Answer:
The break even cost is $0.0063825
Explanation:
Break-even cost is the amount of money, or change in value, which equates to the amount at which an asset must be sold to equal the cost of acquiring it. For easier understanding it can be thought the amount of money for which a product or service must be sold to cover the costs of manufacturing or providing it.
Wattage = W
Cost per kilo watt hour = C
Number of hours per year = H
Price per bulb/CFL = P
Discount rate = 11%
Life of bulb = 2 years
Price of bulb = $0.39
Wattage consumption of bulb per hours = 60
Life of CFL = 24 years
Price of CFL = $3.10
Wattage consumption of CFL per hour = 15
Calculate the Equated Annual Cost (EAC) of bulb
EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 2years)}/ (PVIFA 11%, 2years)
PVIFA 11%, 2years = Annuity PV Factor = [1 – {(1 + r)^(-n)}]/r, where r is the rate per period and n is the number per periods
PVIFA 11%, 2 years = [1 – {(1 + 0.11)^(-2)}]/0.11 = 1.712523 (for 2 years)
PVIFA 11%, 24 years = [1 – {(1 + 0.11)^(-24)}]/0.11 = 8.348136 (for 2 years)
Calculate the EAC of bulb
EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 2 years)}/ (PVIFA 11%, 2 years)
EAC = {- 0.39 - (60/1000 x H x C) x (1.712523)}/ (1.712523)
EAC = {-0.39 – (51.37570 x C)}/ 1.712523, consider this equation 1
Calculate the EAC of CFL
EAC = {- P - (W/1000 x H x C) x (PVIFA 11%, 24 years)}/ (PVIFA 11%, 24 years)
EAC = {- 3.10 - (15/1000 x 500 x C) x (8.348136)}/ (8.348136)
EAC = {-3.10 – (62.61102 x C)}/8.348137, consider this equation 2
Equate 1 and 2 to find the amount of C
{-0.39 – (51.37570 x C)}/ 1.712523 = {-3.10 – (62.61102 x C)}/8.348137
{-0.39 – (51.37570 x C) x 8.348137} = {-3.10 – (62.61102 x C) x 1.712523}
C = $0.0063825
Thus, the break- even cost per kilo – watt is $0.0063825
In an electricity experiment, a 1.20 g plastic ball is suspended on a 59.0 cm long string and given an electric charge. A charged rod brought near the ball exerts a horizontal electrical force F⃗ elec on it, causing the ball to swing out to a 24.0 ∘ angle and remain there.
a. What is the magnitude of Felec ?
b.What is the tension in the string?
Answer:
a.) 5.24 10⁻³ N . b) 0.013 N
Explanation:
a) In absence of other forces, the plastic ball is only subject to the force of gravity (downward) , and to the tension in the string, which are equal each other.
We are told that there exists an horizontal force , of an electric origin, that causes the ball to swing out to a 24º angle (respect the normal) and remain there, so there exists a new equilibrium condition.
In this situation, both the vertical and horizontal components of the external forces acting on the ball (gravity, tension and the electrical force) must be equal to 0.
The only force that has horizontal and vertical components, is the tension in the string.
We can apply Newton's 2nd Law to both directions, as follows:
T cos 24º - mg = 0
-T sin 24º + Fe = 0
where T= Tension in the string.
Fe = Electrical Force
mg = Fg = gravity force
⇒ T = mg/ cos 24º
Replacing in the horizontal forces equation:
-mg/cos 24º . sin 24º = -Fe
∴ Fe = mg. tg 24º = 0.0012 kg. 9.8 m/s². tg 24º = 5.24 10⁻³ N
b) In order to get the value of T, we can simply solve for T the vertical forces component equation , as follows:
T = mg/ cos 24º = 0.0012 kg. 9.8 m/s² / 0.914 = 0.013 N
Final answer:
To find the magnitude of the electrical force (F⃗ elec) exerted on the plastic ball, we can use the fact that the ball is in equilibrium, meaning the net force acting on it is zero. The electrical force is the only horizontal force acting on the ball, so it must be balanced by the horizontal component of the tension in the string.
Explanation:
The horizontal component of the tension (T) in the string can be found using trigonometry. The angle between the string and the vertical is 24.0 degrees, so the horizontal component of the tension is:
T_horizontal = T * cos(24.0 degrees)
Since the ball is in equilibrium, the magnitude of the electrical force is equal to the horizontal component of the tension:
|F⃗ elec| = T_horizontal
To find the tension in the string, we can use the fact that the ball is in equilibrium, meaning the net force acting on it is zero. The gravitational force (F⃗ g) acting on the ball is balanced by the vertical component of the tension in the string.
The vertical component of the tension (T) in the string can be found using trigonometry. The angle between the string and the vertical is 24.0 degrees, so the vertical component of the tension is:
T_vertical = T * sin(24.0 degrees)
Since the ball is in equilibrium, the magnitude of the gravitational force is equal to the vertical component of the tension:
|F⃗ g| = T_vertical
The gravitational force can be calculated using the mass of the ball (m) and the acceleration due to gravity (g):
|F⃗ g| = m * g
Substituting the expression for the vertical component of the tension into the equation for the gravitational force, we get:
m * g = T * sin(24.0 degrees)
Solving for the tension (T), we get:
T = (m * g) / sin(24.0 degrees)
Substituting the given values for the mass of the ball (m) and the acceleration due to gravity (g), we get:
T = (1.20 g * 9.8 m/s^2) / sin(24.0 degrees)
T ≈ 5.10 N
b. The tension in the string is approximately 5.10 N.
Mantles for gas lanterns contain thorium, because it forms an oxide that can survive being heated to incandescence for long periods of time. Natural thorium is almost 100% 232Th, with a half-life of 1.405 ✕ 1010 y. If an average lantern mantle contains 200 mg of thorium, what is its activity (in Bq)?
