Answer:
12.64968 Hz
Explanation:
v = Velocity of sound in seawater = 1522 m/s
u = Velocity of dolphin = 7.2 m/s
f' = Actual frequency = 2674 Hz
From Doppler effect we get the relation
[tex]f=f'\frac{v-u}{v}\\\Rightarrow f=2674\frac{1522-7.2}{1522}\\\Rightarrow f=2661.35032\ Hz[/tex]
The frequency that will be received is 2661.35032 Hz
The difference in the frequency will be
[tex]2674-2661.35032=12.64968\ Hz[/tex]
Final answer:
The difference between the observed frequency and the actual frequency of clicks emitted by the dolphin, calculated using the Doppler Effect, is approximately 12.7 Hz.
Explanation:
The question involves calculating the observed frequency of sound due to the Doppler Effect, which occurs when the source of the sound is moving relative to the observer. In this case, the dolphin is the source of sound waves (clicks), moving away from the marine biologist, who acts as the observer.
The formula for the Doppler Effect for a source moving away from the observer is given by:
f' = f * (v / (v + vs))
Where:
f' is the observed frequency
f is the actual frequency emitted by the source
v is the speed of sound in the medium (1522 m/s in seawater)
vs is the speed of the source relative to the medium (7.2 m/s)
The marine biologist measures the observed frequency (f') as 2674 Hz. We can rearrange the formula to solve for the actual frequency (f):
f = f' * (v + vs) / v
f = 2674 Hz * (1522 m/s + 7.2 m/s) / 1522 m/s
After performing the calculation:
f ≈ 2686.7 Hz
The difference in frequency is then:
Δf = f - f'
Δf ≈ 2686.7 Hz - 2674 Hz
Δf ≈ 12.7 Hz
Therefore, the difference between the observed frequency and the actual frequency of clicks emitted by the dolphin is approximately 12.7 Hz.
A disk with a rotational inertia of 5.0 kg. m2 and a radius of 0.25 m rotates on a frictionless fixed axis perpendicu-
lar to the disk and through its center. A force of 8.0 N is applied tangentially to the rim. If the disk starts at rest,
then after it has turned through half a revolution its angular velocity is :
(1) 0.57 rad/s
(2) 0.64 rad/s
(3) 1.6 rad/s
(4) 3.2 rad/s
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]
A person who weighs 864 N is riding a 85-N mountain bike. Suppose the entire weight of the rider plus bike is supported equally by the two tires. If the gauge pressure in each tire is 7.90 x 105 Pa, what is the area of contact between each tire and the ground?
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 simplest atomic nucleus in nature is molecular formula of hydrogen, which consists of a single proton. Individual protons have charge and can be thought of as small spinning spheres. A charged spinning sphere will generate a magnetic field, whose direction is indicated by the magnetic moment of the object, vector mu. In what direction will the proton rotate, based on the direction of its magnetic moment and the direction of the uniform magnetic field that it is immersed in?
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.
Explanation: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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If the wavelength of a photon in vacuum is the same as the de broglie wavelength of an electron, then the
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 box now rests on a frictionless ramp angled at 15◦ . The mover pulls up on a rope attached to the box to move it up the incline. 33 kg F 29 ◦ 15◦ If the rope makes an angle of 29 ◦ with the horizontal, what is the smallest force the mover would have to exert to move the box up the ramp? The acceleration due to gravity is 9.81 m/s 2 .
Answer:
F = 84.61 N
Explanation:
As in the figure, since there is no friction so if component of Force applied along the incline is greater than the component of weight along the incline, then the object will move up the incline.
component of Force along the incline = F cos(23° - 15°) = F cos(8°)
component of weight along the incline = 33*g*sin(15°) = 33*9.81*sin(15°)
Equating the above two components of forces will give the minimum Force required.
F cos(8°) = 33*9.81*sin(15°)
F = 33*9.81*sin(15°) / cos(8°) (calculate the value using a scientific calculator)
F = 84.61 N
A person runs up several flights of stairs and is exhausted at the top. Later the same person walks up the same stairs and does not feel as tired. Ignoring air resistance, does it take more work to run up the stairs than to walk up?
