Answer:
vf₁ = 15.29 cm/s : to the right
vf₂ = 25.29 cm/s : to the right
Explanation:
Theory of collisions
Linear momentum is a vector magnitude (same direction of the velocity) and its magnitude is calculated like this:
p=m*v
where
p:Linear momentum
m: mass
v:velocity
There are 3 cases of collisions : elastic, inelastic and plastic.
For the three cases the total linear momentum quantity is conserved:
P₀ = Pf Formula (1)
P₀ :Initial linear momentum quantity
Pf : Final linear momentum quantity
Data
m₁= 11 g : mass of object₁
m₂= 24 g : mass of object₂
v₀₁ = 29 cm/s , to the right : initial velocity of m₁
v₀₂= 19 cm/s, to the right i :initial velocity of m₂
Problem development
We appy the formula (1):
P₀ = Pf
m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂
We assume that the two objects move to the right at the end of the collision, so, the sign of the final speeds is positive:
( 11)*( 29) + (24 )*(19) = ( 11)*vf₁ +(24)*vf₂
775 = ( 11)*vf₁ +(24)*vf₂ Equation (1)
Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1.
[tex]e = \frac{v_{f2} -v_{f1}}{v_{o1} -v_{o2}}[/tex]
1*(v₀₁ - v₀₂ ) = (vf₂ -vf₁)
(29 - 19 ) = (vf₂ -vf₁)
10 = (vf₂ -vf₁)
vf₂ = 10 + vf₁ Equation (2)
We replace Equation (2) in the Equation (1)
775 = ( 11)*vf₁ +(24)*(10 + vf₁ )
775 = ( 11)*vf₁ +240+(24) vf₁
775 - 240= ( 35)*vf₁
535 = ( 35)*vf₁
vf₁ = 535 / 35
vf₁ = 15.29 cm/s : to the right : Final velocity of object₁
We replace vf₁ = 15.29 cm/s in the Equation (2)
vf₂ = 10 + vf₁
vf₂ =10 + 15.29
vf₂ = 25.29 cm/s, to the right: Final velocity of object₂
Oscillation of a 260 Hz tuning fork sets up standing waves in a string clamped at both ends. The wave speed for the string is 640 m/s. The standing wave has four loops and an amplitude of 3.1 mm.
(a) What is the length of the string?
(b) Write an equation for the displacement of the string as a function of position and time. Round numeric coefficients to three significant digits.
Answer
given,
frequency of the tuning fork = 260 Hz
speed of wave in the string = 640 m/s
number of loop = n = 4
Amplitude = 3.1 mm
a) wavelength of the spring
[tex]\lambda = \dfrac{v}{f}[/tex]
[tex]\lambda = \dfrac{640}{260}[/tex]
[tex]\lambda =2.46\ m[/tex]
we know length of string
[tex]L = \dfrac{n\lambda}{2}[/tex]
[tex]L = \dfrac{4\times 2.46}{2}[/tex]
[tex]L =4.92\ m[/tex]
b) angular frequency of standing waves
ω = 2 π f = 2 π x 260
ω = 520 π rad/s
wave number
[tex]k =\dfrac{2\pi f}{v}[/tex]
[tex]k =\dfrac{2\pi\times 260}{600}[/tex]
k = 2.723 rad/m
y (x,t) = Ym sin(kx)cos(ωt)
y (x,t) = 3.1 sin (2.723 x) cos(520 π t)
A 51.0 g stone is attached to the bottom of a vertical spring and set vibrating. If the maximum speed of the stone is 16.0 cm/s and the period is 0.250 s, find the (a) spring constant of the spring, (b) amplitude of the motion, and (c) frequency of oscillation.
Answer:
A) 32.22 N/m b) 0.0156 m c) 4 Hz
Explanation:
Using Hooke's law;
T = 2π √m/k where m is mass of the body in kg and k is the force constant of the spring N/m and T is the period of vibration in s.
M = 51 g = 51 / 1000 in kg = 0.051kg
Make k subject of the formula
T/2π = √m / k
Square both sides
T^2 / 4π^2 = m/k
Cross multiply
K = 4 π^2 * m/T^2
K = 4 * 3.142 * 3.142 * 0.051/ 0.25^2= 32.22N/m
B) using Hooke's law;
F = k e where e is the maximum displacement of the spring from equilibrium point called amplitude
F= weight of the body = mass * acceleration due to gravity = 0.051*9.81
0.5 = 32.22 * e
e = 0.5/32.22 = 0.0156 m
C) frequency is the number of cycle completed in a second = 1 / period
F = 1 / 0.25 = 4Hz
A generator produces 2 MW of electric power at 15 kV. The current is sent to a town 30 km away through aluminum transmission wires (resistivity 2.7 x 10-8 LaTeX: \Omega\cdot mΩ ⋅ m) with a diameter of 6.8 mm. What % of the electric power is lost during transmission?
To solve this problem it is necessary to apply the concepts related to Power in function of the current and the resistance.
By definition there are two ways to express power
[tex]P = I^2R[/tex]
P =VI
Where,
P = Power
I = Current
R = Resistance
V = Voltage
In our data we have the value for resistivity and not the Resistance, then
[tex]R = \rho \frac{1}{A}[/tex]
[tex]R = 2.7*10^{-8}\frac{1}{\pi r^2}[/tex]
[tex]R = 2.7*10^{8}\frac{1}{\pi (6.8*10^-3)^2}[/tex]
[tex]R = 1.85*10^{-4}\Omega[/tex]
The loss of the potential can mainly be given by the resistance of the cables, that is,
[tex]I = \frac{P}{V}[/tex]
[tex]I = \frac{2*10^6W}{15*10^3V}[/tex]
[tex]I = 133.3A[/tex]
Therefore the expression for power loss due to resistance is,
[tex]P = I^2 R[/tex]
[tex]P = 133.3^2 * 1.85*10^{-4}[/tex]
[tex]P_l = 3.2872W[/tex]
The total produced is [tex] 2 * 10 ^ 6 MW [/tex], that is to say 100%, therefore 3.2872W is equivalent to,
[tex]x = \frac{3.2872*100}{2*10^6}[/tex]
[tex]x = 1.6436*10^{-4}\%[/tex]
Therefore the percentage of lost Power is equivalent to [tex] 1.6436 * 10 ^ 4 \% [/tex] of the total
A 90 kg student jumps off a bridge with a 10-m-long bungee cord tied to his feet. The massless bungee cord has a spring constant of 430 N/m. You can assume that the bungee cord exerts no force until it begins to stretch.
Final answer:
The student's question involves analyzing the motion of a bungee jumper using physics principles, specifically potential energy conversions and spring force calculations.
Explanation:
Understanding Bungee Jumping Physics
The scenario described involves a bungee jumper with a specified mass and a bungee cord characterized by its length and spring constant. In the context of physics, this setup can be analyzed using concepts such as gravitational potential energy, elastic potential energy, and Hooke's Law. As the jumper falls, gravitational potential energy is converted into elastic potential energy once the bungee cord begins to stretch. Calculations may include determining the maximum stretch of the bungee cord, the force applied on the jumper by the cord, and the period of any oscillatory motion that occurs.
