the mirror.

The situation regards the properties of reflection in a mirror. In a plane mirror, the image of an object appears to be the same distance behind the mirror as the object is in front of the mirror. So the image of the spider in the mirror is 20cm behind the mirror. The eye's distance from the mirror is 65cm. Therefore, the distance from your eye to the image of the spider in the mirror is the sum of these two distances: 20cm (distance from the spider to the mirror) + 65cm (distance from your eye to the mirror) = 85cm.

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Answer:

d) larger than Lily's

Explanation:

If the merry-go-round is rotating at a constant angular speed, this means that any two points, located at different positions along a radius, rotate at the same angular speed, which means that they sweep the same angle at a given time interval.

In order to both points keep aligned along the same radius, we have a single choice (assuming that we are talking about a rigid  body) to meet this premise:

The point farther of the center (Bob) must have a linear speed greater than a point closer to the center (Lily).

Mathematically, we can explain this result as follows:

ω = Δθ / Δt (by definition of angular velocity) (1)

but, by definition of angle, we can say the following:

θ = s/r , where s is the arc along the circumference, and r, the radius.

⇒Δθ = Δs /r

Replacing in (1) we have:

ω = (Δs /Δt) / r

By definition, Δs/Δt = v, so, arranging terms, we get:

v = ω*r

If ω=constant, if r increases, v increases.

So, as Bob is at a distance r from the center larger than Lily's, Bob's speed must be larger than Lily's.

Regardless of their position, Bob and Lily will rotate at the same angular speed. However, due to being further from the center of rotation, Bob's linear speed will be greater than Lily's.

The subject matter of the question revolves around rotational motion, specifically concerning angular speed, and how location on a rotating reference frame, such as a merry-go-round, affects linear speed. Both Bob and Lily, being on the same merry-go-round, share the same angular speed (w), regardless of their location on the platform.

The linear speed of an object in rotational motion is given by: Speed = radius x angular speed. Given that Bob is on the outer edge (larger radius) while Lily is near the center (smaller radius), it's evident that Bob's linear or tangential speed (how fast he's actually moving along a path) will be larger than Lily's for the same angular speed. Therefore, the answer is d) Bob's speed is larger than Lily's.

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Answer:

q = 1,297 10⁻¹⁹ C , n=1

Explanation:

For this problem we will use Newton's second law in the case of equilibrium

        [tex]F_{e}[/tex] + B - W = 0

Where [tex]F_{e}[/tex] is the electrical force up, B the thrust and W the weight of the drop.

Let's look for weight and thrust

oil

      ρ = m / V

      m =  ρ V  

Air

       B =   [tex]\rho _{air}[/tex] g V

Electric force

       [tex]F_{e}[/tex] = qE

       E = V / d

       [tex]F_{e}[/tex] = q V/d

Let's replace

      q V / d + [tex]\rho _{air}[/tex] g V - ρ V g = 0

      qV / d = (4/3 π r³) g (ρ –[tex]\rho _{air}[/tex])

      q = 4/3 π r³ (ρ –[tex]\rho _{air}[/tex])  d / V

Reduce to SI units

      d = 1.87 cm (1m / 100cm) = 1.87 10⁻² m

      ρ= 0.816 r / cm3 (1kg / 1000g) (102cm / 1m)³ = 816 kg / m³

      [tex]\rho_{air}[/tex] = 1.28 kg / m³

Let's calculate the charge

       r = d / 2 = 3.3 10⁻⁶ m

      r = 1.65 10⁻⁶ m

     q = 4/3 π (1.65 10⁻⁶)³  (816 - 1.28)   0.0187 / 2210

     q = 12.9717 10⁻²⁰ C

     q = 1,297 10⁻¹⁹ C

If we assume that the load is

     q = n e

In this case n = 1

Answer:

(D) Va > Vb. The forces between the atoms in a block and the atoms in a surface oppose the motion of the block and are greater on average for block B.

Explanation:

Given that, Two identical blocks, block A and block B, are placed on different horizontal surfaces. Block A is initially at rest on a smooth surface, while block B is initially at rest on a rough surface.

And now, a constant force F₀ is exerted on each block for a time Δt.

Now, speeds of blocks A and B are vA and vB respectively.

In case of block A the surface is smooth, so there is no opposing force.

Whereas as, in case of block B, the surface is rough which means friction opposes the motion

So, some of the applied force is used to overcome this friction which makes the speed of block B vB to be less than that of block A vA.

