The concept used to solve this problem is the conservation of momentum and the conservation of energy.
For conservation of the moment we have the definition:
[tex]m_1v_1 = (m_1+m_2)v_f[/tex]
Where,
m = Mass
[tex]v_1[/tex] = Initial velocity for object 1
[tex]v_f[/tex] = Final velocity
Replacing the values we have to,
[tex]m_1v_1 = (m_1+m_2)v_f[/tex]
[tex]5*12=(5+5)v_f[/tex]
[tex]v_f = 6m/s[/tex]
By conservation of energy we know that the potential energy is equal to the kinetic energy then
[tex]mgh = \frac{1}{2} m(v_f^2-v_i^2)[/tex]
[tex]gh = \frac{1}{2} v_f^2[/tex]
[tex]h = \frac{1}{2} g*v_f^2[/tex]
[tex]h = \frac{1}{2} (9.8)(6)^2[/tex]
[tex]h = 1.837m[/tex]
Therefore after the collision the height when the combined chinks will go is 1.837m
The combined chunks of ice will rise about 3.67 meters above the valley floor after their collision. This is calculated using the principles of conservation of momentum and energy where the initial kinetic energy of the moving ice chunk is conserved and then converted into gravitational potential energy as the combined chunk of ice ascends.
Explanation:This problem involves the concepts of conservation of momentum and energy. The two chunks of ice are initially sliding with a certain momentum and kinetic energy. When they collide and stick together, the total momentum must still be conserved because the system is isolated (i.e., no external forces are acting). Although the speed will be halved (since the total mass doubled), the total kinetic energy is conserved in the collision. This kinetic energy will then be converted to potential energy as the combined chunk of ice ascends. Using the equation for gravitational potential energy (PE = mgh), where m is the mass (10 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height, we can solve for h.
The initial kinetic energy of 5kg chunk of ice is KE = 1/2 mv^2, which equals 1/2 * 5kg * (12m/s)^2 = 360 Joules. After the collision, the kinetic energy is equally shared with the other chunk of ice. Since the initial mass is doubled but the speed is halved, the kinetic energy remains the same. Therefore, we set the gravitational potential energy equal to the initial kinetic energy when it reaches its peak: mgh = KE, 10kg * 9.8 m/s^2 * h = 360 Joules. Solving for h, we get that h is approximately 3.67 meters. Thus, the combined chunks of ice will rise about 3.67 meters above the valley floor after the collision.
Learn more about Conservation of Momentum and Energy here:https://brainly.com/question/3920210
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The radius of the aorta is «10 mm and the blood flowing through it has a speed of about 300 mm/s. A capillary has a radius of about 4ˆ10´3 mm but there are literally billions of them. The average speed of blood through the capillaries is about 5ˆ10´4 m/s. (i) Calculate the effective cross sectional area of the capillaries and (ii) the approximate number of capillaries.
Answer:
(I). The effective cross sectional area of the capillaries is 0.188 m².
(II). The approximate number of capillaries is [tex]3.74\times10^{9}[/tex]
Explanation:
Given that,
Radius of aorta = 10 mm
Speed = 300 mm/s
Radius of capillary [tex]r=4\times10^{-3}\ mm[/tex]
Speed of blood [tex]v=5\times10^{-4}\ m/s[/tex]
(I). We need to calculate the effective cross sectional area of the capillaries
Using continuity equation
[tex]A_{1}v_{1}=A_{2}v_{2}[/tex]
Where. v₁ = speed of blood in capillarity
A₂ = area of cross section of aorta
v₂ =speed of blood in aorta
Put the value into the formula
[tex]A_{1}=A_{2}\times\dfrac{v_{2}}{v_{1}}[/tex]
[tex]A_{1}=\pi\times(10\times10^{-3})^2\times\dfrac{300\times10^{-3}}{5\times10^{-4}}[/tex]
[tex]A_{1}=0.188\ m^2[/tex]
(II). We need to calculate the approximate number of capillaries
Using formula of area of cross section
[tex]A_{1}=N\pi r_{c}^2[/tex]
[tex]N=\dfrac{A_{1}}{\pi\times r_{c}^2}[/tex]
Put the value into the formula
[tex]N=\dfrac{0.188}{\pi\times(4\times10^{-6})^2}[/tex]
[tex]N=3.74\times10^{9}[/tex]
Hence, (I). The effective cross sectional area of the capillaries is 0.188 m².
(II). The approximate number of capillaries is [tex]3.74\times10^{9}[/tex]
A uniform plank is 3.0 m long and has a mass of 10 kg. It is secured at its left end in a horizontal position to be used as a diving platform. To keep the plank in equilibrium, the point of support must supply: a. an upward force and a clockwise torque b. a downward force and a clockwise torque c. an upward force and a counter-clockwise torque d. a downward force and a counter-clockwise torque e. none of these
To develop the problem it is necessary to take into account the concepts related to Torque and sum of moments.
By torque it is understood that
[tex]\tau = F*d[/tex]
Where,
F= Force
d = Distance
The value of the given Torque acts from the center of mass causing it to rotate clockwise.
The Force must then be located at the other end down to make a movement opposite the Torque in the center of mass.
I enclose a graph that allows us to understand the problem in a more didactic way.
The correct answer is D.
A vehicle moves with a velocity, v(t) = exp(0.2t) - 1, 0 ≤ t ≤ 5 s. Peter would like to calculate the displacement of the vehicle as a function of time, x(t), by integrating given velocity over the time from t = 0. Use t = 0.2 s for trapezoidal rule.
Answer:
[tex]x|_0^{0.2}=1.59535[/tex]
Explanation:
Given expression of velocity:
[tex]v(t)=10^{0.2t}-1 ;\ \ 0\leq t\leq 5\ s[/tex]
For getting displacement we need to integrate the above function with respect to t.
Given period of integration:
[tex]t_0=0\ s \to t_f=0.2\ s[/tex]
For trapezoidal rule we break the given interval into two parts of 0.1 s each.
∴take n=2
hence, [tex]\Delta t= 0.1[/tex]
[tex]v(0)=0[/tex]
[tex]v(0.1)=1.0471[/tex]
[tex]v(0.2)=1.0965[/tex]
Now, using trapezoidal rule:
[tex]\int_{0}^{0.2}v(t)\ dt=\Delta x[\frac{1}{2}\times v(0)+v(0.1)+\frac{1}{2}\times v(0.2)][/tex]
[tex]\int_{0}^{0.2}v(t)\ dt=0.1 [\frac{1}{2}\times 0+1.0471+\frac{1}{2}\times 1.0965][/tex]
[tex]x|_0^{0.2}=1.59535[/tex]
Note:Smaller the value of sub-interval better is the accuracy.