It takes the same amount of work to move up the stairs whether running or walking, as work is related to the change in potential energy, which does not depend on the speed of climbing. However, running up the stairs requires additional work to be done against gravity due to the higher kinetic energy, leading to a feeling of greater exhaustion.
Explanation:The question concerns whether it takes more work to run up stairs than to walk up. In physics, the term 'work' refers to the amount of energy transferred by a force through a distance. In the context of climbing stairs, work is done against gravity to elevate a person's body to a higher potential energy level. According to the work-energy principle, work is equal to the change in kinetic energy plus the change in potential energy.
Using the given information, the person does 1764 J of work to move up the stairs, regardless of whether they run or walk.
This is the work done to overcome gravity. Only 120 J of work is used to increase the person's kinetic energy when running, which is additional work that occurs due to the higher speed.
Therefore, the total work done when running will be the sum of these two amounts.
However, the amount of work done to raise the person's body to a higher elevation (the potential energy) does not change whether the person runs or walks.
The difference is in the kinetic energy.
The feeling of exhaustion may be greater when running due to the higher pace requiring more power output (rate of doing work) and possibly higher energy expenditure related to physiological effects, but the work done against gravity, which depends on the change in elevation and mass of the person, remains the same in both cases.
Monochromatic light of a given wavelength is incident on a metal surface. However, no photoelectrons are emitted. If electrons are to be ejected from the surface, how should the incident light be adjusted?
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.
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:
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.
A copper rod of cross-sectional area 11.6 cm2 has one end immersed in boiling water and the other in an ice-water mixture, which is thermally well insulated except for its contact with the copper. The length of the rod between the containers is 19.6 cm, and the rod is covered with a thermal insulator to prevent heat loss from the sides.
How many grams of ice melt each second? (The thermal conductivity of copper is 390 W/m-C°.)
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.
ind the net rate of heat transfer by radiation from a skier standing in the shade, given the following. She is completely clothed in white (head to foot, including a ski mask), the clothes have an emissivity of 0.200 and a surface temperature of 10.0ºC , the surroundings are at −15.0ºC , and her surface area is 1.60 m^2 . (answer in W)
Answer:
Q=116.37 W
Explanation:
Given that
Emissivity ,ε= 0.200
Surface temperature ,T₁ = 10⁰ C = 283 K
Surrounding T₂ = - 15⁰ C = 258 K
Area ,A= 1.6 m²
The net heat transfer given as
[tex]Q=\sigma A \varepsilon (T_1^4-T_2^2)[/tex]
σ = 5.67 x 10⁻⁸
The temperature should be in Kelvin.
Now by putting the values
[tex]Q=5.67\times 10^{-8}\times 1.6\times 0.2\times (283^4-258^2)\ W[/tex]
Q=116.37 W
Therefore answer will be 116.37 W.
A playground merry-go-round with a radius of 1.80 m has a mass of 120 kg and is rotating with an angular speed of 0.400 rev/s. What is its angular speed after a 37.5-kg child gets onto it by grabbing its outer edge? The child is initially at rest.
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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What mass of LP gas is necessary to heat 1.7 L of water from room temperature (25.0 ∘C) to boiling (100.0 ∘C)? Assume that during heating, 16% of the heat emitted by the LP gas combustion goes to heat the water. The rest is lost as heat to the surroundings.
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.
Explanation: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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A __________ 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.
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.
A firecracker breaks up into two pieces , one has a mass of 200 g and files off along the x –axis with a speed of 82.0 m/s and the second has a mass of 300 g and flies off along the y-axis with a speed of 45.0 m/s . What is the total momentum of the two pieces ?
a) 361 kg x m/s at 56.3 degrees from the x-axis .
b) 93.5 kg x m/s at 28.8 degrees from the x-axis .
c) 21.2 kg x m/s at 39.5 degrees from the x-axis .
d) 361 kg d m/s at 0.983 degrees from the x-axis .
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)
Electromagnetic radiation of 8.12×10¹⁸ Hz frequency is applied on a metal surface and caused electron emission. Determine the work function of the metal if the maximum kinetic energy ([tex]E_k[/tex]) of the emitted electron is 4.16×10⁻¹⁷ J.
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
Forces of attraction limit the motion of particles most in
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.