Answer:
The activity is 811.77 Bq
Solution:
As per the question:
Half life of Thorium, [tex]t_{\frac{1}{2}} = 1.405\times 10^{10}\ yrs[/tex]
Mass of Thorium, m = 200 mg = 0.2 g
M = 232 g/mol
Now,
No. of nuclei of Thorium in 200 mg of Thorium:
[tex]N = \frac{N_{o}m}{M}[/tex]
where
[tex]N_{o }[/tex] = Avagadro's number
Thus
[tex]N = \frac{6.02\times 10^{23}\times 0.2}{232} = 5.19\times 10^{20}[/tex]
Also,
Activity is given by:
[tex]\frac{0.693}{t_{\frac{1}{2}}}\times N[/tex]
[tex]A= \frac{0.693}{1.405\times 10^{10}}\times 5.19\times 10^{20} = 2.56\times 10^{10}\ \yr[/tex]
[tex]A = \frac{2.56\times 10^{10}}{365\times 24\times 60\times 60} = 811.77\ Bq[/tex]
A uniform thin circular ring rolls without slipping down an incline making an angle θ with the horizontal. What is its acceleration? (Enter the magnitude. Use any variable or symbol stated above along with the following as necessary: g for the acceleration of gravity.)
Answer:[tex]a=\frac{g\sin \theta }{2}[/tex]
Explanation:
Given
inclination is [tex]\theta [/tex]
let M be the mass and r be the radius of uniform circular ring
Moment of Inertia of ring [tex]I=mr^2[/tex]
Friction will Provide the Torque to ring
[tex]f_r\times r=I\times \alpha [/tex]
[tex]f_r\times r=mr^2\times \alpha [/tex]
in pure Rolling [tex]a=\alpha r[/tex]
[tex]\alpha =\frac{a}{r}[/tex]
[tex]f_r=ma[/tex]
Form FBD [tex]mg\sin \theta -f_r=ma[/tex]
[tex]mg\sin \theta =ma+ma[/tex]
[tex]2ma=mg\sin \theta [/tex]
[tex]a=\frac{g\sin \theta }{2}[/tex]
An object of mass m = 8.0 kg is attached to an ideal spring and allowed to hang in the earth's gravitational field. The spring stretches 2.2 cm before it reaches its equilibrium position.
If it were now allowed to oscillate by this spring, what would be its frequency?
Answer:
Frequency, f = 3.35 Hz
Explanation:
It is given that,
Mass of the object, m = 8 kg
Stretching in the spring, x = 2.2 cm
When the spring is hanged up in the Earth's gravitational field, its weight is balanced by the force in the spring. So,
[tex]mg=kx[/tex]
k is the spring constant
[tex]k=\dfrac{mg}{x}[/tex]
[tex]k=\dfrac{8\times 9.8}{2.2\times 10^{-2}}[/tex]
k = 3563.63 N/m
Let f is the frequency of oscillation. Its expression is given by :
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}[/tex]
[tex]f=\dfrac{1}{2\pi}\sqrt{\dfrac{3563.63}{8}}[/tex]
f = 3.35 Hz
So, the frequency of oscillation by the spring is 3.35 Hz. Hence, this is the required solution.
The earth has a radius of 6.38 × 106 m and turns on its axis once every 23.9 h.
(a) What is the tangential speed (in m/s) of a person living in Ecuador, a country that lies on the equator?
(b) At what latitude (i.e., the angle theta in the drawing) is the tangential speed one-third that of a person living in Ecuador?
Answer:
a) V = 465.9 m/s
b) θ = 70.529°
Explanation:
Let's first calculate angular velocity of earth:
[tex]\omega=\frac{2\pi}{23.9h}*1h/3600s[/tex]
Velocity of a person on Ecuador will be:
[tex]V_E = \omega*R[/tex]
[tex]V_E = 465.9 m/s[/tex]
For part b, since angular velocity is the same:
[tex]\frac{\omega*R}{3}=\omega*(R*cos\theta )[/tex]
Solving for θ:
[tex]\theta=acos(1/3)[/tex]
[tex]\theta=70.529\°[/tex]
(a) speed at the equator: 465.9 m/s
(b) Latitude for 1/3 speed: 50.6°
(a) The tangential speed of a person living in Ecuador is equal to the circumference of the Earth at the equator divided by the period of rotation. The circumference of the Earth at the equator is 2πr, where r is the radius of the Earth. The period of rotation is 23.9 hours, which is equal to 23.9 × 3600 seconds. Therefore, the tangential speed of a person living in Ecuador is:
v = 2πr / T = 2π(6.38 ×[tex]10^6[/tex] m) / (23.9 × 3600 s) ≈ 465.9 m/s
(b) The tangential speed of a person living at a latitude of θ is equal to vr * cos(θ), where vr is the tangential speed of a person living in Ecuador and r is the radius of the Earth. Therefore, we can solve for θ:
cos(θ) = vr / v = 465.9 m/s / v
Solving for θ, we get:
θ = arccos(vr / v) ≈ 50.6°
Therefore, the latitude at which the tangential speed is one-third that of a person living in Ecuador is approximately 50.6°.
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Selena uses a garden hose to fill a bucket of water. The water enters the hose through a faucet with a 6.0 cm diameter.
The speed of the water at the faucet is 5/ms.
If the faucet and nozzlw are at the same height, and the water leaves the nozzle with a speed of 20 m/s, what is the diameter of the nozzle?
Answer:
3 cm
Explanation:
To answer this question the equation of continuity can be used
For us to use the equaition of continuity, we will make a few assumptions:
That the temperature of the water does not change, therefore there is not expansion or contraction (change in volume)That the flow is non-viscousThat there is a single entry and a single exitThat the water is incompressible[tex]A_{1}v_{1} = A_{2} v_{2} \\\pi (3cm)^{2} * 5m/s = \pi r^{2} *20m/s\\\\r = 1.5 cm\\D = 2r = 2*1.5cm=3 cm[/tex]
A thin-walled, hollow sphere of mass M rolls without slipping down a ramp that is inclined at an angle β to the horizontal. Find the magnitude of the acceleration of the sphere along the ramp. Express your answer in terms of β and acceleration due to gravity g.
Answer:
Explanation:
Given
inclination [tex]=\beta [/tex]
Assuming radius of sphere is r
Now from Free Body Diagram
[tex]Mg\sin \theta -f_r=Ma[/tex]
where [tex]f_r=friction\ force[/tex]
[tex]a=acceleration\ of\ system[/tex]
Now friction will Provide the Torque
[tex]f_r\times r=I\cdot \alpha [/tex]
where [tex]I=moment\ of\ inertia[/tex]
[tex]\alpha =angular\ acceleration [/tex]
[tex]f_r\times r=\frac{2}{3}Mr^2\times \frac{a}{r}[/tex]
in pure rolling [tex]a=\alpha r[/tex]
[tex]f_r=\frac{2}{3}Ma[/tex]
[tex]mg\sin \beta -\frac{2}{3}Ma=Ma[/tex]
[tex]Mg\sin \beta =\frac{5}{3}Ma[/tex]
[tex]a=\frac{3g\sin \beta }{5}[/tex]
Vadim bought his wife a "gold" bracelet from a guy on the street for $100. If he drops it in a beaker of water and the volume raises 400 cm3 and the mass is determined to be 0.5 kg, then is the jewelry gold? If it is not gold then what do you think it is made of?
