Final answer:
The colorful effects seen when gasoline creates a thin film on water are a demonstration of thin-film interference, due to light waves reflecting and interfering from the surfaces of the film. The phenomenon causes different colors to appear based on the film's thickness and the light's wavelength.
Explanation:
The colors seen when gasoline forms a thin film on water are due to thin-film interference. This optical phenomenon occurs when light waves reflected from the top and bottom surfaces of a thin film, such as oil or a soap bubble, combine. The interference can be constructive or destructive, depending on the phase difference between the waves. Constructive interference results in bright colors, while destructive interference can cause darkness or cancellation of color. The film's varying thickness results in different colors because each color in the visible spectrum has a different wavelength; thinner or thicker areas interfere with different colors differently.
Early experiments in this area by Agnes Pockels helped to lay the groundwork for our understanding of thin films. Interference effects are most noticeable when light interacts with objects that have sizes similar to the wavelength of light, a principle that is also applied in designing anti-reflection coatings and optical filters.
Final answer:
The colorful effects seen when gasoline or oil creates a thin layer on water are due to thin-film interference, a result of constructive interference of light reflecting off the film's surfaces.
Explanation:
The colors seen when gasoline forms a thin film on water are a demonstration of thin-film interference. This occurs due to the interference of light that is reflected off different surfaces of a thin film such as oil or a soap bubble. The phenomenon is explained by the wave nature of light, which when interacting with objects similar in size to its wavelength, causes some wavelengths of light to interfere constructively, creating bright and vivid colors. The observed colors vary because the thickness of the film affects which wavelengths interfere constructively.
Historical contributions to the study of thin-film interference include the work of Agnes Pockels. She developed a trough for measuring surface films, aiding in the investigation of these colorful effects. Moreover, commercial applications like anti-reflection coatings and optical filters utilize thin film interference principles for effectiveness.
An airplane takes hours to travel a distance of kilometers against the wind. The return trip takes hours with the wind. What is the rate of the plane in still air and what is the rate of the wind?
Answer:
Rate of plane in still air = P = W (t1 +t2)/ (t1-t2)
Rate of wind in still air = W = P (t1 - t2)/(t1 + t2)
Explanation:
Assuming speed of plane are the same on both trips
Rate (D/t1) = (P-W).... EQU 1 going from city a to b
Rate (D/t2) = (P +W)...Equ2 going back to city a
Where t1 is not equal to t2
Where D=distance between two cities
P &W are the speed of plane and wind
t1 &t2 = time taken for travel
Equ 1 & equ 2 becomes
D = ( P - W ) t1.. equ3
D = ( P + W ) t2 equa4
Equating equ 3 and 4
Pt1 - Wt1 = Pt2 +Wt2
P ( t1 - t2) = W ( t1 + t2)
Rate of plane in still air = P = W (t1 +t2)/ (t1- t2)
Rate of wind in still air = W = P (t1 - t2)/(t1 + t2)
A large wooden turntable in the shape of a flat uniform disk has a radius of 2.00 m and a total mass of 120 kg. The turntable is initially rotating at 3.00 rad>s about a vertical axis through its center. Suddenly, a 70.0-kg parachutist makes a soft landing on the turntable at a point near the outer edge. (a) Find the angular speed of the turntable after the parachutist lands. (Assume that you can treat the parachutist as a particle.) (b) Compute the kinetic energy of the system before and after the parachutist lands. Why are these kinetic energies not equal
Answer:
a)1.385 rad/s
b) Before: 1080 J. After 498.46 J
Explanation:
The moments of inertia of the turn table, with the shape of uniform disk is:
[tex] I_1 = 0.5mr^2 = 0.5*120*2^2 = 240 kgm^2[/tex]
The angular momentum of the turn table before the impact is
[tex]A_1 = \omega_1I_1 = 3*240 = 720 radkgm^2[/tex]
The moments of inertia of the system after the impact is (treating the parachute man is a point particle)
[tex]I_2 = I_1 + Mr^2 = 240 + 70*2^2 = 240 + 280 = 520 kgm^2[/tex]
According to angular momentum conservation law:
[tex]A_1 = A_2[/tex]
[tex] 720 = \omega_2I_2[/tex]
[tex]\omega_2 = \frac{720}{I_2} = \frac{720}{520} = 1.385 rad/s[/tex]
(b) Before the impact:
[tex]K_1 = 0.5*I_1*\omega_1^2 = 0.5*240*3^2 = 1080 J[/tex]
After the impct
[tex]K_2 = 0.5*I_2*\omega_2^2 = 0.5*520*1.385^2 = 498.46 J[/tex]
The kinetic energies are not equal because the impact is causing the turn table to lose energy.
A 32-kg child decides to make a raft out of empty 1.0-L soda bottles and duct tape. Neglecting the mass of the duct tape and plastic in the bottles, what minimum number of soda bottles will the child need to be able stay dry on the raft?
Answer:
32 bottles
Explanation:
If we create a free body diagram on the child we have his weight and the bouyant force
W-B=0
They must be equal to mantain equilibrium on the body and he can stay floating, this force is equivalent to the weight of water displaced
W=B=Ww
Mg=mg
32 kg=mass of water displaced
1 kilogram per liter (kg/L) is the density of water, this means that 32 Liters of water are displaced and since the bottles can retain 1 liter, the child needs 32 bottles
The child needs a minimum of 32 soda bottles to stay dry on the raft.
