Answer: [tex]1.955(10)^{13} \frac{Pa.s}{m^{3}}[/tex]
Explanation:
This can be solved by the Poiseuille’s law for a laminar flow:
[tex]R=\frac{8 \eta L}{\pi r^{4}}[/tex]
Where:
[tex]R[/tex] is the resistance of the arteriole
[tex]\eta=3(10)^{-3} Pa.s[/tex] is the viscosity of blood
[tex]L=1000 \mu m=1000(10)^{-6}m[/tex] is the length of the arteriole
[tex]r=25 \mu m=25(10)^{-6}m[/tex] is the radius of the arteriole
[tex]R=\frac{8 (3(10)^{-3} Pa.s)(1000(10)^{-6}m)}{\pi (25(10)^{-6}m)^{4}}[/tex]
[tex]R=1.955(10)^{13} \frac{Pa.s}{m^{3}}[/tex]
The resistance of the arteriole is calculated using the Hagen-Poiseuille equation, resulting in an approximate value of 1.96 × 10¹⁵ Pa·s·m⁻⁴. The formula determines resistance by varying viscosity, length, and radius values, which is essential for understanding blood flow dynamics in physiological settings.
We can solve this problem using the formula for fluid resistance in a cylindrical tube (Hagen-Poiseuille equation):
[tex]R = (8 \times \eta \times L) / (\pi \times r_4)[/tex]
Where:
R is the resistance.η is the viscosity of the fluid (3 × 10⁻³ Pa·s).L is the length of the tube (1000 × 10⁻⁶ m).r is the radius of the tube (25 × 10⁻⁶ m).Substituting the values:
[tex]R = (8 \times 3 \times 10^{-3} Pa\cdot s \times 1000 \times 10^{-6} m) / (\pi \times (25 \times 10^{-6} m))[/tex]
Calculating the values inside the formula:
[tex]R = (24 \times 10^{-3} Pa\cdot s\cdot m) / (\pi \times 3.90625 \times 10^{-18} m^{4})[/tex][tex]R = (24 \times 10^{-3} Pa\cdot s\cdot m) / (1.227 \times 10^{-17} m^{4})[/tex][tex]R \approx 1.96 \times 1015 Pa\cdot s/m^{-4}[/tex]So, the resistance of the arteriole is approximately 1.96 × 10¹⁵ Pa·s·m⁻⁴.
A counterflow double-pipe heat exchanger is used to heat water from 20°C to 80°C at a rate of 1.2 kg/s. The heating is to be completed by geothermal water available at 160°C at a mass flow rate of 2 kg/s. The inner tube is thin-walled and has a diameter of 1.5 cm. If the overall heat transfer coefficient of the heat exchanger is 640 W/m2K, determine the length of the heat exchanger required to achieve the required heating.
Answer:L=109.16 m
Explanation:
Given
initial temperature [tex]=20^{\circ}C[/tex]
Final Temperature [tex]=80^{\circ}C[/tex]
mass flow rate of cold fluid [tex]\dot{m_c}=1.2 kg/s[/tex]
Initial Geothermal water temperature [tex]T_h_i=160^{\circ}C[/tex]
Let final Temperature be T
mass flow rate of geothermal water [tex]\dot{m_h}=2 kg/s[/tex]
diameter of inner wall [tex]d_i=1.5 cm[/tex]
[tex]U_{overall}=640 W/m^2K[/tex]
specific heat of water [tex]c=4.18 kJ/kg-K[/tex]
balancing energy
Heat lost by hot fluid=heat gained by cold Fluid
[tex]\dot{m_c}c(T_h_i-T_h_e)= \dot{m_h}c(80-20)[/tex]
[tex]2\times (160-T)=1.2\times (80-20)[/tex]
[tex]160-T=36[/tex]
[tex]T=124^{\circ}C[/tex]
As heat exchanger is counter flow therefore
[tex]\Delta T_1=160-80=80^{\circ}C[/tex]
[tex]\Delta T_2=124-20=104^{\circ}C[/tex]
[tex]LMTD=\frac{\Delta T_1-\Delta T_2}{\ln (\frac{\Delta T_1}{\Delta T_2})}[/tex]
[tex]LMTD=\frac{80-104}{\ln \frac{80}{104}}[/tex]
[tex]LMTD=91.49^{\circ}C[/tex]
heat lost or gain by Fluid is equal to heat transfer in the heat exchanger
[tex]\dot{m_c}c(80-20)=U\cdot A\cdot[/tex] (LMTD)
[tex]A=\frac{1.2\times 4.184\times 1000\times 60}{640\times 91.49}=5.144 m^2[/tex]
[tex]A=\pi DL=5.144[/tex]
[tex]L=\frac{5.144}{\pi \times 0.015}[/tex]
[tex]L=109.16 m[/tex]
The length of a counterflow double-pipe heat exchanger required can be calculated by first determining the heat needed to raise the water temperature, then calculating the necessary surface area, and finally dividing by the inner circumference of the pipe to find the length.
Explanation:To determine the length of the heat exchanger required to heat water from 20°C to 80°C using geothermal water at 160°C, we can apply the basic principles from thermodynamics and heat transfer. The equation that relates the heat transfer Q with the overall heat transfer coefficient U, the surface area A, and the temperature difference ΔT is Q = U * A * ΔT. We can determine the heat needed to raise the temperature of water by using the specific heat of water and the mass flow rate of water.
First, the amount of heat needed to raise the temperature of water (Q) can be calculated using the specific heat capacity of water (c_p, which is approximately 4.18 kJ/kgK), the mass flow rate of water (m_dot), and the temperature rise (ΔT = T_f - T_i), using the formula Q = m_dot * c_p * ΔT.
Next, we can rearrange the heat transfer equation to solve for the surface area A required for the heat exchanger: A = Q / (U * ΔTLMTD), where ΔTLMTD is the logarithmic mean temperature difference, which can be calculated for a counterflow heat exchanger knowing the inlet and outlet temperatures of both the hot and cold fluids.
Finally, we can find the length of the heat exchanger (L) by dividing the surface area A by the inner circumference of the pipe (π * d), with d being the inner pipe diameter.
By calculating the heat necessary for the temperature rise of the water and applying the given overall heat transfer coefficient, we can determine the required length of the exchanger.
9. A baseball pitcher hurls a 400 g baseball toward a batter with a speed of 30 m/s. The batter hits the pitch directly back toward the pitcher and it travels at 38 m/s. If the collision with the bat required 80 ms (milliseconds), find the force the bat applies to the ball. (Watch signs.)