Answer: It is not gold. The density is 1.25 g/cm3, very little compared with gold.
Explanation:
According to Archimedes’ principle, any body submerged in a liquid, receives an upward force, called buoyant force, which magnitude is equal to the weight of the liquid that displaces.
This weight can be calcultated as follows:
Fb= δH2O . Vbody g
We are told that the volume of the liquid displaced (equal to the volume of the body as it is completely submerged in water) is 400 cm3.
If we know that the mass of the bracelet is 0.5 Kg, we can find out the density of the material, just applying the same relationship between mass, density and volume:
m = δbracelet. V
δbracelet = m / V = 500 g / 400 cm3= 1.25 g/cm3
As the gold density is approximately 19.3 g/cm3, it is clear that the bracelet is not made from gold, but from something less dense.
The density of Vadim's bracelet is calculated to be 1.25 g/cm³, which is significantly lower than the density of pure gold (19.3 g/cm³). Thus, the bracelet is not made of gold and could be made from a lighter metal or alloy.
To determine if Vadim's bracelet is gold, we must calculate its density and compare it to the density of pure gold. The density of a substance is the mass-to-volume ratio, and for gold, it is 19.3 g/cm³. Given that the bracelet has a mass of 0.5 kg (or 500 g) and a volume displacement of 400 cm³ when submerged in water, we can calculate its density using the formula:
Density = Mass / Volume
density = 500 g / 400 cm³ = 1.25 g/cm³
The calculated density of 1.25 g/cm³ is significantly lower than the density of pure gold, which suggests that the bracelet is not made of gold. Since the density is lower than that of gold, the bracelet could be made of a metal or alloy with a lower density, such as aluminium or a mixture that includes materials with lighter densities.
Identify the procedure to determine a formula for self-inductance, or inductance for short. Using the formula derived in the text, find the inductance in henries for a solenoid with 900 loops of wire wound on a rod 9 cm long with radius 3 cm.
Answer:
L = 0.0319 H
Explanation:
Given that,
Number of loops in the solenoid, N = 900
Radius of the wire, r = 3 cm = 0.03 m
Length of the rod, l = 9 cm = 0.09 m
To find,
Self inductance in the solenoid
Solution,
The expression for the self inductance of the solenoid is given by :
[tex]L=\dfrac{\mu_o N^2 A}{l}[/tex]
[tex]L=\dfrac{4\pi\times10^{-7}\times(900)^{2}\times\pi(0.03)^{2}}{0.09}[/tex]
L = 0.0319 H
So, the self inductance of the solenoid is 0.0319 henries.
Nitrogen (N2) undergoes an internally reversible process from 6 bar, 247°C during which pν1.2 = constant. The initial volume is 0.1 m3 and the work for the process is 50 kJ. Assuming ideal gas behavior, and neglecting kinetic and potential energy effects, determine heat transfer, in kJ, and the entropy change, in kJ/K.
The given question involves the thermodynamic process of an internally reversible process of nitrogen gas (N2) at specific pressure and temperature with a constant value of pν1.2. However, the question does not provide enough information to calculate the heat transfer and entropy change accurately.
Explanation:The given question involves the thermodynamic process of an internally reversible process of nitrogen gas (N2) at specific pressure and temperature with a constant value of pν1.2. In order to determine the heat transfer and entropy change, we need to use the first and second laws of thermodynamics. However, the question does not provide enough information to calculate the heat transfer and entropy change accurately.
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A fatigue test was conducted in which the mean stress was 50 MPa (7250 psi) and the stress amplitude was 225 MPa (32,625 psi).
(a) Compute the maximum and minimum stress levels.
(b) Compute the stress ratio.
(c) Compute the magnitude of the stress range.
Answer:
275 MPa, -175 MPa
-0.63636
450 MPa
Explanation:
[tex]\sigma_{max}[/tex] = Maximum stress
[tex]\sigma_{min}[/tex] = Minimum stress
[tex]\sigma_m[/tex] = Mean stress = 50 MPa
[tex]\sigma_a[/tex] = Stress amplitude = 225 MPa
Mean stress is given by
[tex]\sigma_m=\frac{\sigma_{max}+\sigma_{min}}{2}\\\Rightarrow \sigma_{max}+\sigma_{min}=2\sigma_m\\\Rightarrow \sigma_{max}+\sigma_{min}=2\times 50\\\Rightarrow \sigma_{max}+\sigma_{min}=100\ MPa\\\Rightarrow \sigma_{max}=100-\sigma_{min}[/tex]
Stress amplitude is given by
[tex]\sigma_a=\frac{\sigma_{max}-\sigma_{min}}{2}\\\Rightarrow \sigma_{max}-\sigma_{min}=2\sigma_a\\\Rightarrow \sigma_{max}-\sigma_{min}=2\times 225\\\Rightarrow \sigma_{max}-\sigma_{min}=450\ MPa\\\Rightarrow 100-\sigma_{min}-\sigma_{min}=450\\\Rightarrow -2\sigma_{min}=350\\\Rightarrow \sigma_{min}=-175\ MPa[/tex]
[tex]\sigma_{max}=100-\sigma_{min}\\\Rightarrow \sigma_{max}=100-(-175)\\\Rightarrow \sigma_{max}=275\ MPa[/tex]
Maximum stress level is 275 MPa
Minimum stress level is -175 MPa
Stress ratio is given by
[tex]R=\frac{\sigma_{min}}{\sigma_{max}}\\\Rightarrow R=\frac{-175}{275}\\\Rightarrow R=-0.63636[/tex]
The stress ratio is -0.63636
Stress range is given by
[tex]\sigma_{max}-\sigma_{min}=450\ MPa[/tex]
Magnitude of the stress range is 450 MPa
The maximum and minimum stress levels could be calculated as 275 MPa and -175 MPa respectively. The stress ratio comes out to be -0.64, and the magnitude of the stress range is 450 MPa.