Explanation:To determine the minimum number of soda bottles needed for the child to stay dry on the raft, we need to consider the buoyant force exerted by the bottles. The buoyant force is equal to the weight of the displaced water. Since the child wants to stay dry, the buoyant force should be greater than or equal to the weight of the child.
The weight of the child can be calculated using the formula: weight = mass × gravity. Given the mass of the child is 32 kg, and the acceleration due to gravity is 9.8 m/s², we can find that the weight of the child is 32 kg × 9.8 m/s² = 313.6 N.
Next, we need to find the volume of one soda bottle. Since the empty soda bottles have a total volume of 1.0 L, we can assume that each bottle has a volume of 1.0 L ÷ the number of bottles needed. The mass of the water displaced by one bottle can be calculated using the formula: mass = density × volume. Given that the density of water is 1000 kg/m³, and 1 L = 0.001 m³, we can find that the mass of water displaced by one bottle is 1000 kg/m³ × 0.001 m³ = 1 kg.
To find the minimum number of bottles needed, we can set up the equation: buoyant force = weight of child. The buoyant force is equal to the mass of water displaced by one bottle × gravity × the number of bottles needed. Using the values we found earlier, we have: 1 kg × 9.8 m/s² × the number of bottles needed = 313.6 N. Solving for the number of bottles needed, we find that the minimum number of soda bottles the child needs is 313.6 N / (1 kg × 9.8 m/s²) = 32 bottles (rounded up to the nearest whole number).
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What would be your estimate of the age of the universe if you measured a value for Hubble's constant of H0 = 30 km/s/Mly ? You can assume that the expansion rate has remained unchanged during the history of the universe.
Answer:
The age of the universe would be 9.9 billion years
Explanation:
We can calculate an estimate for the age of the Universe from Hubble's Law. Let's suppose the distance between two galaxies is D and the apparent velocity with which they are separating from each other is v. At some point, the galaxies were touching, and we can consider that time the moment of the Big Bang.
Thus, the time it has taken for the galaxies to reach their current separations is:
[tex]\displaystyle{t=D/v}[/tex]
and from Hubble's Law:
[tex]v =H_0D[/tex]
Therefore:
[tex]\displaystyle{t=D/v=D/(H_0\times D)=1/H_0}[/tex]
With the given value for the Hubble's constant we have:
[tex]H_0=(30\ km/s/Mly) \times (1 Mly/ 9.461 \times 10^{18} km) = 3.17\times 10^{-18}\ 1/s[/tex]
and thus,
[tex]t=1/H_0 = 1/(3.17\times 10^{-18} 1/s) = 0.315 \times 10^{18}\ s \approx 9988584474.8858\ years \approx 9.9\ billion\ years[/tex]
Why are unpaired electrons more significant than paired electrons in terms of magnetic properties
Answer:
Opposite spin neutralizes the magnetic fields.
Explanation:
The reason for the significance of the unpaired electrons with respect to the magnetic properties is because electrons have opposite spin and when the electrons are paired then as a result their opposite spins neutralizes the effect of their magnetic field thus resulting in no field effect.
Thus single electrons which are unpaired contributes to the magnetic properties of the material as compared to the paired electrons.
The moon is a satellite that orbits the earth at a radius of 3.85 × 108 m. The mass of the earth is 5.98 × 1024 kg. What is the orbital speed of the moon?
Answer:
The orbital velocity of the moon is, V = 1018 m/s
Explanation:
Given data,
The radius of the moon's orbit, R = 3.85 x 10⁸ m
The mass of the Earth, M = 5.98 x 10²⁴ kg
The formula for orbital velocity is,
V = √(GM/R²)
Substituting the values,
V = √(6.673 x 10⁻¹¹ x 5.98 x 10²⁴ / 3.85 x 10⁸ )
= 1018 m/s
Hence, the orbital velocity of the moon is, V = 1018 m/s
The orbital speed of the Moon can be calculated using a specific formula that takes into account the mass of the Earth, the mass of the Moon, and the radius of the Moon's orbit.
Explanation:The orbital speed of an object in orbit around another object can be calculated using the formula:
v = √[G * (M+E) / r]
Where v is the orbital speed, G is the gravitational constant (approximately 6.67 × 10^-11 N.m^2/kg^2), M is the mass of the larger object (in this case, the Earth), E is the mass of the orbiting object (in this case, the Moon), and r is the radius of the orbit.
Substituting the given values into the formula, we can calculate the orbital speed of the Moon.