Two long parallel wires, each carrying a current of 12 A, lie at a distance of 9 cm from each other. What is the magnetic force per unit length exerted by one wire on the other? The magnetic force per unit length exerted by one wire on the other is ×10-4 N/m.
Answer:
Force per unit length between two conductors will be [tex]3.2\times 10^{-4}N[/tex]
Explanation:
We have given that two long parallel wires each carrying a current of 12 A
So [tex]I_1=I_2=12A[/tex]
Distance between the two conductors d = 9 cm = 0.09 m
We know that magnetic force between two parallel conductors per unit length is given by
[tex]F=\frac{\mu _0I_1I_2}{2\pi d}=\frac{4\times 3.14\times 10^{-7}\times 12\times 12}{2\times 3.14\times 0.09}=3.2\times 10^{-4}N[/tex]
Force per unit length between two conductors will be [tex]3.2\times 10^{-4}N[/tex]
Compute the longitudinal strength of an aligned carbon fiber-epoxy matrix composite having a 0.25 volume fraction of fibers, assuming the following: (1) an average fiber diameter of 10 10-3 mm, (2) an average fiber length of 1 mm, (3) a fiber fracture strength of 2.5 GPa, (4) a fiber-matrix bond strength of 10 MPa, (5) a matrix stress at fiber failure of 10.0 MPa, and (6) a matrix tensile strength of 75 MPa.
Answer:
632.5 MPa
Explanation:
[tex]\sigma_{m}[/tex] = Matrix stress at fiber failure = 10 MPa
[tex]V_f[/tex] = Volume fraction of fiber = 0.25
[tex]\sigma_f[/tex] = Fiber fracture strength = 2.5 GPa
The longitudinal strength of a composite is given by
[tex]\sigma_{cl}=\sigma_{m}(1-V_f)+\sigma_fV_f\\\Rightarrow \sigma_{cl}=10(1-0.25)+(2.5\times 10^3)\times 0.25\\\Rightarrow \sigma_{cl}=632.5\ MPa[/tex]
The longitudinal strength of the aligned carbon fiber-epoxy matrix composite is 632.5 MPa
Final answer:
The given question seeks to calculate the longitudinal strength of a carbon fiber-epoxy composite, but lacks sufficient detail or formulae for a complete answer. Typically, this computation would involve using materials science models that consider fiber orientation and other stress-strain interactions between components.
Explanation:
To compute the longitudinal strength of an aligned carbon fiber-epoxy matrix composite with a 0.25 volume fraction of fibers, we'd need to consider the following given properties: the average fiber diameter, average fiber length, fiber fracture strength, fiber-matrix bond strength, matrix stress at fiber failure, and matrix tensile strength.
Although the exact method for calculating the longitudinal strength would typically involve applying principles from materials science, such as the rule of mixtures, in combination with the given data points, the actual question does not provide enough information or a specific formula to complete the calculation. For a real-life carbon fiber-epoxy composite, the longitudinal strength could be substantially influenced by the alignment of the fibers, bond quality between the fibers and matrix, and the interaction between the stress and strain of the components.
If we had a suitable model or empirical formula, we would proceed by plugging in the given values to determine the longitudinal strength. However, as the provided data from the question is incomplete for this calculation, it is recommended to refer to a textbook or comprehensive resource on composite material mechanics for the detailed step-by-step methodology and equations.
A proton has a speed of 3.50 Ã 105 m/s when at a point where the potential is +100 V. Later, itâs at a point where the potential is â150 V. What is the change in the protonâs electric potential? What is the change in the potential energy of the proton? What is the work done on the proton?
Answer:
(a). The change in the protons electric potential is 0.639 kV.
(b). The change in the potential energy of the proton is [tex]1.022\times10^{-16}\ J[/tex]
(c). The work done on the proton is [tex]-8\times10^{-18}\ J[/tex].
Explanation:
Given that,
Speed [tex]v= 3.50\times10^{5}\ m/s[/tex]
Initial potential V=100 V
Final potential = 150 V
(a). We need to calculate the change in the protons electric potential
Potential energy of the proton is
[tex]U=qV=eV[/tex]
Using conservation of energy
[tex]K_{i}+U_{i}=K_{f}+U_{f}[/tex]
[tex]\dfrac{1}{2}mv_{i}^2+eV_{i}=\dfrac{1}{2}mv_{f}^2+eV_{f}[/tex]
[tex]]\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e(V_{f}-V_{i})[/tex]
[tex]\dfrac{1}{2}mv_{i}^2-]\dfrac{1}{2}mv_{f}^2=e\Delta V[/tex]
[tex]\Delta V=\dfrac{m(v_{i}^2-v_{f}^2)}{2e}[/tex]
Put the value into the formula
[tex]\Delta V=\dfrac{1.67\times10^{-27}(3.50\times10^{5}-0)^2}{2\times1.6\times10^{-19}}[/tex]
[tex]\Delta V=639.2=0.639\ kV[/tex]
(b). We need to calculate the change in the potential energy of the proton
Using formula of potential energy
[tex]\Delta U=q\Delta V[/tex]
Put the value into the formula
[tex]\Delta U=1.6\times10^{-19}\times639.2[/tex]
[tex]\Delta U=1.022\times10^{-16}\ J[/tex]
(c). We need to calculate the work done on the proton
Using formula of work done
[tex]\Delta U=-W[/tex]
[tex]W=q(V_{2}-V_{1})[/tex]
[tex]W=-1.6\times10^{-19}(150-100)[/tex]
[tex]W=-8\times10^{-18}\ J[/tex]
Hence, (a). The change in the protons electric potential is 0.639 kV.
(b). The change in the potential energy of the proton is [tex]1.022\times10^{-16}\ J[/tex]
(c). The work done on the proton is [tex]-8\times10^{-18}\ J[/tex].
Final answer:
The change in electric potential and potential energy of the proton can be calculated based on the provided potentials. The work done on the proton equals the change in potential energy.
Explanation:
The change in the proton's electric potential: The change in electric potential is the final potential minus the initial potential, thus the change is -150 V - 100 V = -250 V.
The change in potential energy of the proton: The potential energy change equals the charge of the proton times the change in potential, giving -proton charge x change in potential.
The work done on the proton: The work done is equal to the change in the potential energy of the proton.
About once every 30 minutes, a geyser known as Old Faceful projects water 18.0 m straight up into the air. Use g = 9.80 m/s2, and take atmospheric pressure to be 101.3 kPa. The density of water is 1000 kg/m3. What is the speed of the water when it emerges from the ground?
