Determine whether each statement below about conduction in semiconductors is true or false.
1. After a photon "kicks" an electron into the upper conduction band, the electron moves to the p-type side of the p-n junction.
2. An n-type semiconductor has missing electrons (or holes) in the lower energy valence band.
3. Electrons cannot move in a full valence band.

Answer:

If we assume that the temperature remains constant. And if the intermolecular forces are increased, then there would be an increase in the boiling point, the viscosity, and surface tension of the substance.

Explanation:

The intermolecular forces are interactions between particles. And decrease as we go from solid >liquid >gas. And on gases we have weak intermolecular forces.

The most common 3 types of intermolecular forces occurs between neutral molecules are known as Van der Walls forces (Dipole-Dipole,Hydrogen bonding, London Dispersion forces).

If the intermodelucar attraction forces increases we have:

1. Decreasing vapor pressure (pressure of the vapor that is in equilibrium with the liquid)

2. Increasing boiling point (temperature at which the vapor pressure is equal to the pressur exerted on the surface of the liquid)

3. Increasing surface tension (resistance of a liquid to spread out and increase the surface area)

4. Increasing viscosity (resistance of a liquid to flow)

If we assume that the temperature remains constant. And if the intermolecular forces are increased, then there would be an increase in the boiling point, the viscosity, and surface tension of the substance.

The speed of the mass when t = 3 s in simple harmonic motion is approximately 6.283 m/s.

The speed of the mass in simple harmonic motion can be determined by using the formula v = ωA sin(ωt), where v is the speed, ω is the angular frequency, A is the amplitude, and t is the time. In this case, the frequency of the motion is given as 4.0 Hz, which corresponds to an angular frequency of ω = 2πf = 2π(4.0 Hz) = 8π rad/s. The amplitude is given as 4.0 cm, which corresponds to A = 0.04 m. Plugging these values into the formula, we get:

v = (8π rad/s)(0.04 m) sin((8π rad/s)(3 s))

v ≈ 6.283 m/s

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The speed of the mass when[tex]\( t = 3 \) s is \( v = 8\pi \)[/tex] cm/s.

To find the speed of the mass at any given time \( t \) during its simple harmonic motion, we can use the following equation for velocity in SHM:

[tex]\[ v(t) = -A\omega \sin(\omega t + \phi) \][/tex]using the relation [tex]\( \omega = 2\pi f \).[/tex]Substituting the given frequency:

[tex]\[ \omega = 2\pi (4.0 \, \text{Hz}) = 8\pi \, \text{rad/s} \][/tex]

 The phase constant \( \phi \) is 0 because the timer is started when the mass is at its maximum displacement, which corresponds to \[tex]( x = A \)[/tex]when [tex]\( t = 0 \).[/tex]

 Now, we can calculate the velocity at [tex]\( t = 3 \)[/tex] s:

[tex]\[ v(3) = -A\omega \sin(\omega \cdot 3 + \phi) \][/tex]

[tex]\[ v(3) = -(4.0 \, \text{cm})(8\pi \, \text{rad/s}) \sin(8\pi \, \text{rad/s} \cdot 3 \, \text{s} + 0) \][/tex]

[tex]\[ v(3) = -32\pi \, \text{cm/s} \sin(24\pi \, \text{rad}) \][/tex]

Since \[tex]( \sin(24\pi \, \text{rad}) = 0 \)[/tex] (because [tex]\( 24\pi \)[/tex] is a multiple of[tex]\( 2\pi \)[/tex]), the velocity at [tex]\( t = 3 \)[/tex] s is:

[tex]\[ v(3) = -32\pi \, \text{cm/s} \cdot 0 = 0 \, \text{cm/s} \][/tex]

 However, this result corresponds to the instantaneous speed at a point where the mass changes direction, and thus its speed is momentarily zero. To find the speed at a point where the mass is actually moving, we need to consider the time just before or after this point of maximum displacement.