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You charge a parallel-plate capacitor, remove it from the battery, and prevent the wires connected to the plates from touching each other. When you pull the plates apart to a larger separation, do the following quantities increase, decrease or stay the same?i) Cii) Qiii) E between the platesiv) delta V
Answer:
i) C decreases
ii) Q remains constant
iii) E remains constant
iv) ΔV increases
Explanation:
i)
We know, capacitance is given by:
[tex]C=\frac{\epsilon_0.A}{d}[/tex]
[tex]\therefore C\propto \frac{1}{d}[/tex]
In this case as the distance between the plates increases the capacitance decreases while area and permittivity of free space remains constant.
ii)
As the amount of charge has nothing to do with the plate separation in case of an open circuit hence the charge Q remains constant.
iii)
Electric field between the plates is given as:
[tex]E=\frac{\sigma}{\epsilon_0}[/tex]
where:
charge density, [tex]\sigma=\frac{Q}{A}[/tex]
As we know that distance of plate separation cannot affect area of the plate. Charge Q and permittivity are also not affected by it, so E remains constant.
iv)
From the basic definition of voltage we know that it is the work done per unit charge to move it through a distance.Here we increase the distance so the work done per unit charge increases.i) Capacitance is decreases
ii) Charge Q remains constant
iii) Electric field E remains constant
iv) Change in potential ΔV is increases
Parallel-plate capacitor:The capacitance is computed as,
[tex]C=\frac{\epsilon A}{d}[/tex]
Where A is area of plates and d is distance between plates.
Following information is to be considered.
Given that the distance between the plates increases, the capacitance decreases while area and permittivity of free space remains constant.the amount of charge is independent on plate separation .Hence the charge Q remains constant.We know that distance of plate separation can not affect area of the plate. So that Charge Q and permittivity are also not affected by it. Thus, electric field E remains constant.Voltage is the work done per unit charge to move it through a distance.Here we increase the distance so the work done per unit charge increases.Learn more about the Parallel plate capacitor here:
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Vector V⃗ 1 points along the z axis and has magnitude V1 = 76. Vector V⃗ 2 lies in the xz plane, has magnitude V2 = 60, and makes a -48 ∘ angle with the x axis (points below x axis).
What is the scalar product of the two vectors? Express answer using two significant figures.
Answer:
R= - 3388.74
Explanation:
Given that
V₁= 76 k ( in z-direction)
θ = 48°
V₂ = 60 cos48° i - 60 sin48° k
The dot product of two vector given as
We know that dot product of two vector is scalar and cross product of two vector is vector.
R= V₁ . V₂
We have to remember
i.i= j.j = k.k = 1
i.j = j.k = k.i = 0
Now
R= V₁ . V₂
R= (76 k ).( 60 cos48° i - 60 sin48° k)
R= 0 - 60 x 76 sin48°
R= - 3388.74
The scalar product of the vectors is - 3388.74
Given information:
V₁= 76k since it is in z-direction
Now vector V₂ makes an angle θ = 48° with x-axis so, it can be resolved as follows:
V₂ = 60 cos48°i - 60 sin48° k
Scalar Product:The scalar product of vectors is the product of the projection of one vector with the other vector.
The scalar product or the dot product of two vectors is given as
V= V₁ . V₂
The dot product of the x,y,and z direction components follow the below mentioned rule:
i.i= j.j = k.k = 1
i.j = j.k = k.i = 0
So, the required scalar product
V = V₁ . V₂
V = (76k ).(60cos48° i - 60sin48° k)
V = 0 -60 x 76sin48°
V = - 3388.74
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A projectile returns to its original height 4.08 s after being launched, during which time it travels 76.2 m horizontally. If air resistance can be neglected, what was the projectile's initial speed?
Answer:27.35 m/s
Explanation:
Given
Time of Flight of Projectile T=4.08 s
Range of Projectile =76.2 m
Time Of Flight of Projectile is given by
[tex]T=\frac{2u\sin \theta }{g}----------1[/tex]
where u=initial Velocity
[tex]\theta =[/tex]Launch angle
g=acceleration due to gravity
Range is given by [tex]R=\frac{u^2\sin 2\theta }{g}------2[/tex]
divide 1 and 2
[tex]\frac{R}{T}=\frac{u^2\sin 2\theta }{g}\times \frac{g}{2u\sin \theta }[/tex]
[tex]\frac{R}{T}=u\cos \theta ------3[/tex]
[tex]u\sin \theta =\frac{Tg}{2}------4[/tex]
squaring and adding 3 & 4 we get
[tex]u^2(\cos ^2\theta +\sin ^2\theta )=(\frac{R}{T})^2+(\frac{Tg}{2})^2[/tex]
[tex]u^2=(19.992)^2+(18.676)^2[/tex]
[tex]u=\sqrt{748.49}[/tex]
[tex]u=27.35 m/s[/tex]
The initial speed is the speed of the object at the beginning of the measurement or the starting speed.
The initial speed of the projectile is 27.35 m/s.
What is the initial speed?
The initial speed is the speed of the object at the beginning of the measurement or the starting speed.
Given information-
The time taken by projectile to returns its original height is 4.08 s.
The distance traveled by it is 76.2 m.
Air resistance can be neglected.
The time of flight of a projectile motion can be given as,
[tex]T=\dfrac{2u\sin \theta}{g}[/tex]
Let the above equation is equation 1.
Here, [tex]u[/tex] is the initial velocity, and [tex]\theta[/tex] is the angle of launch.
Rewrite the above equation as,
[tex]u\sin \theta =\dfrac{Tg}{2}[/tex]
Let the above equation is equation 2.