Answer:
Speed of the water that emerge out of the pipe is 18.8 m/s
Explanation:
Since we know that water drops projected upwards to maximum height of 18 m
So here we can use kinematics equations here
[tex]v_f^2 - v_i^2 = 2 a d[/tex]
here we have
[tex]v_f = 0[/tex]
[tex]d = 18 m[/tex]
[tex]a = -9.80 m/s^2[/tex]
so we will have
[tex]0 - v_i^2 = 2(-9.80)(18)[/tex]
[tex]v_i = 18.8 m/s[/tex]
Two kids are playing on a newly installed slide, which is 3 m long. John, whose mass is 30 kg, slides down into William (20 kg), who is sitting at the very bottom end, and whom he holds onto when he arrives. Laughing, John & William leave the slide horizontally and land in the muddy ground near the foot of the slide. (A) If John starts out 1.8 m above William, and the slide is essentially frictionless, how fast are they going when they leave the slide? (B) Thanks to the mud he acquired, John will now experience an average frictional force of 105 N as he slides down. How much slower is he going when he reaches the bottom than when friction was absent?
Answer:
[tex]v=3.564\ m.s^{-1}[/tex]
[tex]\Delta v =2.16\ m.s^{-1}[/tex]
Explanation:
Given:
mass of John, [tex]m_J=30\ kg[/tex]mass of William, [tex]m_W=30\ kg[/tex]length of slide, [tex]l=3\ m[/tex](A)
height between John and William, [tex]h=1.8\ m[/tex]
Using the equation of motion:
[tex]v_J^2=u_J^2+2 (g.sin\theta).l[/tex]
where:
v_J = final velocity of John at the end of the slide
u_J = initial velocity of John at the top of the slide = 0
Now putting respective :
[tex]v_J^2=0^2+2\times (9.8\times \frac{1.8}{3})\times 3[/tex]
[tex]v_J=5.94\ m.s^{-1}[/tex]
Now using the law of conservation of momentum at the bottom of the slide:
Sum of initial momentum of kids before & after collision must be equal.
[tex]m_J.v_J+m_w.v_w=(m_J+m_w).v[/tex]
where: v = velocity with which they move together after collision
[tex]30\times 5.94+0=(30+20)v[/tex]
[tex]v=3.564\ m.s^{-1}[/tex] is the velocity with which they leave the slide.
(B)
frictional force due to mud, [tex]f=105\ N[/tex]Now we find the force along the slide due to the body weight:
[tex]F=m_J.g.sin\theta[/tex]
[tex]F=30\times 9.8\times \frac{1.8}{3}[/tex]
[tex]F=176.4\ N[/tex]
Hence the net force along the slide:
[tex]F_R=71.4\ N[/tex]
Now the acceleration of John:
[tex]a_j=\frac{F_R}{m_J}[/tex]
[tex]a_j=\frac{71.4}{30}[/tex]
[tex]a_j=2.38\ m.s^{-2}[/tex]
Now the new velocity:
[tex]v_J_n^2=u_J^2+2.(a_j).l[/tex]
[tex]v_J_n^2=0^2+2\times 2.38\times 3[/tex]
[tex]v_J_n=3.78\ m.s^{-1}[/tex]
Hence the new velocity is slower by
[tex]\Delta v =(v_J-v_J_n)[/tex]
[tex]\Delta v =5.94-3.78= 2.16\ m.s^{-1}[/tex]
You’ve been given the challenge of balancing a uniform, rigid meter-stick with mass M = 95 g on a pivot. Stacked on the 0-cm end of the meter stick are n identical coins, each with mass m = 3.1 g, so that the center of mass of the coins is directly over the end of the meter stick. The pivot point is a distance d from the 0-cm end of the meter stick.
Part (a): Determine the distance d = d1, in centimeters, if there is only one coin o the 0 end of the meter stick and the system is in static equilibrium
The distance [tex]\( d_1 \)[/tex] from the pivot point to the center of mass of the meter stick with one coin on the 0-cm end, maintaining static equilibrium, is approximately 46.8 cm.
1. To achieve static equilibrium, the torques on both sides of the pivot point must balance out.
2. The torque due to the meter stick with mass ( M ) is [tex]\( M \times g \times \frac{L}{2} \)[/tex], where ( L ) is the length of the meter stick (100 cm) and \( g \) is the acceleration due to gravity (approximately [tex]\( 9.8 \, \text{m/s}^2 \)[/tex]).
3. The torque due to the coin on the 0-cm end is [tex]\( m \times g \times d_1 \)[/tex], where [tex]\( m \)[/tex] is the mass of the coin.
4. Since the torques balance out, we have the equation: [tex]\( M \times g \times \frac{L}{2} = m \times g \times d_1 \).[/tex]
5. Rearrange the equation to solve for [tex]\( d_1 \): \( d_1 = \frac{M \times \frac{L}{2}}{m} \).[/tex]
6. Substitute the given values: [tex]\( d_1 = \frac{95 \, \text{g} \times \frac{100}{2} \, \text{cm}}{3.1 \, \text{g}} \).[/tex]
7. Calculate [tex]\( d_1 \)[/tex]to find the distance from the pivot point to the center of mass of the meter stick with one coin on the 0-cm end, which is approximately 46.8 cm.
You are given two carts A and B. They look identical and you are told they are made of the same material. You place A at rest at an air track and give B a constant velocity directed to the right so that it collides elastically with A. After the collision cart B moves to the left. What do you conclude?
(A) Cart A is hollow.
(B) The two carts are identical.
(C) Cart B is hollow.
Based on the motion after the collision, cart A is concluded to be hollow.
Let's analyze the situation step by step.
Initial State: Collision: The collision is elastic, meaning both kinetic energy and momentum are conserved.After the Collision:Now, let's consider the possible scenarios:
If the two carts were completely identical (same mass, same structure), and there were no external forces, they would move together after the collision (due to conservation of momentum). The fact that Cart B moves to the left suggests that there might be a difference between the two carts.If Cart A is hollow, it would have less mass than Cart B. After the collision, the two carts would move in the direction of the heavier cart (Cart B) due to conservation of momentum. The fact that Cart B moves to the left supports the idea that there is a mass difference.If Cart B is hollow, it would have less mass than Cart A. After the collision, the two carts would move in the direction of the heavier cart (Cart A) due to the conservation of momentum. However, this contradicts the observed motion of Cart B moving to the left.Therefore, based on the information provided, the conclusion is: (A) Cart A is hollow.
If a 110-W lightbulb emits 2.5 % of the input energy as visible light (average wavelength 550 nm) uniformly in all directions. Part A How many photons per second of visible light will strike the pupil (4.0 mm diameter) of the eye of an observer 2.8 m away? Express your answer using two significant figures.
Answer:
9.7 x 10¹¹ .