Answer:
The force the bat applies to the ball is 340 N
Explanation:
given information:
mass of the ball, m = 400 g = 0.4 kg
initial velocity, [tex]v_{i}[/tex] = 30 m/s
final velocity, [tex]v_{f}[/tex] = 38 m/s
time, t = 80 ms = 0.08 s
the correlation between the change in momentum and the force is shown by the following equation
ΔP = FΔt
F = ΔP/Δt
where,
Δp = mΔv
= m([tex]v_{f} - v_{i}[/tex])
= m([tex]v_{f} + v_{i}[/tex])
[tex]v_{i}[/tex] = - because in opposite direction, thus
F = ΔP/Δt
= m([tex]v_{f} + v_{i}[/tex])/Δt
= (0.4)(38+30)/0.08
= 340 N
During each cycle of operation a refrigerator absorbs 56 cal from the freezer compartment and expels 81 cal tothe room. If one cycle occurs every 10 s, how many minutes will it take to freeze 570 g of water, initially at 0°C?(Lv = 80 cal/g
Answer:
138.18 minutes
Explanation:
[tex]L_v[/tex] = Latent heat of water at 0°C = 80 cal/g
m = Mass of water = 570 g
Heat removed for freezing
[tex]Q=mL_v\\\Rightarrow Q=570\times 80\\\Rightarrow Q=45600\ cal[/tex]
Let N be the number of cycles and each cycle removes 56 cal from the freezer.
So,
[tex]55\times N=45600\\\Rightarrow N=\frac{45600}{55}[/tex]
Each cycle takes 10 seconds so the total time would be
[tex]\frac{45600}{55}\times \frac{10}{60}=138.18\ minutes[/tex]
The total time taken to freeze 138.18 minutes
Please help I have marked the circles with numbers which one is it anybody knows?
Where is the natural light display called aurora borealis located?
Answer:
hi sandra!!
Explanation:
the number 3 is the correct!!
The Aurora is an incredible light show caused by collisions between electrically charged particles released from the sun that enter the earth’s atmosphere and collide with gases such as oxygen and nitrogen. The lights are seen around the magnetic poles of the northern and southern hemispheres.
Auroras that occur in the northern hemisphere are called ‘Aurora Borealis’ or ‘northern lights’ and auroras that occur in the southern hemisphere are called ‘Aurora Australis’ or ‘southern lights’.
Auroral displays can appear in many differents colours, but green is the most common. Colours such as red, yellow, green, blue and violet are also seen occasionally. The auroras can appear in many forms, from small patches of light that appear out of nowhere to streamers, arcs, rippling curtains or shooting rays that light up the sky with an incredible glow.
Answer:
Option-(3): The aurora borealis are the formation of the colored patterns on the pole of the planet, but other celestial bodies also experience such effects due to the heat flares or the sun flares when exposed to them.
Explanation:
Aurora borealis:
The natural phenomenon which is caused in most of the celestial bodies, but when our planet earth is exposed to such electromagnetic rays or flares released from the Sun, when the energy level on the surface increases and the flare is cut off from the environment of the sun. As, starts motion towards the different celestial bodies present in the solar system.
And when it reaches earth, the earth surface reflects the high in energy atoms coming towards its atmosphere. As the magnetic field present on the earth surface interacts with the Sun's flare and then it cut off's into two moving towards the two poles of the planet.
And when the molecules cools down, they are seen are different patterns on the poles.Which are also called as the aurora borealis in scientific terms.A "biconvex" lens is one in which both surfaces of the lens bulge outwards. Suppose you had a biconvex lens with radii of curvature with magnitudes of |R1|=10cm and |R2|=15cm. The lens is made of glass with index of refraction nglass=1.5. We will employ the convention that R1 refers to the radius of curvature of the surface through which light will enter the lens, and R2 refers to the radius of curvature of the surface from which light will exit the lens.Part AIs this lens converging or diverging?Part BWhat is the focal length f of this lens in air (index of refraction for air is nair=1)?Express your answer in centimeters to two significant figures or as a fraction.
A biconvex lens with the given parameters is a converging lens. Using the Lens Maker's Equation with the radii of curvature and index of refraction for glass and air, the focal length of the lens is calculated to be approximately 12 cm.
Explanation:A biconvex lens, where both surfaces of the lens bulge outwards, will bend light rays such that they converge at a focal point. With the parameters given (|R1|=10cm, |R2|=15cm, and nglass=1.5), we can deduce that this lens is a converging lens.
Part A: Since a biconvex lens makes parallel rays of light converge at a point after passing through the lens, it is classified as a converging lens.
Part B: To calculate the focal length (f) of the lens, we use the Lens Maker's Equation:
First, we convert the radii of curvature to the appropriate signs as per the lensmaker's convention (positive for convex surfaces when the outside medium is air). R1 = +10cm and R2 = -15cm, since the light exits from the second surface.Next, we plug the values into the equation (1/f) = (nglass - nair) ((1/R1) - (1/R2)) to get the reciprocal of the focal length.Carrying out the calculation with the data given (nglass=1.5, nair=1, R1=+10cm, and R2=-15cm), we get:
(1/f) = (1.5 - 1) ((1/10cm) - (1/(-15cm)))
(1/f) = 0.5 * (0.1cm⁻¹ + 0.0667cm⁻¹)
(1/f) = 0.5 * 0.1667cm⁻¹
(1/f) = 0.08335cm⁻¹
Therefore, the focal length f is the reciprocal of 0.08335cm⁻¹ which is approximately:
f ≈ 12cm
What is a particulate ? Name a couple of examples.
Answer:Particulates are small, distinct solids suspended in a liquid or gas and example are dust,soot,and salt particles
Explanation:
A planet is in an elliptical orbit around a distant star. At periastron (the point of closest approach to the star), the planet is rp=4.50×108 km from the star and is moving with a speed of vp=18.5 km/s . When the planet is at apastron (the point of greatest distance from the star), it is ra=9.10×108 km from the star. How fast is the planet moving at apastron? va=? km/s
Answer:
9.15 km/s
Explanation:
rp = 4.5 x 10^8 km
vp = 18.5 km/s
ra = 9.10 x 10^8 km
va = ?
According to the conservation of angular momentum constant.
Let m be the mass of planet
m x rp x vp = m x ra x va
4.5 x 10^8 x 18.5 = 9.10 x 10^8 x va
va = 9.15 km/s
To calculate the speed of the planet at apastron, we can use Kepler's second law and the given values of rp, vp, and ra. Plugging in the values, we find that the planet is moving at a speed of 0.92 km/s at apastron.
Explanation:To calculate the speed of the planet at apastron, we can use Kepler's second law, which states that the area swept out by a planet in equal time intervals is constant. At periastron, the planet is moving fastest, so we can use the equation:
A1 = A2
where A1 is the area swept out at periastron and A2 is the area swept out at apastron.
Since the areas are equal, we can set up the following equation:
0.5 * rp * vp = 0.5 * ra * va
where rp is the distance at the periastron, vp is the velocity at the periastron, ra is the distance at the apastron, and VA is the velocity at the apastron. We can rearrange this equation to solve for va:
va = (rp * vp) / ra
Plugging in the given values, we get:
va = (4.50 x 10^8 km * 18.5 km/s) / (9.10 x 10^8 km)
va = 0.92 km/s
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A liquid of density 1290 kg/m 3 1290 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.83 m/s 9.83 m/s and the pipe diameter d 1 d1 is 12.1 cm 12.1 cm . At Location 2, the pipe diameter d 2 d2 is 17.7 cm 17.7 cm . At Location 1, the pipe is 8.35 m higher than it is at location 2. Ignoring viscosity, calculate the difference between fluid pressure at location 2 and the fluid pressure at location 1.