⇒ Va > Vb.

And completely eliminating friction is not possible.

so, even smooth surface has friction which is very very little.

so, The forces between the atoms in a block and the atoms in a surface oppose the motion of the block and are greater on average for block B.

The correct relation between vA and vB is vA > vB because block A encounters less friction on the smooth surface as compared to block B on the rough surface. Consequently, block A will have a larger final velocity vA when the same force is applied to both blocks for the same amount of time.

The correct relation is vA > vB. This is because block A is on a smooth surface, meaning there is less friction to oppose its motion when a force is applied, compared to block B which is on a rough surface. Friction is a force that opposes the motion of an object when it moves across a surface. In this case, the frictional force on block B is greater on average than the force on block A because block B rests on a higher-friction, or rougher, surface.

When we apply a constant force to both blocks, block A, encountering less frictional resistance, moves more easily than block B. Therefore, after force F0 is applied for Δt amount of time, block A on the smooth surface will have moved faster, hence having a larger final velocity vA, than block B on the rough surface (whose final velocity is vB). This falls under the domain of Newton's second law of motion, which states that the acceleration of a body is directly proportional to, and in the same direction as, the net force acting on the body, and inversely proportional to its mass.

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Answer:

distance cover is  = 102.53 m

Explanation:

Given data:

speed of object is 17.1 m/s

[tex]t_1 = 3.32 sec[/tex]

[tex]t_2 = 5.08 sec[/tex]

from equation of motion we know that

[tex]d_1 = vt_1 + \frac{1}{2} gt_1^2[/tex]

where d_1 is distance covered in time t1

so[tex] d_1 = 17.1 \times 3.32 + \frac{1}{2} 9.8 \times 3.32^2 [/tex]=

[tex]d_1 = 110.78 m[/tex]

[tex]d_2 = vt_2 + \frac{1}{2} gt_2^2[/tex]

where d_2 is distance covered in time t2

[tex]d_2 = 17.1 \times 5.08 + \frac{1}{2}\times9.8 \times 5.08^2[/tex]

[tex]d_2 = 213.31 m[/tex]

distance cover is  = 213.31 - 110.78 = 102.53 m

Final answer:

a. The largest angular magnification that Galileo could have obtained with the given lenses is 100x. b. The length of the telescope between the two lenses is 40 cm.

Explanation:

a. Largest angular magnification

The angular magnification of a telescope is equal to the product of the magnification produced by the objective lens and the magnification produced by the eyepiece lens. In this case, we need to find the largest possible angular magnification using the given lenses.

The magnification produced by a lens can be calculated using the formula: magnification = - / , where is the distance of the final image from the eyepiece and is the distance of the object from the objective lens.

Since the angular magnification is the product of the magnifications produced by the two lenses, we can find the largest possible angular magnification by maximizing the product of the two magnifications.

The largest possible angular magnification in this case is 100x.

b. Length of the telescope

The length of the telescope can be calculated by summing the focal lengths of the two lenses and the distances between them. In this case, the length of the telescope between the two lenses is the sum of the focal lengths of the two lenses, which can be calculated using the formula: length = + , where is the focal length of the objective lens and is the focal length of the eyepiece lens.

The length of the telescope between the two lenses is 40 cm.

Answer:

option D.  

Explanation:

The correct answer is option D.                

Diffusion is the particle movement from high concentration to low concentration, such as air, water, etc.                      

Example: if you spray perfume spread throughout the room at one part of the fragrance scent.                                                        

The gas molecule thus spreads out in a gradient of concentration

Diffusion is equivalent to gas molecules spreading out in a concentration gradient. Option D.

Diffusion is the movement of particles (such as gas molecules, liquid molecules, or even ions) from an area of higher concentration to an area of lower concentration.

This process occurs spontaneously and is driven by the natural tendency of particles to move and distribute themselves evenly in a given medium.

Diffusion is essential in various biological, chemical, and physical processes and plays a crucial role in the movement of substances within and between cells, as well as in the mixing of gases and liquids in the environment.

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To solve this problem we will use the kinematic equations of angular motion in relation to those of linear / tangential motion.

We will proceed to find the centripetal acceleration (From the ratio of the radius and angular velocity to the linear velocity) and the tangential acceleration to finally find the total acceleration of the body.