Explanation:
2.5 % of 110 W = 2.75 J/s
energy of one photon
= hc / λ
=[tex]\frac{6.6\times10^{-34}\times3\times10^8}{550\times10^{-9}}[/tex]
= .036 x 10⁻¹⁷ J
No of photons emitted
= 2.75 / .036 x 10⁻¹⁷
= 76.38 x 10¹⁷
Now photons are uniformly distributed in all directions so they will pass through a spherical surface of radius 2.8 m at this distance
photons passing per unit area of this sphere
= 76.38 x 10¹⁷ / 4π ( 2.8)²
Through eye which has surface area of π x ( 2 x 10⁻² )² m² , no of photons passing
= [tex]\frac{76.38\times10^{17}}{4\pi\times(2.8)^2} \times\pi(2\times10^{-3})^2[/tex]
= 9.7 x 10¹¹ .
Consider the air moving over the top of the light bulb. The streamlines near the bulb will be squeezed together as the air goes over the top of the bulb. This leads to a region of _________ on the top of the bulb.
Answer:
low pressure
Explanation:
The streamlines of air particles are squeezed together as the air goes over the top of the bulb. Then, by the law of conservation of mass, the velocity of air particles are increased. And since, the velocity is increase the pressure is bound to decrease. Hence, this leads to region of low pressure on the top of the bulb.
Two UFPD are patrolling the campus on foot. To cover more ground, they split up and begin walking in different directions. Office A is walking at 5 mph directly south while Office B is walking at 6 mph directly west. How long would they need to walk before they are 2 miles away from each other?
Answer:
0.256 hours
Explanation:
Vectors in the plane
We know Office A is walking at 5 mph directly south. Let [tex]X_A[/tex] be its distance. In t hours he has walked
[tex]X_A=5t\ \text{miles}[/tex]
Office B is walking at 6 mph directly west. In t hours his distance is
[tex]X_B=6t\ \text{miles}[/tex]
Since both directions are 90 degrees apart, the distance between them is the hypotenuse of a triangle which sides are the distances of each office
[tex]D=\sqrt{X_A^2+X_B^2}[/tex]
[tex]D=\sqrt{(5t)^2+(6t)^2}[/tex]
[tex]D=\sqrt{61}t[/tex]
This distance is known to be 2 miles, so
[tex]\sqrt{61}t=2[/tex]
[tex]t =\frac{2}{\sqrt{61}}=0.256\ hours[/tex]
t is approximately 15 minutes
Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface. Model a spherically symmetric planet, with the same radius as the earth, as having a density that decreases linearly with distance from the center. Let the density be 1.60 x 10^4 kg/m^3 at the center and 2100 kg/m^3 at the surface.
What is the acceleration due to gravity at the surface of this planet?
Answer:
a = 9.94 m/s²
Explanation:
given,
density at center= 1.6 x 10⁴ kg/m³
density at the surface = 2100 Kg/m³
volume mass density as function of distance
[tex]\rho(r) = ar^2 - br^3[/tex]
r is the radius of the spherical shell
dr is the thickness
volume of shell
[tex]dV = 4 \pi r^2 dr[/tex]
mass of shell
[tex]dM = \rho(r)dV[/tex]
[tex]\rho = \rho_0 - br[/tex]
now,
[tex]dM = (\rho_0 - br)(4 \pi r^2)dr[/tex]
integrating both side
[tex]M = \int_0^{R} (\rho_0 - br)(4 \pi r^2)dr[/tex]
[tex]M = \dfrac{4\pi}{3}R^3\rho_0 - \pi R^4(\dfrac{\rho_0-\rho}{R})[/tex]
[tex]M = \pi R^3(\dfrac{\rho_0}{3}+\rho)[/tex]
we know,
[tex]a = \dfrac{GM}{R^2}[/tex]
[tex]a = \dfrac{G( \pi R^3(\dfrac{\rho_0}{3}+\rho))}{R^2}[/tex]
[tex]a =\pi RG(\dfrac{\rho_0}{3}+\rho)[/tex]
[tex]a =\pi (6.674\times 10^{-11}\times 6.38 \times 10^6)(\dfrac{1.60\times 10^4}{3}+2.1\times 10^3)[/tex]
a = 9.94 m/s²
Air at 207 kPa and 200◦C enters a 2.5-cm-ID tube at 6 m/s. The tube is constructed ofcopper with a thickness of 0.8 mm and a length of 3 m. Atmospheric air at 1 atm and 20◦Cflows normal to the outside of the tube with a free-stream velocity of 12 m/s. Calculate theair temperature at exit from the tube. What would be the effect of reducing the hot-air flowing half?
Answer:
Temperature of air at exit = 24.32 C, After reducing hot air the temperature of the exit air becomes = 20.11 C
Explanation:
ρ = P/R(Ti) where ρ is the density of air at the entry, P is pressure of air at entrance, R is the gas constant, Ti is the temperature at entry
ρ = (2.07 x 10⁵)/(287)(473) = 1.525 kg/m³
Calculate the mass flow rate given by
m (flow rate) = (ρ x u(i) x A(i)) where u(i) is the speed of air, A(i) is the area of the tube (πr²) of the tube
m (flow rate) = 1.525 x (π x 0.0125²) x 6 = 4.491 x 10⁻³ kg/s
The Reynold's Number for the air inside the tube is given by
R(i) = (ρ x u(i) x d)/μ where d is the inner diameter of the tube and μ is the dynamic viscosity of air (found from the table at Temp = 473 K)
R(i) = (1.525) x (6) x 0.025/2.58 x 10⁻⁵ = 8866
Calculate the convection heat transfer Coefficient as
h(i) = (k/d)(R(i)^0.8)(Pr^0.3) where k is the thermal conductivity constant known from table and Pr is the Prandtl's Number which can also be found from the table at Temperature = 473 K
h(i) = (0.0383/0.025) x (8866^0.8) x (0.681^0.3) = 1965.1 W/m². C
The fluid temperature is given by T(f) = (T(i) + T(o))/2 where T(i) is the temperature of entry and T(o) is the temperature of air at exit
T(f) = (200 + 20)/2 = 110 C = 383 K
Now calculate the Reynold's Number and the Convection heat transfer Coefficient for the outside
R(o) = (μ∞ x do)/V(f) where μ∞ is the speed of the air outside, do is the outer diameter of the tube and V(f) is the kinematic viscosity which can be known from the table at temperature = 383 K
R(o) = (12 x 0.0266)/(25.15 x 10⁻⁶) = 12692
h(o) = K(f)/d(o)(0.193 x Ro^0.618)(∛Pr) where K(f) is the Thermal conductivity of air on the outside known from the table along with the Prandtl's Number (Pr) from the table at temperature = 383 K
h(o) = (0.0324/0.0266) x (0.193 x 12692^0.618) x (0.69^1/3) = 71.36 W/m². C
Calculate the overall heat transfer coefficient given by
U = 1/{(1/h(i)) + A(i)/(A(o) x h(o))} simplifying the equation we get
U = 1/{(1/h(i) + (πd(i)L)/(πd(o)L) x h(o)} = 1/{(1/h(i) + di/(d(o) x h(o))}
U = 1/{(1/1965.1) + 0.025/(0.0266 x 71.36)} = 73.1 W/m². C
Find out the minimum capacity rate by
C(min) = m (flow rate) x C(a) where C(a) is the specific heat of air known from the table at temperature = 473 K
C(min) = (4.491 x 10⁻³) x (1030) = 4.626 W/ C
hence the Number of Units Transferred may be calculated by
NTU = U x A(i)/C(min) = (73.1 x π x 0.025 x 3)/4.626 = 3.723
Calculate the effectiveness of heat ex-changer using
∈ = 1 - е^(-NTU) = 1 - e^(-3.723) = 0.976
Use the following equation to find the exit temperature of the air
(Ti - Te) = ∈(Ti - To) where Te is the exit temperature
(200 - Te) = (0.976) x (200 - 20)
Te = 24.32 C
The effect of reducing the hot air flow by half, we need to calculate a new value of Number of Units transferred followed by the new Effectiveness of heat ex-changer and finally the exit temperature under these new conditions.