Answer:
[tex]\Delta P=1060184.8946\ Pa[/tex]
[tex]P_1=124651.2383\ Pa[/tex]
Explanation:
Given:
density of liquid, [tex]\rho=1290\ kg.m^{-3}[/tex]speed of flow at location 1, [tex]v_1=9.83\ m.s^{-1}[/tex]diameter of pipe at location 1, [tex]d_1=0.121\ m[/tex]diameter of pipe at location 2, [tex]d_2=0.177\ m[/tex]height of pipe at location 1, [tex]z_1=8.35\ m[/tex]We know the Bernoulli's equation of in-compressible flow:
[tex]\frac{P}{\rho.g} +\frac{v^2}{2g} + z=constant[/tex] ........................(1)
Cross sectional area of pipe at location 2:
[tex]A_2=\pi \frac{d_2^2}{4}[/tex]
[tex]A_2=\pi\times \frac{0.177^2}{4}[/tex]
[tex]A_2=0.0246\ m^2[/tex]
Cross sectional area of pipe at location 1:
[tex]A_1=\pi \frac{d_1^2}{4}[/tex]
[tex]A_1=\pi\times \frac{0.121^2}{4}[/tex]
[tex]A_1=0.0115\ m^2[/tex]
Using continuity equation:
[tex]A_1.v_1=A_2.v_2[/tex]
[tex]0.0115\times 9.83=0.0246\times v_2[/tex]
[tex]v_2=4.5953\ m.s^{-1}[/tex]
Now apply continuity eq. on both the locations:
[tex]\frac{P_1}{\rho.g} +\frac{v_1^2}{2g} + z_1= \frac{P_2}{\rho.g} +\frac{v_2^2}{2g} + z_2[/tex]
[tex](P_2-P_1) = \rho.g [\frac{v_1^2}{2g} + z_1-\frac{v_2^2}{2g} ][/tex]
[tex]\Delta P=1290\times 9.8 [\frac{9.83^2}{19.6} + 8.35-\frac{4.5953^2}{19.6} ][/tex]
[tex]\Delta P=154266.016\ Pa[/tex]...................................Ans (a)
Now the mass flow rate at location 1:
[tex]\dot{m_1}=\rho\times \dot{V}[/tex]
[tex]\dot{m_1}=1290\times (0.0115\times 9.83)[/tex]
[tex]\dot{m_1}=145.828\ kg.s^{-1}[/tex]
Now pressure at location 1:
[tex]P_1=\frac{\dot{m_1}\times v_1}{A_1}[/tex]
[tex]P_1=\frac{145.828\times 9.83}{0.0115}[/tex]
[tex]P_1=124651.2383\ Pa[/tex] ...................................Ans (b)
The difference between fluid pressure at location 2 and fluid pressure at location 1 is mathematically given as
dP = 114 kPa
What is the difference between fluid pressure at location 2 and fluid pressure at location 1.?Question Parameter(s):
Generally, the Bernoulli's equation is mathematically given as
P + ρ*g*y + v² =pipe constant
Where
A1*v1 = A2*v2
π*(0.105/2)²*9.91 = π*(0.167/2)²*v2
v2 = 3.9 m/s
Therefore
P1 + ρ*g*y1 + v1² = P2 + ρ*g*y2 + v2²
dP = 1290*9.8*9.01 + 9.91² - 3.9²
dP = 114 kPa
In conclusion, difference between fluid pressure is
dP = 114 kPa
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A 20-cm-diameter vertical cylinder is sealed at the top by a frictionless 23 kg piston. The piston is 86 cm above the bottom when the gas temperature is 307°C. The air above the piston is at 1.00 atm pressure.
A) What is the gas pressure inside the cylinder?
B) What will the height of the piston be if the temperature is lowered to 13°C?
Answer:
108507.02596 Pa
42.42 cm
Explanation:
F = Force
m = Mass of piston = 23 kg
[tex]T_1[/tex] = Initial temperature = 307°C
[tex]T_2[/tex] = Final temperature = 13°C
[tex]P_o[/tex] = Outside pressure
[tex]P_i[/tex] = Pressure inside cylinder
A = Area of pistion
h = Height of piston
Change in pressure is given by
[tex]\Delta P=\frac{F}{A}\\\Rightarrow \Delta P=\frac{mg}{\pi r^2}\\\Rightarrow \Delta P=\frac{23\times 9.81}{\pi 0.1^2}\\\Rightarrow \Delta P=7182.02596\ Pa[/tex]
[tex]P_i-P_o=\Delta P\\\Rightarrow P_i=\Delta P+P_0\\\Rightarrow P_i=7182.02596+101325\\\Rightarrow P_i=108507.02596\ Pa[/tex]
Gas pressure inside the cylinder is 108507.02596 Pa
From ideal gas law we have the relation
[tex]\frac{V_1}{T_1}=\frac{V_2}{T_2}\\\Rightarrow \frac{Ah_1}{T_1}=\frac{Ah_2}{T_2}\\\Rightarrow \frac{h_1}{T_1}=\frac{h_2}{T_2}\\\Rightarrow h_2=\frac{h_1}{T_1}\times T_2\\\Rightarrow h_2=\frac{0.86}{307+273.15}\times (13+273.15)\\\Rightarrow h_2=0.42418\ m=42.42\ cm[/tex]
The height of the piston at 13°C will by 42.42 cm
(a) The gauge pressure inside the cylinder is 108,500 Pa.
(b) The height of the piston when the temperature is lowered to 13°C is 42.4 cm.
Change in gas pressure inside the cylinderThe change in the pressure of the gas inside the cylinder is calculated as follows;
ΔP = F/A
ΔP = mg/πr²
ΔP = (23 x 9.8)/(π x0.1²)
ΔP = 7,174.7 Pa
Gauge pressure inside the cylinderThe gauge pressure inside the cylinder is calculated as follows;
Pi = ΔP + Po
Pi = 7,174.7 + 101325
Pi = 108,500 Pa
Height of the pistonThe height of the piston is calculated as follows;
[tex]\frac{V_1}{T_1} = \frac{V_2}{T_2} \\\\\frac{Ah_1}{T_1} = \frac{Ah_2}{T_2} \\\\h_2 = \frac{T_2h_1}{T_1} \\\\h_2 = \frac{(13 + 273) \times 0.86}{(307 + 273)} \\\\h_2 = 0.424 \ m[/tex]
h₂ = 42.4 cm
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How much heat must be removed from 456 g of water at 25.0°C to change it into ice at - 10.0°C ? The specific heat of ice is 2090 J/(kg.K) and the latent heat of fusion of water is 33.5 x 10^4 J/kg .