Explanation: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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A thin spherical shell rolls down an incline without slipping. If the linear acceleration of the center of mass of the shell is 0.23g, what is the angle the incline makes with the horizontal?
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
Two 0.006 Kg bullets are fired with speeds of 20.0 m/s and 50.0 m/s respectively. What are their kinetic energies? Which bullet has more kinetic energy? What is the difference of their kinetic energies?
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
The current and the potential difference in an inductor are in phase. B. The current lags the potential difference by π/2 in an inductor C. The current leads the potential difference by π/2 in an inductor. D. none of the above
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].
A baseball, which has a mass of 0.685 kg., is moving with a velocity of 38.0 m/s when it contacts the baseball bat duringwhich time the velocity of the ball becomes 57.0 m/s in the opposite direction.a. How much impulse has been delivered to the ball by the bat?While in contact with the bat the ball undergoes a maximum compression of approximately 1.0 cm.b. Approximately how long did it take for the ball to be stopped by the bat?c. What will be the average force applied to the ball by the bat while stopping the ball?
Answers:
a) [tex]65.075 kgm/s[/tex]
b) [tex]10.526 s[/tex]
c) [tex]61.82 N[/tex]
Explanation:
a) Impulse delivered to the ballAccording 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)
b) Time
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)
c) Force applied to the ball by the batAccording 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
A proton of mass m moving with a speed of 3.0 × 106 m/s undergoes a head on elastic collision with an alpha particle of mass 4m at rest. What are the velocities of the two particles after the collision?
Answer:6.0×10^5m/s
Explanation:
According to the law of conservation of momentum, sum of the momenta of the bodies before collision is equal to the sum of their momenta after collision.
After their collision, the two bodies will move with a common velocity (v)
Momentum = mass × velocity
Let m1 be the mass of the proton = m
Let m2 be the mass of the alpha particle = m2
Let v1 be the velocity of the proton = 3.0×10^6m/s
Let v2 be the velocity of the alpha particle = 0m/s (since the body is at rest).
Using the law,
m1v1 + m2v2 = (m1 + m2)v
m(3.0×10^6) + 4m(0) = (m + 4m)v
m(3.0×10^6) = 5mv
Canceling 'm' at both sides,
3.0×10^6 = 5v
v = 3.0×10^6/5
The common velocity v = 6.0×10^5m/s
coil of wire is connected to a power supply, and a current runs in the coil. A single loop of wire is located near the coil, with its axis on the same line as the axis of the coil. The radius of the loop is 2 cm.
At time t1 the magnetic field at the center of the loop, due to the coil, is 0.4 T, in the direction shown in the diagram; the current in the coil is constant.
(a) What is the absolute value of the magnetic flux through the loop at time t1?
Φ mag = T m2
(b) What approximations or assumptions did you make in calculating your answer to part (a)?
Check all that apply.
1.The magnitude of the magnetic field due to the coil is uniform over the area of the loop.
2.The magnetic field outside the loop is zero.
3. The magnetic field due to the coil is uniform in direction over the area of the loop.
(c) What is the direction of the "curly" electric field inside the wire of the loop at time t1? (Remember that at this time the current in the coil is constant.)
E = 0
At a later time t2, the current in the coil begins to decrease.
(d) Now what is the direction of the "curly" electric field in the loop?
counter-clockwise
At time t2 the rate of change of the magnetic field at the center of the loop, due to the coil, is -0.36 T/s.
(e) At this time, what is the absolute value of the rate of change of the magnetic flux through the loop?
|dΦmag/dt| = T m2/s
(f) At this time, what is the absolute value of the emf in the loop?
|emf| = V
(g) What is the magnitude of the electric field at location P, which is inside the wire?
|E| = V/m
(h) Now the wire loop is removed. Everything else remains as it was at time t2; the magnetic field is still changing at the same rate. What is the magnitude of the electric field at location P?
|E| = V/m
a) The magnetic flux is 0.0005Tm².
b) 1. The magnitude of the magnetic field due to the coil is uniform over the area of the loop.3. The magnetic field due to the coil is uniform in direction over the area of the loop.
c) The direction of the electric field inside the wire of the loop at time t1 is zero
d) The direction of the "curly" electric field in the loop at time when the current in the coil t₂ is counter clockwise
e) The absolute value of the rate of change of the magnetic flux through the loop is [tex]0.000045T^2[/tex]
f) The absolute value of the emf in the loop at previous instant is 0.000045V
g) The magnitude of the electric field at location P, which is inside the wire is 0.0036V/m
h) Now the wire loop is removed.