Since the new NTU is half of the previous NTU we can say that
NTU (new) = 2 x NTU = 2 x 3.723 = 7.446
∈(new) = 1 - e^(-7.446) = 0.999
(200 - Te (new)) = (0.999) x (200 - 20)
Te (new) = 20.11 C
Complete the following statement: The interior of a thermos bottle is silvered to minimize heat transfer due to
A. conduction and convection
B. conduction
C. conduction, convection and radiation
D. conduction and radiation
E. radiation.
Answer:
E. radiation.
Explanation:
As we know that heat transfer due to conduction depends on thermal conductivity of the materials and heat transfer due to convection depends on the velocity of the fluid.But on the other hand heat transfer due to radiation depends on the surface properties like emmisivity .So when bottle is silvered then it will leads to minimize the radiation heat transfer.
Therefore answer is --
E. radiation.
The answer is option C.
The interior of a thermos is silvered to minimize heat transfer by conduction, convection, and radiation. The silvering acts like a mirror, reflecting heat, and the vacuum between the thermos walls almost eliminates conduction and convection.
Explanation:The interior of a thermos bottle is silvered to minimize heat transfer due to C. conduction, convection and radiation.
This is because the silvering on the inner surface of the thermos acts like a mirror, reducing the amount of heat that can be transferred by all three modes: conduction, convection, and especially radiation.
Conduction is the transfer of heat through direct contact of molecules, minimized in a thermos by the vacuum between its double walls. Convection is the transfer of heat in a fluid (like air or liquid) through the motion of the fluid itself, which is also nearly eliminated by the vacuum. Radiation is the transfer of heat through electromagnetic waves, which the silvering reflects back, greatly reducing heat loss this way.
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A 20-foot ladder is leaning against the wall. If the base of the ladder is sliding away from the wall at the rate of 3 feet per second, find the rate at which the top of the ladder is sliding down when the top of the ladder is 8 feet from the ground.
Answer:
6.87 ft/s is the rate at which the top of ladder slides down.
Explanation:
Given:
Length of the ladder is, [tex]L=20\ ft[/tex]
Let the top of ladder be at height of 'h' and the bottom of the ladder be at a distance of 'b' from the wall.
Now, from triangle ABC,
AB² + BC² = AC²
[tex]h^2+b^2=L^2\\h^2+b^2=20^2\\h^2+b^2=400----1[/tex]
Differentiating the above equation with respect to time, 't'. This gives,
[tex]\frac{d}{dt}(h^2+b^2)=\frac{d}{dt}(400)\\\\\frac{d}{dt}(h^2)+\frac{d}{dt}(b^2)=0\\\\2h\frac{dh}{dt}+2b\frac{db}{dt}=0\\\\h\frac{dh}{dt}+b\frac{db}{dt}=0--------2[/tex]
In the above equation the term [tex]\frac{dh}{dt}[/tex] is the rate at which top of ladder slides down and [tex]\frac{db}{dt}[/tex] is the rate at which bottom of ladder slides away.
Now, as per question, [tex]h=8\ ft, \frac{db}{dt}=3\ ft/s[/tex]
Plug in [tex]h=8[/tex] in equation (1) and solve for [tex]b[/tex]. This gives,
[tex]8^2+b^2=400\\64+b^2=400\\b^2=400-64\\b^2=336\\b=\sqrt{336}=18.33\ ft[/tex]
Now, plug in all the given values in equation (2) and solve for [tex]\frac{dh}{dt}[/tex]
[tex]8\times \frac{dh}{dt}+18.33\times 3=0\\8\times \frac{dh}{dt}+54.99=0\\8\times \frac{dh}{dt}=-54.99\\ \frac{dh}{dt}=-\frac{54.99}{8}=-6.87\ ft/s[/tex]
Therefore, the rate at which the top of ladder slide down is 6.87 ft/s. The negative sign implies that the height is reducing with time which is true because it is sliding down.
A 0.500-kg block, starting at rest, slides down a 30.0° incline with static and kinetic friction coefficients of 0.350 and 0.250, respectively. After sliding 77.3 cm along the incline, the block slides across a frictionless horizontal surface and encounters a spring (k = 35.0 N/m).What is the maximum compression of the spring?
Answer:x=23.4 cm
Explanation:
Given
mass of block [tex]m=0.5 kg[/tex]
inclination [tex]\theta =30[/tex]
coefficient of static friction [tex]\mu =0.35[/tex]
coefficient of kinetic friction [tex]\mu _k=0.25[/tex]
distance traveled [tex]d=77.3 cm[/tex]
spring constant [tex]k=35 N/m [/tex]
work done by gravity+work done by friction=Energy stored in Spring
[tex]mg\sin \theta d-\mu _kmg\cos \theta d=\frac{kx^2}{2}[/tex]
[tex]mgd\left ( \sin \theta -\mu _k\cos \theta \right )=\frac{kx^2}{2}[/tex]
[tex]0.5\times 9.8\times 0.773\left ( \sin 30-0.25\cos 30\right )=\frac{35\times x^2}{2}[/tex]
[tex]x=\sqrt{\frac{2\times 0.5\times 9.8\times 0.773(\sin 30-0.25\times \cos 30)}{35}}[/tex]
[tex]x=0.234 m[/tex]
[tex]x=23.4 cm[/tex]
A puck of mass 0.5100.510kg is attached to the end of a cord 0.827m long. The puck moves in a horizontal circle without friction. If the cord can withstand a maximum tension of 126N, what is the highest frequency at which the puck can go around the circle without the cord breaking?