Now the range of the projectile motion can be given as,
[tex]R=\dfrac{u^2\sin (2\theta) }{g}\\R=\dfrac{u^22\cos \theta\sin \theta }{g}\\[/tex]
Divide this equation by the equation 1 as,
[tex]\dfrac{R}{T}=u\cos \theta[/tex]
Square and add the the above equation and equation 2 as,
[tex]u^2(\cos^2 \theta+\sin^2 \theta)=\dfrac{R}{T}+\dfrac{Tg}{2}\\u^2(1)=\dfrac{76.2}{4.08}+\dfrac{4.08\times9.81}{2}\\u=27.35\rm m/s[/tex]
Hence, the initial speed of the projectile is 27.35 m/s.
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At one point in time, the price of gold was about $1400 per ounce, while that of silver was about $20 an ounce. The "ounce" in this case is the troy ounce, which is equal to 31.1035 g. (The more familiar avoirdupois ounce is equal to 28.35 g.) The density of gold is 19.3 g/cm3 and that of silver is 10.5 g/cm3.
A. If you found a spherical gold nugget worth $5 million, what would be its diameter?
ANSWER: ______________ cm
B. How much would a silver nugget of that size be worth?
ANSWER: $ ____________
Answer:
Gold nugget diameter 22,24 cm
Silver nugget price 38860 $
Explanation:
Gold price 1400 $ /ou 1 ounce = 31.1035 grs
so 1400 /31.1035
Gold price is 45,01 $/grs.
If a nugget worth 5000000 $ then 5000000/ 45.01
a nugget mass : 111086,43 grs
Now gold density is d = 19.3 grs/cm³
And she volume is V = 4/*π*r³
d = m/V V = m/d V = 111086.43/19.3 cm³
V = 5755,77 cm³
Now V of the sphere is V = 5755,77 = 4/3*π*r³
r³ =3*5755,77 / 4π r³ = 1374,79
r = 11.12 cm 2r = 22,24 = Φ (sphere diameter)
B) d (silver) = m/V m = d*V
V = 5755,77 cm³ The same size the same volume
m = 10,5 * 5755,77 [grs/cm³ * cm³] m = 60435,59 grs
Silver nugget worth :
20 $ /ou 20/31.1035 = 0.643 $ /grs
Price 0,643 * 60435,59 = 38860 $
Calculate the final speed (in m/s) of a 106 kg rugby player who is initially running at 7.75 m/s but collides head-on with a padded goalpost and experiences a backward force of 1.85 ✕ 10⁴ N for 7.50 ✕ 10⁻²s.
Answer:
Final velocity will be -5.33 m/sec
Explanation:
We have given mass of the rugby player m = 106 kg
Initial speed = 7.75 m/sec
Backward force [tex]F=1.85\times 10^4N[/tex]
Time is given as [tex]t=7.5\times 10^{-2}sec[/tex]
Impulse is given by impulse = force × time
So impulse [tex]=-1.85\times 10^4\times 7.5\times 10^{-2}=-1387.5N-s[/tex] ( as force is backward )
We know that impulse is given by change in momentum
So [tex]m(v_f-v_i)=-1387.5[/tex]
[tex]106\times (v_f-7.75)=-1387.5[/tex]
[tex]v_f=-5.33m/sec[/tex]
Calculate the energy that is required to change 50.0 g ice at -30.0°C to a liquid at 73.0°C. The heat of fusion = 333 J/g, the heat of vaporization = 2256 J/g, and the specific heat capacities of ice = 2.06 J/gK and liquid water = 4.184 J/gK
Answer:
[tex]Q=35011.6\ J[/tex]
Explanation:
Given:
mass of ice, [tex]m=50\ g[/tex]initial temperature of ice, [tex]T_i=-30^{\circ}C[/tex]final temperature of liquid water, [tex]T_f=73^{\circ}C[/tex]heat of fusion of ice, [tex]L=333\ J.g^{-1}[/tex]specific heat capacity of ice, [tex]c_i=2.06\ J.g^{-1}[/tex]specific heat capacity of liquid water, [tex]c_w=4.184\ J.g^{-1}.K^{-1}[/tex]Now, total heat energy required get to the final state:
[tex]Q=m(c_i.\Delta T_i+L+c_w.\Delta T_w)[/tex]
where:
[tex]\Delta T_w=[/tex] change in temperature of water from 0 to 73 degree C
[tex]\Delta T_i=[/tex] change in temperature of ice from -30 to 0 degree C
[tex]\therefore Q=50(2.06\times 30+333+4.184\times 73)[/tex]
[tex]Q=35011.6\ J[/tex]
Answer:
The heat energy required is 350.1J
Explanation:
Given data
Mass of ice =50g---kg =50/1000= 0.05kg
Temperature of ice T1= - 30°c
Temperature of ice T2=0°c
Temperature of liquid T3=0°c
Temperature of water T4= 73°c
The heat of fusion = 333 J/g
heat of vaporization = 2256 J/g, and the specific heat capacities of ice = 2.06 J/gK
and liquid water = 4.184 J/gK
It will be a good idea to first understand the path this process will follow
Ice at - 30°c - - ice at 0°c
Ice at 0°c - - - - liquid at 0°c fusion
Liquid at 0°c -- - liquid at 73°c
First, you have to calculate the heat absorbed by ice going from -30 C to 0 C. Use the equation:
q = m c (T2-T1)
q= 0.5*2.06(0-(-30))
q= 30.9J
Then, calculate the heat required to melt that ice at 0C. Use the equation:
q = m *(heat of fusion)
q=0.5*333
q=166.5J
Then, calculate the heat required to raise the temperature of water from 0°C to 73°C. Use
q = m c (T4-T3)
q= 0.5*4.184(73-0)
q= 152.7J
Finally, we will sum up the heat required
Total heat energy required = 30.9+166.5+152.7= 350.1J
A box slides with uniform acceleration up an incline. The box has an initial speed of 9.0 m/s and rises vertically 2.60 m before coming to rest. If the angle of the incline is 30°, what is the coefficient of kinetic friction between the box and the incline?