Our data is given as:

[tex]\omega = 1.25 rad/s \rightarrow[/tex] The angular speed

[tex]\alpha = 0.745 rad/s2 \rightarrow[/tex] The angular acceleration

[tex]r = 4.65 m \rightarrow[/tex] The distance

The relation between the linear velocity and angular velocity is

[tex]v = r\omega[/tex]

Where,

r = Radius

[tex]\omega =[/tex] Angular velocity

At the same time we have that the centripetal acceleration is

[tex]a_c = \frac{v^2}{r}[/tex]

[tex]a_c = \frac{(r\omega)^2}{r}[/tex]

[tex]a_c = \frac{r^2\omega^2}{r}[/tex]

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

[tex]a_c = (4.65 )(1.25 rad/s)^2[/tex]

[tex]a_c = 7.265625 m/s^2[/tex]

Now the tangential acceleration is given as,

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

Here,

[tex]\alpha =[/tex] Angular acceleration

r = Radius

[tex]\alpha = (0.745)(4.65)[/tex]

[tex]\alpha = 3.46425 m/s^2[/tex]

Finally using the properties of the vectors, we will have that the resulting component of the acceleration would be

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

[tex]|a| = \sqrt{(7.265625)^2+(3.46425)^2}[/tex]

[tex]|a| = 8.049 m/s^2 \approx 8.05 m/s2[/tex]

Therefore the correct answer is C.

Sound intensity (energy) falls inversely proportional to the square of the distance  from the sound:

[tex]I \propto \frac{1}{r^2}[/tex]

Therefore if we have two values of intensities we have

[tex]\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}[/tex]

As we have that

[tex]r_1 = 3 r_2[/tex]

Then we have that

[tex]\frac{I_1}{I_2} = \frac{r_2^2}{(3r_2)^2}[/tex]

[tex]\frac{I_1}{I_2} = \frac{1}{9}[/tex]

Therefore the correct answer is D. The sound intensity drops to 1 / 9 of its original value.

Answer

given,

frequency of sound (f)= 120 Hz

speed of sound (v)= 343 m/s

a) speed of approaching = 3 % of speed of sound

                          = 0.03 x 343 = 10.3 m/s

     frequency you would hear

     [tex]f' = f\dfrac{v+v_0}{v}[/tex]

     [tex]f' = 120 \times \dfrac{343+10.3}{343}[/tex]

            f' = 123.60 Hz

b)  speed of going away = 3 % of speed of sound

                          = 0.03 x 343 = 10.3 m/s

     frequency you would hear

     [tex]f' = f\dfrac{v-v_0}{v}[/tex]

     [tex]f' = 120 \times \dfrac{343-10.3}{343}[/tex]

            f' = 116.39 Hz

Answer:

289.714 times bigger

Explanation:

[tex]\Delta x[/tex] = Uncertainty in position

[tex]\Delta p[/tex] = Uncertainty in momentum = [tex]\Delta v m[/tex]

[tex]\Delta v[/tex] = Uncertainty in velocity = [tex]2\times 10^3\ m/s[/tex]

h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]

m = Mass of electron = [tex]9.1\times 10^{-31}\ kg[/tex]

From the Heisenberg uncertainty principle we have

[tex]\Delta x\Delta p=\dfrac{h}{4\pi}\\\Rightarrow \Delta x\Delta v m=\dfrac{h}{4\pi}\\\Rightarrow \Delta x=\dfrac{h}{4\pi\Delta v m}\\\Rightarrow \Delta x=\dfrac{6.626\times 10^{-34}}{4\pi \times 2\times 10^3\times 9.1\times 10^{-31}}\\\Rightarrow \Delta x=2.89714\times 10^{-8}\ m[/tex]

Comparing with 0.1 nm size atom

[tex]\dfrac{\Delta x}{x}=\dfrac{2.89714\times 10^{-8}}{0.1\times 10^{-9}}\\\Rightarrow \dfrac{\Delta x}{x}=289.714[/tex]

So, the electron’s minimum uncertainty in position is 289.714 times bigger than an atom of size 0.1 nm

The Heisenberg Uncertainty Principle in quantum physics states a limit to how precisely the position and momentum of a particle, such as an electron, can be known simultaneously. In the given context, though there's a large uncertainty in the electron's velocity, the uncertainty in its position within the atom remains significantly small, thus reflecting the inverse relationship between the precision of these two measurements.

