Answer:
The speed of the car was 28 m/s.
Explanation:
Hi there!
The initial kinetic energy of the car, KE, is equal to the negative work, W, done by friction to bring the car to stop. Let´s write the work-energy theorem:
W = ΔKE = final kinetic energy - initial kinetic energy
In this case, the final kinetic energy is zero, then:
W = - initial kinetic energy
Since the work done by friction is negative (the work is done in opposite direction to the movement of the car), then:
Wfr = initial kinetic energy
The work done by friction is calculated as follows:
Wfr = Fr · d
Where:
Fr = friction force.
d = distance.
The friction force is calculated as follows:
Fr = N · μ
Where:
N = normal force.
μ = coefficient of friction
Since the only vertical forces acting on the car are the weight of the car and the normal force, and the car is not being accelerated in the vertical direction, the normal force has to be equal to the weight of the car (with opposite sign).
Then the friction force can be written as follows:
Fr = m · g · μ
Where:
m = mass of the car.
g = acceleration due to gravity (9.8 m/s²)
The work done by friction will be:
W = m · g · μ · d
The equation of kinetic energy is the following:
KE = 1/2 · m · v²
Where
m = mass of the car.
v = speed.
Then:
W = KE
m · g · μ · d = 1/2 · m · v²
2 · g · μ · d = v²
2 · 9.8 m/s² · 0.40 · 98 m = v²
v = 28 m/s
The speed of the car was 28 m/s.
You come across an open container that is filled with two liquids. Since the two liquids have different densities, there is a distinct separation between them. Water, which has a density of rho w = 1.00 × 10 3 kg/m 3 , fills the lower portion of the container to a depth of 0.221 m . The fluid that is floating on top of the water is 0.335 m deep. If the absolute pressure on the bottom of the container is 1.049 × 10 5 Pa , what is the density, rho l , of the unknown fluid? The acceleration due to gravity is g = 9.81 m/s 2 and atmospheric pressure is P 0 = 1.013 × 10 5 Pa .
Answer:
496.57492 kg/m³
Explanation:
[tex]P_a[/tex] = Atmospheric pressure = 101300 Pa
[tex]\rho_w[/tex] = Density of water = [tex]1000 kg/m^3[/tex]
[tex]h_w[/tex] = Height of water = 0.221 m
[tex]h_l[/tex] = Height of fluid = 0.335 m
g = Acceleration due to gravity = 9.81 m/s²
[tex]\rho_l[/tex] = Density of the unknown fluid
Absolute pressure at the bottom
[tex]P_{abs}=P_a+\rho_wgh_w+\rho_lgh_l\\\Rightarrow \rho_l=\frac{P_{abs}-P_a-\rho_wgh_w}{gh_l}\\\Rightarrow \rho_l=\frac{104900-101300-1000\times 9.81\times 0.221}{9.81\times 0.335}\\\Rightarrow \rho_l=435.73873\ kg/m^3[/tex]
The density of the unknown fluid is 496.57492 kg/m³
To find the density of the unknown fluid in the container, we subtract the atmospheric pressure and the pressure due to water from the absolute pressure at the bottom of the container and divide the result by the product of gravity and the height of the unknown fluid.
Explanation:The question is asking about the determination of the density of an unknown fluid in an open container where you have two separated fluids due to their different densities. In this setup, water fills the lower portion of the container and the unknown fluid fills the upper part.
The pressure at the bottom of this container (P) can be expressed as the sum of the atmospheric pressure (P0), pressure due to water (ρw * g * h1), and pressure due to the unknown fluid (ρl * g * h2) where 'g' is the acceleration due to gravity and 'h' represents the depth of each fluid. Given in the problem are P, P0, ρw, g, h1, and h2 and to find the density of the unknown fluid (ρl), we rearrange the equation into ρl = (P - P0 - ρw * g * h1) / (g * h2). By substituting the given values into this equation, we can find the density of the unknown fluid.