Answer: 0.58
Explanation:
First we need to get the acceleration of the body using equation of motion
v²=u²-2as
v is the final velocity
u is the initial velocity
a is acceleration
s is the distance moved
0²=9²-2a(2.6)
-81=-5.2a
a=81/5.2
a= 15.6m/s²
Angle of inclination =30°
To get the coefficient of friction, we use the formula
Ff =nR
Ff is frictional force
n is coefficient of friction
R is normal reaction
n = Ff/R = Wsin30°/Wcos30°
n = tan30°
n = 0.58
Yellow light travels through a certain glass block at a speed of 1.97 × 108 m/s. The wavelength of the light in this particular type of glass is 3.81 × 10−7 m (381 nm). What is the frequency of the yellow light in the glass block? Answer in units of H
The frequency of the yellow light is [tex]5.17\cdot 10^{14}Hz[/tex]
Explanation:
The relationship between wavelength, frequency and speed of a wave is given by the wave equation:
[tex]v = f \lambda[/tex]
where
v is the speed of the wave
f is the frequency
[tex]\lambda[/tex] is the wavelength
For the yellow light in this problem, we have:
[tex]v=1.97\cdot 10^8 m/s[/tex] is the speed
[tex]\lambda=3.81\cdot 10^{-7} m[/tex] is the wavelength
Solving for f, we find its frequency:
[tex]f=\frac{v}{\lambda}=\frac{1.97\cdot 10^8}{3.81\cdot 10^{-7}}=5.17\cdot 10^{14}Hz[/tex]
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The frequency of the yellow light in the glass block is [tex]5.17 \times 10^{14}Hz[/tex]
Given information:Yellow light travels through a certain glass block at a speed of 1.97 × 108 m/s. The wavelength of the light in this particular type of glass is 3.81 × 10−7 m (381 nm).
Calculation of the frequency:[tex]= 1.97 \times 10^8 \div 3.81 \times 10^{-7}\\\\= 5.17 \times 10^{14}Hz[/tex]
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Two horizontal forces act on a 1.4 kg chopping block that can slide over a friction-less kitchen counter, which lies in an xy plane. One force is [tex]\vec{F}_1 = (3.9 N)\hat{i} + (3.3 N)\hat{j}[/tex]. Find the acceleration of the chopping block in unit-vector notation for each of the following second forces.[tex]a) \vec{F}_2= (-3.0N)\hat{i} + (-4.0N)\hat{j}\\b) \vec{F}_2= (-3.0N)\hat{i} + (4.0N)\hat{j}\\c) \vec{F}_2=(3.0N)\hat{i} + (-4.0N)\hat{j}[/tex]
Answer:
Part a)
[tex]a = (0.64\hat i - 0.5 \hat j)m/s^2[/tex]
Part b)
[tex]a = (0.64\hat i + 5.21 \hat j)m/s^2[/tex]
Part c)
[tex]a = (4.92\hat i - 0.5 \hat j)m/s^2[/tex]
Explanation:
As per Newton's II law we know that
F = ma
so we will have
[tex]a = \frac{F}{m}[/tex]
so we will have
[tex]a = \frac{F_1 + F_2}{m}[/tex]
Part a)
[tex]a = \frac{(3.9 \hat i + 3.3 \hat j) + (-3\hat i - 4\hat j)}{1.4}[/tex]
[tex]a = \frac{0.9 \hat i - 0.7 \hat j}{1.4}[/tex]
[tex]a = (0.64\hat i - 0.5 \hat j)m/s^2[/tex]
Part b)
[tex]a = \frac{(3.9 \hat i + 3.3 \hat j) + (-3\hat i + 4\hat j)}{1.4}[/tex]
[tex]a = \frac{0.9 \hat i + 7.3 \hat j}{1.4}[/tex]
[tex]a = (0.64\hat i + 5.21 \hat j)m/s^2[/tex]
Part c)
[tex]a = \frac{(3.9 \hat i + 3.3 \hat j) + (3\hat i - 4\hat j)}{1.4}[/tex]
[tex]a = \frac{6.9 \hat i - 0.7 \hat j}{1.4}[/tex]
[tex]a = (4.92\hat i - 0.5 \hat j)m/s^2[/tex]
If the atoms that share electrons have an unequal attraction for the electrons is called
If the atoms that share electrons have an unequal attraction for electrons, the bond is called a Polar covalent bond.