Answer:209.98 kJ
Explanation:
mass of water [tex]m=456 gm[/tex]
Initial Temperature of Water [tex]T_i=25^{\circ}C[/tex]
Final Temperature of water [tex]T_f=-10^{\circ}C[/tex]
specific heat of ice [tex]c=2090 J/kg-K[/tex]
Latent heat [tex]L=33.5\times 10^4 J/kg[/tex]
specific heat of water [tex]c_{water}=4.184 KJ/kg-K[/tex]
Heat require to convert water at [tex]T=25^{\circ}C[/tex] to [tex]T=0^{\circ}C[/tex]
[tex]Q_1=0.456\times 4.184\times (25-0)=47.69 kJ[/tex]
Heat require to convert water at [tex]T=0^{\circ}[/tex] to ice at [tex]T=0^{\circ}[/tex]
[tex]Q_2=m\times L=0.456\times 33.5\times 10^4=152.76 kJ[/tex]
heat require to convert ice at [tex]T=0^{\circ} C\ to\ T=-10^{\circ} C[/tex]
[tex]Q_3=0.456\times 2090\times (0-(-10))=9.53 kJ[/tex]
Total heat [tex]Q=Q_1+Q_2+Q_3[/tex]
[tex]Q=47.69+152.76+9.53=209.98 kJ[/tex]
(7%) Problem 5: A thermos contains m1 = 0.73 kg of tea at T1 = 31° C. Ice (m2 = 0.095 kg, T2 = 0° C) is added to it. The heat capacity of both water and tea is c = 4186 J/(kg⋅K), and the latent heat of fusion for water is Lf = 33.5 × 104 J/kg. dho32@student.mtsac.edu
The final temperature of the mixture is approximately 29.91°C.
The heat exchange between the tea and the ice. The temperature of the final mixture will be somewhere between 0°C and 31°C, and we need to determine that final temperature. The heat transfer can be calculated using the principle of conservation of energy:
Qin = Qout
The heat gained by the ice as it melts is given by:
Qice = m2 ⋅ Lf
The heat gained by the tea as it cools down is given by:
Qtea = m1 ⋅ c ⋅ (T1 - Tfinal)
The negative sign is used because the tea is losing heat. Setting these equal to each other and solving for Tfinal, we get:
m1 ⋅ c ⋅ (T1 - Tfinal) = m2 ⋅ Lf
Now, let's plug in the given values:
0.73 kg ⋅ 4186 J/(kg ⋅ K) ⋅ (31°C - Tfinal) = 0.095 kg ⋅ 33.5 × 10^4 J/kg
Now, solve for Tfinal:
0.73 ⋅ 4186 ⋅ (31 - Tfinal) = 0.095 ⋅ (33.5 × 10^4)
3050.78 ⋅ (31 - Tfinal) = 3182.5
94602.78 - 3050.78 ⋅ Tfinal = 3182.5
-3050.78 ⋅ Tfinal = -91420.28
Tfinal = 91420.28/3050.78
Tfinal ≈ 29.91°C
So, the final temperature of the mixture is approximately 29.91°C.
Light emitting diode (LEDs) light bulbs have become required in recent years, but do they make financial sense? Suppose a typical 60-watt incandescent light bulb costs $.39 and lasts for 1,000 hours. A 15-watt LED, which provides the same light, costs $3.10 and lasts for 12,000 hours. A kilowatt hour of electricity costs $.115. A kilowatt-hour is 1,000 watts for 1 hour. However, electricity costs actually vary quite a bit depending on location and user type. An industrial user in West Virginia might pay $.04 per kilowatt-hour whereas a residential user in Hawaii might pay $.25. You require a return of 11 percent and use a light fixture 500 hours per year. What is the break-even cost per kilowatt-hour?
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
A 60-kg woman stands on the very end of a uniform board, which is supported one-quarter of the way from one end and is balanced. What is the mass of the board?
a. 60 kg
b. 30 kg
c. 20 kg
d. 15 kg
e. 120 kg
The correct option can be seen in Option A.
The diagrammatic expression of the question can be seen in the image attached below.
From the given question, we are being informed that the uniform board is balanced. As a result, the torque(i.e. a measurement about how significantly a force acts on a body for it to spin about an axis) acting on the right-hand side of the balance point should be equal to that of the left-hand side.
Mathematically;
[tex]\mathbf{\tau_{_{right}}= \tau_{_{left}}}[/tex]
Given that the mass of the woman = 60 kg
[tex]\mathbf{\tau =\dfrac{m\times g \times l}{\mu}}[/tex]
[tex]\mathbf{\tau_{left} =\dfrac{m\times g \times l}{\mu}}---(1)[/tex]
[tex]\mathbf{\tau_{_{right}} =\dfrac{60 \times g \times l}{\mu}}---(2)[/tex]
Equating both (1) and (2) together, we have:
[tex]\mathbf{\dfrac{m\times g \times l}{\mu} =\dfrac{60 \times g \times l}{\mu} }[/tex]
Dividing like terms on both side
mass (m) = 60 kg
As such, the correct option can be seen in Option A.
Thus, we can conclude that from the 60-kg woman who stands on the very end of a uniform board, the mass of the board on the other end is also 60 kg.
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With the switch open, roughly what must be the resistance of the resistor on the right for the current out of the battery to be the same as when the switch is closed (and the resistances of the two resistors are 20 Ω and 10 Ω)?
A) 7ΩB) 15ΩC) 30ΩD) 5Ω
Answer:
The resistance must be 6.67[tex]\Omega[/tex]
Solution:
Resistance, [tex]R_{1} = 20\Omega[/tex]
Resistance, [tex]R_{2} = 10\Omega[/tex]
For the current to be the same when the switch is open or closed, the resistances must be connected in parallel as current is distributed in parallel with the same voltage across the circuit:
Thus in parallel:
[tex]\frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}[/tex]
[tex]\frac{1}{R_{eq}} = \frac{1}{20} + \frac{1}{10}[/tex]
[tex]R_{eq} = 6.67\ \Omega[/tex]
A photographer uses his camera, whose lens has a 50 mm focal length, to focus on an object 2.5 m away. He then wants to take a picture of an object that is 50 cm away.
Part A
In which direction must the lens move to focus on this second object?
Part B
How far must the lens move to focus on this second object?
Answer:
0.004 m away from the film
Explanation:
u = Object distance
v = Image distance
f = Focal length = 50 mm
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{2.5}\\\Rightarrow \frac{1}{v}=\frac{98}{5} \\\Rightarrow v=\frac{5}{98}=0.051\ m[/tex]
The image distance is 0.051 m
When u = 50 cm
[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}-\frac{1}{u}=\frac{1}{v}\\\Rightarrow \frac{1}{v}=\frac{1}{0.05}-\frac{1}{0.5}\\\Rightarrow \frac{1}{v}=18\\\Rightarrow v=\frac{1}{18}=0.055\ m[/tex]
The image distance is 0.055 m
The lens has moved 0.055-0.051 = 0.004 m away from the film
To focus on the closer object, the lens must move away from the object. By applying the lens formula, we calculate that the lens must move exactly 7mm away from the initial position.
Explanation:This problem comes down to the concept of the lens formula in optical physics. The lens formula is 1/f = 1/v - 1/u where f is the focal length, v is the image distance, and u is the object distance.
Part A: To focus on the second object which is closer, the lens must move away from the object.
