During heavy exercise, the body pumps 2.00 L of blood per minute to the surface, where it is cooled by 2.00ºC . What is the rate of heat transfer from this forced convection alone, assuming blood has the same specific heat as water and its density is 1050 kg/m

Answer:

(a) 0.05 Am^2

(b) 1.85 x 10^-3 Nm

Explanation:

width, w = 10 cm = 0.1 m

length, l = 20 cm = 0.2 m

Current, i = 2.5 A

Magnetic field, B = 0.037 T

(A) Magnetic moment, M = i x A

Where, A be the area of loop

M = 2.5 x 0.1 x 0.2 = 0.05 Am^2

(B) Torque, τ = M x B x Sin 90

τ = 0.05 x 0.037 x 1

τ = 1.85 x 10^-3 Nm

Answer:

Orbital

Explanation:

The orbitals is the area or region outside the nucleus where an electron can most likely be found

Answer:

s sublevel

Explanation:

IS a orbital

Answer:

1. True

2. False

3. True

Explanation:

Semiconductors are the materials whose conduction is intermediary to that of the conductor and insulator and can be made more conducting by mixing impurities or we call it as doping.

In semiconductors, the concentration of the holes in the valence band and the electrons in the conduction band is usually very small and can be varied noticeably by doping.

(1) After being excited by a photon, an electron can move to the conduction band as a hole is created in the valence band band and this hole moves to the p-side of the p-n junction.

(2) It is not the n-type but the p-type semiconductor with missing electrons or holes in the valence band of lower energy.

(3) Electrons are free to move in a partially filled conduction band and valence band.

There is no movement of electrons along the band which are completely full or totally vacant.

Answer:

Circuit breaker

Explanation:

Circuit breaker is the devise designed to protect the circuit from over current by opening the circuit automatically. Breaker can also be off manually by toggle switch. Earlier fuses were used but circuit breakers have replaced them. Fuse and circuit breakers operates differently. in case of overloading fuses blown off and opens the circuit while circuit breaker opens the circuit automatically without being blown off.

Final answer:

A circuit breaker is a safety device in a circuit that opens the circuit when an overcurrent occurs. It is faster than fuses and can be reset.

Explanation:

A circuit breaker is a device designed to open and close a circuit by non-automatic means and to open the circuit automatically on a predetermined overcurrent without damage to itself when properly applied within its rating. It acts as a safety device that switches off an appliance if the current in the circuit is too strong.

Circuit breakers are rated for a maximum current and can be reset, reacting much faster than fuses. They consist of components like bimetallic strips that respond to heat to break the electrical connection in the circuit.

Fnet = ma = m*0.23g = mgsinΘ - Ff

where Ff is the friction force upslope.

net torque τ = Ff * r,

but also τ = I*α =(2/3)mr² * a/r = 2mra/3 = 2mr(0.23g)/3 = 0.46mrg/3

Then Ff * r = 0.46mrg / 3

Ff = 0.46mg/3 → put this into net force equation:

m*0.23g = mgsinΘ - 0.46mg/3 → mg cancels; multiply through by 3

0.69 = 3sinΘ - 0.46

3sinΘ = 1.15

sinΘ = 1.15/3

Θ = arctan(1.15/3) = 22.54 º

The tangential velocity of the car is 124.5 m/s

Explanation:

The centripetal acceleration of an object in uniform circular motion is given by:

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

where

a is the acceleration

v is the tangential velocity

r is the radius of the circle

For the car in this problem, we have:

[tex]a=100 m/s^2[/tex] is the centripetal acceleration

r = 155 m is the radius of the turn

Solving for v, we find the tangential velocity of the car:

[tex]v=\sqrt{ar}=\sqrt{(100)(155)}=124.5 m/s[/tex]

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The angular speed of the merry-go-round after the child gets on is approximately 1.28 rev/s.

To find the angular speed of the merry-go-round after the child gets onto it, we can use the principle of conservation of angular momentum. The total angular momentum before the child gets on should be equal to the total angular momentum after.

The angular momentum (L) of the merry-go-round before the child gets on is given by:

L_initial = I_initial * ω_initial

Where:

I_initial = Moment of inertia of the merry-go-round

ω_initial = Initial angular speed

The angular momentum (L) of the child after getting on is given by:

L_child = I_child * ω_final

Where:

I_child = Moment of inertia of the child

ω_final = Final angular speed

Since angular momentum is conserved, L_initial = L_child.