Answer:2.74 Hz
Explanation:
Given
mass Puck [tex]m=0.51 kg[/tex]
length of cord [tex]L=0.827 m[/tex]
Maximum Tension in chord [tex]T=126 N[/tex]
as the Puck is moving in a horizontal circle so maximum Tension in the string will be equal to centripetal force
[tex]F_c=m\omega ^2L=T[/tex]
[tex]126=0.51\times (\omega )^2\times 0.827[/tex]
[tex]\omega =\sqrt{298.74}[/tex]
[tex]\omega =17.28 rad/s[/tex]
[tex]\omega =2\pi f[/tex]
[tex]f=\frac{2\pi }{\omega }[/tex]
[tex]f=2.74 Hz[/tex]
To find the highest frequency at which the puck can go around the circle without the cord breaking, we use the formula for tension in a circular motion and solve for velocity. Then we use the velocity to find the frequency. The highest frequency is approximately 2.18 Hz.
To determine the highest frequency at which the puck can go around the circle without the cord breaking, we need to find the maximum tension in the cord.
Since the tension in the cord is equal to the centripetal force required to keep the puck moving in a circle, we can use the formula:
Tension = mass × velocity² / radius
Substituting the given values, we get:
126N = 0.51kg × v² / 0.827m
Now, solving for v, we find:
v² = (126N × 0.827m) / 0.51kg
v² = 204.2 m²/s²
v = √(204.2 m²/s²) = 14.29 m/s
Since the frequency of an object moving in a circle is equal to its velocity divided by the circumference of the circle, we can calculate the highest frequency as:
Frequency = v / (2πr)
Frequency = 14.29 m/s / (2π × 0.827m)
Frequency ≈ 2.18 Hz
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Sound level B in decibels is defined as
B= 10 log (i/i)
where i = 1 × 10-12 W/m2 . The decibel
scale intensity for busy traffic is 80 dB. Two
people having a loud conversation have a deci-
bel intensity of 70 dB.
What is the approximate combined sound
intensity?
Answer in units of W/m2
Answer:
The approximate combined sound intensity is [tex]I_{T}=1.1\times10^{-4}W/m^{2}[/tex]
Explanation:
The decibel scale intensity for busy traffic is 80 dB. so intensity will be
[tex]10log(\frac{I_{1}}{I_{0}} )=80[/tex], therefore [tex]I_{1}=1\times10^{8}I_{0}=1\times10^{8} * 1\times10^{-12}W/m^{2}=1\times10^{-4}W/m^{2}[/tex]
In the same way for the loud conversation having a decibel intensity of 70 dB.
[tex]10log(\frac{I_{2}}{I_{0}} )=70[/tex], therefore [tex]I_{2}=1\times10^{7}I_{0}=1\times10^{7} * 1\times10^{-12}W/m^{2}=1\times10^{-5}W/m^{2}[/tex]
Finally we add both of them [tex]I_{T}=I_{1}+I_{2}=1\times10^{-4}W/m^{2}+1\times10^{-5}W/m^{2}=1.1\times10^{-4}W/m^{2}[/tex], is the approximate combined sound intensity.
Which of the following statements is false?
• The energy of electromagnetic radiation increases as its frequency increases.
• An excited atom can return to its ground state by absorbing electromagnetic radiation.
• An electron in the n = 4 state in the hydrogen atom can go to the n = 2 state by emitting electromagnetic radiation at the appropriate frequency.
• The frequency and the wavelength of electromagnetic radiation are inversely proportional to each other.
An excited atom can return to its ground state by absorbing electromagnetic radiation is false about the electromagnetic radiation.
Option B
Explanation:
In the scope of modern quantum theory, the term Electromagnetic radiation is identified as the movement of photons through space. Almost all the sources of energy that we utilize today such as coal, oil, etc are a product of electromagnetic radiation which was absorbed from the sun millions of years ago.
Various properties of electromagnetic radiations are a directly proportional relationship between the energy and the frequency, Inverse proportionality between frequency and the wavelength, etc. Hence, we can conclude that an "excited atom" can never return to its ground state by assimilating electromagnetic radiation and the 2nd statement is false.
The statement 'An excited atom can return to its ground state by absorbing electromagnetic radiation' is false, as the process actually involves emitting radiation. Electromagnetic radiation's energy increases with frequency, and frequency and wavelength have an inverse relationship.
Explanation:The false statement among the ones provided is: An excited atom can return to its ground state by absorbing electromagnetic radiation. An excited atom returns to its ground state by emitting radiation, not absorbing it. When it comes to electromagnetic radiation, increasing frequency indeed results in increasing energy. This is because the energy of electromagnetic radiation is directly proportional to its frequency. Additionally, when an electron in the n = 4 state in the hydrogen atom transitions to the n = 2 state, it emits radiation at an appropriate frequency. Lastly, the frequency and the wavelength of electromagnetic radiation are inversely proportional, meaning as one increases, the other decreases.
A toy balloon, which has a mass of 2.90 g before it is inflated, is filled with helium (with a density of 0.180 kg/m^3) to a volume of 8400 cm^3. What is the minimum mass that should be hung from the balloon to prevent it from rising up into the air? Assume the air has a density of 1.29 kg/m^3.
Answer:
[tex]M=6.4243\ g[/tex]
Explanation:
Given:
mass of deflated balloon, [tex]m_b=2.9\ g=0.0029\ kg[/tex]density of helium, [tex]\rho_h=0.180\ kg.m^{-3}[/tex]volume of inflation, [tex]V=8400\ cm^3=0.0084\ m^3[/tex]density of air, [tex]\rho_a=1.29\ kg.m^{-3}[/tex]To stop this balloon from rising up we need to counter the buoyant force.
mass of balloon after inflation:
[tex]m=m_h+m_b[/tex]
[tex]m=0.0084\times 0.180+0.0029[/tex]
[tex]m=0.004412\ kg[/tex]
Now the density of inflated balloon:
[tex]\rho_b=\frac{m}{V}[/tex]
[tex]\rho_b=\frac{0.004412}{0.0084}[/tex]
[tex]\rho_b=0.5252\ kg.m^{-3}[/tex]
Now the buoyant force on balloon
[tex]F_B=V(\rho_a-\rho_b).g[/tex]
[tex]F_B=0.0084(1.29-0.5252)\times 9.8[/tex]
[tex]F_B=0.063\ N[/tex]
∴Mass to be hung:
[tex]M=\frac{F_B}{g}[/tex]
[tex]M=0.00642432\ kg[/tex]
[tex]M=6.4243\ g[/tex]
Two waves traveling on a string in the same direction both have a frequency of 135 Hz, a wavelength of 2 cm, and an amplitude of 0.04 m. What is the amplitude of the resultant wave if the original waves differ in phase by each of the following values?