The concept involved in this question is called the Heisenberg Uncertainty Principle which is a fundamental theory in quantum mechanics. This theory describes a limit to how precisely we can know both the simultaneous position of an object (such as an electron) and its momentum.

Using the principles of uncertainty (ΔxΔp ≥ h/4π), where h is the Planck constant, the uncertainty in position (Δx) is given. If the given velocity (v) of the electron is 2.0×103m/s with accuracy or uncertainty (Δv), the minimum uncertainty in position can be calculated, where the product of the uncertainties in position and velocity equal to or greater than Planck's constant divided by 4π, i.e. Δx ≥ h / (4πmΔv).

After calculation, it shows that the uncertainty in the electron's position within the atom is very small. Even though the uncertainty in velocity is large, the uncertainty in position remains smaller compared to the size of the atom. Hence, this represents the principle that increasing precision in measuring one quantity leads to greater uncertainty in the measurement of the other quantity.

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Answer:

The velocity is  [tex]4.6 m/s^2[/tex]

Explanation:

Given:

Force = 500N

Distance  s= 0

To find :

Its velocity at s = 0.5 m

Solution:

[tex]\sum F_{x}=m a[/tex]

[tex]F\left(\frac{4}{5}\right)-F_{S}=13 a[/tex]

[tex]500\left(\frac{4}{5}\right)-\left(k_{s}\right)=13 a[/tex]

[tex]400-(500 s)=13 a[/tex]

[tex]a = \frac{400 -(500s)}{13}[/tex]

[tex]a = (30.77 -38.46s) m/s^2[/tex]

Using the relation,

[tex]a=\frac{d v}{d t}=\frac{d v}{d s} \times \frac{d s}{d t}[/tex]

[tex]a=v \frac{d v}{d s}[/tex]

[tex]v d v=a d s[/tex]

Now integrating on both sides

[tex]\int_{0}^{v} v d v=\int_{0}^{0.5} a d s[/tex]

[tex]\int_{0}^{v} v d v=\int_{0}^{0.5}(30.77-38.46 s) d s[/tex]

[tex]\left[\frac{v^{2}}{2}\right]_{0}^{2}=\left[\left(30.77 s-19.23 s^{2}\right)\right]_{0}^{0.5}[/tex]

[tex]\left[\frac{v^{2}}{2}\right]=\left[\left(30.77(0.5)-19.23(0.5)^{2}\right)\right][/tex]

[tex]\left[\frac{v^{2}}{2}\right]=[15.385-4.807][/tex]

[tex]\left[\frac{v^{2}}{2}\right]=10.578[/tex]

[tex]v^{2}=10.578 \times 2[/tex]

[tex]v^{2}=21.15[/tex]

[tex]v = \sqrt{21.15}[/tex]

[tex]v = 4.6 m/s^2[/tex]

Answer:

0.243

Explanation:

Step 1: Identify the given parameters

Force (f) = 5kN, length of pitch (L) = 5mm, diameter (d) = 25mm,

collar coefficient of friction = 0.06 and thread coefficient of friction = 0.09

Frictional diameter =45mm

Step 2: calculate the torque required to raise the load

[tex]T_{R} = \frac{5(25)}{2} [\frac{5+\pi(0.09)(25)}{\pi(25)-0.09(5)}]+\frac{5(0.06)(45)}{2}[/tex]

[tex]T_{R}[/tex] = (9.66 + 6.75)N.m

[tex]T_{R}[/tex] = 16.41 N.m

Step 3: calculate the torque required to lower the load

[tex]T_{L} = \frac{5(25)}{2} [\frac{\pi(0.09)(25) -5}{\pi(25)+0.09(5)}]+\frac{5(0.06)(45)}{2}[/tex]

[tex]T_{L}[/tex] = (1.64 + 6.75)N.m

[tex]T_{L}[/tex] = 8.39 N.m

Since the torque required to lower the thread is positive, the thread is self-locking.

The overall efficiency = [tex]\frac{F(L)}{2\pi(T_{R})}[/tex]

                        = [tex]\frac{5(5)}{2\pi(16.41})}[/tex]

                        = 0.243

The overall efficiency of the power screw is approximately 38.6%. The torque required to raise the load is 46.8 N·m, while the torque to lower the load is about 22.53 N·m.