The moment of inertia (I) of a solid disk like a merry-go-round can be calculated using the formula:

I = (1/2) * m * r^2

Where:

m = Mass

r = Radius

For the merry-go-round:

I_initial = (1/2) * 120 kg * (1.80 m)^2

I_child = (1/2) * 37.5 kg * (1.80 m)^2

Now, we can set up the equation for conservation of angular momentum:

I_initial * ω_initial = I_child * ω_final

Solve for ω_final:

(1/2) * 120 kg * (1.80 m)^2 * ω_initial = (1/2) * 37.5 kg * (1.80 m)^2 * ω_final

Now, we can cancel out some terms:

120 * ω_initial = 37.5 * ω_final

Now, solve for ω_final:

ω_final = (120 * ω_initial) / 37.5

Substitute the given value for ω_initial (0.400 rev/s):

ω_final = (120 * 0.400 rev/s) / 37.5

ω_final ≈ 1.28 rev/s

So, the angular speed of the merry-go-round after the child gets on is approximately 1.28 rev/s.

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Answer:

 [tex]v = 11.29\ m/s[/tex]

Explanation:

given,

bead is released from the height = 24.5 m from bottom

radius of the loop = 9 m

acceleration due to gravity = 9.8 m/s

A be the top most point of the loop

Difference of elevation from the top

     H = 24.5 - 2 x r

     H = 24.5 - 2 x 9

     H = 24.5 - 18

     H = 6.5 m

now using conservation of energy

   KE = PE

 [tex]\dfrac{1}{2}mv^2 = m g H[/tex]

 [tex]v^2 = 2 g H[/tex]

 [tex]v = \sqrt{2 g H}[/tex]

 [tex]v = \sqrt{2 \times 9.8 \times 6.5}[/tex]

 [tex]v = \sqrt{127.4}[/tex]

 [tex]v = 11.29\ m/s[/tex]

speed at point A is equal to  [tex]v = 11.29\ m/s[/tex]

Answer:Decrease Wavelength

Explanation:

To remove the Electrons from a metal surface the minimum amount of energy is provided i.e. threshold Energy is Provided to remove the electrons.

If no photoelectrons are emitted then it can be Concluded that Energy is less than threshold Energy.

Energy is inversely Proportional to the wavelength of light thus to increase Energy we have to decrease the wavelength of monochromatic emission.

Answer:

The current lags the potential difference by π/2 in an inductor

Explanation:

The potential difference leads to the current by  [tex]\frac{\pi}{2}[/tex]. Alternate signals such as current and voltage -in this case- are periodic, this means that this signals are repeated at fixed spaces of time. Thus, In an inductor the current lags the potential difference by [tex]\frac{\pi}{2}[/tex].

Answer:

Explanation:

The photons travel faster faster through space because photons always travel through space faster than electrons, in fact when an electron gets hitted by a photon this boost its speed

The direction of rotation of a proton can be determined using the right-hand rule in physics when considering the direction of its magnetic moment and the uniform magnetic field it's immersed in. The resultant rotation direction works to facilitate energy minimization in the system.

In the presence of an external magnetic field, the principle of physics concerning the interaction between magnetic fields and magnetic moments state that the magnetic moment of an object, in this case a proton, will align itself along the direction of the external magnetic field. This means, if the magnetic moment vector mu of the proton is directed along the positive z-axis, and the magnetic field is also along the positive z-axis, the proton's rotation will be in the direction opposite to that of the magnetic field following the right-hand rule. The proton's rotation and the magnetic field directions facilitate energy minimization in the system. Hence, the direction of rotation of a proton, based on the direction of its magnetic moment and the direction of the uniform magnetic field it is immersed in, can be determined by the right-hand rule; if your thumb points in the direction of the proton's magnetic moment (as opposed to the magnetic field), your fingers will curl in the direction of its rotation.

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Answer:

Height of slide = 0.936 m

Explanation:

Gravitational potential = Mass x Acceleration due to gravity x Height

P.E = mgh

Mass, m = 49 kg

Acceleration due to gravity, g = 9.81 m/s²

Height, h = ?

Potential energy, P.E = 450 J

Substituting

              P.E = mgh

              450 = 49 x 9.81 x h

                 h = 0.936 m

Height of slide = 0.936 m

Answer:

a) Kinetic energies

K₁ = 1.2 J

K₂ = 7.5 J

b) The bullet that has the highest kinetic energy is the one with the highest speed , v = 50 m/s , K₂ = 7.5 J

c) K₂ -K₁  = 6.3 J

Explanation:

The kinetic energy (K) is that due to the movement of a body and is calculated as follows:

K = (1/2) m*v²  (J)

Where :

m : the mass of the body ( kg)

v is the speed of the body (m/s)

Data

m₁ = m₂ = 0.006 Kg

v₁ = 20 m/s

v₂ = 50 m/s

a)Calculation of the kinetic energy

K₁ = (1/2) (m₁)*(v₁)²

K₁ = (1/2) (0.006)*(20)²

K₁ = 1.2 J

K₂= (1/2) (m₂)*(v₂)²

K₂ = (1/2) (0.006)*(50)²

K₂ = 7.5 J

b) K₂ ˃ K₁

The bullet that has the highest kinetic energy is the one with the highest speed , v = 50 m/s, K₂ = 7.5 J

c) Difference of their kinetic energies (K₂ -K₁)