The question is investigating the interactions of current, magnetic fields, magnetic flux, and electromagnetic forces.
(a) The magnetic flux is the magnetic field times the area it is flowing through. So we have
[tex]\phi_{mag}= B * A = 0.4T * \pi*(0.02m)^2 = 0.0005 T m^2[/tex]
(b) The assumptions made are: 1. The magnetic field is uniform over the loop's area, 2. Magnetic field outside the loop is zero, 3. The magnetic field is uniform in direction over the loop's area.
(c) The direction of the electric field inside the wire of the loop at time t1 is zero, as the magnetic field and the current are constant and hence there’s no change in magnetic flux.
(d) If the current in the coil begins to decrease, Lenz's law tells us that an induced current will always counteract the change in magnetic flux that caused it. Therefore, the direction of the "curly" electric field in the loop will be counter-clockwise.
(e) At the later time the rate of change of the magnetic flux, |dΦmag/dt|, would be equal to the rate of change of the magnetic field times the area of the loop [tex]= |0.36 T/s * \pi*(0.02m)^2| = 0.000045 T m^2/s.[/tex]
(f) The absolute value of the emf in the loop would be equal to the absolute value of the rate of change of magnetic flux. Based on Faraday's law this means [tex]|emf| = |d\phi_{mag}/dt| = 0.000045 V.[/tex]
(g) [tex]\|E| = \frac{|\text{emf}|}{2 \pi r} = \frac{0.000452 \, \text{V}}{2 \pi (0.02 \, \text{m})} \approx 0.0036 \, \text{V/m}[/tex]
(h ) The magnitude of the electric field at location P, which is inside the wire, would depend on specifics not given in the question.
A 1-kilogram parcel of air is at 35°c and contains 7 grams of water vapor. What is the relative humidity?
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.
A rectangular loop of wire of width 10 cm and length 20 cm has a current of 2.5 A flowing through it. Two sides of the loop are oriented parallel to a uniform magnetic field of strength 0.037 T, the other two sides being perpendicular to the magnetic field.
A)What is the magnitude of the magnetic moment of the loop?
B)What torque does the magnetic field exert on the loop?
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
Consider the following statements:
A) The entropy of an isolated system never decreases.
B) Heat never flows spontaneously from cold to hot.
C) The total thermal energy of an isolated system is constant.
Which of these express the second law of thermodynamics?
1. A only
2. B only
3. C only
4. Both A and B
5. Both B and C
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
The main greenhouse gases in the atmospheres of the terrestrial planets are
Answer:
Answered
Explanation:
Terrestrial Planets are the planets that have solid surface and are smaller in size. Mercury, Venus, Earth and Mars are the terrestrial planets of our solar system.
The main greenhouse gases in the atmospheres of the terrestrial planets are, Carbon dioxide( the atmosphere of Venus has mainly CO_2 in it making it hottest planet of the solar system because of greenhouse effect). Water vapor and Cloroflorocarbons in the Earth atmosphere. Traces of methane and nitrous oxide are also present.
Steam at 4 MPa and 400C enters a nozzle steadily with a velocity of 60 m/s, and it leaves at 2 MPa and 300C. The inlet area of the nozzle is 50 cm2, and heatis being lost at a rate of 75 kJ/s. Determine(a)the mass flow rate of the steam,(b)the exit velocity of the steam, and(c)the exit area of the nozzle.
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.
Explanation: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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Calculate the partial pressure of ozone at 441 ppb if the atmospheric pressure is 0.67 atm.
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]
Explanation: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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A popular classroom demonstration is to place a gas can on a burner and boil water in it. Left unchecked this has the potential to be a very boring demo. However the can is removed from the flame and the lid is screwed on tightly. After it cools down the can will
A) still be boring and not change.
B) shrivel up, since colder gases have less pressure.
C) bulge and expand, since colder gases are denser and exert more pressure
D) shrivel up, since the atmosphere exerts more force on the can as it cools.
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.
The wavelength of light mainly affects our perception of
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.