Explanation:In material fatigue studies, the stress levels in a material are often evaluated. In your case:
The mean stress (σm) is given as 50 MPa, and the stress amplitude (σa) is 225 MPa.(a) The maximum stress (σmax) can be calculated as σmax = σm + σa = 50 + 225 = 275 MPa, and the minimum stress (σmin) can be calculated as σmin = σm - σa = 50 - 225 = -175 MPa.(b) The stress ratio (R) is derived as the ratio of the minimum to the maximum stress. So R = σmin / σmax = -175 / 275 = -0.64(rounded to two decimal places).(c) The stress range (Δσ) is defined as the difference between the maximum and minimum stresses in a cycle. So Δσ = σmax - σmin = 275 - (-175) = 450 MPa.Learn more about Material Fatigue here:https://brainly.com/question/33293027
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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]
A 2.8-carat diamond is grown under a high pressure of 58 × 10 9 N / m 2 .
(a) By how much does the volume of a spherical 2.8-carat diamond expand once it is removed from the chamber and exposed to atmospheric pressure?
(b) What is the increase in the diamond’s radius? One carat is 0.200 g, and you can use 3.52 g/cm3 for the density of diamond, and 4.43 × 10 11 N / m 2 for the bulk modulus of diamond.
Final answer:
To calculate the change in volume of a diamond exposed to atmospheric pressure, we can use the formula for bulk modulus. The increase in the diamond's radius can be found using the formula for increase in volume.
Explanation:
(a) To calculate the change in volume, we can use the formula for bulk modulus:
ΔV = V * (Pf - Pi) / B
Where ΔV is the change in volume, V is the initial volume, Pf is the final pressure, Pi is the initial pressure, and B is the bulk modulus.
Substituting the given values, we get:
ΔV = (4/3) * π * r^3 * (Pf - Pi) / B
Since the sphere is symmetrical, the change in radius (Δr) is the same in all directions. So, we can calculate it as:
Δr = ΔV / ((4/3) * π * (r^2) * ΔP)
(b) To find the increase in the diamond's radius, we can use the formula for increase in volume:
ΔV = (4/3) * π * (Rf^3 - Ri^3)
Substituting the given values, we get:
Rf = (3 * ΔV + 4 * π * Ri^3) / (4 * π * Ri^2)
The diamond's volume expands by 2.08 x 10⁻⁹ m³ and its radius increases by 10 micrometers upon being exposed to atmospheric pressure. These calculations utilize the mass, density, and bulk modulus of the diamond.
First, we convert the mass of the diamond into kilograms:
2.8 carats * 0.200 g/carat = 0.56 g = 0.00056 kg
Next, calculate the initial volume ([tex]V_i[/tex]) of the diamond using the density formula:
Density = Mass / Volume
So, the initial volume:
[tex]V_i[/tex] = Mass / Density
[tex]V_i[/tex] = 0.56 g / 3.52 g/cm³
[tex]V_i[/tex] = 0.1591 cm³ = 1.591 × 10⁻⁷ m³
The bulk modulus (B) of diamond is given as 4.43 × 10¹¹ N/m² and the pressure change (ΔP) is:
ΔP = 58 × 10⁹ N/m²
Using the formula for volume change (ΔV), where:
ΔV / V = -ΔP / B
we get:
ΔV = -[tex]V_i[/tex] * ΔP / B
ΔV = -1.591 × 10⁻⁷ m³ * 58 × 109 N/m² / 4.43 × 10¹¹ N/m²
ΔV ≈ -2.08 × 10⁻⁹ m³
Therefore, the volume expansion when exposed to atmospheric pressure is:
ΔV = 2.08 × 10⁻⁹ m³
(b) For the increase in radius (Δr), use the formula for the volume of a sphere:
V = (4/3)πr³
The new volume ([tex]V_f[/tex]) is:
[tex]V_f[/tex] = Vi + ΔV
[tex]V_f[/tex] ≈ 1.591 × 10⁻⁷ m³ + 2.08 × 10⁻⁹ m³
[tex]V_f[/tex] ≈ 1.611 × 10⁻⁷ m³
Setting the initial and final volumes equal to the sphere volume formula, we solve for the radii:
[tex]r_{i[/tex] = (3[tex]V_i[/tex] / 4π)^(1/3)
[tex]r_{i[/tex] ≈ (3 * 1.591 × 10⁻⁷ m³ / 4π)^(1/3)
[tex]r_{i[/tex] ≈ 3.37 × 10-3 m
[tex]r_f[/tex] = (3[tex]V_f[/tex] / 4π)^(1/3)
[tex]r_f[/tex] ≈ (3 * 1.611 × 10⁻⁷ m³ / 4π)^(1/3)
[tex]r_f[/tex] ≈ 3.38 × 10⁻³ m
Thus the increase in radius Δr is:
Δr ≈ [tex]r_f[/tex] - [tex]r_{i[/tex]
Δr ≈ 3.38 × 10⁻³ m - 3.37 × 10⁻³ m
Δr = 0.01 × 10⁻³ m
Δr = 10 × 10⁻⁶ m = 10 μm
Halogen lightbulbs allow their filaments to operate at a higher temperature than the filaments in standard incandescent bulbs. For comparison, the filament in a standard lightbulb operates at about 2900K, whereas the filament in a halogen bulb may operate at 3400K. Which bulb has the higher peak frequency? Calculate the ratio of the peak frequencies. The human eye is most sensitive to a frequency around 5.5x10^14 Hz. Which bulb produces a peak frequency close to this value?