K₂ -K₁  = 7.5 J - 1.2 J = 6,3 J

Answer:1.6 rad/s

Explanation:

Given

moment of Inertia  of disk [tex]I=5 kg-m^2[/tex]

radius of disc [tex]r=0.25 m[/tex]

Force [tex]F=8 N[/tex]

Torque [tex]T=I\alpha =F\cdot r[/tex]

[tex]5\times \alpha =8\times 0.25[/tex]

[tex]\alpha =0.4 rad/s^2[/tex]

using

[tex]\theta =\omega _0\times t+\frac{\alpha t^2}{2}[/tex]

[tex]\pi =0+\frac{0.4t^2}{2}[/tex]

[tex]2\pi =0.4t^2[/tex]

[tex]t^2=5\pi [/tex]

[tex]t=\sqrt{5\pi }[/tex]

[tex]t=3.96 s[/tex]

[tex]\omega =\omega _0+\alpha t[/tex]

[tex]\omega =0+0.4\times 3.96[/tex]

[tex]\omega =1.58 rad/s\approx 1.6 rad/s[/tex]

The correct answer is (3) 1.6 rad/s. The angular velocity of the disk after it has turned through half a revolution is approximately [tex]\(1.6 \, \text{rad/s}\)[/tex] .

To solve this problem, we can use the relationship between torque, rotational inertia, and angular acceleration, which is given by the equation:

[tex]\[\tau = I \alpha\][/tex]

 where \(\tau\) is the torque, \(I\) is the rotational inertia, and \(\alpha\) is the angular acceleration.

 First, we calculate the torque \(\tau\) caused by the force \(F\) applied tangentially to the rim of the disk at a distance \(r\) from the axis of rotation:

[tex]\[\tau = F \times r\][/tex]

Given that [tex]\(F = 8.0 \, \text{N}\) and \(r = 0.25 \, \text{m}\),[/tex]we have:

[tex]\[\tau = 8.0 \, \text{N} \times 0.25 \, \text{m} = 2.0 \, \text{Nm}\][/tex]

Now, we know the rotational inertia \(I\) of the disk is [tex]\(5.0 \, \text{kg} \cdot \text{m}^2\)[/tex]. Using the torque and rotational inertia, we can find the angular acceleration \(\alpha\):

[tex]\[2.0 \, \text{Nm} = 5.0 \, \text{kg} \cdot \text{m}^2 \times \alpha\][/tex]

[tex]\[\alpha = \frac{2.0 \, \text{Nm}}{5.0 \, \text{kg} \cdot \text{m}^2} = 0.4 \, \text{rad/s}^2\][/tex]

 Next, we use the equation for angular displacement \(\theta\) when an object starts from rest and has a constant angular acceleration:

[tex]\[\theta = \frac{1}{2} \alpha t^2\][/tex]

We are given that the disk has turned through half a revolution, which is \(\pi\) radians (since one full revolution is \(2\pi\) radians). We can now solve for the time \(t\) it takes to achieve this angular displacement:

[tex]\[\pi = \frac{1}{2} \times 0.4 \, \text{rad/s}^2 \times t^2\[/tex]

[tex]\[t^2 = \frac{\pi}{0.2 \, \text{rad/s}^2}\][/tex]

[tex]\[t = \sqrt{\frac{\pi}{0.2 \, \text{rad/s}^2}}\][/tex]

[tex]\[t \approx \sqrt{\frac{3.14159}{0.2}} \approx \sqrt{15.70795} \approx 3.96 \, \text{s}\][/tex]

 Finally, we can find the angular velocity \(\omega\) after this time using the equation:

[tex]\[\omega = \alpha t\][/tex]

[tex]\[\omega = 0.4 \, \text{rad/s}^2 \times 3.96 \, \text{s}\][/tex]

[tex]\[\omega \approx 1.584 \, \text{rad/s}\][/tex]

 Rounding to two significant figures, we get:

[tex]\[\omega \approx 1.6 \, \text{rad/s}\][/tex]

 Therefore, the angular velocity of the disk after it has turned through half a revolution is approximately[tex]\(1.6 \, \text{rad/s}\)[/tex], which corresponds to option (3).[tex]\[\omega = \alpha t\][/tex]

Answer:

292.3254055 W/m²

469.26267 V/m

[tex]1.56421\times 10^{-6}\ T[/tex]

Explanation:

P = Power of bulb = 90 W

d = Diameter of bulb = 7 cm

r = Radius = [tex]\frac{d}{2}=\frac{7}{2}=3.5\ cm[/tex]

[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

The intensity is given by

[tex]I=\frac{P}{A}\\\Rightarrow I=\frac{90}{4\pi 0.035^2}\\\Rightarrow I=5846.50811\ W/m^2[/tex]