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A student holds a bike wheel and starts it spinning with an initial angular speed of 9.0 rotations per second. The wheel is subject to some friction, so it gradually slows down. In the 10-s period following the inital spin, the bike wheel undergoes 65.0 complete rotations. Assuming the frictional torque remains constant, how much more time delta ts will it take the bike wheel to come to a complete stop? The bike wheel has a mass of 0.725 kg and a radius of 0.315 m. If all the mass of the wheel is assumed to be located on the rim, find the magnitude of the frictional torque τf that was acting on the spinning wheel.
After 10 seconds, the bike wheel will take roughly 15.98 seconds to come to a complete halt. The frictional torque operating on the spinning bike wheel is -0.112 N*m.
Explanation:The process to solve this problem involves several steps. The first one is to determine the initial and final angular velocities of the wheel. Angular velocities are usually measured in radians per second, but we have them in rotations per second. So, firstly these values need to be converted by multiplying the number of rotations by 2π radians. The initial angular velocity, ωi, hence becomes
9 * 2π = 56.55 rad/s,
and after 10 seconds, the final angular velocity, ωf, is
65 / 10 * 2π = 40.84 rad/s.
Now, we already know 10 seconds.
Using these values, the angular acceleration (α) can be calculated as
(ωf - ωi) / Δt = (40.84 rad/s - 56.55 rad/s) / 10 s = -1.571 rad/s².
Note that this is negative because it's a deceleration. We're also given the wheel's mass (m) and radius (r), so the moment of inertia (I), which can be calculated as
I = m*r², will be 0.725 kg * (0.315 m)² = 0.071 kg*m².
From here, frictional torque (τf), which is the product of the moment of inertia and angular acceleration, can be found as
I * α = 0.071 kg*m² * -1.571 rad/s² = -0.112 N*m.
The negative sign just indicates the torque acts in the direction to slow down the wheel.
Finally, to find the time Δts it takes to come to rest, we can use the equation for angular velocity:
ω = ωi + α * t.
Setting ω to 0 (as the wheel stops) and solving for t gives us:
t = (0 - ωi) / α = - (40.84 rad/s) / -1.571 rad/s² = 25.98 s.
But our Δts is the time after the first 10 seconds, so Δts = t - 10 s = 15.98 s.
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Because of its metabolic processes, your body continually emits thermal energy. Suppose that the air in your bedroom absorbs all of this thermal energy during the time you sleep at night (8.0 hours). Assume your metabolic rate to be P = 120 J/s. Estimate the temperature change you expect in this air. Let the volume of your room be 36 m2 and Heat capacity of air is 1000 J/kg C.
Answer:
77.83783 °C
Explanation:
V = Volume of room = 36 m²
c = Heat capacity = 1000 J/kg C
P = Power = 120 J/s = 120 W
t = Time = 8 hours
[tex]\rho[/tex] = Density of air = 1.225 kg/m³
[tex]\Delta T[/tex] = Change in temperature
Density is given by
[tex]\rho=\frac{m}{V}\\\Rightarrow m=\rhoV\\\Rightarrow m=1.225\times 36\\\Rightarrow m=44.1\ kg[/tex]
Heat is given by
[tex]Q=mc\Delta T\\\Rightarrow Pt=mc\Delta T\\\Rightarrow \Delta T=\frac{Pt}{mc}\\\Rightarrow \Delta T=\frac{120\times 8\times 3600}{44.4\times 1000}\\\Rightarrow \Delta T=77.83783^{\circ}C[/tex]
The temperature change you expected in this air 77.83783 °C
Final answer:
The estimated temperature change in the air of your bedroom while you sleep is approximately 78.3 °C.
Explanation:
In order to estimate the temperature change in the air of your bedroom while you sleep, we need to calculate the amount of thermal energy emitted by your body and the amount of energy absorbed by the air. Your metabolic rate is given as P = 120 J/s. Assuming you sleep for 8.0 hours, the total energy emitted is 120 J/s * 8 hours * 3600 seconds/hour = 3456000 J.
The specific heat capacity of air is given as 1000 J/kg C and the volume of your room is 36 m³. Assuming the density of air is 1.225 kg/m³, the mass of the air in your room is 1.225 kg/m³ * 36 m³ = 44.1 kg.