Explanation:A covalent chemical bond is formed in case of two different non-metals when one or more electron pairs are shared between bonding atoms. A difference in electronegativity of subsequent atoms of a covalent bond leads to formation of a small net charge around nucleus of each atom, pulling the shared electrons to one side of the bond, to the nucleus which has higher electronegativity.
HCl is an example of polar covalent bond and the HCl bond has Chlorine more electronegative. The bonding electrons are more close to Cl than H and hence Cl is partially negatively charged than H which has partial positive charge (HCl bond : [tex]H^{+} - Cl^{-}[/tex]). When electrons shared in a covalent bond have equal attraction, the bond is a Non-Polar covalent bond.
A motorcyclist is traveling along a road and accelerates for 4.50s to pass another cyclist. The angular acceleration of each wheelis +6.70 rad/s^2, and, just after passing, the angular velocity ofeach is +74.5 rad/s, where the plus signs indicate counterclockwisedirections. What is the angular displacement of each wheel duringthis time?
a. +221 rad
b. +131 rad
c. +335 rad
d. +355 rad
e. +267 rad
Answer:
Angular displacement of the wheel, [tex]\theta=267.41\ rad[/tex]
Explanation:
It is given that,
Angular acceleration of the wheel, [tex]\alpha =6.7\ rad/s^2[/tex]
Final speed of the wheel, [tex]\omega_f=74.5\ rad/s[/tex]
Time taken, t = 4.5 s
Initially, it is required to find the initial angular velocity of the wheel. Using the first equation of rotational kinematics as :
[tex]\omega_f=\omega_o+\alpha t[/tex]
[tex]\omega_o[/tex] is the initial speed of the wheel
[tex]\omega_o=\omega_f-\alpha t[/tex]
[tex]\omega_o=74.5-6.7\times 4.5[/tex]
[tex]\omega_o=44.35\ rad/s[/tex]
Let [tex]\theta[/tex] is the angular displacement of each wheel during this time. Using the second equation of motion as :
[tex]\theta=\omega_o t+\dfrac{1}{2}\alpha t^2[/tex]
[tex]\theta=44.35\times 4.5+\dfrac{1}{2}\times 6.7\times (4.5)^2[/tex]
[tex]\theta=267.41\ rad[/tex]
So, the angular displacement of each wheel during this time is 267 radian.
During fabrication processes, polymeric materials are generally subjected to which of the following conditions? The ambient temperature. Elevated temperatures. The ambient pressure. Elevated pressures.
Answer:
- Elevated temperatures.
- Elevated pressures.
Explanation:
In the manufacture of plastic products are used:
Raw material (pellets) which are the monomers that promote the chemical reaction.
To these are added the charges, in order to reduce the cost of the final product and improve some properties. These charges can be fiberglass, paper, metal structures.
Additives are also added whose mission is to improve or achieve certain properties, such as reducing friction, reducing chemical degradation, increasing electrical conductivity, coloring the product, and all this happens in the presence of a catalyst that is responsible for initiating and accelerating the chemical reaction process.
There are different methods of production of plastics, one of the most frequent is by injection.
Injection is a process that is carried out on machines similar to extrusion machines, in which the spindle, in addition to rotating, has an axial displacement.
In the injection, once the mold is filled, it is separated from the nozzle of the machine, breaking the feed channel. After a certain time, the piece already cooled is demoulded.
High pressures and temperatures are necessary, but parts of good finish and at high production speeds are obtained
The correct answer is: Elevated temperatures.
During fabrication processes, polymeric materials are often subjected to elevated temperatures. This is because many polymers are thermoplastic, meaning they soften and can be reshaped when heated. The elevated temperatures allow the polymer chains to move more freely, enabling the material to be molded, extruded, or otherwise formed into the desired shape or structure.
While ambient temperature and pressure can also affect polymeric materials, the key condition for shaping and fabricating these materials is typically elevated temperature. Ambient temperature is usually not sufficient to induce the necessary molecular mobility for fabrication. Ambient pressure is generally constant and does not play a significant role in the fabrication process unless specific pressure conditions are required for certain molding techniques, such as injection molding or compression molding, where controlled pressure is applied to ensure the polymer fills the mold properly.
A beam of electrons passes through a single slit, and a beam of protons passes through a second, but identical, slit. The electrons and the protons have the same speed. Which one of the following correctly describes the beam that experiences the greatest amount of diffraction?
(a) The electrons, because they have the smaller momentum and, hence, the smaller de Broglie wavelength
(b) The electrons, because they have the smaller momentum and, hence, the larger de Broglie wavelength
(c) The protons, because they have the smaller momentum and, hence, the smaller de Broglie wavelength
(d) The protons, because they have the larger momentum and, hence, the smaller de Broglie wavelength
(e) Both beams experience the same amount of diffraction, because the electrons and protons have the same de Broglie wavelength.
Answer:
(b) The electrons, because they have the smaller momentum and, hence, the larger de Broglie wavelength
Explanation:
de Broglie wavelength λ = h / m v
Since both electrons and protons have same velocity , momentum mv will be less for electrons because mass of electron is less .
for electron , momentum is less so . Therefore de Broglie wavelength λ will be more for electrons .