Answer:
Halogen
0.85294
Explanation:
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
b = Wien's displacement constant = [tex]2.897\times 10^{-3}\ mK[/tex]
T = Temperature
From Wien's law we have
[tex]\lambda_m=\frac{b}{T}\\\Rightarrow \lambda_m=\frac{2.897\times 10^{-3}}{2900}\\\Rightarrow \lambda_m=9.98966\times 10^{-7}\ m[/tex]
Frequency is given by
[tex]\nu=\frac{c}{\lambda_m}\\\Rightarrow \nu=\frac{3\times 10^8}{9.98966\times 10^{-7}}\\\Rightarrow \nu=3.00311\times 10^{14}\ Hz[/tex]
For Halogen
[tex]\lambda_m=\frac{b}{T}\\\Rightarrow \lambda_m=\frac{2.897\times 10^{-3}}{3400}\\\Rightarrow \lambda_m=8.52059\times 10^{-7}\ m[/tex]
Frequency is given by
[tex]\nu=\frac{c}{\lambda_m}\\\Rightarrow \nu=\frac{3\times 10^8}{8.52059\times 10^{-7}}\\\Rightarrow \nu=3.52088\times 10^{14}\ Hz[/tex]
The maximum frequency is produced by Halogen bulbs which is closest to the value of [tex]5.5\times 10^{14}\ Hz[/tex]
Ratio
[tex]\frac{3.00311\times 10^{14}}{3.52088\times 10^{14}}=0.85294[/tex]
The ratio of Incandescent to halogen peak frequency is 0.85294
Two generators use the same magnetic field and operate at the same frequency. Each has a single-turn circular coil. One generator has a coil radius of 5.1 cm and a peak emf of 1.8 V. The other generator has a peak emf of 3.9 V. Find the coil radius r of this other generator.
Answer:
The radius of the second coil will be 7.5 cm
Explanation:
We have given that radius of the first coil [tex]r_1=5.1cm=0.051m[/tex]
Peak emf of the first generator [tex]e_1=1.8Volt[/tex]
Peak emf of the second generator [tex]e_2=3.9Volt[/tex]
We know that induced emf is rate of change of magnetic flux , that is equal to magnetic field multiply by change in area
So emf will be proportional to the change in area
So [tex]\frac{e_1}{e_2}=\frac{A_1}{A_2}[/tex]
[tex]\frac{1.8}{3.9}=\frac{0.051^2}{r_2^2}[/tex]
[tex]r_2=0.075m = 7.5cm[/tex]
So the radius of the second coil will be 7.5 cm
A sled of mass m is being pulled horizontally by a constant horizontal force of magnitude F. The coefficient of kinetic friction is mu_k. During time interval t, the sled moves a distance s, starting from rest.Find the average velocity vavg of the sled during that time interval.Express your answer in terms of the given quantities and, if necessary, appropriate constants. You may or may not use all of the given quantities.
Answer:
The average velocity of the sled is vavg = s/t.
Explanation:
Hi there!
The average velocity is calculated as the traveled distance over time:
vavg = Δx/Δt
Where:
vavg = average velocity.
Δx = traveled distance.
Δt = elapsed time.
We already know the traveled distance (s) and also know the time it takes the sled to travel that distance (t). Then, the average velocity can be calculated as follows:
vavg = s/t
Have a nice day!
The average velocity of the sled, given by the formula v_avg = s / t, evaluates the distance covered over a certain duration of time. The sled has moved a distance 's' during a time interval 't', and so its average velocity over this interval is simply 's' divided by 't'. The presence of friction does not affect the calculation of average velocity.
Explanation:The average velocity of the sled is simply the distance traveled over a certain time interval. In this case, the sled starts from rest and is being pulled by a horizontal force F against a frictional force. Average Velocity, v_avg, is given by the formula:
v_avg = s / t
In physics, velocity is a measure of the rate of change of position concerning time, so the average velocity is simply the total distance (or displacement) divided by the total time. Given the fact that the sled has moved a distance, s, during the time interval t, the average velocity is simply s divided by t.
With this definition, the sled's average velocity can be determined without needing to tabulate its speed or direction throughout the time interval even in the presence of friction.
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A solid brass (bulk modulus 6.7 x 1010 N/m2) sphere is subjected to a pressure of 1.0 x 105 Pa due to the earth's atmosphere. On Venus the pressure due to the atmosphere is 9.0 x 106 Pa. By what fraction r/r0 (including the algebraic sign) does the radius of the sphere change when it is exposed to the Venusian atmosphere? Assume that the change in radius is very small relative to the initial radius.
Answer:
[tex]\frac{r}{r_0}=-4.4x10^{-5}[/tex]
Explanation:
Using the equation to find the fraction does the radius of the sphere change when it is exposed to Venusian atmosphere.
[tex]\frac{r}{r_0}=\frac{1}{3}*\frac{V}{V_0}[/tex]
Replacing to find the relation between the radius
[tex]\frac{V}{V_0}=-\frac{P}{B}[/tex]
Replacing numeric to find the relation
[tex]\frac{r}{r_0}=\frac{1}{3}*\frac{8.9x10^6pa}{6.7x10^{10}pa}[/tex]
[tex]\frac{r}{r_0}=\frac{1}{3}*-1.33x10^{-4}[/tex]
So the relation is
[tex]\frac{r}{r_0}=-4.4x10^{-5}[/tex]
A truck with a heavy load has a total mass of 7100 kg. It is climbing a 15∘ incline at a steady 15 m/s when, unfortunately, the poorly secured load falls off! Immediately after losing the load, the truck begins to accelerate at 1.5 m/s2. Part A What was the mass of the load? Ignore rolling friction.
Answer:
The load has a mass of 2636.8 kg
Explanation:
Step 1 : Data given
Mass of the truck = 7100 kg
Angle = 15°
velocity = 15m/s
Acceleration = 1.5 m/s²
Mass of truck = m1 kg
Mass of load = m2 kg
Thrust from engine = T
Step 2:
⇒ Before the load falls off, thrust (T) balances the component of total weight downhill:
T = (m1+m2)*g*sinθ
⇒ After the load falls off, thrust (T) remains the same but downhill component of weight becomes m1*gsinθ .