Using the equation Q = mcΔθ, where Q is the thermal energy absorbed/lost, m is the mass, c is the specific heat capacity, and Δθ is the change in temperature, we can calculate the temperature change as follows:
Q = mcΔθ
Δθ = Q / mc
Δθ = 3456000 J / (44.1 kg * 1000 J/kg C)
Δθ ≈ 78.3 °C
Therefore, we can estimate that the temperature change in the air of your bedroom is approximately 78.3 °C.
A combination lock has a 1.3-cm-diameter knob that is part of the dial you turn to unlock the lock. To turn that knob, you twist your wrist as you grip it between your thumb and forefinger with a force of 5.0 N . Suppose the coefficient of static friction between the knob and your fingers is 0.65.
What is the most torque that you can exert on the knob without having it slip between your fingers?
Answer:
0.04225 Nm
Explanation:
N = Force applied = 5 N
[tex]\mu[/tex] = Coefficient of static friction = 0.65
d = Diameter of knob = 1.3 cm
r = Radius of knob = [tex]\frac{d}{2}=\frac{1.3}{2}=0.65\ cm[/tex]
Force is given by
[tex]F=N\mu\\\Rightarrow F=5\times 0.65\\\Rightarrow F=3.25\ N[/tex]
When we multiply force and radius we get torque
Torque on thumb
[tex]\tau_t=F\times r\\\Rightarrow \tau_t=3.25\times 0.0065\\\Rightarrow \tau_t=0.021125\ Nm[/tex]
Torque on forefinger
[tex]\tau_f=F\times r\\\Rightarrow \tau_f=3.25\times 0.0065\\\Rightarrow \tau_f=0.021125\ Nm[/tex]
The total torque is given by
[tex]\tau=\tau_t+\tau_f\\\Rightarrow \tau=0.021125+0.021125\\\Rightarrow \tau=0.04225\ Nm[/tex]
The most torque that exerted on the knob is 0.04225 Nm
The torque is dependent on the force applied to the system. The torque exerted on the knob is 0.04225 Nm.
What is torque?
The torque is defined as the measure of the force that can cause an object to rotate about its axis.
Given that the applied force F is 5 N and the coefficient of static friction [tex]\mu[/tex] between the knob and your fingers is 0.65. The diameter d of the knob is 1.3 cm.
The radius r of the knob is given as,
[tex]r = \dfrac {d}{2}[/tex]
[tex]r = \dfrac {1.3}{2}[/tex]
[tex]r = 0.65 \;\rm cm[/tex]
The force F' applied to the system is calculated as given below.
[tex]F' = F\mu[/tex]
[tex]F' = 5\times 0.65[/tex]
[tex]F'=3.25 \;\rm N[/tex]
Now the torque on the thumb is given as,
[tex]\tau_t = F'\times r[/tex]
[tex]\tau_t = 3.25\times 0.65\times 10^{-2}[/tex]
[tex]\tau_t = 0.021125 \;\rm Nm[/tex]
The torque on the forefinger is,
[tex]\tau_f = F'\times r[/tex]
[tex]\tau_f= 3.25\times 0.65\times 10^{-2}[/tex]
[tex]\tau_f = 0.021125 \;\rm Nm[/tex]
The total torque of the system is calculated as,
[tex]\tau = \tau_t + \tau_f[/tex]
[tex]\tau= 0.021125 + 0.021125[/tex]
[tex]\tau = 0.04225 \;\rm Nm[/tex]
Hence we can conclude that the torque exerted on the knob is 0.04225 Nm.