5% of this energy goes to the visible light so the intensity is

[tex]I=0.05\times 5846.50811\\\Rightarrow I=292.3254055\ W/m^2[/tex]

The visible light intensity at the surface of the bulb is 292.3254055 W/m²

Energy density of the wave is

[tex]u=\frac{1}{2}\epsilon_0E^2[/tex]

Energy density is also given by

[tex]\frac{I}{c}=\frac{1}{2}\epsilon_0E^2\\\Rightarrow E=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E=\sqrt{\frac{2\times 292.3254055}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E=469.26267\ V/m[/tex]

The amplitude of the electric field at this surface is 469.26267 V/m

Amplitude of a magnetic field is given by

[tex]B=\frac{E}{c}\\\Rightarrow B=\frac{469.26267}{3\times 10^8}\\\Rightarrow B=1.56421\times 10^{-6}\ T[/tex]

The amplitude of the magnetic field at this surface is [tex]1.56421\times 10^{-6}\ T[/tex]

The intensity of visible light at the surface of the bulb is approximately 920 W/m². The amplitude of the electric field at this surface, for a sinusoidal wave with this intensity, is approximately 600 V/m. The amplitude of the magnetic field at this surface, for a sinusoidal wave with this intensity, is approximately 2 x 10^-6 T.

A 90-W incandescent light bulb radiates only about 5% of the energy as visible light, this gives 0.05 * 90 = 4.5 W as the power of the visible light. The intensity in watts per square meter is given by power divided by surface area. The surface area of a sphere is 4πR², where R represents the radius of the sphere (3.5 cm = 0.035 m in this case). Thus, the visible light intensity I is given by: I = 4.5W / (4π*0.035²) ≈ 920 W/m².

For a sinusoidal wave, the amplitude E0 of the electric field is related to the intensity I by E0 = sqrt (2µ0cI), where µ0 is the permittivity of free space (8.85x10^-12) and c is the speed of light (3x10^8 m/s). Using the calculated intensity, we find E0 ≈ 600 V/m.

The amplitude B0 of the magnetic field at the light bulb surface related to the electric field's amplitude through the relation B0 = E0 / c, so B0 ≈ 600V/m / 3x10^8 m/s ≈ 2x10^-6 T (Tesla).

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Answer:

D) shrivel up, since the atmosphere exerts more force on the can as it cools.

Explanation:

As the water in the can is boiled the can gets heated up and contains hot vapour and gases which are rare in density and are in their expanded state. In this state when the can is sealed tightly such that no air leaks in or out of the can. When the temperature of the can drops, the gases shrink in volume and the pressure inside the can become less than the pressure of the atmosphere which leads to shriveling of the can.

Answer:

72.25 g

Explanation:

mass of 1.7 L water = 1.7 x 10⁻³ x 10³ kg

= 1.7 kg

heat required to raise its temperature from 25 degree to 100 degree

= mass x specific heat x rise in temperature

= 1.7 x 4.18 x 10³ x 75 J

= 532.95 kJ

Now calorific value of LP gas = 46.1 x 10⁶J / kg

Let required mass of LP gas be m kg

heat evolved from this amount of gas

=  46.1 x 10⁶ m

Heat utilized in warming water  

= 46.1 x 10⁶ m  x .16 J

So

46.1 x 10⁶ m  x .16  =  532.95 x 10³

7376 x 10³ m = 532.95 x 10³

m = 532.95 / 7376 kg

= 72.25 g

In order to heat 1.7 L of water from room temperature to boiling, approximately 66.7 kg of LP gas, assuming that during heating only 16% of the heat emitted by LP gas combustion goes to heat the water, would be required.

To calculate the mass of LP gas needed to heat the water, we firstly need to determine the amount of energy required to heat the water from room temperature to the boiling point. This can be determined using the formula for heat Q=m*c* ΔT, where m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature. Considering water's specific heat capacity is 4.184 J/g °C and the change in temperature is 75 °C, the heat needed to warm the water would be (1.7 kg * 4.184 kJ/kg °C * 75 °C) = 532.722 kJ.

Since only 16% of the heat from the LP gas goes to heat the water, the total energy provided by the combustion of LP gas would be obtained by dividing the heat to warm water (532.722 kJ) by 0.16, which equals to 3.33 * 10^3 kJ.

Considering that the heat of combustion of LP gas is about 50 kJ/g approx, the mass of LP gas needed would finally be found by dividing the total energy provided by the LP gas (3.33 * 10^3 kJ) by the heat of combustion of LP gas, which results in approximately 66.7 kg.

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Answer:

20%

Explanation:

Relative Humidity (%) = (water vapor content÷water vapor capacity) × 100

=(7÷35)×100

=(0.2)×100

=20%

According to the Temperature-Water Vapor Capacity Table, the water capacity at 35 °C is 35 grams.