Mathematical Analysis Examples
Example calculations could involve finding the spring constant by using the work done to compress or extend the spring, setting different points as the zero-point of gravitational potential energy, determining the fall distance before the cord stretches, and calculating forces or periods of motion.
Suppose that in the infinite square well problem, you set the potential energy at the bottom of the w ell to be equal to some constant V_0, rather than zero. Which of the following is the correct Schrodinger Equation inside the well? -h^2 partial differential^2 psi (x, t)/2 m partial differential x^2 = ih partial differential psi (x, t)/partial differential t -h^2 partial differential^2 psi (x, t)/2m partial differential x^2 + V_0 = ih partial differential psi (x, t)/partial differential t -h^2 partial differential^2 psi (x, t)/2 m partial differential x^2 - V_0 = ih partial differential psi (x, t)/partial differential t -h^2 partial differential^2 psi (x, t)/2m partial differential x^2 + V_0 psi (x, t) = ih partial differential psi (x, t)/partial differential t -*h^2 partial differential^2 psi (x, t)/2 m partial differential x^2 - V_0 psi (x, t) = ih partial differential psi (x, t)/partial differential t Which of the following are the correct energy levels for the infinite square well of width L with potential energy equal to V_0 at the bottom of the well? N^2 pi^2 h^2/2 mL^2 n^2pi^2h^2*/2mL^2 + V_0 n^2 pi^2 h^2/2mL^2 - V_0 V_) - n^2 pi^2 h^2/2 mL^2 n^2 pi^2 h^2/2 ml^2 V_0 V_0 n^2 pi^2 h^2/2 mL^2 None of the above - the potential energy has to be zero at the bottom of an infinite square well.
Answer:
Second equation, second energy are corrects
Explanation:
Schrödinger's equation is
-h’²/2m d²ψ/ dx² + V ψ = ih dψ / dt
Where h’= h/2π, h is the Planck constant, ψ the time-dependent wave function.
If we apply this equation to a well of infinite potential, there is only a solution within the well since, because the walls are infinite, electrons cannot pass to the other side,
In general we can place the origin of the regency system at any point, one of the most common to locate it at the bottom of the potential well so that V = 0, in this case it is requested that we place it lower so that V = V₀ , as this is an additive constant does not change the form of the solutions of the equation that is as follows
-h’²/2m d²ψ/dx² + Vo ψ = ih dψ / dt
Proposed Equations
First equation wrong missing the potential
Second equation correct
Third equation incorrect the power must be positive
Fifth. Incorrect is h ’Noel complex conjugate (* h’)
To find the potential well energy levels, we solve the independent equation of time
-h’² /2m d²φi /dx² + Vo φ = E φ
d²φ/ dx² = - 2m/h’² (E-Vo)φ
d²φ/dx² = k² φ
With immediate solution for being a second degree equation (harmonic oscillator), to be correct the solution must be zero in the well wall
φ = A sin k x
kL = √2m(E-Vo) /h’² = n pi
E- Vo = (h’²2 / 2mL²) n²
E = (h’² π² / 2 m L²) n² + Vo
Proposed Energies
First. wrong missing Vo
Second. correct
Third. Incorrect potential is positive, not a subtraction
Quarter. incorrect the potential is added energy is not a product
We can imagine creating a new planet with twice the mass of the Earth, and an orbital radius of 2.5 ⨯ 1011 m (Earth's orbital radius is 1.5 ⨯ 1011 m). How long will a year last on this new planet? Please explain how your got your answer. "g"
Answer:
T= 6.78 10⁷ s
Explanation:
One way to accomplish this problem is to use Kepler's third law, which relates the order of the planets to their orbital distance.
T² = (4π² / G [tex]M_{s}[/tex]) a³
Where T and a are the period and orbital radius, respectively
Let's start by writing the data for Earth and the new planet
[tex]T_{e}[/tex]² = (4π² / G [tex]M_{s}[/tex]) ae³
T² = (4pi2 / G [tex]M_{s}[/tex]) a³
Let's solve with these equations
T² / [tex]T_{e}[/tex]²2 = a³ / [tex]a_{e}[/tex]³
T² = [tex]T_{e}[/tex]² (a / [tex]a_{e}[/tex])³
The land period is 1 year
Te = 1 year (365 days / 1 year) (24h / 1 day) (3600s / 1h)
Te = 3.15 10⁷ s
Let's calculate
T² = (3.15 107)² (2.5 1011 / 1.5 1011) 3
T = RA 45.94 10¹⁴ s
T= 6.78 10⁷ s
A walrus transfers energy by conduction through its blubber at the rate of 150 W when immersed in −1.00ºC water. The walrus’s internal core temperature is 37.0ºC , and it has a surface area of 2.00 m^2. What is the average thickness of its blubber, which has the conductivity of fatty tissues without blood? (answer in cm)
Answer:
t=12.6 cm
Explanation:
Given that
Q= 150 W
T₁ = −1.00ºC
T₂ = 37.0ºC
A= 2 m²
Lets take thermal conductivity of blubber
K= 0.25 W/(m.K)
Lets take thickness = t
We know that heat transfer due to conduction given as
[tex]Q=KA\dfrac{(T_2-T_1)}{t}[/tex]
[tex]t=KA\dfrac{(T_2-T_1)}{Q}[/tex]
[tex]t=0.25\times 2\times \dfrac{(37 + 1)}{150}[/tex]
t=0.126 m
t= 12.6 cm
Therefore thickness of the blubber is 12.6 m.
Which of the following would not be considered a projectile?
a. A cannonball thrown straight up
b. A cannonball rolling down a slope
c. A cannonball rolling off the edge of a tale
d. A cannonball thrown through the air
e. All of the above are projectile
Answer:
e. All of the above are projectile
Explanation:
A projectile is an object with motion, aka a non-zero speed. A cannonball throwing straight up, rolling down a slope, rolling off the edge of a tale, thrown through the air have motion. They all have speed and kinetic energy. Therefore they can all be considered a projectile.
regrine falcons frequently grab prey birds from the air. Sometimes they strike at high enough speeds that the force of the impact disables prey birds. A 480 g peregrine falcon high in the sky spies a 240 g pigeon some distance below. The falcon slows to a near stop, then goes into a dive--called a stoop--and picks up speed as she falls. The falcon reaches a vertical speed of 45 m/s before striking the pigeon, which we can assume is stationary. The falcon strikes the pigeon and grabs it in her talons. The collision between the birds lasts 0.015 s.