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A geothermal plant uses geothermal water (yes a liquid) extracted from beneath the Earth using drilling apparatus. The geothermal water (a condensed liquid) comes up at a temperature of 200 C at a rate of 200 kg/s. It is used to drive a turbine that extracts 6000 kw of power. The water leaves the turbine at 80 C and enters a lake at 25 C. What is the actual thermal efficiency of the turbine in percent to three significant digits
Answer:
4.1%
Explanation:
Given that water specific heat c = 4186 J/kgC. If water is flowing with mass rate of 200kg/s and temperature is dropping from 200C to 25C. Then the thermal energy rate from the water should be
[tex]P_i = mc\Delta t = 200*4186*(200 - 25) = 146510000 J/s[/tex] or 146510kW
Since the turbine is only able to extract 6000kW of power, then the thermal efficiency is:
[tex]Eff = \frac{P_o}{P_i} = \frac{6000}{146510} = 0.041[/tex] or 4.1%
Consider a 145 gram baseball being thrown by a pitcher. The ball approaches the batter with a speed of 44 m/s. The batter swings and sends the ball flying back at a speed of 90 m/s into the outfield.
If the ball was in contact with the bat for a time of 3 ms, what is the average force exerted by the bat during the collision?
For the development of this problem it is necessary to apply the concepts related to Force, momentum and time.
By definition we know that momentum can be expressed as
[tex]P = F*t[/tex]
Where
P = Momentum
F = Force
t = Time
At the same time, another way of expressing the momentum is through mass and speed, that is,
P = mv
Where
m = Mass
v = Velocity
Equating both equations
F*t = mv
[tex]F = \frac{mv}{t}[/tex]
Our values are given as
[tex]m = 145*10^{-3}Kg[/tex]
[tex]t = 3*10^{-3}s[/tex]
[tex]v = 90-44 = 46m/s \rightarrow[/tex] Net velocity
Replacing we have then,
[tex]F = \frac{mv}{t}[/tex]
[tex]F = \frac{3*10^{-3}*46}{3*10^{-3}}[/tex]
[tex]F = 2223.33N[/tex]
Therefore the average force exerted by the bat during the collision is 2223.3N
Steel is very stiff, and the Young's modulus for steel is unusually large, 2.0×1011 N/m2. A cube of steel 25 cm on a side supports a load of 85 kg that has the same horizontal cross section as the steel cube. What is the magnitude of the "normal" force that the steel cube exerts on the load? By how much does the steel cube compress, due to the load that it supports? (Give your answer as a positive number.)
Answer:
Force (normal) = 833.85 N Compression = 1.67 x 10⁻⁸ m
Explanation:
Given data Young's Modulus (Y) = 2 x 10¹¹ N/m², Length of one side of cube = 25 cm = 0.25 m, mass of load = 85 kg
Normal force is the force exerted upon an object that is in contact with another stable object. This force would be applied by the surface onto the object in the same vector and is used to keep the object stable while it rests on a surface.
We know from Newton's Second Law that
F = ma where m is the mass and a is the acceleration (in this case due to gravity) hence, the normal opposing force to the load applied by the surface would be equal to the force applied on the surface by the weight of the load on the surface, So
F (normal) = M (load) x a = 85 x 9.81 = 833.85 N
Compression is the change in length of an object by the exertion of force upon it. Using the Young's Modulus formula we can find this change in the cube of steel. The Young's Modulus is given by
Y = (F/A)/(ΔL/L), where Y is the Young's Modulus, F is the Force being applied on the object, A is the cross sectional area on which the said force is applied, ΔL is the change in length due to said force being applied and L is the original Length of the side of the cross sectional area.
Solving this for ΔL, we can re- arrange the equation
ΔL = (F x L)/(Y x A) since area of square is L x L we can simplify the equation to get
ΔL = (F)/(Y x L), substitute the values
ΔL = (833.85)/(2 x 10¹¹ x 0.25) = 1.67 x 10⁻⁸m
Two spheres have the same radius and equal mass. One sphere is solid, and the other is hollow and made of a denser material. Which one has the bigger moment of inertia about an axis through its center?
a. Both the same.
b. The solid one.
c. The hollow one.
Answer:
Explanation:
Two sphere have same mass and radius but one is solid and another is hollow
Moment of inertia is the distribution of mass about the axis of rotation
For Hollow sphere more mass is at outside therefore its moment of inertia is more than solid sphere
[tex]I_{hollow}=\frac{2}{3}mr^2[/tex]
[tex]I_{solid}=\frac{2}{5}mr^2[/tex]
Blood takes about 1.65 s to pass through a 2.00 mm long capillary. If the diameter of the capillary is 5.00 μm and the pressure drop is 2.85 kPa , calculate the viscosity of blood. Assume laminar flow.