Water Vapor Capacity: The amount of water (grams) which air can hold at a given temperature.

Water Vapor Content: The amount of water vapor actually present in the air.

Answer:

I would think the answer is color, if the wavelength is within the visible light spectrum. This could be answered in different ways but I'm pretty sure the answer you are looking for is hue/color.

Answer:

A=60.06 x 10⁻⁵ m²

Explanation:

Given that

Weight of the person = 864 N

Weight of the bike = 85 N

The total weight ,car + bike ,Wt= 864 + 85 N

 Wt = 949 N

2 p = 949 N

The weight on the each tire(F) = 474.5  N

P = 7.9 x 10 ⁵  Pa

We know that

Force = Pressure x Area

F= P A

By putting the values

474.5 =  7.9 x 10 ⁵  A

A=60.06 x 10⁻⁵ m²

Therefore area of contact A

A=60.06 x 10⁻⁵ m²

The partial pressure of ozone at 441 ppb with an atmospheric pressure of 0.67 atm is [tex]2.947 x 10^-7 atm.[/tex]

Atmospheric pressure=0.67 atm

We have to calculate the partial pressure of ozone at 441 ppb .

To calculate the partial pressure of ozone at 441 ppb, we need to use the formula:

Partial Pressure = Concentration x Total Pressure

Given that the atmospheric pressure is 0.67 atm and the concentration of ozone is 441 ppb (441 parts per billion), we can calculate the partial pressure as follows:

Partial Pressure of Ozone

= (441/1,000,000,000) x 0.67 atm

[tex]= 2.947 x 10-7[/tex]

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Answer:

0.686 g of ice melts each second.

Solution:

As per the question:

Cross-sectional Area of the Copper Rod, A = [tex]11.6\ cm^{2} = 11.6\times 10^{- 4}\ m^{2}[/tex]

Length of the rod, L = 19.6 cm = 0.196 m

Thermal conductivity of Copper, K = [tex]390\ W/m.^{\circ}C[/tex]

Conduction of heat from the rod per second is given by:

[tex]q = \frac{KA\Delta T}{L}[/tex]

where

[tex]\Delta T = 100^{\circ} - 0^{\circ} = 100^{\circ}C[/tex] = temperature difference between the two ends of the rod.

Thus

[tex]q = \frac{390\times 11.6\times 10^{- 4}\times 100}{0.196} = 228.48\ J/s[/tex]

Now,

To calculate the mass, M of the ice melted per sec:

[tex]M = \frac{q}{L_{w}}[/tex]

where

[tex]L_{w}[/tex] = Latent heat of fusion of water = 333 kJ/kg

[tex]M = \frac{228.48}{333\times 10^{3}} = 6.86\times 10^{- 4}\ kg = 0.686\ g[/tex]

To find the number of grams of ice melted per second, we can calculate the rate of heat transfer through the copper rod from the boiling water to the ice-water mixture using Fourier's Law of Heat Conduction. The formula is:

[tex]\[ Q = kA \frac{\Delta T}{L} \][/tex]

Where:

- Q is the rate of heat transfer (in watts, W)

- k  is the thermal conductivity of the material (in W/m°C)

-  A  is the cross-sectional area of the rod (in m²)

- [tex]\( \Delta T \)[/tex] is the temperature difference between the hot and cold ends (in °C)

- L is the length of the rod (in meters)

First, we need to find the temperature difference [tex]\( \Delta T \)[/tex]between the hot end (boiling water) and the cold end (ice-water mixture). We know water boils at 100°C and ice-water mixture is at 0°C. Therefore,[tex]\( \Delta T = 100°C - 0°C = 100°C \).[/tex]

Given:

- Cross-sectional area [tex]\( A = 11.6 \, \text{cm}^2 = 0.00116 \, \text{m}^2 \)[/tex]

- Length of the rod [tex]\( L = 19.6 \, \text{cm} = 0.196 \, \text{m} \)[/tex]

- Thermal conductivity of copper [tex]\( k = 390 \, \text{W/m°C} \)[/tex]

- Latent heat of fusion of ice [tex]\( L_f = 334 \, \text{J/g} \)[/tex]

Now, let's calculate the rate of heat transfer  Q :

[tex]\[ Q = (390 \times 0.00116) \times \frac{100}{0.196} \][/tex]

[tex]\[ Q \approx 236.8 \, \text{W} \][/tex]

Next, we need to convert this rate of heat transfer into grams of ice melted per second using the latent heat of fusion of ice:

[tex]\[ \text{Grams of ice melted per second} = \frac{Q}{L_f} \][/tex]

[tex]\[ \text{Grams of ice melted per second} = \frac{236.8 \, \text{J/s}}{334 \, \text{J/g}} \][/tex]

[tex]\[ \text{Grams of ice melted per second} \approx 0.708 \, \text{g/s} \][/tex]

Therefore, approximately [tex]\( 0.708 \)[/tex] grams of ice melt each second. This means that [tex]\( 0.708 \)[/tex] grams of ice in the ice-water mixture will melt every second due to the heat transferred from the boiling water through the copper rod.