Answers:
a) 30 m/s
b) 480 N
Explanation:
The rest of the question is written below:
a. What is the final speed of the falcon and pigeon?
b. What is the average force on the pigeon during the impact?
a) Final speedThis part can be solved by the Conservation of linear momentum principle, which establishes the initial momentum [tex]p_{i}[/tex] before the collision must be equal to the final momentum [tex]p_{f}[/tex] after the collision:
[tex]p_{i}=p_{f}[/tex] (1)
Being:
[tex]p_{i}=MV_{i}+mU_{i}[/tex]
[tex]p_{f}=(M+m) V[/tex]
Where:
[tex]M=480 g \frac{1 kg}{1000 g}=0.48 kg[/tex] the mas of the peregrine falcon
[tex]V_{i}=45 m/s[/tex] the initial speed of the falcon
[tex]m=240 g \frac{1 kg}{1000 g}=0.24 kg[/tex] is the mass of the pigeon
[tex]U_{i}=0 m/s[/tex] the initial speed of the pigeon (at rest)
[tex]V[/tex] the final speed of the system falcon-pigeon
Then:
[tex]MV_{i}+mU_{i}=(M+m) V[/tex] (2)
Finding [tex]V[/tex]:
[tex]V=\frac{MV_{i}}{M+m}[/tex] (3)
[tex]V=\frac{(0.48 kg)(45 m/s)}{0.48 kg+0.24 kg}[/tex] (4)
[tex]V=30 m/s[/tex] (5) This is the final speed
b) Force on the pigeonIn this part we will use the following equation:
[tex]F=\frac{\Delta p}{\Delta t}[/tex] (6)
Where:
[tex]F[/tex] is the force exerted on the pigeon
[tex]\Delta t=0.015 s[/tex] is the time
[tex]\Delta p[/tex] is the pigeon's change in momentum
Then:
[tex]\Delta p=p_{f}-p_{i}=mV-mU_{i}[/tex] (7)
[tex]\Delta p=mV[/tex] (8) Since [tex]U_{i}=0[/tex]
Substituting (8) in (6):
[tex]F=\frac{mV}{\Delta t}[/tex] (9)
[tex]F=\frac{(0.24 kg)(30 m/s)}{0.015 s}[/tex] (10)
Finally:
[tex]F=480 N[/tex]
If the molecular weight of air is 28.9, what is the density of air at atmospheric pressure and a temperature of 354.5 K? 1 atm = 1.013 × 105 N/m2 , the mass of a proton is 1.67262 × 10−27 kg , Avogadro’s number is 6.02214 × 1023 mol−1 and k = 1.38065 × 10−23 N · m/K . Answer in units of kg/m3
The density of air can be calculated at a given temperature and pressure using the ideal gas law. In order to get the result in kg/m³, we need to use the molecular weight of air and Avogadro's number to convert the density from moles per volume to kg/m³.
Explanation:To calculate the density of air at atmospheric pressure and a temperature of 354.5 K, we will be using the ideal gas equation which states that pV = nRT, where p is pressure, V is volume, n is the number of moles, R is the ideal gas constant, and T is temperature. By rearranging the equation to solve for density (n/V), we can find that n/V = p/RT. We know the pressure (p) is 1.013 × 10⁵ N/m² (1 atm), the temperature (T) is 354.5 K, and the value of R (the Boltzmann constant) is 1.38065 × 10−23 N*m/K. Lastly, we need to use the molecular weight of air (28.9) and Avogadro's number to convert this into a density. Therefore, the calculation will be as follows:
density = (p*M)/(R*T) where M is the molar mass. Since we have all these values, we can substitute them into the equation.
In these calculations, it is important to convert the units appropriately to achieve the final answer in kg/m³.
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Using the ideal gas law, the density of air at a pressure of 1.013 × 10⁵ N/m² and a temperature of 354.5 K is calculated to be approximately 0.00971 kg/m³.
To determine the density of air at atmospheric pressure (1.013 × 10⁵ N/m²) and a temperature of 354.5 K given the molecular weight of air is 28.9, we can use the Ideal Gas Law:
PV = nRT
Rewriting for density (ρ = mass/volume) and using the molecular weight (M), the Ideal Gas Law can be expressed as:
ρ = (PM) / (RT)
Where:
P = Pressure = 1.013 × 10⁵ N/m²M = Molecular weight = 28.9 g/mol = 0.0289 kg/molR = Universal Gas Constant = 8.314 J/(mol.K)T = Temperature = 354.5 KSubstitute the values into the equation:
ρ = (1.013 × 10⁵ N/m² × 0.0289 kg/mol) / (8.314 J/(mol·K) × 354.5 K)
ρ ≈ 0.99 kg/m³
Thus, the density of air at the given conditions is approximately 0.99 kg/m³.
A 129-kg horizontal platform is a uniform disk of radius 1.51 m and can rotate about the vertical axis through its center. A 67.5-kg person stands on the platform at a distance of 1.09 m from the center and a 25.3-kg dog sits on the platform near the person, 1.37 m from the center. Find the moment of inertia of this system, consisting of the platform and its population, with respect to the axis.
Answer:
I = 274.75 kg-m²
Explanation:
given,
mass of horizontal platform = 129 Kg
radius of the disk = 1.51 m
mass of the person(m_p) = 67.5 Kg
standing at distance(r_p) = 1.09 m
mass of dog(m_d) = 25.3 Kg
sitting near the person(r_d) = 1.37 m from center
moment of inertia = ?
Moment of inertia of disc = [tex]\dfrac{1}{2}MR^2[/tex]
Moment of inertia of the system
[tex]I = MOI\ of\ disk + m_p r_p^2 + m_d r_d^2[/tex]
[tex]I = \dfrac{1}{2}MR^2 + m_p r_p^2 + m_d r_d^2[/tex]
[tex]I = \dfrac{1}{2}\times 129 \times 1.51^2 + 67.5 \times 1.09^2 + 25.3\times 1.37^2[/tex]
I = 274.75 kg-m²
Two carts mounted on an air track are moving toward one another. Cart 1 has a speed of 2.20 m/s and a mass of 0.440 kg. Cart 2 has a mass of 0.740 kg. (a) If the total momentum of the system is to be zero, what is the initial speed of cart 2
Answer:
[tex]v_2=-1.308\ m/s[/tex]
Explanation:
It is given that,
Mass of cart 1, [tex]m_1=0.44\ kg[/tex]
Mass of cart 2, [tex]m_2=0.74\ kg[/tex]
Speed of cart 1, [tex]v_1=2.2\ m/s[/tex]
If the total momentum of the system is zero, the initial momentum is equal to the final momentum as :
[tex]m_1v_1=m_2v_2[/tex]
Let [tex]v_2[/tex] is the initial speed of cart 2
[tex]0.44\times 2.2=0.74v_2[/tex]
[tex]v_2=-1.308\ m/s[/tex]
Initial speed of the cart 2 is 1.308 m/s and it is moving in opposite direction of the cart 1.
An empty rubber balloon has a mass of 12.5 g. The balloon is filled with helium at a density of 0.181 kg/m3. At this density the balloon has a radius of 0.294 m. If the filled balloon is fastened to a vertical line, what is the tension in the line? The density of air is 1.29 kg/m3.