Part B: To calculate how far the lens must move, we need to apply the lens formula. Let's calculate the image distances for the two object distances.
For an object at 2.5 m, 1/v = 1/f + 1/u = 1/0.05 + 1/2.5 = 20 + 0.4 = 20.4. So, v = 0.049 m or 49 mm.
For an object at 0.5 m, 1/v = 1/f + 1/u = 1/0.05 - 1/0.5 = 20 - 2 = 18. So, v = 0.056 m or 56 mm.
Hence, the lens must move 56mm - 49mm = 7mm away from the initial position.
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A 0.20 kg particle moves along the x axis under the influence of a stationary object. The potential energy is given by U(x) = 8x2 + 2x4, where U is in joules and x is in meters. If the particle has a speed of 5.0 m/s when it is at x = 1.0 m, its speed when it is at the origin is:
Answer:11.18 m/s
Explanation:
Given
mass of particle m=0.2 kg
Potential Energy U(x) is given by
[tex]U(x)=8x^2+2x^4[/tex]
at x=1 m
[tex]U(1)=8+2=10 J[/tex]
kinetic energy at x=1 m
[tex]K.E.=\frac{1}{2}mv^2=\frac{1}{2}\times 0.2\times 5^2[/tex]
[tex]K.E.=2.5 J[/tex]
Total Energy =U+K.E.
[tex]Total=10+2.5=12.5 J[/tex]
at x=0, U(0)=0
as total Energy is conserved therefore K.E. at x=0 is equal to Total Energy
[tex]\frac{1}{2}\times 0.2\times v^2=12.5[/tex]
[tex]v^2=125[/tex]
[tex]v=\sqrt{125}[/tex]
[tex]v=11.18 m/s[/tex]
By conserving energy, we find that the speed of the particle at the origin is 25 m/s.
Explanation:The question is asking for the speed of the particle at the origin given its potential energy function and speed at x = 1.0 m. This can be solved using the principle of energy conservation which states that the total energy (kinetic energy + potential energy) of the particle is conserved unless acted upon by an outside force.
In this case, we first find the total energy of the particle at x = 1.0 m. The kinetic energy is (1/2)mv² = (1/2)*0.2 kg*(5 m/s)² = 2.5 J. The potential energy at x = 1.0 m, according to the given function, is U(1) = 8(1)² + 2(1)⁴ = 10 J. So, the total energy at x = 1.0 m is 2.5 J + 10 J = 12.5 J.
At the origin (x = 0), the potential energy is U(0) = 0. So, the kinetic energy at the origin is equal to the total energy, which is 12.5 J. From the kinetic energy, we can find the speed using the equation KE = (1/2)mv², which gives v = sqrt((2*KE)/m) = sqrt((2*12.5 J)/0.20 kg) = 25 m/s.
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Suppose you are selling apple cider for two dollars a gallon when the temperature is 3.3 degree C. The coefficient of volume expansion of the cider is 280*10^-6(C degree)^-1. How much more money (in pennies) would you make per gallon be refilling the container on a day when the temperature is 32 degrees C? Ignore the expansion of the container. Round your answer to 0.1 penny.
Answer:
1.6 penny
Explanation:
[tex]V_0[/tex] = Original volume = 1 gal (Assumed)
[tex]\Delta T[/tex] = Change in temperature
[tex]\beta[/tex] = Coefficient of volume expansion = [tex]280\times 10^{-6}\ /^{\circ}[/tex]
Change in volume is given by
[tex]\Delta_V=\beta V_0\Delta T\\\Rightarrow \Delta_V=280\times 10^{-6}\times 1\times (32-3.3)\\\Rightarrow \Delta_V=0.008036[/tex]
New volume would be
[tex]1+0.008036=1.008036\ gal[/tex]
The amount of money earned extra would be
[tex]0.008036\times 2=0.016072\ \$[/tex]
1.6 penny more would be earned if the temperature is 32°C
Final answer:
By refilling a container of apple cider at 32 degrees C instead of 3.3 degrees C, you would make approximately 1.6 pennies more per gallon due to thermal expansion of the cider.
Explanation:
To calculate how much more money you would make per gallon by refilling the container of apple cider when the temperature is 32 degrees C, as opposed to 3.3 degrees C, you need to determine the change in volume due to thermal expansion.
The formula for volume expansion is ΔV = βV₀ΔT, where ΔV is the change in volume, β is the coefficient of volume expansion, V₀ is the initial volume, and ΔT is the change in temperature.
The initial temperature T1 is 3.3°C, and the final temperature T2 is 32°C, thus ΔT = T2 - T1 = 32°C - 3.3°C = 28.7°C. The coefficient of volume expansion of the cider, given as β, is 280 x 10^-6 (C°)^-1.
Assuming that the initial volume V₀ of the cider is 1 gallon, the change in volume ΔV would be:
ΔV = 280 x 10^-6 x 1 x 28.7 = 0.008036 gallons
To convert gallons to liters, we use the fact that 1 gallon is approximately 3.78541 liters. So, the increase in volume in liters would be:
ΔV (liters) = 0.008036 x 3.78541 = 0.0304 liters
Since there are approximately 3.78541 liters in a gallon, and knowing that the price for one gallon is two dollars, we can calculate the additional revenue (in pennies) as follows:
Extra revenue = ΔV (liters) / 3.78541 x 200 pennies = 0.0304 / 3.78541 x 200 ≈ 1.6 pennies
Therefore, you would make approximately 1.6 pennies more per gallon by refilling the container at 32°C compared to 3.3°C.
A 2.95 g sample of an unknown chlorofluorocarbon (an organic compound containing chlorine and fluorine) is decomposed and produces 581 mL of chlorine gas at a pressure of 752 mmHg and a temperature of 298 K. What is the percent by mass of chlorine in the chlorofluorocarbon?
Answer:
mass percent of chlorine is 56.5%
Explanation:
Given data:
Volume V = 581 mL = 0.581 L
Pressure P = 752 mm of Hg = 752/760 = 0.9894 atm
Temperature T = 298 K
Molar gas constant R = 0.08206 atm.L/mol.K
from Ideal gas equation we have following relation
PV = nRT solving for n
[tex]n = \frac{PV}{RT}[/tex]
[tex]n = \frac{0.9894 \times 0.581}{0.08206 \times 298}[/tex]
n = 0.0235 mole
mas of [tex]Cl_2[/tex] = moles × molar mass of [tex]Cl_2[/tex]
[tex]= 0.0235 \times 70.906 = 1.66 g[/tex]
Mass percent of chlorine [tex]= \frac{mass\ of\ Cl}{mass\ of\ sample}[/tex]
[tex]=\frac{1.66}{2.95} = 0.565 = 56.5 %[/tex]
The volume of water in the Pacific Ocean is about 7.00 × 108 km3. The density of seawater is about 1030 kg/m3. For the sake of the calculations, treat the Pacific Ocean as a point like object (obviously a very rough approximation). 1) Determine the gravitational potential energy of the Moon–Pacific Ocean system when the Pacific is facing away from the Moon. (Express your answer to three significant figures.) Answer in Joules 2) Repeat the calculation when Earth has rotated so that the Pacific Ocean faces toward the Moon. (Express your answer to three significant figures.) Answer in Joules
The concepts used to solve this exercise are given through the calculation of distances (from the Moon to the earth and vice versa) as well as the gravitational potential energy.