 Now, we calculate the velocity at [tex]( t = 3 - \Delta t \):[/tex]

[tex]\[ v(3 - \Delta t) = -A\omega \sin(\omega (3 - \Delta t) + \phi) \][/tex]

[tex]\[ v(3 - \Delta t) = -(4.0 \, \text{cm})(8\pi \, \text{rad/s}) \sin(8\pi (3 - \frac{1}{8\pi}) \, \text{rad}) \][/tex]

[tex]\[ v(3 - \Delta t) = -32\pi \, \text{cm/s} \sin(24\pi - 1) \, \text{rad}) \][/tex]

[tex]\[ v(3 - \Delta t) = -32\pi \, \text{cm/s} \sin(-1) \, \text{rad}) \][/tex]

 Since [tex]\( \sin(-1) \approx -1 \)[/tex] for a very small angle in radians:

[tex]\[ v(3 - \Delta t) = -32\pi \, \text{cm/s} \cdot (-1) \][/tex]

[tex]\[ v(3 - \Delta t) = 32\pi \, \text{cm/s} \][/tex]

 Therefore, just before the mass reaches its maximum displacement at [tex]\( t = 3 \) s[/tex], its speed is [tex]\( 32\pi \)[/tex] cm/s.

Answer:

166 666 666.7 years

Explanation:

We start the question by making the units uniform. We are told that the continents move at 3 cm/year = 0.03 m/year.

We are also told that the continents are now 5000 km = 5 000 000 m apart

So to calculate the time it took for them to be this far apart

t = distance/speed

t = 5 000 000 m/(0.03 m/year) = 166 666 666.7 years

Answer:

[tex]Re=1992.24[/tex]

Explanation:

Given:

vertical height of oil coming out of pipe, [tex]h=25\ m[/tex]

diameter of pipe, [tex]d=0.1\ m[/tex]

length of pipe, [tex]l=50\ m[/tex]

density of oil, [tex]\rho = 900\ kg.m^{-3}[/tex]

viscosity of oil, [tex]\mu=1\ Pa.s[/tex]

Now, since the oil is being shot verically upwards it will have some initial velocity and will have zero final velocity at the top.

Using the equation of motion:

[tex]v^2=u^2-2gh[/tex]

where:

v = final velocity

u = initial velocity

Putting the respective values:

[tex]0^2=u^2-2\times 9.8\times 25[/tex]

[tex]u=22.136\ m.s^{-1}[/tex]

For Reynold's no. we have the relation as:

[tex]Re=\frac{\rho.u.d}{\mu}[/tex]

[tex]Re=\frac{900\times 22.136\times 0.1}{1}[/tex]

[tex]Re=1992.24[/tex]

Answer:

[tex]fraction = 0.11[/tex]

Explanation:

Linear kinetic energy of the bicycle is given as

[tex]KE = \frac{1}{2}mv^2[/tex]

[tex]K_1 = \frac{1}{2}(13) v^2[/tex]

[tex]K_1 = 6.5 v^2[/tex]

Now rotational kinetic energy of the wheels

[tex]K_2 = 2(\frac{1}{2}(I)(\omega^2))[/tex]

[tex]K_2 = (mR^2)(\frac{v^2}{R^2})[/tex]

[tex]K_2 = mv^2[/tex]

[tex]K_2 = 3.1 v^2[/tex]

now kinetic energy of the rider is given as

[tex]K_3 = \frac{1}{2}Mv^2[/tex]

[tex]K_3 = \frac{1}{2}(38) v^2 [/tex]

[tex]K_3 = 19 v^2[/tex]

So we have

[tex]fraction = \frac{K_2}{K_1 + K_2 + K_3}[/tex]

[tex]fraction = \frac{3.1 v^2}{6.5 v^2 + 3.1 v^2 + 19 v^2}[/tex]

[tex]fraction = 0.11[/tex]

Answer:

Radius, r = 1.5 meters

Explanation:

It is given that,

When an object is moving in an amusement park it will exert centripetal force. The centripetal force acting on the person, F = 560 N

Mass of the person, m = 83 kg

Speed of the wall, v = 3.2 m/s

The centripetal force acting on the person is given by :

[tex]F=\dfrac{mv^2}{r}[/tex]

[tex]r=\dfrac{mv^2}{F}[/tex]

[tex]r=\dfrac{83\times (3.2)^2}{560}[/tex]

r = 1.51 meters

or

r = 1.5 meters

So, the radius of the chamber is 1.5 meters.

Final answer:

Using the given formula and solving for the constant using the temperature 10 minutes after pouring, the temperature of the coffee 30 minutes after it was poured is found to be 60 degrees Fahrenheit.