(a) p/6 cm(b) p/3 cm
Answer:
The amplitude of the resultant wave are
(a). 0.0772 m
(b). 0.0692 m
Explanation:
Given that,
Frequency = 135 Hz
Wavelength = 2 cm
Amplitude = 0.04 m
We need to calculate the angular frequency
[tex]\omega=2\pi f[/tex]
[tex]\omega=2\times\pi\times135[/tex]
[tex]\omega=848.23\ rad/s[/tex]
As the two waves are identical except in their phase,
The amplitude of the resultant wave is given by
[tex]y+y=A\sin(kx-\omega t)+Asin(kx-\omega t+\phi)[/tex]
[tex]y+y=A[2\sin(kx-\omega t+\dfrac{\phi}{2})\cos\phi\dfrac{\phi}{2}[/tex]
[tex]y'=2A\cos(\dfrac{\phi}{2})\sin(kx-\omega t+\dfrac{\phi}{2})[/tex]
(a). We need to calculate the amplitude of the resultant wave
For [tex]\phi =\dfrac{\pi}{6}[/tex]
The amplitude of the resultant wave is
[tex]A'=2A\cos(\dfrac{\phi}{2})[/tex]
Put the value into the formula
[tex]A'=2\times0.04\cos(\dfrac{\pi}{12})[/tex]
[tex]A'=0.0772\ m[/tex]
(b), We need to calculate the amplitude of the resultant wave
For [tex]\phi =\dfrac{\pi}{3}[/tex]
[tex]A'=2\times0.04\cos(\dfrac{\pi}{6})[/tex]
[tex]A'=0.0692\ m[/tex]
Hence, The amplitude of the resultant wave are
(a). 0.0772 m
(b). 0.0692 m
Astronomers discover an exoplanet, a planet obriting a star other than the Sun, that has an orbital period of 3.27 Earth years in a circular orbit around its star, which has a measured mass of 3.03×1030 kg . Find the radius of the exoplanet's orbit.
Answer:
r = 3.787 10¹¹ m
Explanation:
We can solve this exercise using Newton's second law, where force is the force of universal attraction and centripetal acceleration
F = ma
G m M / r² = m a
The centripetal acceleration is given by
a = v² / r
For the case of an orbit the speed circulates (velocity module is constant), let's use the relationship
v = d / t
The distance traveled Esla orbits, in a circle the distance is
d = 2 π r
Time in time to complete the orbit, called period
v = 2π r / T
Let's replace
G m M / r² = m a
G M / r² = (2π r / T)² / r
G M / r² = 4π² r / T²
G M T² = 4π² r3
r = ∛ (G M T² / 4π²)
Let's reduce the magnitudes to the SI system
T = 3.27 and (365 d / 1 y) (24 h / 1 day) (3600s / 1h)
T = 1.03 10⁸ s
Let's calculate
r = ∛[6.67 10⁻¹¹ 3.03 10³⁰ (1.03 10⁸) 2) / 4π²2]
r = ∛ (21.44 10³⁵ / 39.478)
r = ∛(0.0543087 10 36)
r = 0.3787 10¹² m
r = 3.787 10¹¹ m
You have a two-wheel trailer that you pull behind your ATV. Two children with a combined mass of 76.2 kg hop on board for a ride through the woods and the springs (one for each wheel) each compress by 6.17 cm. When you pull the trailer over a tree root in the trail, it oscillates with a period of 2.09 s. Determine the following. (a) force constant of the springs? N/m
(b) mass of the trailer? kg
(c) frequency of the oscillation? Hz
(d) time it takes for the trailer to bounce up and down 10 times? s
a) The spring constant is 12,103 N/m
b) The mass of the trailer 2,678 kg
c) The frequency of oscillation is 0.478 Hz
d) The time taken for 10 oscillations is 20.9 s
Explanation:
a)
When the two children jumps on board of the trailer, the two springs compresses by a certain amount
[tex]\Delta x = 6.17 cm = 0.0617 m[/tex]
Since the system is then in equilibrium, the restoring force of the two-spring system must be equal to the weight of the children, so we can write:
[tex]2mg = k'\Delta x[/tex] (1)
where
m = 76.2 kg is the mass of each children
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
[tex]k'[/tex] is the equivalent spring constant of the 2-spring system
For two springs in parallel each with constant k,
[tex]k'=k+k=2k[/tex]
Substituting into (1) and solving for k, we find:
[tex]2mg=2k\Delta x\\k=\frac{mg}{\Delta x}=\frac{(76.2)(9.8)}{0.0617}=12,103 N/m[/tex]
b)
The period of the oscillating system is given by
[tex]T=2\pi \sqrt{\frac{m}{k'}}[/tex]
where
And for the system in the problem, we know that
T = 2.09 s is the period of oscillation
m is the mass of the trailer
[tex]k'=2k=2(12,103)=24,206 N/m[/tex] is the equivalent spring constant of the system
Solving the equation for m, we find the mass of the trailer:
[tex]m=(\frac{T}{2\pi})^2 k'=(\frac{2.09}{2\pi})^2 (24,206)=2,678 kg[/tex]
c)
The frequency of oscillation of a spring-mass system is equal to the reciprocal of the period, therefore:
[tex]f=\frac{1}{T}[/tex]
where
f is the frequency
T is the period
In this problem, we have
T = 2.09 s is the period
Therefore, the frequency of oscillation is
[tex]f=\frac{1}{2.09}=0.478 Hz[/tex]
d)
The period of the system is
T = 2.09 s
And this time is the time it takes for the trailer to complete one oscillation.
In this case, we want to find the time it takes for the trailer to complete 10 oscillations (bouncing up and down 10 times). Therefore, the time taken will be the period of oscillation multiplied by 10.
Therefore, the time needed for 10 oscillations is:
[tex]t=10T=10(2.09)=20.9 s[/tex]
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A 2.0-kg block travels around a 0.40-m radius circle with an angular speed of 16 rad/s. The circle is parallel to the xy plane and is centered on the z axis, 0.60 m from the origin. What is the magnitude of the component in the xy plane of the angular momentum around the origin?
The magnitude of the component in the xy plane of the angular momentum around the origin is 7.68 kg・m²/s. This is calculated using the formula for angular momentum, with the velocity determined from the product of the radius and angular speed.
Explanation:The magnitude of the component in the xy plane of the angular momentum around the origin can be calculated using the formula for angular momentum, L = mvr, where m is the mass, v is the velocity (which can be obtained from v = rw where r is the radius, and w is the angular speed), and r is the perpendicular distance from the center of the circular path to the origin. In this case, m = 2.0 kg, r = 0.60 m (the distance from the z-axis, not the radius of the circle), and the velocity v = (0.40 m)(16 rad/s) = 6.4 m/s.
Plugging these values into the angular momentum formula gives us, L = mvr = (2.0 kg)(6.4 m/s)(0.60 m) = 7.68 kg・m²/s as the magnitude of the component in the xy plane of the angular momentum around the origin.