By definition the gravitational potential energy is given by,
[tex]PE=\frac{GMm}{r}[/tex]
Where,
m = Mass of Moon
G = Gravitational Universal Constant
M = Mass of Ocean
r = Radius
First we calculate the mass through the ratio given by density.
[tex]m = \rho V[/tex]
[tex]m = (1030Kg/m^3)(7*10^8m^3)[/tex]
[tex]m = 7.210*10^{11}Kg[/tex]
PART A) Gravitational potential energy of the Moon–Pacific Ocean system when the Pacific is facing away from the Moon
Now we define the radius at the most distant point
[tex]r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m[/tex]
Then the potential energy at this point would be,
[tex]PE_1 = \frac{GMm}{r_1}[/tex]
[tex]PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}[/tex]
[tex]PE_1 = 9.05*10^{15}J[/tex]
PART B) when Earth has rotated so that the Pacific Ocean faces toward the Moon.
At the nearest point we perform the same as the previous process, we calculate the radius
[tex]r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m[/tex]
The we calculate the Potential gravitational energy,
[tex]PE_2 = \frac{GMm}{r_2}[/tex]
[tex]PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}[/tex]
[tex]PE_2 = 9.361*10^{15}J[/tex]
A 45.0-kg girl is standing on a 166-kg plank. The plank, originally at rest, is free to slide on a frozen lake, which is a flat, frictionless surface. The girl begins to walk along the plank at a constant velocity of 1.48 m/s to the right relative to the plank. (Let the direction the girl is moving in be positive. Indicate the direction with the sign of your answer.)
1. What is her velocity relative to the surface of the ice?
2. What is the velocity of the plank relative to the surface of ice?
Answer:
-0.31563 m/s
1.16437 m/s
Explanation:
[tex]m_1[/tex] = Mass of girl = 45 kg
[tex]m_2[/tex] = Mass of plank = 166 kg
[tex]v_1[/tex] = Velocity of girl relative to plank = 1.48 m/s
[tex]v_2[/tex] = Velocity of the plank relative to ice surface
In this system the linear momentum is conserved
[tex](m_1+m_2)v_2+m_1v_1=0\\\Rightarrow v_2=-\frac{m_1v_1}{m_1+m_2}\\\Rightarrow v_2=-\frac{45\times 1.48}{45+166}\\\Rightarrow v_2=-0.31563\ m/s[/tex]
Velocity of the plank relative to ice surface is -0.31563 m/s
Velocity of the girl relative to the ice surface is
[tex]v_1+v_2=1.48-0.31563=1.16437\ m/s[/tex]
Part APart complete If the CD rotates clockwise at 500 rpm (revolutions per minute) while the last song is playing, and then spins down to zero angular speed in 2.60 s with constant angular acceleration, what is α, the magnitude of the angular acceleration of the CD, as it spins to a stop?
Answer:
20.13841 rad/s²
Explanation:
[tex]\omega_i[/tex] = Initial angular velocity = [tex]500\times \frac{2\pi}{60}\ rad/s[/tex]
[tex]\omega_f[/tex] = Final angular velocity = 0
t = Time taken = 2.6 s
[tex]\alpha[/tex] = Angular acceleration
Equation of rotational motion
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-500\times \frac{2\pi}{60}}{2.6}\\\Rightarrow \alpha=-20.13841\ rad/s^2[/tex]
The magnitude of the angular acceleration of the CD, as it spins to a stop is 20.13841 rad/s²
Consider a particle in a square well. It has mass m and the length of the box is I.Consider the following situations (1) the particle of mass m in the ground state (2) the particle of mass m in the first excited state (3) a particle of mass m in the ground state (4) a particle of mass m in a box of length 2 (5) a particle of mass m in a box of length 21 in the first excited state Which is the proper order from lowest energy to highest energy? (Use«. > to reflect the relative values; for example (1) (3)く(4) 5). 2) Select the correct answer O 1)(2) (3)(4)(5)
Answer:
Explanation:
The energy of the particle in a square power well is given by the Schrödinger equation, with
E₁ = RA (h²/ 8m L²) n²
With h the Planck constant the mass of the particle, L the length of the box and n an integer starting 1
We find the energy for each situation presented
1) For the fundamental state n = 1
E₁ = RA (h/ 8m L²)
2) the first excited state corresponds to n = 2
E₂ = RA (h/ 8m L²) 2²
E₂ = E₁ 4
3) E₁ = RA (h/ 8m L²)
4) The length of the box is L = 2L
E’= RA (h’ / 8m (2L)²)
E’= RA (h’ / 8m L²) Ra ¼
E’= E₁ / 2
5) L = 2l first excited state
n = 2
E ’’ = RA (h/ 8m (2L)²2) 2²
E ’’ = RA (h ’/ 8m (L)²) 4/2
E ’’ = E₁ 2
All energies are in relation to the fundamental state (E1), milking from least to greatest
E1 / 2 <E1 = E1 <2E1 <4E1
4 <3 = 1 <5 <2
Final answer:
The correct order from lowest to highest energy for a particle in a square well, considering both quantum state and box length, is: (5) < (4) < (3) < (2) < (1). This order is based on the quantum energy level formula E_n = n^2 * pi^2 * h-bar^2 / (2 * m * L^2), where energy increases with the square of the principal quantum number and decreases with the square of the box length.
Explanation:
To determine the proper order from lowest energy to highest energy for the particle in a square well, we must consider the principles of quantum mechanics, particularly the energy levels of a particle in a one-dimensional box. The energy levels are quantized and are given by the expression En = n2π2ħ2/(2mL2), where ħ is the reduced Planck constant, m is the mass of the particle, L is the length of the box, and n is the principal quantum number (n = 1 for the ground state, n = 2 for the first excited state, and so on). Given this, here's the reasoning for the order:
A particle in the ground state (n = 1) in a box of length L has the lowest energy.A particle in the first excited state (n = 2) in a box of the same length L will have a higher energy due to the n2 dependency.A particle in the ground state in a box of length 2L will have a lower energy than the particle in the ground state in a box of length L, because the energy is inversely proportional to the square of the box length.A particle in the box of length 2L in the first excited state will still have lower energy than the first excited state in a box of length L, but higher than the ground state in the box of length 2L.Finally, a particle in the first excited state in a box of length 21L will have the lowest energy of all, due to the large increase in box length.Therefore, the order from lowest energy to highest energy is as follows:
(5) < (4) < (3) < (2) < (1)
The diffusion constant for oxygen diffusing through tissue is 1.0 × 10-11 m2/s. In a certain sample oxygen flows through the tissue at 2.0 × 10-6 kg/s. If the thickness of the tissue is doubled, then what is the rate of oxygen flow through the tissue?