Amount of diffraction that is angle of diffraction is proportional to λ
Therefore electrons having greater de Broglie wavelength will show greater diffraction.
A physics student looks into a microscope and observes that small particles suspended in water are moving about in an irregular motion. Which of the following statements is the best explanation for this observation?
a. water molecules strike the particles giving them the same average kinetic energy as the water
b. the particles are carried by convection currents in the water
c. the small particles may be considered a fluid; and thus, move about randomly
d. the actual motion is regular, but the speeds of particles are too large to observe the regular motion
e. the particles are moving to be uniformly distributed throughout the volume of the water
Answer:
d. the actual motion is regular, but the speeds of particles are too large to observe the regular motion
Explanation:
The speeds of the particles are very large and comparatively the average free path is very small . Therefore time taken in covering the free path ( path between two consecutive collision with medium particles ) is very small . Hence the st line path covered by particles between two collision is less likely to be visible. Hence motion appears irregular or zig-zag.
A 70.0-kg person throws a 0.0420-kg snowball forward with a ground speed of 35.0 m/s. A second person, with a mass of 57.0 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.20 m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged?
Answer:
so throwers new velocity = 2.18032m/s
so catchers new velocity = 0.02577m/s
Explanation:
Directly by conservation of momentum we can write
[tex]m_1u_1+m_2u_2= m_1v_1+m_2v_2[/tex]
let x be the thrower's new velocity
(70+0.042)×2.2 + 57×0 = 70× x +0.042×35 +57×0
x = 2.18032m/s
so the velocity of 70 kg man = 2.18032m/s
so throwers new velocity = 2.18032m/s
now again by conservation of momentum
0.042×35 = (57+0.042) ×y
y = 0.02577m/s
so catchers new velocity = 0.02577m/s
A space telescope travels about Earth in a circular orbit at a distance of 380 miles from Earth's surface. It makes one orbit every 95 minutes. Find its linear velocity in miles per hour. (The radius of Earth is approximately 3960 miles.) Round to the nearest tenth place.
Final answer:
The linear velocity of the space telescope is about 17,193.2 miles per hour when rounded to the nearest tenth. This is calculated using the orbit radius of 4340 miles and the orbital period of 1.5833 hours.
Explanation:
To calculate the linear velocity of a space telescope traveling around Earth, we need to use the formula v = 2πr / T, where v is the linear velocity, r is the radius of the orbit, and T is the orbital period. To find the orbit radius, we add Earth's radius and the altitude of the telescope above Earth's surface: r = 3960 miles + 380 miles = 4340 miles.
Next, we convert the orbital period from minutes to hours. There are 60 minutes in an hour, so T = 95 minutes / 60 minutes/hour = 1.5833 hours. We can then plug the radius and period into the formula to find the linear velocity: v = 2 x π x 4340 miles / 1.5833 hours = 17,193 miles/hour. When rounded to the nearest tenth, the linear velocity is approximately 17,193.2 miles per hour.
Calculate the work required to move a planet’s satellite of mass 1820 kg from a circular orbit of radius 2R to one of radius 3R, where 7.37×106 m is the radius of the planet. The mass of the planet is 7.51 × 1024 kg. Answer in units of J.
Answer:
The work required to move a planet's satellite is W = 2854.61 J
Explanation:
Given data,
The mass of the satellite, m = 1820 kg
The radius of the circular orbit, r = 2R
The radius of the planet, r = 5.37 x 10⁶ m
The mass of the planet, M = 7.5 x 10²⁴ kg
The formula for work done from the 2R to 3R is,
W = [tex]\int_{2R}^{3R}\frac{GMm}{r^{2}}dr[/tex]
W = GMm/3R - GMm/2R
W = (-0.17)GMm/R J
The negative sign indicates that the energy stored in the satellite as the potential energy.
Substituting the values
W = (-0.17) 6.673 x 10⁻¹¹ X 7.51 x 10²⁴ X 1820 / (7.37 x 10⁶)²
= -2854.61 J
Hence, the work required to move a planet's satellite is W = 2854.61 J
Light with a wavelength of 587.5 nm illuminates a single slit 0.750 mm in width.
(a.) At what distance from the slit should a screen be located if the first minimum in the diffraction pattern is to be 0.850 mm from the center of the screen?
b.) What is the width of the central maximum?
(c.) Sketch the intensity distribution for the diffraction pattern observed on the viewing screen.
Answer:
a) The screen should be located at 1.08 meters
b) The width of the central maximum is 1.7 mm
c) See figure below.
Explanation:
a) This is a single slit diffraction problem, the equation that describes this kind of phenomenon is:
[tex]a\sin\theta=m\lambda [/tex] (1)
Because we’re interested in a minimum near the center of the screen, we can use the approximation [tex] \sin\theta\approx\tan\theta=\frac{y}{x} [/tex]
So equation (1) is now:
[tex]a\frac{y}{x}=m\lambda [/tex] (2)
Solving (2) for x:
[tex] x=\frac{ay}{m\lambda}=\frac{(0.75\times10^{-3})(0.85\times10^{-3})}{1(587.5\times10^{-9})}\approx1.08m [/tex]
b) As you can see on the figure below a maximum is approximately between the two adjacent minimums, because the diffraction pattern is approximately symmetric respect the center of the screen the width of the central maximum is 2*0.850mm = 1.7 mm.