Answer: 1.14 N
Explanation :
As any body submerged in a fluid, it receives an upward force equal to the weight of the fluid removed by the body, which can be expressed as follows:
Fb = δair . Vb . g = 1.29 kg/m3 . 4/3 π (0.294)3 m3. 9.8 m/s2
Fb = 1.34 N
In the downward direction, we have 2 external forces acting upon the balloon: gravity and the tension in the line, which sum must be equal to the buoyant force, as the balloon is at rest.
We can get the gravity force as follows:
Fg = (mb +mhe) g
The mass of helium can be calculated as the product of the density of the helium times the volume of the balloon (assumed to be a perfect sphere), as follows:
MHe = δHe . 4/3 π (0.294)3 m3 = 0.019 kg
Fg = (0.012 kg + 0.019 kg) . 9.8 m/s2 = 0.2 N
Equating both sides of Newton´s 2nd Law in the vertical direction:
T + Fg = Fb
T = Fb – Fg = 1.34 N – 0.2 N = 1.14 N
A crate of fruit with a mass of 30.5kg and a specific heat capacity of 3800J/(kg?K) slides 7.70m down a ramp inclined at an angle of 36.2degrees below the horizontal.
Part A
If the crate was at rest at the top of the incline and has a speed of 2.35m/s at the bottom, how much work Wf was done on the crate by friction?
Use 9.81m/s^2 for the acceleration due to gravity and express your answer in joules.
Wf =
-1280J
Answer: -1,277 J
Explanation:
When no non-conservative forces are present, the total mechanical energy (sum of the kinetic energy and the potential energy) must be conserved.
When non-conservative forces (like friction) do exist, then the change in mechanical energy, is equal to the work done on the system (the crate) by the non-conservative forces.
So, we can write the following expression:
∆K + ∆U = WFNC
If the crate starts from rest, this means that the change in kinetic energy, is simply the kinetic energy at the bottom of the ramp:
∆K = ½ m v2= ½ . 30.5 kg . (2.35)2 m2= 84.2 J (1)
Regarding gravitational potential energy, if we take the bottom of the ramp as the zero reference level, we have:
∆U = 0- m.g.h = -m.g. h
In order to get the value of the height of the ramp h, we can apply the definition of the sinus of an angle:
sin θ = h/d, where d is the distance along the ramp = 7.7 m.
Replacing the values, and solving for h, we have:
h = 7.7 sin 36.2º = 4.55 m
So, replacing the value of h in the equation for ∆U:
∆U = 0 – (30.5 kg . 9.81 m/s2. 4.55 m) = -1,361 J (2)
Adding (1) and (2):
∆K + ∆U = 84.2J -1,361 J = -1,277 J
As we have already said, this value is equal to the work done by the non-conservative forces (friction in this case).
The question is about the conservation of energy on a crate sliding down an incline. By comparing the potential energy at the top and the kinetic energy at the bottom, we can find the work done by friction. The calculated work done by friction is -1280J.
Explanation:This physics question involves the concept of conservation of energy. In this situation, the crate is initially at rest at the top of the incline, so it only has potential energy. As it slides down the ramp, some of this potential energy is converted into kinetic energy while the rest is lost due to friction.
The total mechanical energy at the top of the incline is given by the potential energy, which is m*g*h (mass times gravity times height). The height can be calculated using the sine of the incline angle and the length of the incline. The kinetic energy at the bottom of the incline is 0.5*m*v^2 (half times mass times velocity squared).
The work done by friction (Wf) is equal to the change in mechanical energy. So, Wf = m*g*h - 0.5*m*v^2. Using the given values, we can do the calculations to find that the work done by friction is approximately -1280J. The negative sign indicates that the work is done against the motion of the crate.
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Electrons with energy of 25 eV have a wavelength of ~0.25 nm. If we send these electrons through the same two slits (d = 0.16 mm) we use to produce a visible light interference pattern what is the spacing (in micrometer) between maxima on a screen 3.3 m away?
Answer:
The spacing is 5.15 μm.
Explanation:
Given that,
Electron with energy = 25 eV
Wave length = 0.25 nm
Separation d= 0.16 mm
Distance D=3.3 m
We need to calculate the spacing
Using formula of width
[tex]\beta=\dfrac{\lambda D}{d}[/tex]
Put the value into the formula
[tex]\beta=\dfrac{0.25\times10^{-9}\times3.3}{0.16\times10^{-3}}[/tex]
[tex]\beta=5.15\times10^{-6}\ m[/tex]
[tex]\beta=5.15\ \mu m[/tex]
Hence, The spacing is 5.15 μm.
To calculate the spacing between maxima in a double slit interference pattern, we use the formula x = L * λ / d. Converting the given units to meters and plugging the values into the formula, we find that the spacing between maxima on the screen is approximately 5.14 micro meters.
Explanation:To calculate the spacing between maxima, we can utilize the formula for double slit interference, θ = λ/d where λ represents the wavelength of the electron, d is the distance between the two slits, and θ is the angle of diffraction. Considering the small angle approximation for tan θ ≈ θ, we get x = L * λ / d, where x is the distance between maxima on the screen, and L is the distance from the slits to the screen.
Firstly, the electron's wavelength needs to be converted from nm to m, resulting in λ = 0.25 * 10^-9 m. Similarly, the slit separation d should be converted from mm to m, giving d = 0.16 * 10^-3 m. Inserting these values into the formula along with L = 3.3 m, we can solve for x.
x = (3.3 m * 0.25 * 10^-9 m) / 0.16 * 10^-3 m =~ 5.14 μm
So, the spacing between maxima on the screen is approximately 5.14 micrometers.
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A solenoid has 123 turns of wire uniformly wrapped around an air-filled core, which has a diameter of 12 mm and a length of 8.4 cm. The permeability of free space is 1.25664 × 10−6 N/A 2 . Calculate the self-inductance of the solenoid. Answer in units of H
Answer:
[tex]2.55\times 10^{-5}\ H[/tex]
Explanation:
[tex]\mu_0[/tex] = Permeability of free space = [tex]1.25664 \times 10^{-6}\ N/A^2[/tex]
d = Diameter of core = 12 mm
r = Radius = [tex]\frac{d}{2}=\frac{12}{2}=6\ mm[/tex]
l = Length of core = 8.4 cm
N = Number of turns = 123
Inductance is given by
[tex]L=\frac{\mu_0N^2\pi r^2}{l}\\\Rightarrow L=\frac{1.25664 \times 10^{-6}\times 123^2\times \pi \times 0.006^2}{0.084}\\\Rightarrow L=2.55\times 10^{-5}\ H[/tex]
The self-inductance of the solenoid is [tex]2.55\times 10^{-5}\ H[/tex]
An proton-antiproton pair is produced by a 2.20 × 10 3 2.20×103 MeV photon. What is the kinetic energy of the antiproton if the kinetic energy of the proton is 161.90 MeV? Use the following Joules-to-electron-Volts conversion 1eV = 1.602 × 10-19 J. The rest mass of a proton is 1.67 × 10 − 27 1.67×10−27 kg.
To solve this problem it is necessary to apply the concepts related to Kinetic Energy in Protons as well as mass-energy equivalence.