Answer:

The work function ϕ of the metal = 53.4196 x 10⁻¹⁶ J      

Explanation:

When light is incident on a photoelectric material like metal, photoelectrons are emitted from the surface of the metal. This process is called photoelectric effect.

The relationship between the maximum kinetic energy ([tex]E_{k}[/tex]) of the photoelectrons to the frequency of the absorbed photons (f) and the threshold frequency (f₀) of the photoemissive metal surface is:

                                        [tex]E_{k}[/tex] = h(f − f₀)

                                        [tex]E_{k}[/tex] = hf - hf₀

E is the energy of the absorbed photons:  E = hf

ϕ is the work function of the surface:  ϕ = hf₀

                                        [tex]E_{k}[/tex] = E - ϕ

Frequency f = 8.12×10¹⁸ Hz

Maximum kinetic energy [tex]E_{k}[/tex] = 4.16×10⁻¹⁷ J  

Speed of light  c = 3 x 10⁸ m/s

Planck's constant h = 6.63 × 10⁻³⁴ Js                                

                                        E = hf = 6.63 × 10⁻³⁴ x 8.12×10¹⁸

                                        E = 53.8356 x 10⁻¹⁶ J

from [tex]E_{k}[/tex] = E - ϕ ;

                                        ϕ = E - [tex]E_{k}[/tex]

                                        ϕ = 53.8356 x 10⁻¹⁶ - 4.16×10⁻¹⁷

                                        ϕ = 53.4196 x 10⁻¹⁶ J

The work function of the metal ϕ = 53.4196 x 10⁻¹⁶ J      

Answer:

Solids

Explanation:

Some of the properties that limit the motion of particles in a solid:

1. The molecules of a solid are very closely packed together thereby limiting their motion compare to molecules of liquid which are free to move  and gas with loosely bound molecules.

2. The molecules of a solid are held together by strong inter-molecular force due to very small inter-molecular spaces.

3. The motion of molecules in a solid vibrates only in their mean or fixed positions.

4. The particles are arranged in definite pattern and shape. A solid neither takes on the shape of its container like liquid, nor does it fill the entire volume available like a gas.

5. A solid is rigid and does not flow like liquid and gas.

6. A solid has a definite volume and can not be easily compressed.

Forces of attraction most limit the motion of particles in solids. This is because particles in a solid are tightly packed together and held in place by strong intermolecular forces. As solids change to liquids or gases, these forces lessen, allowing particles more freedom.

Forces of attraction most limit the motion of particles in solids. A solid's particles are tightly packed together and held in place by strong intermolecular forces. For example, atoms in a solid are always in close contact with neighboring atoms, held in place by forces (Figure 14.2 (a)). In contrast, particles in a liquid, while also in close contact, are able to slide over one another (b), and particles in a gas move about freely (c).

As solids are heated and change state to a liquid and then a gas, these forces of attraction lessen, allowing particles more freedom of movement. In summary, the state of a substance (solid, liquid, or gas) is largely dependent upon the behavior of its particles, which, in turn, is determined by the attractive forces between them. The stronger the forces of attraction, the less motion the particles will have, as is the case with solids.

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Answers:

a) [tex]65.075 kgm/s[/tex]

b) [tex]10.526 s[/tex]

c) [tex]61.82 N[/tex]

Explanation:

According to the Impulse-Momentum theorem we have the following:

[tex]I=\Delta p=p_{2}-p_{1}[/tex] (1)

Where:

[tex]I[/tex] is the impulse

[tex]\Delta p[/tex] is the change in momentum

[tex]p_{2}=mV_{2}[/tex] is the final momentum of the ball with mass [tex]m=0.685 kg[/tex] and final velocity (to the right) [tex]V_{2}=57 m/s[/tex]

[tex]p_{1}=mV_{1}[/tex] is the initial momentum of the ball with initial velocity (to the left) [tex]V_{1}=-38 m/s[/tex]

So:

[tex]I=\Delta p=mV_{2}-mV_{1}[/tex] (2)

[tex]I=\Delta p=m(V_{2}-V_{1})[/tex] (3)

[tex]I=\Delta p=0.685 kg (57 m/s-(-38 m/s))[/tex] (4)

[tex]I=\Delta p=65.075 kg m/s[/tex] (5)

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately [tex]1.0 cm=0.01 m[/tex]:

[tex]V_{2}=V_{1}+at[/tex] (6)

[tex]V_{2}^{2}=V_{1}^{2}+2ad[/tex] (7)

Where:

[tex]a[/tex] is the acceleration

[tex]d=0.01 m[/tex] is the length the ball was compressed

[tex]t[/tex] is the time

Finding [tex]a[/tex] from (7):

[tex]a=\frac{V_{2}^{2}-V_{1}^{2}}{2d}[/tex] (8)

[tex]a=\frac{(57 m/s)^{2}-(-38 m/s)^{2}}{2(0.01 m)}[/tex] (9)

[tex]a=90.25 m/s^{2}[/tex] (10)

Substituting (10) in (6):

[tex]57 m/s=-38 m/s+(90.25 m/s^{2})t[/tex] (11)

Finding [tex]t[/tex]:

[tex]t=1.052 s[/tex] (12)

According to Newton's second law of motion, the force [tex]F[/tex] is proportional to the variation of momentum  [tex]\Delta p[/tex] in time  [tex]\Delta t[/tex]:

[tex]F=\frac{\Delta p}{\Delta t}[/tex] (13)

[tex]F=\frac{65.075 kgm/s}{1.052 s}[/tex] (14)

Finally:

[tex]F=61.82 N[/tex]

Answers:

a) 65.125 Ns

b) 5.263 * 10^(-4) s

c) 123737.5 N

Explanation:

a) Impulse delivered to the ball F.dt

According to the Impulse-Momentum we have the following:

[tex]F*dt = m*(V_{2} - V_{1})[/tex]

Using the given data we insert in equation above:

[tex]Impulse = 0.685 kg (57 - (-38))\\\\Impulse = 65.1225 Ns[/tex]

Answer: 65.125 Ns

b)

This time can be calculated by the following equations, taking into account the ball undergoes a maximum compression of approximately 0.01m :

Using the kinematic equations for constant acceleration:

[tex](v_{f})^2 = (v_{i})^2 + 2*a*s[/tex]

Where:

vf = 0 (ball stops on the bat)

vi = 38 m/s

s = compression = 0.01 m

Using the equation above we compute acceleration:

[tex]a = \frac{(v_{f})^2 - (v_{i})^2}{2*s} \\\\a = \frac{0^2 - 38^2}{2*0.01} \\\\a = -72,200 m/s^2[/tex]

Using the acceleration to compute time:

[tex]v_{f} = v_{i} + a*t\\\\t = \frac{v_{f} - v_{i}}{a}\\\\t = \frac{0 - 38}{-72,200}\\\\t = 5.263*10^(-4) s[/tex]

Answer: 5.263*10^(-4) s

c)

According to Newton's second law of motion:

[tex]F_{avg} * dt = Impulse[/tex]

Using answer from part a and b

[tex]F_{avg} = \frac{Impulse}{dt} \\\\F_{avg} = \frac{65.125}{5.263*10^(-4)} \\\\F_{avg} = 123737.5 N[/tex]

Answer: 123737.5 N

Answer:

(A) 4.09 kg/s

(B) 589.9 m/s

(C)   0.0008707 m^{3} =  8.71 cm^{2}

Explanation:

inlet pressure of steam (P1) = 4 MPa

inlet temperature of steam (T1) = 400 degree celcius

inlet velocity (V1) = 60 m/s

outlet pressure (P2) = 2 MPa

outlet temperature (T2) = 300 degree celcius

inlet area (A1) = 50 cm^{2} = 0.005 m^{2}

rate of heat loss (Q) = 75 kJ/s

(A) mass flow rate (m) = \frac{A1 x V1}{α1}

where the initial specific volume (α1) for the given temperature and pressure is gotten from tables A-6 = 0.07343 m^3/kg

m = \frac{0.005 x 60}{0.07343}

m = 4.09 kg/s

(B) we can get the outlet velocity using the energy balance equation

  E in = E out

   m(h1 + \frac{(V1)^{2}}{2}) =  m(h2 + \frac{(V2)^{2}}{2})

V2 = [tex]\sqrt{2(h1 - h2) +(V1)^{2} - 2\frac{Q}{m}[/tex]

where h1 and h2 are the enthalpies and are gotten from table A-6

V2 = [tex]\sqrt{2 x 1000 x(3214.5 - 3024.2) +(60)^{2} - 2\frac{75 x 1000}{4.09}[/tex]

V2 = 589.9 m/s  

(C) the outlet area is gotten from mass flow rate (m) = \frac{A2 x V2}{α}

  A2 = (α2 x m) / V2

where the initial specific volume (α2) for the given temperature and pressure is gotten from tables A-6 = 0.12552 m^3/kg

A2 = (0.12552 x 4.09) / 589.5 = 0.0008707 m^{3} =  8.71 cm^{2}

To solve this problem on thermodynamics and fluid dynamics, apply the principles of conservation of mass and energy. First, calculate the mass flow rate using specific volume and continuity equation. Then, considering kinetic and internal energy, calculate the exit velocity. Finally, using mass flow rate and exit velocity, determine the exit area of the nozzle.