Explanation:

The question asks us to determine the temperature of coffee 30 minutes after it was poured, given the formula F = 120(2–at) + 60, where F is in degrees Fahrenheit and t is the time in minutes since the coffee was poured. Since the temperature of the coffee 10 minutes after it was poured was 120 degrees Fahrenheit, we first need to find the value of the constant a using the given formula. Substituting F with 120 and t with 10 gives us 120 = 120(2-10a) + 60. Solving for a, we find that a is equal to 0.1. Now, to find the temperature of the coffee 30 minutes after it was poured, we substitute t with 30 in the formula, getting F = 120(2-0.1*30) + 60, which simplifies to F = 60 degrees Fahrenheit.

Answer:

the volume is 0.253 cm³

Explanation:

The pressure underwater is related with the pressure in the surface through Pascal's law:

P(h)= Po + ρgh

where Po= pressure at a depth h under the surface (we assume = 1atm=101325 Pa) , ρ= density of water ,g= gravity , h= depth at h meters)

replacing values

P(h)= Po + ρgh = 101325 Pa + 1025 Kg/m³ * 9.8 m/s² * 20 m = 302225 Pa

Also assuming that the bubble behaves as an ideal gas

PV=nRT

where

P= absolute pressure, V= gas volume ,n= number of moles of gas, R= ideal gas constant , T= absolute temperature

therefore assuming that the mass of the bubble is the same ( it does not absorb other bubbles, divides into smaller ones or allow significant diffusion over its surface) we have

at the surface) PoVo=nRTo

at the depth h) PV=nRT

dividing both equations

(P/Po)(V/Vo)=(T/To)

or

V=Vo*(Po/P)(T/To) = 0.80 cm³ * (101325 Pa/302225 Pa)*(277K/293K) = 0.253 cm³

V = 0.253 cm³

Answer:

Explanation:

Let the velocity of rocket case and payload after the separation be v₁ and v₂ respectively. v₂ will be greater because payload has less mass so it will be fired with greater speed .

v₂ - v₁ = 910

Applying law of conservation of momentum

( 250 + 100 ) x 7700 = 250 v₁ + 100 v₂

2695000 = 250 v₁ + 100 v₂

2695000 = 250 v₁ + 100 ( 910 +v₁ )

v₁ = 7440 m /s

v₂ = 8350 m /s

Total kinetic energy before firing

= 1/2 ( 250 + 100 ) x 7700²

= 1.037575 x 10¹⁰ J

Total kinetic energy after firing

= 1/2 ( 250 x 7440² + 100 x 8350² )

= 1.0405325 x 10¹⁰ J

The kinetic energy has been increased due to addition of energy generated in firing or explosion which separated the parts or due to release of energy from compressed spring.

Answer:

Equilibrium temperature will be [tex]T=52.2684^{\circ}C[/tex]

Explanation:

We have given weight of the lead m = 2.61 gram

Let the final temperature is T

Specific heat of the lead c = 0.128

Initial temperature of the lead = 11°C

So heat gain by the lead = 2.61×0.128×(T-11°C)

Mass of the water m = 7.67 gram

Specific heat = 4.184

Temperature of the water = 52.6°C

So heat lost by water = 7.67×4.184×(T-52.6)

We know that heat lost = heat gained

So [tex]2.61\times 0.128\times (T-11)=7.67\times 4.184\times (52.6-T)[/tex]

[tex]0.334T-3.67=1688-32.031T[/tex]

[tex]T=52.2684^{\circ}C[/tex]

By equating the heat gained by the lead to the heat lost by the water, the final temperature is calculated to be approximately 41.5°C.

To determine the final temperature of the lead weight and water, we can use the principle of calorimetry, which states that the heat gained by one object is equal to the heat lost by another object in a closed system.

Given:

Mass of lead[tex](m_l_e_a_d) = 2.61 g[/tex]

Specific heat capacity of lead [tex](c_l_e_a_d)[/tex] = 0.128 J/g°C

Initial temperature of lead[tex](T_l_e_a_d_i_n_i_t_i_a_l)[/tex]) = 11.1°C

Mass of water [tex](m_w_a_t_e_r)[/tex]= 7.67 g

Specific heat capacity of water[tex](c_w_a_t_e_r)[/tex] = 4.184 J/g°C

Initial temperature of water [tex](T_w_a_t_e_r_i_n_i_t_i_a_l)[/tex] = 52.6°C

Let T_final be the final equilibrium temperature.