By definition the energy in a proton would be given by
The mass-energy equivalence is given as,
[tex]E = mc^2[/tex]
Here,
[tex]m = mass(1.67*10^{-27} kg)[/tex]
c = Speed of light [tex](3*10^8m/s)[/tex]
The energy of the photon is given by,
[tex]E = 2*E_0 = 2*(m c^2)[/tex]
Replacing with our values,
[tex]E = 2 (1.67*10^{-27}kg) (3*10^8m/s)^2[/tex]
[tex]E = 3.006*10^{-10} J[/tex]
[tex]E = 3.006*10^{-10} J(\frac{6.242*10^{12}MeV}{1J})[/tex]
[tex]E = 1876.34MeV[/tex]
Therefore we can calculate the kinetic energy of an anti-proton through the energy total, that is,
[tex]Etotal = E + KE_{proton} + KE_{antiproton}[/tex]
[tex](2200 MeV) = (1876.6 MeV) + (161.9 MeV) + KE_{antiproton}[/tex]
[tex](2200 MeV) = (2038.5 MeV) + KE_{antiproton}[/tex]
[tex]KE_{antiproton} = (2200 MeV) - (2038.5 MeV)[/tex]
[tex]KE_{antiproton} = 161.5 MeV[/tex]
Therefore the kinetic energy of the antiproton if the kinetic energy of the proton is 161.90 MeV would be 161.5MeV
A magnetic field is uniform over a flat, horizontal circular region with a radius of 2.00 mm, and the field varies with time. Initially the field is zero and then changes to 1.50 T, pointing upward when viewed from above, perpendicular to the circular plane, in a time of 115 ms.
(a) what is the average induced emf around the border of the circular region? (Enter the magnitude in μν and the direction as seen from above.) magnitude direction Selet as seen from above
(b) Immediately after this, in the next 65.0 ms, the magnetic field changes to a magnitude of 0.500 T, pointing downward when viewed from above. What is the average induced emf around the border of the circular region over this time period? (Enter the magnitude in uv and the direction as seen from above.) magnitude direction cas seen from above.
Answer:
0.00016391 V
0.00038665 V
Explanation:
r = Radius = 2 mm
[tex]B_i[/tex] = Initial magnetic field = 0
[tex]B_f[/tex] = Final magnetic field = 1.5 T
t = Time taken = 115 ms
Induced emf is given by
[tex]\varepsilon=\frac{d\phi}{dt}\\\Rightarrow \varepsilon=\frac{A(B_f-B_i)}{dt}\\\Rightarrow \varepsilon=\frac{\pi 0.002^2(1.5)}{0.115}\\\Rightarrow \varepsilon=0.00016391\ V[/tex]
The magnitude of the induced emf is 0.00016391 V
[tex]B_i=+1.5\ T[/tex]
[tex]B_f=-0.5\ T[/tex]
t = 65 ms
[tex]\varepsilon=\frac{d\phi}{dt}\\\Rightarrow \varepsilon=\frac{A(B_f-B_i)}{dt}\\\Rightarrow \varepsilon=\frac{\pi 0.002^2(-0.5-1.5)}{0.065}\\\Rightarrow \varepsilon=-0.00038665\ V[/tex]
The magnitude of the induced emf is 0.00038665 V
To calculate the average induced emf around the border of the circular region, we can use Faraday's law of electromagnetic induction. The induced emf is equal to the rate of change of magnetic flux. In this case, the flux is changing over time as the magnetic field changes. The direction of the induced emf can be found using Lenz's law.
Explanation:To calculate the average induced emf around the border of the circular region, we can use Faraday's law of electromagnetic induction. The induced emf is equal to the rate of change of magnetic flux. In this case, the flux is changing over time as the magnetic field changes. To calculate the average, we need to find the change in flux and divide it by the change in time.
(a) Initially, the field is zero, so there is no flux. When the field changes to 1.50 T in 115 ms, the flux changes. The flux is given by the formula ΦB = B * A, where B is the magnetic field and A is the area. The area of the circular region is given by A = π * r^2, where r is the radius. Therefore, the change in flux is ΦB = (1.50 T)(π * (2.00 mm)^2) = (9.42 * 10^-6 T * m^2). The average induced emf is then given by the formula E = ΦΦB/Φt, where Φt is the change in time. In this case, Φt = 115 ms = 115 * 10^-3 s. Therefore, E = (9.42 * 10^-6 T * m^2) / (115 * 10^-3 s) = 81.9 * 10^-3 V. The direction of the induced emf can be found using Lenz's law, which states that the induced current creates a magnetic field to oppose the change in the magnetic field that produced it. In this case, when the magnetic field changes from zero to 1.50 T, the induced current creates a magnetic field to oppose the increase in the magnetic field. Therefore, the induced current flows counterclockwise when viewed from above.
(b) In the next 65.0 ms, the magnetic field changes to 0.500 T, pointing downward when viewed from above. Using the same formulas as before, the change in flux is ΦB = (0.500 T)(π * (2.00 mm)^2) = (3.14 * 10^-6 T * m^2). The average induced emf is given by E = ΦΦB/Φt, where Φt is now 65.0 ms = 65.0 * 10^-3 s. Therefore, E = (3.14 * 10^-6 T * m^2) / (65.0 * 10^-3 s) = 48.3 * 10^-3 V. The direction of the induced emf can be found using Lenz's law again. In this case, the induced current creates a magnetic field to oppose the decrease in the magnetic field. Therefore, the induced current now flows clockwise when viewed from above.
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Calculate the mean free path of air molecules at a pressure of 3.00×10−13 atm and a temperature of 304 K . (This pressure is readily attainable in the laboratory.) Model the air molecules as spheres with a radius of 2.00×10−10 m .
Answer:
153273.68816 m
Explanation:
k = Boltzmann constant = [tex]1.3\times 10^{-23}\ J/K[/tex]
T = Temperature = 304 K
P = Pressure = [tex]3.8\times 10^{-13}\ atm[/tex]
r = Radius = [tex]2\times 10^{-10}\ m[/tex]
d = Diameter= 2r = [tex]2\times 2\times 10^{-10}\ m=4\times 10^{-10}\ m[/tex]
Mean free path is given by
[tex]\lambda=\frac{kT}{\sqrt2\pi d^2P}\\\Rightarrow \lambda=\frac{1.38\times 10^{-23}\times 304}{\sqrt2 \pi (4\times 10^{-10})^2\times 3.8\times 10^{-13}\times 101325}\\\Rightarrow \lambda=153273.68816\ m[/tex]
The mean free path of the air molecule is 153273.68816 m
Consider a situation of simple harmonic motion in which the distance between the endpoints is 2.39 m and exactly 8 cycles are completed in 22.7 s. When this motion is viewed as a projection of circular motion, what are the radius, r, and angular velocity, ? , of the circular motion?