Resultant force on truck is F = T – m1*gsinθ
F causes the acceleration of the truck: F= m*a
This gives the equation:
T – m1*gsinθ = m1*a
T = m1(a + gsinθ)
Combining both equations gives:
(m1+m2)*g*sinθ = m1*(a + gsinθ)
m1*g*sinθ + m2*g*sinθ =m1*a + m1*g*sinθ
m2*g*sinθ = m1*a
Since m1+m2 = 7100kg, m1= 7100 – m2. This we can plug into the previous equation:
m2*g*sinθ = (7100 – m2)*a
m2*g*sinθ = 7100a – m2a
m2*gsinθ + m2*a = 7100a
m2* (gsinθ + a) = 7100a
m2 = 7100a/(gsinθ + a)
m2 = (7100 * 1.5) / (9.8sin(15°) + 1.5)
m2 = 2636.8 kg
The load has a mass of 2636.8 kg
The mass of the load is 2636.8 kg
given information:Mass of the truck m = 7100 kg
Angle θ = 15°
velocity v = 15m/s
Acceleration a = 1.5 m/s²
Now let us assume that the mass of the truck is m₁ and the mass of the load is m₂. The thrust from the engine be T.
resolving forces:Initially, the thrust (T) balances the total weight:
[tex]T = (m_1+m_2) g sin\theta[/tex]
When the load falls off, the thrust (T) remains the same
so the net force on the truck
[tex]F = T -m_1gsin\theta[/tex]
[tex]T-m_1gsin\theta = m_1a \\\\ T = m_1(a + gsin\theta)[/tex]
comparing both the equation for T we get:
[tex](m_1+m_2)gsin\theta = m_1(a + gsin\theta)\\\\m_1gsin\theta + m_2gsin\theta =m_1a + m_1gsin\theta\\\\m_2gsin\theta = m_1a[/tex]
Now, m₁+m₂ = 7100kg
m₁= 7100 – m₂
[tex]m_2gsin\theta = (7100-m_2)a\\\\m_2gsin\theta = 7100a-m_2a\\\\m_2gsin\theta + m_2a = 7100a\\\\m_2 (gsin\theta + a) = 7100a\\\\m_2 = 7100a/(gsin\theta + a)\\\\m_2 = (7100 \times1.5) / (9.8sin(15) + 1.5)\\\\m_2 = 2636.8 kg[/tex]
The load has a mass of 2636.8 kg
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A 25 - kg television set rests on four rubber pads , each having a height of 1 . 0 cm and a radius of 0 . 60 cm . A 200 - N horizontal force is applied to the television set . How far does it move sideways ? The shear modulus of rubber is 2 . 6 x 106 N / m2
The television set moves sideways approximately 0.15 µm.
Explanation:
To calculate the distance the television set moves sideways, we need to use the equation for shear deformation. The equation is Ax = (F * L0 * L2) / (S * A), where Ax is the displacement, F is the force applied, L0 is the original length, L2 is the height of the rubber pads, S is the shear modulus, and A is the cross-sectional area. Substituting the known values, we have Ax = (200 N * 1.0 cm) / (2.6 x 10^6 N/m^2 * (3.14 x (0.0060 m)^2)). Solving for Ax, we find Ax ≈ 1.50 x 10^-7 m, or 0.15 µm.
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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.
An object at rest begins to rotate with a constant angular acceleration. If this object rotates through an angle θ in time t, through what angle did it rotate in the time ½t?
a. 4θ
b. ¼θ
c. ½θ
d. 2θ
e. θ
Answer:
Angular displacement will be [tex]\frac{1}{4}\Theta[/tex]
So option (b) will be the correct option
Explanation:
We have given that firstly object is at rest
So [tex]\omega _i=0rad/sec[/tex]
From law of motion we know that angular displacement is given by
[tex]\Theta =\omega _it+\frac{1}{2}\alpha t^2=0\times t+\frac{1}{2}\alpha t^2=\frac{1}{2}\alpha t^2[/tex]
Now angular displacement by the object in [tex]\frac{t}{2}sec[/tex]
[tex]\Theta =0\times t+\frac{1}{2}\alpha (\frac{t}{2})^2=\frac{1}{4}(\frac{1}{2}\alpha t^2)=\frac{1}{4}\Theta[/tex]
So option (b) will be the correct option
The angle the object rotate through in the time [tex]\frac{1}{2} t[/tex] is [tex]\frac{1}{4} (\theta)[/tex]
Given the following data:
Initial angular speed = 0 m/s (since it starts from rest).Angle = [tex]\theta[/tex]Time = tTo determine the angle the object rotate through in the time [tex]\frac{1}{2} t[/tex]:
How to calculate angular displacement.Mathematically, angular displacement is given by this formula:
[tex]\theta = \omega_i t +\frac{1}{2} \alpha t^2[/tex]
Where:
[tex]\theta[/tex] is the angular displacement.[tex]\omega[/tex] is the initial angular speed.[tex]\alpha[/tex] is the angular acceleration.t is the time.Substituting the given parameters into the formula, we have;
[tex]\theta = 0( t )+\frac{1}{2} \alpha t^2\\\\\theta = \frac{1}{2} \alpha t^2[/tex]
when t = [tex]\frac{1}{2} t[/tex]:
[tex]\theta = \frac{1}{2} \alpha (\frac{t}{2} )^2\\\\\theta = \frac{1}{2} \alpha (\frac{t^2}{4} )\\\\\theta =\frac{1}{4} (\frac{1}{2} \alpha t^2)\\\\\theta =\frac{1}{4} (\theta)[/tex]
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A toy car having mass m = 1.10 kg collides inelastically with a toy train of mass M = 3.55 kg. Before the collision, the toy train is moving in the positive x-direction with a velocity of Vi = 2.20 m/s and the toy car is also moving in the positive x-direction with a velocity of vi = 4.95 m/s. Immediately after the collision, the toy car is observed moving in the positive x-direction with a velocity of 1.80 m/s.
(a) Determine the final velocity of the toy train. cm/s
(b) Determine the change ake in the total kinetic energy.
Answer:
[tex]V_{ft}= 317 cm/s[/tex]
ΔK = 2.45 J
Explanation:
a) Using the law of the conservation of the linear momentum:
[tex]P_i = P_f[/tex]
Where:
[tex]P_i=M_cV_{ic} + M_tV_{it}[/tex]
[tex]P_f = M_cV_{fc} + M_tV_{ft}[/tex]
Now:
[tex]M_cV_{ic} + M_tV_{it} = M_cV_{fc} + M_tV_{ft}[/tex]
Where [tex]M_c[/tex] is the mass of the car, [tex]V_{ic}[/tex] is the initial velocity of the car, [tex]M_t[/tex] is the mass of train, [tex]V_{fc}[/tex] is the final velocity of the car and [tex]V_{ft}[/tex] is the final velocity of the train.