Heat gained by lead [tex](Q_lead) = m_lead * c_lead * (T_final - T_lead_initia[/tex]l)

Heat lost by water [tex](Q_water) = m_water * c_water * (T_water_initial - T_final)[/tex]

Since the system is thermally insulated, Q_lead = -Q_water

[tex]m_lead * c_lead * (T_final - T_lead_initial) = - m_water * c_water * (T_water_initial - T_final)[/tex]

Solving for T_final:

[tex]T_final = (m_lead * c_lead * T_lead_initial + m_water * c_water * T_water_initial) / (m_lead * c_lead + m_water * c_water)[/tex]

Plugging in the values:

T_final = (2.61 g * 0.128 J/g°C * 11.1°C + 7.67 g * 4.184 J/g°C * 52.6°C) / (2.61 g * 0.128 J/g°C + 7.67 g * 4.184 J/g°C)

T_final ≈ 41.5°C

Therefore, the final temperature of both the lead weight and the water at thermal equilibrium is approximately 41.5°C.

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Answer:

-No

Explanation:

Given that

Mass of box = 100 kg

Coefficient of friction ,μ= 0.23

We know that friction force depends on the normal force acts on the box

Fr= μ N

When we pull the box then :

Normal force N= mg

Friction force Fr=  μ mg                  

When we push the box :

Lets take pushing force = F

θ =Angle make by pushing force from the vertical line

Normal force

N = mg + F cosθ

Fr= μ ( mg + F cosθ )

The friction force is more when we push the box.That is why this is not easier to push the box.

Therefore answer is ---No

Answer:

[tex]\theta = 20.98 degree[/tex]

Explanation:

As we know that the speed of the sound is given as

[tex]v = 332 + 0.6 t[/tex]

now at t = 273 k = 0 degree

[tex]v = 332 m/s[/tex]

so we have

[tex]a sin\theta = N\lambda[/tex]

[tex]a sin\theta = N(\frac{v_1}{f})[/tex]

now when temperature is changed to 313 K we have

[tex]t = 313 - 273 = 40 degree[/tex]

now we have

[tex]v = 332 + (0.6)(40)[/tex]

[tex]v_2 = 356 m/s[/tex]

[tex]a sin\theta' = N(\frac{v_2}{f})[/tex]

now from two equations we have

[tex]\frac{sin19.5}{sin\theta} = \frac{332}{356}[/tex]

so we have

[tex]sin\theta = 0.358[/tex]

[tex]\theta = 20.98 degree[/tex]

The diffraction angle of sound from a loudspeaker changes with temperature because the speed of sound varies with temperature. The higher the temperature, the faster the sound travels, resulting in a longer wavelength and, consequently, a larger diffraction angle.

The question involves the concept of wave diffraction, specifically as it pertains to sound waves exiting a diffraction horn loudspeaker. In this scenario, the diffraction angle θ of sound is observed to change with temperature, due to the dependence of the speed of sound on air temperature. To find the new diffraction angle when the temperature rises from 273 K to 313 K, we need to understand that the speed of sound in air increases with temperature. The formula v = 331.4 + 0.6T (with T in ℃) can be used to calculate the speed of sound at different temperatures, but here we'll use Kelvin and recall that v = 331.4 + 0.6(T - 273) to scale it correctly.

At 273 K, the speed of sound is approximately 331.4 m/s, and at 313 K, it is higher. Assuming the frequency of the sound remains constant, and using the relationship v = fλ (where v is the speed of sound, f is the frequency, and λ is the wavelength), the wavelength will increase with the speed of sound. According to the diffraction formula for a single slit, sin(θ) = mλ/d (where m is the order of the minimum, λ is the wavelength, and d is the width of the slit), if wavelength λ increases while d stays the same, the angle θ must also increase to satisfy the equation for the first minimum (m=1). Thus, the diffraction angle on a summer day when the temperature is higher would be larger than 19.5°.

The solution to this problem requires the understanding of the concepts of specific heat and heat of fusion. It involves three parts: heating up the ice to the melting point, melting the ice, and as the melted ice is absorbed by the water, it cools down the water's temperature. The heat absorbed by the ice is equal to the heat lost by the water.

The subject of this question falls under Physics, specifically in the topic of Thermodynamics. The problem involves the concept of specific heat, and heat of fusion. We need to account for three parts: heating the ice from -5°C to 0°C, melting the ice at 0°C, and the absorbed heat by the water from the melting ice.

1. Firstly, calculate the energy needed to heat the ice from -5°C to 0°C (phase change has not yet happened). Use Q = m*c*ΔT, where m is the mass of ice, c is the specific heat of ice, and ΔT is the change in temperature.