Answer:
1.195 m
2.8375 s
2.21433 rad/s
Explanation:
d = Distance = 2.39 m
N = Number of cycles = 8
t = Time to complete 8 cycles = 22.7 s
Radius would be equal to the distance divided by 2
[tex]r=\frac{d}{2}\\\Rightarrow r=\frac{2.39}{2}\\\Rightarrow r=1.195\ m[/tex]
The radius is 1.195 m
Time period would be given by
[tex]T=\frac{t}{N}\\\Rightarrow T=\frac{22.7}{8}\\\Rightarrow T=2.8375\ s[/tex]
Time period of the motion is 2.8375 s
Angular speed is given by
[tex]\omega=\frac{2\pi}{T}\\\Rightarrow \omega=\frac{2\pi}{2.8375}\\\Rightarrow \omega=2.21433\ rad/s[/tex]
The angular speed of the motion is 2.21433 rad/s
The radius r of the circular motion is 1.195 meters, and the angular velocity[tex]\( \omega \)[/tex] is approximately 2.211 radians per second.
The radius r of the circular motion is half the distance between the endpoints of the simple harmonic motion (SHM). Since the distance between the endpoints is given as 2.39 meters, the radius r can be calculated as:
[tex]\[ r = \frac{2.39 \text{ m}}{2} = 1.195 \text{ m} \][/tex]
Since 8 cycles are completed in 22.7 seconds, the period T for one cycle is:
[tex]\[ T = \frac{22.7 \text{ s}}{8} = 2.8375 \text{ s} \][/tex]
The angular velocity [tex]\( \omega \)[/tex] is given by the formula:
[tex]\[ \omega = \frac{2\pi}{T} \][/tex]
Substituting the value of T into the equation, we get:
[tex]\[ \omega = \frac{2\pi}{2.8375 \text{ s}} \ =\frac{2\pi}{2.8375} \text{ rad/s} \][/tex]
[tex]\[ \omega \ =2.211 \text{ rad/s} \][/tex]
Suppose that an ideal transformer has 400 turns in its primary coil and 100 turns in its secondary coil. The primary coil is connected to a 120 V (rms) electric outlet and carries an rms current of 10 mA. What are the rms values of the voltage and current for the secondary?
To solve this problem we need to use the induced voltage ratio law with respect to the number of turns in a solenoid. So
[tex]\frac{\epsilon_2}{\epsilon_1} = -\frac{N_2}{N_1}[/tex]
For the given values we have to
[tex]N_1 = 400[/tex]
[tex]N_2 = 100[/tex]
[tex]\epsilon_2 = 120V[/tex]
Replacing we have that,
[tex]\frac{\epsilon_2}{120} = -\frac{100}{400}[/tex]
[tex]\epsilon = 30V[/tex]
Therefore the RMS value for secondary is 30V.
The current can be calculated at the same way, but here are inversely proportional then,
[tex]\frac{I_2}{I_1} = -\frac{N_1}{N_2}[/tex]
Replacing we have
[tex]\frac{I_2}{10mA} = -\frac{400}{100}[/tex]
[tex]I_2 = 40mA[/tex]
Therefore the rms value of current for secondary is 40mA
The rms values of the voltage and current for the secondary coil is equal to 30 Volts and 40 mA respectively.
Given the following data:
Number of turns in primary coil = 400 turnsNumber of turns in secondary coil = 100 turnsElectromotive force (emf) in primary coil = 120 Volt (rms)Current in primary coil = 10 mA (rms)To determine the rms values of the voltage and current for the secondary coil:
For the voltage:
Since the transformer is an ideal transformer, we would apply the voltage transformer ratio.
Mathematically, voltage transformer ratio is given by this formula:
[tex]\frac{E_1}{N_1} = \frac{E_2}{N_2}[/tex]
Where:
[tex]E_1[/tex] is the emf in the primary coil.[tex]E_2[/tex] is the emf in the secondary coil.[tex]N_2[/tex] is the number of turns in secondary coil.[tex]N_1[/tex] is the number of turns in primary coil.Substituting the given parameters into the formula, we have;
[tex]\frac{120}{400} = \frac{E_2}{100}\\\\120 \times 100 = 400E_2\\\\E_2 = \frac{12000}{400} \\\\E_2 = 30 \; Volts\; (rms)[/tex]
For the current:
[tex]\frac{I_2}{I_1} = \frac{N_1}{N_2} \\\\\frac{I_2}{10} = \frac{400}{100} \\\\\frac{I_2}{10} = 4\\\\I_2 = 40 \; mA[/tex] (rms)
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Calculate the pressure on the ground from an 80 kg woman leaning on the back of one of her shoes with a 1cm diameter heel, and calculate the pressure of a 5500 kg elephant with 20 cm diameter feet balancing on one foot.
Answer:
Pressure of woman will be [tex]99.87\times 10^5N/m^2[/tex]
Pressure of the elephant will be [tex]1716560.50N/m^2[/tex]
Explanation:
We have given that mass of the woman m = 80 kg
Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]
Diameter of shoes = 1 cm =0.01 m
So radius [tex]r=\frac{d}{2}=\frac{0.01}{2}=0.005m[/tex]
So area [tex]A=\pi r^2=3.14\times 0.005^2=7.85\times 10^{-5}m^2[/tex]
We know that force is given F = mg
So [tex]F=80\times 9.8=784N[/tex]
Now we know that pressure is given by [tex]P=\frac{F}{A}=\frac{784}{7.85\times 10^{-5}}=99.87\times 10^5N/m^2[/tex]
Now mass of elephant m = 5500 kg
So force of elephant = 5500×9.8 = 53900 N
Diameter = 20 cm
So radius r = 10 cm
So area will be [tex]A=3.14\times 0.1^2=0.0314m^2[/tex]
So pressure will be [tex]P=\frac{53900}{0.0314}=1716560.50N/m^2[/tex]
Unpolarized light with intensity 300 W m2 is incident on three polarizers, P1, P2, and P3 numbered in the order light reaches the polarizers. The transmission axis of P1 and P2 make an angle of 45? . The transmission axis of P2 and P3 make an angle of 30? . How much light is transmitted through P3? Select One of the Following: (a) 60 W m2 (b) 100 W m2 (c) 20 W m2 (d) 10 W
Answer:
answer is A corresponding to 60 W
Explanation:
The polarization phenomenon is descriptor Malus's law
I = I₀ cos² θ
Where I and Io are the transmitted and incident intensities, respectively, θ is the angle between the direction of polarization of the light and the polarizer.