Answer:
m' = 1 x 10⁻⁶ kg/s
Explanation:
Given that
Diffussion constant = 1 x 10⁻¹¹
Mass flow rate ,m = 2 x 10⁻⁶ kg/s
The diffusion is inversely proportional to the thickness of the membrane and therefore when the thickness is doubled, the mass flow rate would become half.
So new flow rate m'
[tex]m'=\dfrac{m}{2}[/tex]
[tex]m'=\dfrac{2\times 10^{-6}}{2}\ kg/s[/tex]
m' = 1 x 10⁻⁶ kg/s
Which one of the following phrases best describes the electric potential of a charged particle?
A) the total force exerted on or by the charged particle
B) the force per unit charge
C) the potential energy of the particle relative to infinity
D) the potential energy per unit charge
E) the potential energy per unit force on the particle
Answer:
D.The potential energy per unit charge
Explanation:
Electric potential of a charged particle:
It is scalar quantity because it has magnitude but it does not have direction.
It is the amount of work done required to move a unit positive charge from reference point to specific point in the electric field without producing any acceleration.
Mathematical representation:
[tex]V=\frac{W}{Q_0}[/tex]
Where W= Work done
[tex]Q_0[/tex]= Unit positive charge
Other formula to calculate electric field:
[tex]V=\frac{KQ}{r}[/tex]
Where K=[tex]\frac{1}{4\pi \epsilon_0}[/tex]
It can be defined as potential energy per unit charge.
Hence, option D is true.
A horizontal rod 0.300 m long carries a current through a uniform horizontal magnetic field of magnitude 6.40×10−2 T that points perpendicular to the rod. Part A If the magnetic force on this rod is measured to be 0.140 N , what is the current flowing through the rod?
Answer:
Current, 7.29 A
Explanation:
It is given that,
Length of the horizontal rod, L = 0.3 m
Magnetic field through a horizontal rod, [tex]B=6.4\times 10^{-2}\ T[/tex]
The magnetic force acting on the rod, F = 0.14 N
Let the current flowing through the rod is given by I. The magnetic force acting on an object in the uniform magnetic field is given by :
[tex]F=ILB\ sin\theta[/tex]
Here, [tex]\theta=90^{\circ}[/tex]
[tex]F=ILB[/tex]
[tex]I=\dfrac{F}{LB}[/tex]
[tex]I=\dfrac{0.14\ N}{0.3\ m\times 6.4\times 10^{-2}\ T}[/tex]
I = 7.29 A
So, the current flowing through the rod 7.29 A.
Final answer:
The current flowing in the rod is approximately 7.292 A when it is subject to a 0.140 N magnetic force within a 6.40 x [tex]10^-^2[/tex] T magnetic field. The calculation uses the formula F = I * L * B, and since the rod is perpendicular to the magnetic field, the angle θ is 90°, which simplifies the calculation.
Explanation:
The question asks for the current flowing through a rod that is experiencing a magnetic force due to an external magnetic field.
The force on a current-carrying conductor in a magnetic field is given by the equation F = I * L * B * sin(θ), where F is the force in newtons, I is the current in amperes, L is the length of the conductor in meters, B is the magnetic field in teslas, and θ is the angle between the direction of the current and the magnetic field. In this case, the rod is perpendicular to the magnetic field, so the angle θ is 90°, making sin(θ) equal to 1.
To find the current I, we rearrange the formula to be I = F / (L * B). Substituting the given values:
Force F = 0.140 N
Length L = 0.300 m
Magnetic field B = 6.40×[tex]10^-^2[/tex] T
The current I can thus be calculated as I = 0.140 N / (0.300 m * 6.40×[tex]10^-^2[/tex] T).
Performing the calculation, I equals approximately 7.292 A.
A 10.0 cm long wire is pulled along a U-shaped conducting rail in a perpendicular magnetic field. The total resistance of the wire and rail is 0.330 ohms . Pulling the wire at a steady speed of 4.00 m/s causes 4.40 W of power to be dissipated in the circuit.a) How big is the pulling force?b) What is the strength of the magnetic field?
Answer
given,
length of wire = 10 cm = 0.1 m
resistance of the wire = 0.330 ohms
speed of pulling = 4 m/s
Power = 4.40 W
a) Force of pull = ?
[tex]P = F_{pull} v[/tex]
[tex]F_{pull} =\dfrac{P}{v}[/tex]
[tex]F_{pull} =\dfrac{4.40}{4}[/tex]
[tex]F_{pull} =1.1\ N[/tex]
b) using formula
[tex]P = \dfrac{B^2l^2v^2}{R}[/tex]
where B is the magnetic field
v is the pulling velocity
R is the resistance of the wire
[tex]B =\sqrt{\dfrac{PR}{l^2v^2}}[/tex]
[tex]B =\sqrt{\dfrac{F_{pull} \times v R}{l^2v^2}}[/tex]
[tex]B =\sqrt{\dfrac{F_{pull} \times R}{l^2v}}[/tex]
[tex]B =\sqrt{\dfrac{1.1 \times 0.33}{0.1^2\times 4}}[/tex]
[tex]B =\sqrt{9.075}[/tex]
B = 3.01 T
The pulling force in this magnetic field is equal to 1.1 Newton.
Given the following data:
Length of wire = 10.0 cm.Total resistance = 0.330 Ohms.Speed = 4.00 m/s.Power = 4.40 Watts.How to calculate the pulling force.Mathematically, the pulling force in a magnetic field is given by this formula:
[tex]F = \frac{Power}{Speed} \\\\F=\frac{4.40}{4.00}[/tex]
F = 1.1 Newton.
To determine the strength of the magnetic field, we would apply this formula:
[tex]B=\sqrt{\frac{PR}{L^2V^2} }\\\\B=\sqrt{\frac{FVR}{L^2V^2} }\\\\B=\sqrt{\frac{FR}{L^2V} }[/tex]
Where:
B is the magnetic field.P is the power.V is the velocity.L is the length.R is the resistance.F is the pulling force.Substituting the given parameters into the formula, we have;
[tex]B=\sqrt{\frac{1.1 \times 0.330}{0.1^2 \times 4.00} } \\\\B=\sqrt{\frac{0.363}{0.01 \times 4.00} }\\\\B=\sqrt{9.075}[/tex]
B = 3.012 T.
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Case 1: A 0.780-kg silver pellet with a temperature of 85 oC is added to 0.150 kg of water in a copper cup of unknown mass. The initial temperature of the water and the copper cup is 14 oC. The equilibrium temperature of the system (silver water copper cup) is measured to be 26.0 °C. Assume no heat is exchanged with the surroundings. The specific heats of silver, water and copper are: 234 J/(kg oC), 4186 J/(kg oC) and 387 J/(kg oC) , respectively. (a) Which substance releases heat
Answer: The silver pellet will release heat
Explanation:
Based on the case scenario, the silver pellet has a higher temperature that the system of water and copper cup and is thereby added to the system. Because of the higher kinetic energy of the molecules of silver in the silver pellet, some of energy will be released to the water and copper cup system because the system will aim to achieve thermal equilibrium.