2. Secondly, calculate the energy required to melt the ice. This time, use Q = m*Lf, where m is the mass of ice and Lf is the heat of fusion.

3. Lastly, as the melted ice (now at 0°C) is absorbed by the water, it will cool down the water's temperature. The heat absorbed by the ice is equal to the heat lost by the water.

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Answer:

seismograph or seismometer.

Explanation:

Seismograph is an instrument used to detect and record earthquakes.

It has a mass attached to a fixed base. During an earthquake, the base moves and the mass does not. The motion of the base with respect to the mass is commonly transformed into an electrical voltage. The electrical voltage is recorded. This record is proportional to the motion of the seismometer mass relative to the earth, but it can be mathematically converted to a record of the absolute motion of the ground.

Answer:

Total load = 2999.126 kg

Explanation:

Let the spring constant of the shock absorber be k.

We know that the force applied on a spring is directly proportional to elongated length and the constant of proportionality is called spring constant.

Thus

Force, F = kx

where,

x = elongation = 9.1 cm 0.091 m

mass of the people, m = 127 kg

F = weight of the people = mg = 127 x 9.8 = 1244.6 N

substituting these values in the first equation,

1244.6 = k x 0.091

thus, k = 13,676.923 N/m

Now we know that the time period, T of an oscillating spring with a load of mass m is

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

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

thus,

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

T = 1.66s

substituting these values in the equation,

[tex]m =13,676.923\frac{1.66^{2} }{4\pi ^{2}  }[/tex]

m = 2999.126 kg

The mass attached to the spring must be 0.72 kg

Explanation:

The frequency of vibration of a spring-mass system is given by:

[tex]f=\frac{1}{2\pi} \sqrt{\frac{k}{m}}[/tex] (1)

where

k is the spring constant

m is the mass attached to the spring

We can find the spring constant by using Hookes' law:

[tex]F=kx[/tex]

where

F is the force applied on the spring

x is the stretching of the spring

When a mass of m = 2.82 kg is applied to the spring, the force applied is the weight of the mass, so we have

[tex]mg=kx[/tex]

and using [tex]g=9.8 m/s^2[/tex] and [tex]x=0.0177 m[/tex], we find

[tex]k=\frac{mg}{x}=\frac{(2.82)(9.8)}{0.0177}=1561.3 N/m[/tex]

Now we want the frequency of vibration to be

f = 7.42 Hz

So we can rearrange eq.(1) to find the mass m that we need to attach to the spring:

[tex]m=\frac{k}{(2\pi f)^2}=\frac{1561.3}{(2\pi (7.42))^2}=0.72 kg[/tex]

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Answer:

1. Rotating 2. Revolving

Explanation:

Just did it on edge 2021

Answer:

The answer is to insulate Earth's surface temperature.

Explanation:

Clouds inhibit the outflow of terrestrial radiation. This acts to insulate Earth's surface temperature, keeping it warmer at night and cooler in the day.

Answer:

Neither Technician A nor Technician B.

Explanation:

chrome is very expensive and we do need to check the ring before reassembling engine.

Answer:

T= 210.15 N

F= 75.31 N

Explanation:

Let the tension in string be T newton.

According to the question

⇒T×cos21°= mg

⇒T= mg/cos21°

⇒T=20×9.81/cos21

⇒T= 210.15 N

now, the magnitude of horizontal force

F= Tsin21°

⇒F= 210.15×sin21°

=75.31 N

Answer:

D = 71.20 km

Explanation:

In this case, let's get the exercise by parts.

Part 1 would be the first displacement which is 30 km. However, it's displacing north to east 30°, so in this case, if we trace a line of this, it should be a diagonal. So we need to get the value of this displacement in the x and y axis.

Dx1 = 30 cos 30° = 25.98 km

Dy1 = 30 sin 30° = 15 km

Doing the same thing but in the second travel we have:

Dx2 = 70 cos 50° = 45 km

Dy2 = 70 sin 50° = 53.62 km

Now we need to know how was the displacement in both axis. In travel 2, the plane is moving north to west, so, in the x axis, is moving in the opposite direction, therefore:

Dx = -45 + 25.98 = -19.02 km

In the y axis, is moving upward so:

Dy = 53.62 + 15 = 68.62 km

Finally to get the resultant displacement:

D = √Dx² + Dy²

Replacing:

D = √(-19.02)² + (68.62)²

D = √5070.46

D = 71.20 km

Answer:[tex]24.70 ^{\circ}C[/tex]

Explanation:

Given

mass of lead piece [tex]m_l=234 gm\approx 0.234 kg[/tex]

mass of water in calorimeter [tex]m_w=611 gm\approx 0.611 kg[/tex]

Initial temperature of water [tex]T_w=24^{\circ}C[/tex]

Initial temperature of lead piece [tex]T_l=24^{\circ}C[/tex]

we know heat capacity of lead and water are [tex]125.604 J/kg-k[/tex] and [tex]4.184 kJ/kg-k[/tex] respectively

Let us take [tex]T ^{\circ}C[/tex] be the final temperature of the system

Conserving energy

heat lost by lead=heat gained by water

[tex]m_lc_l(T_l-T)=m_wc_w(T-T_w)[/tex]

[tex]0.234\times 125.604(86-T)=0.611\times 4.184\times 1000(T-24)[/tex]

[tex]86-T=\frac{0.611\times 4.184\times 1000}{29.391}(T-24)[/tex]

[tex]86-T=86.97T-2087.49[/tex]

[tex]T=\frac{2173.491}{87.97}=24.70^{\circ}C[/tex]

Answer and Explanation:

Tropical storm:

The point at which the tropical depression intensifies and can sustain a maximum wind speed in the range of 39-73 mph, it is termed as a tropical storm.

Hurricane:

A hurricane is that type of tropical cyclone which includes with it high speed winds and thunderstorms and the intensity of the sustained wind speed is 74 mph or more than this.

Hurricanes are the most intense tropical cyclones.

The only difference between the tropical storm and hurricane is in the intensity.

A tropical storm becomes a hurricane when its sustained wind speeds exceed 74 miles per hour, with both systems featuring low pressure centers and heavy rains, but hurricanes pose a greater threat due to their higher wind speeds.

The primary difference between a tropical storm and a hurricane lies in their wind speeds. When a storm's sustained winds reach between 39 to 73 miles per hour, it is classified as a tropical storm. Once those winds exceed 74 miles per hour, the storm is then classified as a hurricane. Both types of storm systems form over warm ocean waters, typically warmer than 80 °F and involve low pressure centers, strong winds, and heavy rains. However, a hurricane's increased wind speed is what gives it the potential to cause significantly more damage upon making landfall.

The Coriolis force is responsible for the cyclonic rotation of these storms—with hurricanes in the Northern Hemisphere rotating counterclockwise and those in the Southern Hemisphere rotating clockwise. The terms hurricane, typhoon, and tropical storm are regional names for these cyclones. Regardless of the name, these storms are dangerous due to the combination of high winds, heavy rains, and consequential risks such as flooding and structural damage.

The required height of the tin can is given by the height at the minimum value of the surface area function of the tin can

Reason:

The given information on the tin can are;

Volume of the tin can = 16·π in.³

The amount of tin to be used = Minimum amount

The height of the can in inches required

Solution;

Let h, represent the height of the can, we have;

The surface area of the can, S.A. = 2·π·r² + 2·π·r·h

The volume of the can, V = π·r²·h

Where;

r = The radius of the tin can

h = The height of the tin can

Which gives;

16·π = π·r²·h

16 = r²·h

[tex]h = \dfrac{16}{r^2}[/tex]

Which gives;

[tex]S.A. = 2 \cdot \pi \cdot r^2 + 2 \cdot \pi \cdot r \cdot \dfrac{16}{r^2} = 2 \cdot \pi \cdot r^2 + \pi \cdot \dfrac{32}{r}[/tex]

When the minimum amount of tin is used, we have;

[tex]\dfrac{d(S.A.)}{dx} =0 = \dfrac{d}{dx} \left( 2 \cdot \pi \cdot r^2 + \pi \cdot \dfrac{32}{r} \right) = \dfrac{4 \cdot \pi \cdot \left (r^3-8 \right)}{r^2}[/tex]

Therefore;

[tex]\dfrac{4 \cdot \pi \cdot \left (r^3-8 \right)}{r^2} = 0[/tex]

4·π·(r³ - 8) = r² × 0

r³ = 8

r = 2

The radius of the tin can, r = 2 inches

The

[tex]h = \dfrac{16}{2^2} = 4[/tex]

The height of the tin can, h = 4 inches

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The required height of can is 4 inches.

Given data:

The volume of cylindrical tin with top and bottom is, [tex]V = 16 \pi \;\rm in^{3}[/tex].