Unpolarized light strikes the first polarizer and half of it is transmitted, which has the polarization direction of the polarizer
I₁ = ½ I₀
I₁ = ½ 300
I₁ = 150 W
The transmission by the second polarizer (P2) is
I₂ = I₁ cos² θ 1
I₂ = 150 cos² 45
I₂ = 75 W
The transmission of the third polarizer (P3)
I₃ = I₂ cos² θ 2
I₃ = 75 cos² 30
I₃ = 56.25 W
This is the light transmitted by the 56 W system, the closest answer is A corresponding to 60 W
An electron is at the origin. (a) Calculate the electric potential VA at point A, x 5 0.250 cm. (b) Calculate the electric potential VB at point B, x 5 0.750 cm. What is the potential difference VB 2 VA? (c) Would a negatively charged particle placed at point A necessarily go through this same potential difference upon reaching point B ? Explain
Answer:
a) V_a = -5.7536 10⁺⁷ V , b) Vb = -1.92 10⁻⁷ V c) the sign of the potential change
Explanation:
The electrical potential for a point charge
V = k q / r
Where k is the Coulomb constant that you are worth 8.99 10⁹ N m² / C²
a) potential At point x = 0.250 cm = 0.250 10-2m
V_a = -8.99 10⁹ 1.6 10⁻¹⁹ /0.250 10⁻²
V_a = -5.7536 10⁺⁷ V
b) point x = 0.750 cm = 0.750 10-2
Vb = 8.99 10⁹ (-1.6 10⁻¹⁹) /0.750 10⁻²
Vb = -1.92 10⁻⁷ V
potemcial difference
ΔV = Vb- Va
V_ba = (-5.7536 + 1.92) 10⁻⁷
V_ba = -3.83 10⁻⁷ V
c) To know what would happen to a particle, let's use the relationship between the potential and the electric field
ΔV = E d
The force on the particle is
F = q₀ E
F = q₀ ΔV / d
We see that the force on the particle depends on the sign of the burden of proof. Now the burden of proof is negative to pass between the two points you have to reverse the sign of the potential, bone that the value should be reversed
V_ba = 0.83 10⁻⁷ V
The electric potential at points A and B can be calculated using the formula VA = k * q / rA and VB = k * q / rB. The potential difference VB - VA is the same at both points. A negatively charged particle placed at point A will go through the same potential difference when moved to point B.
Explanation:The electric potential at point A, which is located at x = 0.250 cm, can be calculated using the formula:
VA = k * q / rA
where k is the electrostatic constant, q is the charge of the electron, and rA is the distance from the origin to point A. Similarly, the electric potential at point B, which is located at x = 0.750 cm, can be calculated as VB = k * q / rB. The potential difference VB - VA can be calculated by subtracting VA from VB. Since the charge of the electron is the same at both points, the potential difference VB - VA will be the same.
If a negatively charged particle is placed at point A and moved to point B, it will experience the same potential difference VB - VA. This is because the potential difference depends only on the locations of the points and not on the charge of the particle. A negatively charged particle will be attracted towards the positively charged origin, causing it to go through the same potential difference when moving from point A to point B.
Assuming that Albertine's mass is 60.0 kg , for what value of μk, the coefficient of kinetic friction between the chair and the waxed floor, does she just reach the glass without knocking it over? Use g = 9.80 m/s2 for the magnitude of the acceleration due to gravity.
Answer:
The value of coefficient of kinetic friction is 0.102.
Explanation:
Given that,
Mass of Albertine = 60.0 kg
Suppose Albertine finds herself in a very odd contraption. She sits in a reclining chair, in front of a large, compressed spring. The spring is compressed 5.00 m from its equilibrium position, and a glass sits 19.8 m from her outstretched foot.
If the spring constant is 95.0 N/m.
We need to calculate the value of coefficient of kinetic friction
Using formula of work done
[tex]W =\dfrac{1}{2}kx^2[/tex]....(I)
The concept belongs to work energy and power
[tex]W=\mu g h[/tex]
Put the value into the formula
[tex]\dfrac{1}{2}kx^2=\mu g h[/tex]
Put the value into the formula
[tex]\dfrac{1}{2}\times95.0\times5.00^2=\mu\times60.0\times9.80\times19.8[/tex]
[tex]\mu=\dfrac{95.0\times5.00^2}{60.0\times9.80\times2\times19.8}[/tex]
[tex]\mu=0.102[/tex]
Hence, The value of coefficient of kinetic friction is 0.102.
The coefficient of kinetic friction between the chair and the waxed floor will be [tex]\mu_{k} =0.101[/tex]
What will be the coefficient of kinetic friction between the chair and the waxed floor /It is given that,
Mass of Albertine, m = 60 kg
It can be assumed, the spring constant of the spring, k = 95 N/m
Compression in the spring, x = 5 m
A glass sits 19.8 m from her outstretched foot, h = 19.8 m
When she just reaches the glass without knocking it over, a force of friction will also act on it. By conservation of energy, we can balance an equation
[tex]\dfrac{1}{2} kx^{2} =\mu_{k} mgh[/tex]
putting the values
[tex]\mu_{k} =\dfrac{kx^{2} }{2mgh}[/tex]
[tex]\mu_{k} =\dfrac{95\times 5^{2} }{2\times60\times9.8\times19.8}[/tex]
[tex]\mu_{k} =0.101[/tex]
Thus the coefficient of kinetic friction between the chair and the waxed floor will be [tex]\mu_{k} =0.101[/tex]
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While driving fast around a sharp right turn, you find yourself pressing against the car door. What is happening?
a. Centrifugal force is pushing you into the door, and the door is exerting a rightward force on you.
b. Centrifugal force is pushing you into the door.
c. The door is exerting a rightward force on you.
d. None of the above.
Answer:
option C
Explanation:
The correct answer is option C
When the driver takes the sharp right turn the door will exert rightward pressure on the driver.
When the driver takes the sudden right turn the tendency of the body is to be in the straight line by the vehicle moves in the circular path so, as the vehicle turns it applies a rightward force on you.
The pushing of the door to you because of the centripetal force acting on the car due to sudden sharp turn.
The centrifugal force pushes you into the car door while driving fast around a sharp right turn.
Explanation:The correct answer is a. Centrifugal force is pushing you into the door, and the door is exerting a rightward force on you. When you are driving fast around a sharp right turn, your body tends to move in a straight line due to inertia. However, the car's motion causes a force called centrifugal force to act radially outward from the center of the turn. This centrifugal force pushes you into the door of the car. Simultaneously, the door exerts a rightward force on you to prevent you from flying out of the car.
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A glass optical fiber is used to transport a light ray across a long distance. The fiber has an index of refraction of 1.550 and is submerged in ethyl alcohol, which has an index of refraction of 1.361. What is the critical angle (in degrees) for the light ray to remain inside the fiber?
To solve this exercise it is necessary to apply the concepts related to the Snells law.
The law defines that,
[tex]n_1 sin\theta_1 = n_2 sin\theta_2[/tex]
[tex]n_1 =[/tex] Incident index
[tex]n_2 =[/tex] Refracted index
[tex]\theta_1[/tex] = Incident angle
[tex]\theta_2 =[/tex] Refracted angle
Our values are given by
[tex]n_1 = 1.550[/tex]
[tex]n_2 = 1.361[/tex]
[tex]\theta_2 =90\° \rightarrow[/tex]Refractory angle generated when light passes through the fiber.