Replacing data:
[tex](1.1 kg)(4.95 m/s) + (3.55 kg)(2.2 m/s) = (1.1 kg)(1.8 m/s) + (3.55 kg)V_{ft}[/tex]
Solving for [tex]V_{ft}[/tex]:
[tex]V_{ft}= 3.17 m/s[/tex]
Changed to cm/s, we get:
[tex]V_{ft}= 3.17*100 = 317 cm/s[/tex]
b) The kinetic energy K is calculated as:
K = [tex]\frac{1}{2}MV^2[/tex]
where M is the mass and V is the velocity.
So, the initial K is:
[tex]K_i = \frac{1}{2}M_cV_{ic}^2+\frac{1}{2}M_tV_{it}^2[/tex]
[tex]K_i = \frac{1}{2}(1.1)(4.95)^2+\frac{1}{2}(3.55)(2.2)^2[/tex]
[tex]K_i = 22.06 J[/tex]
And the final K is:
[tex]K_f = \frac{1}{2}M_cV_{fc}^2+\frac{1}{2}M_tV_{ft}^2[/tex]
[tex]K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2[/tex]
[tex]K_f = \frac{1}{2}(1.1)(1.8)^2+\frac{1}{2}(3.55)(3.17)^2[/tex]
[tex]K_f = 19.61 J[/tex]
Finally, the change in the total kinetic energy is:
ΔK = Kf - Ki = 22.06 - 19.61 = 2.45 J
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
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.
A 3 GHz line-of-sight microwave communication link consists of two lossless parabolic dish antennas, each1 m in diameter. If the receive antenna requires 10 nW of receive power for good reception and the distance between the antennas is 40 km, how much power should be transmitted?
Answer:
0.25938 W
Explanation:
c = Speed of light = [tex]3\times 10^8\ m/s[/tex]
[tex]\nu[/tex] = Frequency = 3 GHz
d = Diameter of lossless antenna = 1 m
r = Radius = [tex]\frac{d}{2}=\frac{1}{2}=0.5\ m[/tex]
[tex]A_t[/tex] = Area of transmitter
[tex]A_r[/tex] = Area of receiver
R = Distance between the antennae = 40 km
[tex]P_r[/tex] = Power of receiver = [tex]10\times 10^{-9}\ W[/tex]
[tex]P_t[/tex] = Power of Transmitter
Wavelength
[tex]\lambda=\frac{c}{\nu}\\\Rightarrow \lambda=\frac{3\times 10^8}{3\times 10^9}\\\Rightarrow \lambda=0.1\ m[/tex]
From Friis transmission formula we have
[tex]\frac{P_t}{P_r}=\frac{\lambda^2R^2}{A_tA_r}\\\Rightarrow P_t=P_r\frac{\lambda^2R^2}{A_tA_r}\\\Rightarrow P_t=10\times 10^{-9}\frac{0.1^2\times (40\times 10^3)^2}{\pi 0.5^2\times \pi 0.5^2}\\\Rightarrow P_t=0.25938\ W[/tex]
The power that should be transmitted is 0.25938 W
At one particular moment, a 15.0 kg toboggan is moving over a horizontal surface of snow at 4.80 m/s. After 7.00 s have elapsed, the toboggan stops. Use a momentum approach to find the magnitude of the average friction force (in N) acting on the toboggan while it was moving
Answer:
10.28571 N
Explanation:
m = Mass of toboggan = 15 kg
u = Initial velocity = 4.8 m/s
v = Final velocity = 0
t = Time taken = 7 seconds
Friction force is given by the change in momentum over time
[tex]F=\frac{m(v-u)}{t}\\\Rightarrow F=\frac{15(0-4.8)}{7}\\\Rightarrow F=-10.28571\ N[/tex]
The magnitude of the average friction force acting on the toboggan while it was moving is 10.28571 N
Using the principle of conservation of momentum, the magnitude of the average friction force acting on the toboggan can be found. The initial momentum of the toboggan is equal to the change in momentum, which is equal to the mass of the toboggan multiplied by the change in velocity. Dividing the change in momentum by the time interval gives the magnitude of the average friction force as 10.29 N.
Explanation:To find the magnitude of the average friction force acting on the toboggan, we can use the principle of conservation of momentum. The initial momentum of the toboggan is given by P = m * v, where m is the mass (15.0 kg) and v is the velocity (4.80 m/s). The final momentum is zero, as the toboggan comes to a stop. Therefore, the change in momentum is equal to the initial momentum.
The change in momentum is given by δP = m * δv, where δv is the change in velocity. Since the velocity changes from 4.80 m/s to 0 m/s, the change in velocity is -4.80 m/s. Therefore, the change in momentum is -15.0 kg * 4.80 m/s = -72.0 kg*m/s.
The average friction force is equal to the change in momentum divided by the time interval. The time interval is given as 7.00 s. Therefore, the magnitude of the average friction force is |-72.0 kg*m/s / 7.00 s| = 10.29 N.
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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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An automobile starter motor has an equivalent resistance of 0.0500Ω and is supplied by a 12.0-V battery with a 0.0100-Ω internal resistance.
(a) What is the current to the motor?
(b) What voltage is applied to it?
(c) What power is supplied to the motor?
(d) Repeat these calculations for when the battery connections are corroded and add 0.0900Ω to the circuit. (Significant problems are caused by even small amounts of unwanted resistance in low-voltage, high-current applications.)
Answer
given,
resistance = 0.05 Ω
internal resistance of battery = 0.01 Ω
electromotive force = 12 V
a) ohm's law
V = IR
and volage
[tex]V = \epsilon - Ir[/tex]
now,
[tex]IR = \epsilon - Ir[/tex]
[tex]I(R+r) = \epsilon[/tex]
[tex]I= \dfrac{\epsilon}{R+r}[/tex]
inserting the values
[tex]I= \dfrac{12}{0.05+0.01}[/tex]
I = 200 A
b) Voltage
V = I R
V = 200 x 0.05
V = 10 V
c) Power
P = I V
P = 200 x 10 = 2000 W
d) total resistance = 0.05 + 0.09 = 0.14 Ω
[tex]I= \dfrac{\epsilon}{R+r}[/tex]
[tex]I= \dfrac{12}{0.14+0.01}[/tex]
I = 80 A
V = 80 x 0.05 = 4 V
P = 4 x 80 = 320 W
Answer:
Explanation:
Resistance of motor, R = 0.05 ohm
internal resistance of battery, r = 0.01 ohm
Voltage of battery, V = 12 V
(a) Total resistance, R' = R + r = 0.05 + 0.01 = 0.06 ohm
Let the current be i.
use Ohm's law
i = V / R'
i = 12 / 0.06 = 200 A
(b) Voltage across motor, V' = i x R = 200 x 0.05 = 10 V
(c) Power, P = i²R = 200 x 200 x 0.05 = 2000 Watt.