Answer:
Viscosity of blood will be [tex]918.54\times 10^{-6}Pa-sec[/tex]
Explanation:
We have given time to pass the capillary = 1.65 sec
Length of capillary L = 2 mm = 0.002 m
Pressure drop [tex]\Delta P=2.85kPa=2.85\times 10^3Pa[/tex]
Diameter [tex]d=5\mu m=5\times 10^{-6}m[/tex]
So radius [tex]r=\frac{5\times 10^{-6}}{2}=2.5\times 10^{-6}m[/tex]
Velocity will be [tex]v=\frac{L}{t}=\frac{0.002}{1.65}=1.212\times 10^{-3}m/sec[/tex]
We know that viscosity is given by [tex]\eta =\frac{r^2\Delta P}{8Lv}=\frac{(2.5\times 10^{-6})^2\times 2.85\times 10^3}{8\times 0.002\times 1.212\times 10^{-3}}=918.54\times 10^{-6}Pa-sec[/tex]
Viscosity of blood will be [tex]918.54\times 10^{-6}Pa-sec[/tex]
The viscosity of the blood flowing in the given capillary is [tex]9.27 \times 10^{-4} \ Pa.s[/tex]
The given parameters;
time taken for the blood to pass, t = 1.65 sdiameter of the capillary, d = 5 μmlength of the capillary, l = 2.00 mmpressure drop, ΔP = 2.85 kPaThe velocity of the blood is calculated as follows;
[tex]v = \frac{L}{t} = \frac{2\times 10^{-3} \ m}{1.65 \ s} = 0.0012 \ m/s[/tex]
The radius of the capillary is calculated as follows;
[tex]r = \frac{d}{2} \\\\r = \frac{5\times 10^{-6}}{2} = 2.5 \times 10^{-6} \ m[/tex]
Assuming laminar flow, the viscosity of the blood is calculated as;
[tex]\mu = \frac{r^2 \times \Delta P}{8lv} \\\\\mu = \frac{(2.5\times 10^{-6})^2 \times 2850}{8 \times 2\times 10^{-3} \times 0.0012} \\\\\mu= 9.27 \times 10^{-4} \ Pa.s[/tex]
Thus, the viscosity of the blood flowing in the given capillary is [tex]9.27 \times 10^{-4} \ Pa.s[/tex]
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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 m/s. If the faucet and the nozzle 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?
I keep coming up with 4 cm. Please someone confirm my answer. If incorrect please explain why.
Answer:
D₂ = 2.738 cm
Explanation:
Continuity equation
The continuity equation is nothing more than a particular case of the principle of conservation of mass. It is based on the flow rate (Q) of the fluid must remain constant throughout the entire pipeline.
Since the flow rate is the product of the surface of a section of the duct because of the speed with which the fluid flows, we will have to comply with two points of the same pipeline:
Q = v*A : Flow Equation
where:
Q = Flow in (m³/s)
A is the surface of the cross sections of points 1 and 2 of the duct.
v is the flow velocity at points 1 and 2 of the pipe.
It can be concluded that since the flow rate must be kept constant throughout the entire duct, when the section decreases, the flow rate increases in the same proportion and vice versa.
Data
D₁= 6.0 cm : faucet diameter
v₁ = 5 m/s : speed of fluid in the faucet
v₂ = 20 m/s : speed of fluid in the nozzle
Area calculation
A = (π*D²)/4
A₁ = (π*D₁²)/4
A₂ = (π*D₂²)/4
Continuity equation
Q₁ = Q₂
v₁A₁ = v₂A₂
v₁(π*D₁²)/4 = v₂(π*D₂²)/4 : We divide by (π/4) both sides of the equation
v₁ (D₁)² = v₂(D₂)²
We replace data
6 *(5)² = 20*(D₂)²
150 = 20*(D₂)²
(150 /20) = (D₂)²
7.5 = (D₂)²
[tex]D_{2} = \sqrt{7.5}[/tex]
D₂ = 2.738 cm : nozzle diameter
At a height of fourteen meters above the surface of a freshwater lake, a sound pulse is generated. The echo from the bottom of the lake returns to the point of origin 0.107 s later. The air and water temperature is 20 °C. How deep is the lake?