The subject of this question falls within the field of Engineering, particularly it relates to thermodynamics and fluid dynamics.

Starting with the given situation, we can see that we have the following known quantities: Initial pressure (P1) = 4 MPa, initial temperature (T1) = 400°C, initial velocity (V1) = 60 m/s, final pressure (P2) = 2 MPa, final temperature (T2) = 300°C, and heat loss Q = 75 kJ/s.

To answer part (a), you would first calculate the specific volume at the given initial state using a steam table, then apply the equation of continuity (ρ1A1V1=ρ2A2V2) to find the mass flow rate.

For part (b), you would follow a similar method but this time utilizing the equation of energy conservation considering both kinetic energy and internal energy of the steam, since the mechanical work is zero and there is heat loss. Considering the energy balance, you can find the exit velocity of the steam.

For part (c), you would use the mass flow rate you found in part (a) and the exit velocity from part (b) in the continuity equation to find the exit area of the nozzle.

Keep in mind that these calculations involve the use of particular thermodynamic values, which might be provided in course materials or general engineering tables.

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The time Δt it will take for the bike wheel to come to a complete stop is approximately 23.6 seconds.  

Let's break down the problem step by step:

1. Calculate the angular deceleration (α):

  - We're given the initial angular speed (ω0) as 9.0 rotations per second and the time it takes for the wheel to slow down as 10.0 seconds.

  - Using the formula α = (ωf - ω0) / t, where ωf is the final angular speed and t is the time, we get α = (0 - 9.0) / 10.0 = -0.9 rotations per second squared.

2. Calculate the total number of rotations during deceleration:

  - Given that the wheel undergoes 65.0 complete rotations during this time, we divide the rotations by the initial angular speed to find the time it takes to complete these rotations:

  - Time = Rotations / Initial angular speed = 65.0 / 9.0 = 7.22 seconds.

3. Use the equation to find the time to stop (tstop):

  - We can use the formula ωf = ω0 + α * tstop, where ωf = 0 (as the wheel comes to a stop), ω0 = 9.0 rotations per second, and α = -0.9 rotations per second squared.

  - Substituting these values, we get 0 = 9.0 + (-0.9) * tstop.

  - Solving for tstop gives us tstop = 9.0 / 0.9 = 10 seconds.

So, it will take approximately 23.6 seconds for the bike wheel to come to a complete stop.

The problem involves angular motion and the effects of friction on a spinning bike wheel. To solve for the time it takes for the wheel to stop, we first determine the angular deceleration using the given initial angular speed and the time it takes for the wheel to slow down. With this deceleration, we calculate the time it takes for the wheel to complete the given number of rotations. Finally, using the formula for angular motion with constant acceleration, we find the additional time needed for the wheel to stop completely.

Complete Question:
A student holds a bike wheel and starts it spinning with an initial angular speed of 9.0 rotations per second. The wheel is subject to some friction, so it gradually slows down. In the 10.0 s period following the inital spin, the bike wheel undergoes 65.0 complete rotations. Assuming the frictional torque remains constant, how much more time Δ t s will it take the bike wheel to come to a complete stop?

The wheel will take approximately 7.22 seconds to come to a complete stop.

To determine how much more time Δt it will take for the bike wheel to come to a complete stop, let's follow these steps:

Given:

Assuming constant angular deceleration (α),

We can find α using the formula:

Solving for α:

so

Hence,

Now, calculate the time (Δt) it will take to stop from 13π rad/s:

So, the additional time required for the wheel to stop, Δt, is approximately 7.22 seconds.

Answer:

1. A only

Explanation:

First law of thermodynamics it sates that energy of the universe is constant.It is also known as energy conservation law.

Therefore according to the first law of thermodynamics ,the thermal energy of the system is unchanged.

Second law of thermodynamics states that ,it is impossible to make a device which take heat from higher temperature energy resource and give 100 % work without heat rejecting.

Therefore only option A is correct.

1. A only

Answer:

A) 21.2 kg.m/s at 39.5 degrees from the x-axis

Explanation:

Mass of the smaller piece = 200g = 200/1000 = 0.2 kg

Mass of the bigger piece = 300g = 300/1000 = 0.3 kg

Velocity of the small piece = 82 m/s

Velocity of the bigger piece = 45 m/s

Final momentum of smaller piece = 0.2 × 82 = 16.4 kg.m/s

Final momentum of bigger piece = 0.3 × 45 = 13.5 kg.m/s

since they acted at 90oc to each other (x and y axis) and also momentum is vector quantity; then we can use Pythagoras theorems

Resultant momentum² = 16.4² + 13.5² = 451.21

Resultant momentum = √451.21 = 21.2 kg.m/s at angle 39.5 degrees to the x-axis  ( tan^-1 (13.5 / 16.4)