The volume of tin is,

[tex]V = \pi r^{2}h\\16 \pi = \pi r^{2}h\\[/tex]

here, r is the radius of can and h is the height of can.

[tex]16 =r^{2}h\\\\h=\dfrac{16}{r^{2}}[/tex]

The surface area of tin is,

[tex]SA = 2\pi r(r+h)\\SA = 2\pi r(r+\dfrac{16}{r^{2}})\\SA = 2\pi r^{2}+\dfrac{32 \pi}{r})[/tex]

For minimum amount of tin used, we have,

[tex]\dfrac{d(SA)}{dr} =0\\\dfrac{ d(2\pi r^{2}+\dfrac{32 \pi}{r})}{dr} = 0\\4\pi r-\dfrac{32 \pi}{r^{2}}=0\\4\pi r=\dfrac{32 \pi}{r^{2}}\\r^{3}=8\\r =2[/tex]

So, height of can is,

[tex]h=\dfrac{16}{2^{2}}\\h=4[/tex]

Thus, the required height of can is 4 inches.

Learn more about the curved surface area here:

https://brainly.com/question/16140623?referrer=searchResults

Answer:12.12 m/s

Explanation:

Given

mass of child [tex]m=49 kg[/tex]

height of hill [tex]h=7.5 m[/tex]

inclination [tex]\theta =26^{\circ}[/tex]

Conserving Energy at top and bottom Point of hill

Potential Energy at Top =Kinetic Energy at bottom

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

[tex]v=\sqrt{2gh}[/tex]

[tex]v=\sqrt{2\times 9.8\times 7.5}[/tex]

[tex]v=\sqrt{147}[/tex]

[tex]v=12.12 m/s[/tex]

Answer:

[tex]v\approx 12.129\,\frac{m}{s}[/tex]

Explanation:

The final speed of the child-sled system is determined by means of the Principle of Energy Conservation:

[tex]U_{1} + K_{1} = U_{2} + K_{2}[/tex]

[tex]K_{2} = (U_{1}-U_{2})+K_{1}[/tex]

[tex]\frac{1}{2}\cdot m \cdot v^{2} = m\cdot g \cdot \Delta h[/tex]

[tex]v = \sqrt{2\cdot g \cdot \Delta h}[/tex]

[tex]v = \sqrt{2\cdot\left(9.807\,\frac{m}{s^{2}} \right)\cdot (7.50\,m)}[/tex]

[tex]v\approx 12.129\,\frac{m}{s}[/tex]

Explanation:

Given that, Speed of the automobile, v = 55 km/h = 15.27 m/s

Diameter of the tire, d = 77 cm

Radius, r = 0.385 m

(a) Let [tex]\omega[/tex] is the angular speed of the tires about their axles.

The relation between the linear speed and the angular speed is given by :

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

[tex]\omega=\dfrac{v}{r}[/tex]

[tex]\omega=\dfrac{15.27\ m/s}{0.385\ m}[/tex]

[tex]\omega=39.66\ rad/s[/tex]

(b) Number of revolution,

[tex]\theta=38\ rev=238.76\ radian[/tex]

Final angular speed of the car, [tex]\omega_f=0[/tex]

Initial angular speed, [tex]\omega_i=39.66\ rad/s[/tex]

Let [tex]\alpha[/tex] is the angular acceleration of the car. Using third equation of rotational kinematics as :

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

[tex]\alpha =\dfrac{-\omega_i^2}{2\theta}[/tex]

[tex]\alpha =\dfrac{-(39.66)^2}{2\times 238.76}[/tex]

[tex]\theta=-3.29\ rad/s^2[/tex]

(c) Let d is the distance covered by the car during the braking. It is given by :

[tex]d=\theta\times r[/tex]

[tex]d=238.76\times 0.385[/tex]

d = 91.92 meters

Answer:

See explanation

Explanation:

To do this, you need to use energy conservation.  The sum of kinetic and potential energies is the same at all points along the path so, you can write the expression like this:

1/2mv1² + mgh1 = (1/2)mv2² + mgh2

Where:

v1 = 0 because it's released from rest

h1 = 18.6 m

v2 = speed we want to solve.

h2 = height at point A. In this case, you are not providing the picture or data, so, I'm going to suppose a theorical data to solve this. Let's say h2 it's 12 m.