Steam is accelerated by a nozzle steadily from zero velocity to a velocity of 280 m/s at a rate of 2.5 kg/s. If the temperature and pressure of the steam at the nozzle exit are 400°C and 2 MPa, determine the exit area of the nozzle. Solve using appropriate software.
Final answer:
To determine the exit area of the nozzle, use the principle of conservation of mass and the equation for mass flow rate. Calculate the density using the Ideal Gas Law and substitute it into the equation for area.
Explanation:
To determine the exit area of the nozzle, we can use the principle of conservation of mass and the equation for mass flow rate:
Mass flow rate = density x velocity x area
Given that the mass flow rate is 2.5 kg/s and the velocity is 280 m/s, we can rearrange the equation to solve for the area:
Area = mass flow rate / (density x velocity)
However, we need to find the density of the steam at the nozzle exit. To do this, we can use the Ideal Gas Law:
Pressure x Volume = n x R x Temperature
Where pressure = 2 MPa, volume can be assumed to be the volume of the nozzle exit, R is the gas constant, and temperature is 400°C converted to Kelvin.
Once we have the density, we can substitute it into the equation for the area to find the exit area of the nozzle.
The exit area of the nozzle is approximately [tex]\( 0.00140 \text{ m}^2 \) or \( 1.40 \text{ mm}^2 \)[/tex].
The continuity equation for a steady-state flow is given by:
[tex]\[ \dot{m} = \rho \cdot A \cdot v \][/tex]
To find the density [tex]\( \rho \)[/tex], we need to use the ideal gas law, which is a good approximation for steam under these conditions:
[tex]\[ P = \rho \cdot R \cdot T \][/tex]
where:
- P is the absolute pressure at the nozzle exit (2 MPa or 2000 kPa),
- R is the specific gas constant for steam (0.4615 kJ/kg·K),
- T is the absolute temperature at the nozzle exit (400°C + 273.15 = 673.15 K).
Rearranging the ideal gas law to solve for [tex]\( \rho \)[/tex]:
[tex]\[ \rho = \frac{P}{R \cdot T} \][/tex]
Now, we can substitute the density [tex]\( \rho \)[/tex] back into the continuity equation to solve for the exit area A:
[tex]\[ A = \frac{\dot{m}}{\rho \cdot v} \][/tex]
Substituting the values we have:
[tex]\[ \rho = \frac{2000 \text{ kPa}}{0.4615 \text{ kJ/kg·K} \cdot 673.15 \text{ K}} \] \[ \rho = \frac{2000}{310.56} \text{ kg/m}^3 \] \[ \rho \approx 6.44 \text{ kg/m}^3 \][/tex]
Now, we can find the exit area A:
[tex]\[ A = \frac{2.5 \text{ kg/s}}{6.44 \text{ kg/m}^3 \cdot 280 \text{ m/s}} \] \[ A = \frac{2.5}{1787.2} \text{ m}^2 \] \[ A \approx 0.00140 \text{ m}^2 \][/tex]
The hydraulic oil in a car lift has a density of 8.81 x 102 kg/m3. The weight of the input piston is negligible. The radii of the input piston and output plunger are 5.07 x 10-3 m and 0.150 m, respectively. What input force F is needed to support the 27800-N combined weight of a car and the output plunger, when (a) the bottom surfaces of the piston and plunger are at the same level, and (b) the bottom surface of the output plunger is 1.20 m above that of the input plunger?
Answer:
a. [tex]F_2=31.76N[/tex]
b. [tex]F_2=185.86N[/tex]
Explanation:
Given:
[tex]F_1=27800N[/tex]
[tex]r_1=5.07x10^{-3}m[/tex]
[tex]r_2=0.150 m[/tex]
[tex]p=8.81x10^2 kg/m^3[/tex]
Using the equation to find the force so replacing
a.
[tex]F_1*A_2=F_2*A_1[/tex]
[tex]A=\pi*r^2[/tex]
[tex]F_2=F_1*\frac{A_2}{A_1}=27800*\frac{\pi*(5.07x10^{-3}m)^2}{\pi*(0.150m)^2}[/tex]
[tex]F_2=31.76N[/tex]
b.
[tex]F_2=F_1+F_p[/tex]
[tex]F_2=27800*\frac{\pi*(5.07x10^{-3}m)^2}{\pi*(0.150m)^2}+(8.81x10^2kg/m^3*9.8m/s^2*1.20m*\pi*(5.07x10^{-3})m^2)[/tex]
[tex]F_2=185.86N[/tex]
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 1.30×104 kg/m3 at the center and 2100 kg/m3 at the surface. Part A What is the acceleration due to gravity at the surface of this planet?
The acceleration due to gravity at the surface of a planet depends on its mass and radius, and assumes a uniform density. Since your model has a density that decreases linearly from the center to the surface, the exact value for gravity would require integration over the volume of the planet to account for mass distribution. This arrangement involves advanced calculus.
Explanation:The acceleration due to gravity at the surface of any planet, including Earth, is determined by a constant (G), the mass of the planet (M), and the radius of the planet (R). The formula is g = GM/R². However, this calculation assumes a uniform density throughout the planet, which is often not the case. In reality, like in your model where the density decreases linearly from the center to the surface, finding the precise acceleration due to gravity at the surface becomes more complicated and involves integration over the entire volume of the planet to account for how the mass is distributed.
Given that you provided the densities at the center and surface of the modeled planet, and these densities decrease linearly, one can utilize the formula for the linear density ρ(r) = ρ_center - r(ρ_center - ρ_surface)/R, where R is the radius of the planet, r is the distance from the center, and ρ_center and ρ_surface are the density at the center and surface, respectively. Then, integrate over the volume of the planet to find the total mass.
Once you have the mass, you can use the formula g = GM/R² again to find the acceleration due to gravity at the surface. However, this calculation goes beyond a basic understanding of gravity and requires knowledge of calculus. Without specific numbers for the mass and the integration result, I cannot provide the exact value for surface gravity in this case.
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The acceleration due to gravity at a planet's surface depends on the planet's radius, mass and the linear decrease of density from center to surface. The formula of this acceleration is G×M/r², considering that M is the planet's mass obtained by the product of volume and average density. However, as the density changes linearly, the force of gravity also decreases linearly from the center to the surface.
Explanation:To calculate the acceleration due to gravity at the surface of the planet, we have to consider the planet's radius, mass and density. Given the density at the center and surface, we can calculate the average density which is the total mass of the planet divided by the total volume. In this spherically symmetric planet model, we can use the formula for the volume of a sphere, which is 4/3πr³, with r being the Earth's radius. We consider that mass (M) equals density (ρ) times volume (V), and the force of gravity (F) is G×(M1×M2)/r², where G is the gravitational constant. In this case, M1 is the mass of the planet and M2 is the mass of the object where we want to know the acceleration, and r is the distance between the centers of the two masses, or in this case the radius of the planet. As force is also mass times acceleration, we can replace F in the formula with M2 times a (acceleration), and find that acceleration is G×M1/r². However, as the density changes linearly from the center to the surface, the force of gravity will also decrease linearly, affecting the acceleration.
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