Calculating Lead and Mean Diameter:Lead (L) = Pitch (p) = 5 mm

Mean diameter (dm) = 25 mm

Calculating Torque to Raise the Load ([tex]T_{up[/tex]):

Using the formula: [tex]T_{up[/tex] = (W dm/2) (L + π μt dm)/(π dm - μt L)

Where:

W = 5 kN = 5000 N

μt = 0.09

dm = 25 mm = 0.025 m

L = 5 mm = 0.005 m

Substituting the values:

[tex]T_{up[/tex] = (5000 × 0.025/2) (0.005 + π × 0.09 × 0.025)/(π × 0.025 - 0.09 × 0.005)

[tex]T_{up[/tex] ≈ 46.8 N·m

Calculating Torque to Lower the Load ([tex]T_{down[/tex]):

Using the formula: [tex]T_{down[/tex] = (W dm/2) (L - π μt dm)/(π dm + μt L)

Substituting the values:

[tex]T_{down[/tex] = (5000 × 0.025/2) (0.005 - π × 0.09 × 0.025)/(π × 0.025 + 0.09 × 0.005)

[tex]T_{down[/tex] ≈ 22.53 N·m

Calculating Overall Efficiency (η):

Efficiency η = (tan θ / (tan(θ + φ)))

Where:

θ = tan-1(L/π dm)

φ = tan-1(μt)

θ ≈ 0.064

φ ≈ 0.09

η ≈ (tan 0.064) / (tan(0.064 + 0.09))

η ≈ 38.6%

Therefore, the overall efficiency of the power screw is approximately 38.6%, the torque to raise the load is 46.8 N·m, and the torque to lower the load is around 22.53 N·m.

Answer:

I think d is the answer haha

Galaxy collisions were more common in the past as C. the universe was smaller, hence galaxies were closer together.

Galaxy collisions were more common in the past than they are today primarily because galaxies were closer together in the past due to the smaller size of the universe. As the universe expanded, it carried galaxies along with it, increasing the distances between them and making collisions less frequent. When the universe was younger, galaxy interactions and collisions were frequent events that significantly influenced galaxy evolution, increased star-formation rates, and contributed to the prevalence of quasars during that time.

Observations indicate that when the universe was about 20% of its current age, interactions such as galaxy mergers happened most frequently, likely accounting for the active quasar populations observed from that epoch. These collisions often resulted in starburst galaxies, and the debris from these encounters could fuel the supermassive black holes at the centers of galaxies.

Answer:

given,

mass of the block = m₁

mass of the another block = 3 m₁

initial Amplitude, A = 6 cm

final amplitude = 6 cm

total mechanical energy = 12 J

total energy of the block spring

   [tex]E = \dfrac{1}{2}kA^2[/tex]

A is the amplitude and k is spring constant

initial energy is equal to 12 J

from the above expression we can say that

Energy of the given system depends up on the magnitude of spring constant and the amplitude.

so, energy of both the system will be same.

we know,

[tex]E = \dfrac{1}{2}mv^2[/tex]

[tex]12= \dfrac{1}{2}\times 3 m_1 v^2[/tex]

  [tex]v^2 = \dfrac{8}{m_1}[/tex]

  [tex]v = \sqrt{ \dfrac{8}{m_1}}[/tex]

Answer:

a)    t = 0.113 s , b)     W = 0.1756 N , c)   # λ = 49

Explanation:

a) Let's use the relationship

         v = λ f

Of the equation

         y = 8.55 10-3 cos (172 x + 2730 t)

When comparing this with the general equation

         y = A cos (kx - wt)

The wave number k = 172

       k = 2π /λ

       λ = 2π / 172

       λ = 0.03653 m

The angular velocity w = 2730

     w = 2π f

      f = w / 2π

     f = 2730 / 2π

     f = 434.49 Hz

The speed of the wave is

      v = 0.03653    434.49

      v = 15.87 m / s

The speed the wave on a string is constant, so

      v = d / t

      t = d / v

      t = 1.80 /15.87

      t = 0.113 s

b) The weight applied to the rope

      v = √ T /μ

The density

     μ = m / l

     μ = (0.0123 / 9.8) /1.80

     μ = 6.97 10-4 kg / m

The tension equal to the applied weight

    T = v² μ

    W = T = 15.87²   6.97 10⁻⁴

    W = 0.1756 N

c) let's use a rule of proportions

    # λ = 1.8 /0.03653

      # λ = 49

Answer:

correct answer is b    Z --->Z+1

Explanation:

In the processes of radioactive decay there are three basic processes the emission of alpha particles and the emission of beta rays and the emission of ayos range

The emission of a beta ray implies the transformation of a neutral into a proton, which implies the increase of the atomic number in a unit

          Z   ---->  Z +1

the atomic mass does not change since the mass of the two particles is practically the same, to balance the reaction antineutrino must also be emitted

The daughter particle is in an execrated state and passes to its base state with the emission of a gamma ray that does not change its atomic number or its atomic mass.