To solve this problem it is necessary to resort to concepts related to the kinematic equations of movement description which define speed as
[tex]v = \frac{x}{t}[/tex]
Where,
x = Distance
t=Time
Re-arrange to find the time we have,
[tex]t = \frac{x}{v}[/tex]
At 20°C the speed of sound in air is 343.2m/s and the speed of sound in water is 1484.3m/s
The time in which the sound lasts in traveling in the air would be given by the shipping and return (even when it touches the water) which is 28m (14 way and 14 back), in other words
[tex]t_a = \frac{28}{343.2} = 0.08158s[/tex]
In the case of the lake, the same logic is followed, only that there is no net distance here, therefore
[tex]t_w = \frac{2d}{1484.3}[/tex]
The total time would be given by
[tex]t_T = 0.08158 + \frac{2d}{1484.3}[/tex]
The total time is given in the statement so we clear to find d
[tex]0.108-0.08158= \frac{2d}{1484.3}[/tex]
[tex]d = \frac{1484.3(0.108-0.08158)}{2}[/tex]
[tex]d = 18.86m[/tex]
Therefore the lake has a depth of 18.86 m
A cylinder is closed by a piston connected to a spring of constant 2.40 103 N/m. With the spring relaxed, the cylinder is filled with 5.00 L of gas at a pressure of 1.00 atm and a temperature of 20.0°C.
(a) If the piston has a cross-sectional area of 0.0110 m2 and negligible mass, how high will it rise when the temperature is raised to 250°C? m
(b) What is the pressure of the gas at 250°C?
Answer:
a) The correct solution would be x=0.1544m.
b) [tex]P_f = P_i +\frac{Kx}{A}=101325Pa +\frac{2400N/m (0.1544m)}{0.011m^2}=135012.27 Pa[/tex]
Explanation:
Initial states
[tex]P_i =1atm=101325 Pa[/tex] represent the initial pressure
[tex]V_i =0.005m^3 = 5L[/tex] represent the initial volume
N represent the number of moles
[tex]T_i = 20+273.15=293.15K[/tex] represent the initial temperature
Final states
[tex]P_f[/tex] represent the final pressure
[tex]V_f[/tex] represent the final volume
[tex]T_f = 250+273.15=523.15K[/tex]
Part a
From the initial states and with the law of ideal gases we can find the moles like this
[tex]n=\frac{P_iV_i}{RT_i}=\frac{101325Pa(0.005m^3)}{8.314\frac{J}{mol K} 293.15K}=0.2079mol[/tex]
In order to find the final pressure we need to take in count the atmospheric pressure and the pressure related to the force that the springs applies to the piston, like this:
[tex]P_f =P_i +\frac{F}{A}= P_i +\frac{Kx}{A}[/tex]
Where x is the distance that the piston is displaced upward, since increase the temperature and the gas inside tends to expand.
And we can find the final volume on this way [tex]V_f = V_i + Ax[/tex]
For the final state we have the following equation:
[tex]P_f V_f=nR T_f [/tex]
And if we replace [tex]V_f[/tex] we got:
[tex](P_i +\frac{Kx}{A})(V_i +Ax)=nRT_f[/tex]
[tex]P_i V_i +P_i A x +\frac{KxV_i}{A}+kx^2=nRT_f[/tex]
And x can be solved with a quadratic equation:
[tex]kx^2+(P_i A +\frac{V_i K}{A})x+(P_i V_i-nRT_f)=0[/tex]
[tex]2400 k^2 +(101325Pa)(0.0110m^2)+\frac{0.005m^3 x2400N/m}{0.0110m^2})x+(101325Pa(0.005m^3)-0.2079mol(8.314J/molK)(523.15K))=0[/tex]
[tex]2400 k^2 +(2205.484)x-397.63=0[/tex]
And solving for x we got x=0.1544 or x=-1.07331 m. The correct solution would be x=0.1544m.