(d) Total resistance, R' = 0.05 + 0.1 + 0.09 = 0.15 ohm
i = V / R' = 12 / 0.15 = 80 A
V' = i x R = 80 x 0.05 = 4 V
P' = i²R = 80 x 80 x 0.05 = 320 Watt
A rectangular loop of wire with width w = 5 cm, length L = 10cm, mass m = 40 g, and resistance R = 20 mΩ has an initial velocity v0 = 1 m/s to the right. It crosses from a region with zero magnetic field to a region with B = 2T pointing out of the page. How far does the loop penetrate into the magnetic field?
Answer:
The loop penetrate 4 cm into the magnetic field.
Explanation:
Given that,
Width w= 5 cm
Length L= 10 cm
mass m = 40 g
Resistance R = 20 mΩ
Initial velocity = 1 m/s
Magnetic field = 2 T
We need to calculate the induced emf
Using formula of emf
[tex]\epsilon=v_{0}Bw[/tex]
Put the value into the formula
[tex]\epsilon =1\times2\times5\times10^{-2}[/tex]
[tex]\epsilon =10\times10^{-2}\ volt[/tex]
We need to calculate the current
Using Lenz's formula
[tex]i=\dfrac{\epsilon}{R}[/tex]
[tex]i=\dfrac{10\times10^{-2}}{20\times10^{-3}}[/tex]
[tex]i=5\ A[/tex]
We need to calculate the force
Using formula of force
[tex]F=i(\vec{w}\times\vec{B})[/tex]
[tex]F=iwB[/tex]
Put the value into the formula
[tex]F=5\times5\times10^{-2}\times2[/tex]
[tex]F=0.5\ N[/tex]
We need to calculate the acceleration
Using formula of acceleration
[tex]a=\dfrac{F}{m}[/tex]
Put the value in to the formula
[tex]a=\dfrac{0.5}{40\times10^{-3}}[/tex]
[tex]a=12.5\ m/s^2[/tex]
We need to calculate the distance
Using equation of motion
[tex]v^2=u^2+2as[/tex]
[tex]s=\dfrac{v^2-u^2}{2a}[/tex]
[tex]s=\dfrac{0-1^2}{2\times(-12.5)}[/tex]
[tex]s=0.04\ m[/tex]
[tex]s=4\ cm[/tex]
Hence, The loop penetrate 4 cm into the magnetic field.
A wheel with a weight of 395 N comes off a moving truck and rolls without slipping along a highway. At the bottom of a hill it is rotating at an angular velocity of 23.1 rad/s . The radius of the wheel is 0.652 m and its moment of inertia about its rotation axis is 0.800 MR2. Friction does work on the wheel as it rolls up the hill to a stop, at a height of h above the bottom of the hill; this work has a magnitude of 3470 J .
Answer:
h=12.04m
Explanation:
1) Notation and some important concepts
[tex]\omega_i[/tex] represent the initial angular velocity
[tex]v_i[/tex] represent the initial velocity
[tex]h[/tex] represent the final height reached by the mass
[tex]M[/tex] represent the mass of the object
[tex]W_f[/tex] represent the work due the friction force (variable of interest)
[tex]KE_{rot}[/tex] represent the rotational energy
[tex]KE_{tra}[/tex] represent the transitional kinetic energy
[tex]PE=mgh[/tex] represent the potential energy
[tex]I=0.8MR^2[/tex] represent the rotational inertia
W= 395 N is the weight of the object
For this problem we can use the principle of energy conservation, this principle states that "the total energy of an isolated system remains constant; it is said to be conserved over time".
At the begin the wheel had rotational energy defined as "The kinetic energy due to rotational motion. Is a scalar quantity and have units of energy usually Joules". And this energy is represented by the following formula: [tex]KE_{rot}=\frac{1}{2}I\omega^2_i[/tex]
At the starting point the wheel also had kinetic energy defined as "The energy of mass in motion" and is given by the formula : [tex]KE_{tran}=\frac{1}{2}mv_i^2_i[/tex]
At the end of the movement we have potential energy since the mass is at height h the potential energy is defined as "The energy stored within an object, due to the object's position, arrangement or state" and is given by the formula [tex]PE=mgh[/tex].
Since we have friction acting we have a work related to the force of friction and we need to subtract this from the formula of conservation of energy
2) Formulas to use
The figure attached is an schematic draw for the problem
Using the principle of energy conservation we have:
[tex]KE_{rot}+KE_{tran}-W_f =PE[/tex]
Replacing the formulas for each energy w ehave:
[tex]\frac{1}{2}I\omega^2_i+\frac{1}{2}mv_i^2_i-W_f =Mgh[/tex] (1)
We also know that [tex]v_i =\omega_i R[/tex] and [tex]I=0.8MR^2[/tex] so if we replace this into equation (1) we got:
[tex]0.8(\frac{1}{2})M(R\omega_i)^2 +\frac{1}{2}M(\omega_i R)^2-W_f=Mgh[/tex] (2)
We also know that the weight is defined as [tex]W=mg[/tex] so then [tex]M=\frac{W}{g}[/tex], so if we replace this into equation (2) we have:
[tex]0.8(\frac{1}{2})\frac{W}{g}(R\omega_i)^2 +\frac{1}{2}\frac{W}{g}(\omega_i R)^2-W_f=Wh[/tex] (3)
So then if we solve for h we got:
[tex]h=\frac{0.8(\frac{1}{2})\frac{W}{g}(R\omega_i)^2 +\frac{1}{2}\frac{W}{g}(\omega_i R)^2-W_f}{W}[/tex] (4)
3) Solution for the problem
Now we can replace the values given into equation (4):
[tex]h=\frac{0.8(\frac{1}{2})\frac{395N}{9.8\frac{m}{s^2}}(0.652m(23.1\frac{rad}{s}))^2 +\frac{1}{2}\frac{395}{9.8\frac{m}{s^2}}(23.1\frac{rad}{s}(0.652m))^2-3470J}{395N}=12.04m[/tex] (4)
So then our final answer would be h=12.04m