Now, let's replace the data in the above expression (assuming h2 = 12 m). Also, remember that we don't have the mass of the bead, but we don't need it to solve it, because it's simplified by the equation, therefore the final expression is:

1/2v1² + gh1 = 1/2v2² + gh2  

Replacing the data we have:

1/2*(0) + 9.8*18.6 = 1/2v² + 9.8*12

182.28 = 117.6 + 1/2v²

182.28 - 117.6 = v²/2

64.68 * 2 = v²

v = √129.36

v = 11.37 m/s

Now, remember that you are not providing the picture to see exactly the value of height at point A. With that picture, just replace the value in this procedure, and you'll get an accurate result.

Answer:

positive reinforcer; negative reinforcer

Explanation:

positive reinforcer; negative reinforcer

Receiving delicious food can be seen as a positive reinforcer while escaping electric shock can be seen as a negative reinforcer.

Answer:

Explanation:

The tendency of the automobile running on a circular path at high speed to turn towards left or right around one of its wheels , is due to torque by centripetal force acting at its centre of mass about that wheel .

Suppose the automobile tends to turn in clockwise direction about its wheel on the outer edge . A rotational angular momentum is created . To counter this effect , we can take the help of a rotating wheel or flywheel . We shall have to keep its rotation in anticlockwise direction so that it can create rotational angular momentum in direction opposite to that created by centrifugal force on fast moving automobile. Each of them will nullify the effect of the other . In this way , rotating flywheel will save the automobile from turning upside down .

Final answer:

To counteract the tendency for a car to roll over when rounding curves at high speed, a flywheel should be mounted horizontally and perpendicular to the travel direction, with a sense of rotation that produces gyroscopic precession force downward on the inside wheels.

Explanation:

When an automobile rounds a curve at high speeds, the loading on the wheels changes, with the tendency to lighten the load on the inside wheels, which can lead to the car rolling over. To combat this, a large spinning flywheel can help equalize the loading on the wheels. The flywheel should be mounted such that its axis of rotation is horizontal and perpendicular to the direction of the car's travel. The sense of rotation should be such that when the car turns, the gyroscopic precession of the flywheel produces a force that pushes down on the inside wheels. This force counteracts the effect of the weight transfer to the outside wheels, reducing the risk of the car tipping over.

This concept exploits the conservation of angular momentum and the phenomenon known as gyroscopic precession. In a closed system, any attempt to change the direction of the axis of rotation of the spinning flywheel produces a force perpendicular to the direction of the applied force (precession). If the top of the flywheel is tilted outward, the precession force acts downward on the side of the flywheel closest to the inside of the turn, creating a stabilizing effect on the inside wheels.

Answer:

[tex]x=0.0049\ m= 4.9\ mm[/tex]

[tex]d=0.01153\ m=11.53\ mm[/tex]

Explanation:

Given:

(a)

Let x be the maximum distance of stretch without moving the mass.

The spring can be stretched up to the limiting frictional force 'f' till the body is stationary.

[tex]f=k.x[/tex]

[tex]\mu_s.N=k.x[/tex]

where:

N = m.g = the normal reaction force acting on the body under steady state.

[tex]0.1\times (9.8\times 0.75)=150\times x[/tex]

[tex]x=0.0049\ m= 4.9\ mm[/tex]

(b)

Now, according to the question:

Let d be the total distance the object travels before stopping.

Now, the energy stored in the spring due to vibration of amplitude:

[tex]U=\frac{1}{2} k.A^2[/tex]

This energy will be equal to the work done by the kinetic friction to stop it.

[tex]U=F_k.d[/tex]

[tex]\frac{1}{2} k.A^2=\mu_k.N.d[/tex]

[tex]0.5\times 150\times 0.0098^2=0.0850 \times 0.75\times 9.8\times d[/tex]

[tex]d=0.01153\ m=11.53\ mm[/tex]

is the total distance does it travel before stopping.

To find the maximum distance the spring can be stretched without moving the mass, calculate the force of static friction and equate it to the spring force.

(a) How far can the spring be stretched without moving the mass?

To find the maximum distance the spring can be stretched without moving the mass, we need to calculate the force of static friction. The force of static friction can be calculated using the equation:

Fs = μs * m * g

where μs is the coefficient of static friction, m is the mass, and g is the acceleration due to gravity.

Once we have the force of static friction, we can equate it to the spring force:

Fs = k * x

where k is the force constant of the spring and x is the maximum distance the spring is stretched without moving the mass.