Does the photoelectric effect support the wave theory of light? The particle theory of light?
Final answer:
The photoelectric effect supports the particle theory of light as it shows that light's ability to eject electrons depends on its frequency, indicative of light's particle-like behavior through photons.
Explanation:
The photoelectric effect does not support the wave theory of light, but rather supports the particle theory of light. According to classical wave theory, energy of light is related to its intensity or amplitude. However, the photoelectric effect demonstrates that the kinetic energy of ejected electrons from a metal surface depends on the light's frequency rather than its intensity.
This phenomenon can be explained by considering light to consist of particles called photons. Each photon carries a quantized amount of energy determined by the equation E = hv, where E is energy, h is Planck's constant, and v is frequency. If the frequency of light is above a certain threshold, it can dislodge electrons from the metal because the photons have sufficient energy, showing light's particle-like nature.
A 5.0 g coin is placed 15 cm from the center of a turntable. The coin has static and kinetic coefficients of friction with the turntable surface of μs = 0.70 and μk = 0.50. The turntable very slowly speeds up.
Answer:
Angular speed will reach 6.833rad/s before the coin starts slipping
Explanation:
There is no question but I'll asume the common one: Calculate the speed of the turntable before the coin starts slipping.
With a sum of forces:
[tex]Ff = m*a[/tex]
[tex]Ff=m*V^2/R[/tex]
At this point, friction force is maximum, so:
[tex]\mu*N=m*V^2/R[/tex]
[tex]\mu*m*g=m*V^2/R[/tex]
Solving for V:
[tex]V=\sqrt{\mu*g*R}[/tex]
V=1.025 m/s
The angular speed of the turntable will be:
ω = V/R = 6.833 rad/s This is the maximum speed it can reach before the coin starts slipping.
Given the atomic radius of argon, 0.97 Å, and knowing that a sphere has a volume of 4πr3/3, calculate the fraction of space that Ar atoms occupy in a sample of argon at STP. Express your answer using two significant figures.
Answer:
1.0x10^-4
Explanation:
First, in order to do this, we need to calculate the volume of 1 simple atom of Ar. Using the formula of the volume of a sphere we have the following
Converting A to cm:
0.97 * 1x10^-8 = 9.7x10^-9 cm
Now the volume:
V = 4/3π(9.7x10^-9)³
V = 3.82x10^-24 cm³
We know that 1 cm³ is 1 mL, and 1 L is 1000 mL so:
V = 3.82x10^-24 mL / 1000 = 3.82x10^-27 L
Now, using avogadro's number, we should get the total volume of all atoms of Ar so:
3.82x10^-27 * 6.02x10^23 = 2.3x10^-3 L
Finally, at STP the volume of an ideal gas is 22.4 L so:
2.3x10^-3 / 22.4 = 1.03x10^-4
With two significant figure, it would be 1.0x10^-4
As heat is added to water, is it possible for the temperature measured by a thermometer in the water to remain constant?
a. Maybe; it depends on the rate at which the heat is added.
b. No, adding heat will always change the temperature.
c. Yes, the water could be changing the phase.
d. Maybe; it depends on the initial water temperature.
Answer:
C. Yes, the water could be changing the phase.
Explanation:
Stars of spectral type A and F are considered ________.
a. reasonably likely to have Earth-like planets with complex plant- and animal-like life
b. reasonably likely to have habitable planets but much less likely to have planets with complex plant- or animal-like life
c. unlikely to have planets of any kind
d. unlikely to have habitable planets
Answer:
B. the stars of spectral type A and F are considered reasonably to have habitable planets but much less likely to have planets with complex plant - or animal - like life.
Explanation:
The appropriate spectral range for habitable stars is considered to be "late F" or "G", to "mid-K" or even late "A". This corresponds to temperatures of a little more than 7,000 K down to a little less than 4,000 K (6,700 °C to 3,700 °C); the Sun, a G2 star at 5,777 K, is well within these bounds. "Middle-class" stars (late A, late F, G , mid K )of this sort have a number of characteristics considered important to planetary habitability:
• They live at least a few billion years, allowing life a chance to evolve. More luminous main-sequence stars of the "O", "B", and "A" classes usually live less than a billion years and in exceptional cases less than 10 million.
• They emit enough high-frequency ultraviolet radiation to trigger important atmospheric dynamics such as ozone formation, but not so much that ionisation destroys incipient life.
• They emit sufficient radiation at wavelengths conducive to photosynthesis.
• Liquid water may exist on the surface of planets orbiting them at a distance that does not induce tidal locking.
Thus , the stars of spectral type A and F are considered reasonably to have habitable planets but much less likely to have planets with complex plant - or animak - like life.
Adding or removing protons from an atom does what to the atom
Answer:
Changes the element of the atom.
Explanation:
The elements are classified by the number of protons they have in their nucleus, so if the number of protons is changed (added or removed), that atom will become one of a different element.
For example, hydrogen has only 1 proton in its nucleus, and helium has 2. So if a proton is added to a hydrogen atom, it becomes a helium atom, and consequently its atomic number, wich is determided by the protons in an element, will also change.