Consequently, from the above the correct answer is b

Answer:

t = 47.62 sec

Explanation:

Given data;

[tex]A_1 = 4.62 \times 10^4 m^2[/tex]

[tex]A_2 = 2.52 \times 10^4 m^2[/tex]

h = 2.50 cm

volume 10 L

from

[tex]A_1 v_1 = A_2 v_2[/tex]

[tex]4.62 \times 10^4 v_1 = 2.52 \times 10^4 v_2[/tex]

[tex]4.62 v_1 = 2.52 v_2[/tex] ......1

from bernoulli eq

[tex]P_1 + \frac{1}{2} \rho v_1^2 + \rho g h = P_2 + \frac{1}{2} \rho v_2^2 [/tex]

[tex]P_1 =P_2 = P_{atm}[/tex]

[tex]v_2^2 = v_1^2 +2gh[/tex] ... 2

from 1 and 2 equation

[tex]v_1 = 0.46 m/s[/tex]

volume flow rate is

[tex]Q = A_1 \times v_1 = 4.62 \times 10^[-4} v_1 = 2.1 \times 10^{-4} m^3/s[/tex]

[tex]t  = \frac{v}{Q} [/tex]

[tex]t =\frac{10\times 10^{-3}}{2.1 \times 10^{-4}} = 47.62 s[/tex]

Answer:

The time is [tex]4.76\times10^{-7}\ sec[/tex]

Explanation:

Given that,

Area [tex]A_{1}=4.62\times10^{4}\ m^2[/tex]

Area [tex]A_{2}=2.52\times10^{4}\ m^2[/tex]

Height = 2.50 cm

Volume = 10.0 L

We need to calculate the speed

Using equation of continuity

[tex]A_{1}v_{1}=A_{2}v_{2}[/tex]

Put the value into the formula

[tex]4.62\times10^{4}\times v_{1}=2.52\times10^{4}\times v_{1}[/tex]

[tex]4.62v_{1}=2.52v_{2}[/tex].....(I)

[tex]v_{1}=\dfrac{2.52}{4.62}v_{2}[/tex]

[tex]v_{1}=0.545v_{2}[/tex]

Now, using Bernoulli equation

[tex]P_{1}+\dfrac{1}{2}\rhi\times v_{1}^2+\rho gh=P_{2}+\dfrac{1}{2}\rhi\times v_{2}^2[/tex]

Here, [tex]P_{1}=P_{2}=P_{atm}[/tex]

[tex]v_{2}^2=v_{1}^2+2gh[/tex].....(II)

Put the value [tex]v_{1}[/tex] into the formula

[tex]v_{2}^2=(0.545v_{2})^2+2\times9.8\times2.50\times10^{-2}[/tex]

[tex]v_{2}^2=0.297v_{2}^2+0.49[/tex]

[tex]v_{2}^2(1-0.297)=0.49[/tex]

[tex]v_{2}=\sqrt{\dfrac{0.49}{0.703}}[/tex]

[tex]v_{2}=0.835\ m/s[/tex]

Put the value of [tex]v_{2}[/tex] in the equation (I)

[tex]v_{1}=0.545\times0.835[/tex]

[tex]v_{1}=0.46\ m/s[/tex]

We need to calculate the flow rate

Using formula of flow rate

[tex]Q=A_{1}v_{1}[/tex]

[tex]Q=(4.62\times10^{4})\times0.46[/tex]

[tex]Q=2.1\times10^{4}\ m^3/s[/tex]

We need to calculate the time

Using formula of time

[tex]t = \dfrac{V}{Q}[/tex]

Put the value into the formula

[tex]t=\dfrac{10.0\times10^{-3}}{2.1\times10^{4}}[/tex]

[tex]t=4.76\times10^{-7}\ sec[/tex]

Hence, The time is [tex]4.76\times10^{-7}\ sec[/tex]