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Which of the following statements concerning the electric field inside a conductor is true?
A) The electric field inside a conductor is never zero.
B) The electric field inside a conductor is always zero.
C) The electric field inside a conductor is always zero if charges inside the conductor are not moving.
D) The electric field inside a conductor is always zero unless there are excess charges inside the conductor.
Answer:
C
Explanation:
The electric field inside a conductor is always zero if the charges inside the conductor are not moving.
Since the electron are not moving then they must be in electrostatic equilibrium which means the electric field inside the conductor is zero. if the electric field existed inside the conductor then there will be net force on all the electrons and the electrons will accelerate.
The electric field inside a conductor is always zero if charges inside the conductor are not moving. Under electrostatic conditions, free electrons in the conductor rearrange to cancel any internal electric fields. Thus, the correct answer is Option C.
1. In a conductor in electrostatic equilibrium, the electric field inside the conductor is zero. This is because any free electrons within the conductor will move in response to any electric field until they reach a state where there is no net force acting on them. This movement of electrons cancels out any existing electric field.
2. Properties of conductors in electrostatic equilibrium include that any excess charge resides on the surface of the conductor, and the electric field just outside the surface is perpendicular to the surface.
3. Therefore, in the absence of moving charges (static conditions), the electric field inside a conductor must be zero.
Suppose the Earth's magnetic field at the equator has magnitude 0.00005 T and a northerly direction at all points. How fast must a singly ionized uranium atom (m=238u, q=e) move so as to circle the Earth 1.44 km above the equator? Give your answer in meters/second.
Answer:
Velocity will be [tex]v=1.291\times 10^8m/sec[/tex]
Explanation:
We have given magnetic field B = 0.00005 T
Mass m = 238 U
We know that [tex]1u=1.66\times 10^{-27}kg[/tex]
So 238 U [tex]=238\times 1.66\times 10^{-27}=395.08\times 10^{-27}kg[/tex]
Radius [tex]=R+1.44=6378+1.44=6379.44KM[/tex]
We know that magnetic force is given by
[tex]F=qvB[/tex] which is equal to the centripetal force
So [tex]qvB=\frac{mv^2}{r}[/tex]
[tex]1.6\times 10^{-19}\times v\times 0.00005=\frac{395.08\times 10^{-27}v^2}{6379.44}[/tex]
[tex]v=1.291\times 10^8m/sec[/tex]
A soup can is in the shape of a right cylinder. The can has a volume of 16 fluid ounces. The height is three times its radius. The metal used to make the lateral surface of the can costs $0.01 per square inch. The metal used to make the top and bottom costs $0.02 per square inch. If one fluid ounce is approximately 1.805 cubic inches, what is the total cost to make one empty soup can? Use 3.14 for straight pi
Answer:
$0.662772
Explanation:
v = Volume of can = 16 fl oz.
[tex]1\ floz.=1.805\ in^3[/tex]
r = Radius of can
h = Height of can = 3r
Volume of cylinder is given by
[tex]\pi r^2h=16\times 1.805\\\Rightarrow \pi r^23r=16\times 1.805\\\Rightarrow 3\pi r^3=16\times 1.805\\\Rightarrow r=\left(\frac{16\times 1.805}{3\times 3.14}\right)^{\frac{1}{3}}\\\Rightarrow r=1.45247\ in[/tex]
h=3r\\\Rightarrow h=3\times 1.45247\\\Rightarrow h=4.35741\ in[/tex]
Surface area of sides is given by
[tex]2\pi rh\\ =2\times 3.14\times 1.45247\times 4.35741\\ =39.76632\ in^2[/tex]
Surface area of top and bottom is given by
[tex]2\pi r^2\\ =2\times 3.14\times 1.45247^2\\ =13.25544\ in^2[/tex]
Cost of making the can will be
[tex]39.76632\times 0.01+13.25544\times 0.02=\$0.662772[/tex]
The cost to make the can is $0.662772
An electric motor can drive grinding wheel at two different speeds. When set to high the angular speed is 2000 rpm. The wheel turns at 1000 rpm when set to low. When the switch is changed from high to low, it takes the wheel 60 sec to slow down. A) ( 5 points) What is the initial angular speed of the high setting in rad/sec B) (5 points) What is the angular acceleration in rad/s2 of the wheel? C) (5 points) What is the angular speed in rad/s 40 seconds after the setting is changed? D) (5 points) How many revolutions did it make as it changes speed?
a) The initial angular speed is 209.3 m/s
b) The angular acceleration is [tex]-1.74 rad/s^2[/tex]
c) The angular speed after 40 s is 139.7 rad/s
d) The wheel makes 1501 revolutions
Explanation:
a)
The initial angular speed of the wheel is
[tex]\omega_i = 2000 rpm[/tex]
which means 2000 revolutions per minute.
We have to convert it into rad/s. Keeping in mind that:
[tex]1 rev = 2\pi rad[/tex]
[tex]1 min = 60 s[/tex]
We find:
[tex]\omega_i = 2000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=209.3 rad/s[/tex]
b)
To find the angular acceleration, we have to convert the final angular speed also from rev/min to rad/s.
Using the same procedure used in part a),
[tex]\omega_f = 1000 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=104.7 rad/s[/tex]
Now we can find the angular acceleration, given by
[tex]\alpha = \frac{\omega_f - \omega_i}{t}[/tex]
where
[tex]\omega_i = 209.3 rad/s[/tex] is the initial angular speed
[tex]\omega_f = 104.7 rad/s[/tex] is the final angular speed
t = 60 s is the time interval
Substituting,
[tex]\alpha = \frac{104.7-209.3}{60}=-1.74 rad/s^2[/tex]
c)
To find the angular speed 40 seconds after the initial moment, we use the equivalent of the suvat equations for circular motion:
[tex]\omega' = \omega_i + \alpha t[/tex]
where we have
[tex]\omega_i = 209.3 rad/s[/tex]
[tex]\alpha = -1.74 rad/s^2[/tex]
And substituting t = 40 s, we find
[tex]\omega' = 209.3 + (-1.74)(40)=139.7 rad/s[/tex]
d)
The angular displacement of the wheel in a certain time interval t is given by
[tex]\theta=\omega_i t + \frac{1}{2}\alpha t^2[/tex]
where
[tex]\omega_i = 209.3 rad/s[/tex]
[tex]\alpha = -1.74 rad/s^2[/tex]
And substituting t = 60 s, we find:
[tex]\theta=(209.3)(60) + \frac{1}{2}(-1.74)(60)^2=9426 rad[/tex]
So, the wheel turns 9426 radians in the 60 seconds of slowing down. Converting this value into revolutions,
[tex]\theta = \frac{9426 rad}{2\pi rad/rev}=1501 rev[/tex]
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