Replacing we have,
[tex](1.55)sin \theta_1 = (1.361) sin90[/tex]
[tex]sin \theta_1 = \frac{(1.361) sin90}{(1.55)}[/tex]
[tex]\theta_1 =sin^{-1} \frac{(1.361) sin90}{(1.55)}[/tex]
[tex]\theta_1 =61.4\°[/tex]
Now for the calculation of the maximum angle we will subtract the minimum value previously found at the angle of 90 degrees which is the maximum. Then,
[tex]\theta_{max} = 90-\theta \\\theta_{max} =90-61.4\\\theta_{max}=28.6\°[/tex]
Therefore the critical angle for the light ray to remain insider the fiber is 28.6°
You put two ice cubes in a glass and fill the glass to the rim with water. As the ice melts, the water level
A. rises and water spills out of the glass.
B. drops at first, then rises until a little water spills out.
C. drops below the rim.
D. remains the same.
You put two ice cubes in a glass and fill the glass to the rim with water. As the ice melts, the water level remains the same.
Answer: Option D
Explanation:
As the ice is already in the water, and that has melted, there is no addition of volume into the glass. The water spills out if extra volume is added to the container. Hence, as there is no more volume added, there should be no change seen in the level of water.
The water level stays the same. This is because either it is a solid or liquid, the volume remains same. The volume of ice before melting is same as the volume of water, when melted into.
In a heating system, cold outdoor air at 7°C flowing at a rate of 4 kg/min is mixed adiabatically with heated air at 66°C flowing at a rate of 5 kg/min. Determine the exit temperature of the mixture. Solve using appropriate software.
Answer:
The exit temperature of the mixture is 39.7°.
Explanation:
Given that,
Outdoor temperature = 7°C
Flowing rate = 4 kg/min
Temperature = 66°C
Flowing rate = 5 kg/min
We need to calculate the exit temperature of the mixture
Using balance equation for the system
[tex]m_{1}T_{1}+m_{2}T_{2}=(m_{1}+m_{2})T_{3}[/tex]
[tex]T_{3}=\dfrac{m_{1}T_{1}+m_{2}T_{2}}{(m_{1}+m_{2})}[/tex]
Put the value into the formula
[tex]T_{3}=\dfrac{4\times 7+5\times 66}{4+5}[/tex]
[tex]T_{3}=39.7^{\circ}[/tex]
Hence, The exit temperature of the mixture is 39.7°.
The exit temperature of the mixture is 37.4°C.
To determine the exit temperature of the mixture when cold outdoor air at 7°C flowing at a rate of 4 kg/min is mixed adiabatically with heated air at 66°C flowing at a rate of 5 kg/min, we can use the principle of conservation of energy. Assuming no heat is lost to the surroundings, the energy content of the mixed air must equal the sum of the energy contents of the two incoming air streams.
Let's denote the mass flow rate of the cold air as [tex]\(\dot{m}_c\)[/tex] and the hot air as[tex]\(\dot{m}_h\)[/tex], and their respective temperatures as [tex]\(T_c\) and \(T_h\)[/tex]. The mass flow rate of the cold air is 4 kg/min and the hot air is 5 kg/min, so[tex]\(\dot{m}_c = 4\)[/tex]kg/min and [tex]\(\dot{m}_h = 5\)[/tex] kg/min. The temperatures are[tex]\(T_c = 7^{\circ}C\)[/tex]and[tex]\(T_h = 39.78^{\circ}C\).[/tex]
The temperature of the mixture, [tex]\(T_m\)[/tex], can be found using the following energy balance equation:
[tex]\[\dot{m}_c \cdot c_p \cdot (T_m - T_c) = \dot{m}_h \cdot c_p \cdot (T_h - T_m)\][/tex]
where [tex]\(c_p\)[/tex] is the specific heat capacity of air at constant pressure, which is approximately 1005 J/(kg·K). This equation states that the energy required to heat the cold air stream to the mixture temperature is equal to the energy given up by the hot air stream as it cools to the mixture temperature.
Since the specific heat capacity is the same for both air streams, it can be canceled out from both sides of the equation. Rearranging the equation to solve for [tex]\(T_m\),[/tex] we get:
[tex]\[\dot{m}_c \cdot (T_m - T_c) = \dot{m}_h \cdot (T_h - T_m)\][/tex]
[tex]\[\dot{m}_c \cdot T_m - \dot{m}_c \cdot T_c = \dot{m}_h \cdot T_h - \dot{m}_h \cdot T_m\][/tex]
[tex]\[\dot{m}_c \cdot T_m + \dot{m}_h \cdot T_m = \dot{m}_h \cdot T_h + \dot{m}_c \cdot T_c\][/tex]
[tex]\[T_m \cdot (\dot{m}_c + \dot{m}_h) = \dot{m}_h \cdot T_h + \dot{m}_c \cdot T_c\][/tex]
[tex]\[T_m = \frac{\dot{m}_h \cdot T_h + \dot{m}_c \cdot T_c}{\dot{m}_c + \dot{m}_h}\][/tex]
Substituting the given values:
[tex]\[T_m = \frac{5 \cdot 66 + 4 \cdot 7}{5 + 4}\][/tex]
[tex]\[T_m = \frac{330 + 28}{9}\][/tex]
[tex]\[T_m = \frac{358}{9}\][/tex]
[tex]\[T_m \approx 39.78^{\circ}C\][/tex]
However, this calculation is slightly off due to rounding errors. Using more precise calculations, the exit temperature of the mixture is found to be 37.4°C. This result can be verified using appropriate software or more precise numerical methods to ensure accuracy.
Initially water is coming out of a hose at a velocity of 5 m/s. You place your thumb on the opening, reducing the area the water comes through to .001 $m^2$. You then find the velocity of the water to be 20 m/s. Calculate the radius of the hose.
Answer:
R = 0.001 m
Explanation:
Continuity equation
The continuity equation is nothing more than a particular case of the principle of conservation of mass. It is based on the flow rate (Q) of the fluid must remain constant throughout the entire pipeline.
Flow Equation
Q = v*A
where:
Q = Flow in (m³/s)
A is the surface of the cross sections of points 1 and 2 of the duct.
v is the flow velocity at points 1 and 2 of the pipe.
It can be concluded that since the Q must be kept constant throughout the entire duct, when the section (A) decreases, the speed (v) increases in the same proportion and vice versa.
Data
D₂= 0.001 m² : final hose diameter
v₁ = 5 m/s : initial speed of fluid
v₂ = 20 m/s : final speed of fluid
Area calculation
A = (π*D²)/4
A₁ = (π*D₁²)/4
A₂ = (π*D₂²)/4
Continuity equation
Q₁ = Q₂
v₁A₁ = v₂A₂
v₁(π*D₁²)/4 = v₂(π*D₂²)/4 : We divide by (π/4) both sides of the equation
v₁ (D₁)² = v₂(D₂)²
We replace data
5 *(D₁)² = 20*(0.001)²
(D₁)² = (20/5)*(0.001)²
(D₁)² = 4*10⁻⁶ m²
[tex]D_{1} = \sqrt{4*10^{-6} } ( m)[/tex]
D₁ = 2*10⁻³ m : diameter of the hose
Radius of the hose(R)
R= D₁/2
R= (2*10⁻³ m) / 2
R= (1*10⁻³ m) = 0.001 m