Part b
And the final pressure would be given by:
[tex]P_f = P_i +\frac{Kx}{A}=101325Pa +\frac{2400N/m (0.1544m)}{0.011m^2}=135012.27 Pa[/tex]
A person with a surface area of 1.20 m2, and a skin temperature of 27 oC, is in a room that is at a temperature of 18.8 °C. The emissivity of the skin is 0.895. The Stefan-Boltzmann constant is 5.67 x 10-8 W/(m2·K4).(a) How much power (power is the energy per second) is radiated by the person? Keep 2 decimal places.b. How much power is absorbed by the person from the surroundings? Keep 2 decimal places.c. What is his NET Power loss? Report as a Positive value. Keep 2 decimal places.
Answer:
494.24 W
442.41 W
51.83 W
Explanation:
A = Area of person = 1.2 m²
[tex]T_1[/tex] = Person's temperature = 27+273.15 = 300.15 K
[tex]T_2[/tex] = Room temperature = 18.8+273.15 = 291.95 K
[tex]\epsilon[/tex] = Emissivity of skin = 0.895
[tex]\sigma[/tex] = Stefan-Boltzmann constant = [tex]5.67\times 10^{-8}\ W/m^2K^4[/tex]
Radiated power of the person
[tex]P_1=\sigma\epsilon AT_1^4\\\Rightarrow P_1=5.67\times 10^{-8}\times 0.895\times 1.2\times 300.15^4\\\Rightarrow P_1=494.24\ W[/tex]
Power radiated by the person is 494.24 W
Radiated power absorbed by the person
[tex]P_2=\sigma\epsilon AT_2^4\\\Rightarrow P_2=5.67\times 10^{-8}\times 0.895\times 1.2\times 291.95^4\\\Rightarrow P_2=442.41\ W[/tex]
Radiated power absorbed by the person is 442.41 W
Net power would be
[tex]\Delta P=P_1-P_2\\\Rightarrow \Delta P=494.24-442.41\\\Rightarrow \Delta P=51.83\ W[/tex]
The net power loss is 51.83 W
Answer:
(a) 493.256 watt
(b) 441.5 Watt
(c) 51.76 Watt
Explanation:
Area, A = 1.20 m^2
T = 27°C = 300 K
To = 18.8°C = 291.8 K
e = 0.895
Stefan's constant, σ = 5.67 x 10^-8 W/m^2K^4
(a) Power emitted by the person
P1 = A e σ T^4
P1 = 1.2 x 0.895 x 5.67 x 10^-8 x (300)^4
P1 = 493.256 watt
(b) Pwer absorbed by the person, P2
P2 = A e σ To^4
P2 = 1.2 x 0.895 x 5.67 x 10^-8 x (291.8)^4
P2 = 441.5 watt
(c) Net power loss, P = P1 - P2
P = 493.256 - 441.5 = 51.76 Watt
g An object with mass m=2 kg is completely submerged, and tethered, to the bottom of a large body of water. If the density of the water is rhow = 1000 kg/m3and the density of the object is rhoob j=500 kg/m3, find the tension in the rope. Take g=10 m/s2and assume the object has a uniform mass density
Answer:
The tension in the rope is 20 N
Solution:
As per the question:
Mass of the object, M = 2 kg
Density of water, [tex]\rho_{w} = 1000\ kg/m^{3}[/tex]
Density of the object, [tex]\rho_{ob} = 500\kg/m^{3}[/tex]
Acceleration due to gravity, g = [tex]10\ m/s^{2}[/tex]
Now,
From the fig.1:
'N' represents the Bouyant force and T represents tension in the rope.
Suppose, the volume of the block be V:
V = [tex]\frac{M}{\rho_{ob}}[/tex] (1)
Also, we know that Bouyant force is given by:
[tex]N = \rho_{w}Vg[/tex]
Using eqn (1):
[tex]N = \rho_{w}\frac{M}{\rho_{ob}}g[/tex]
[tex]N = 1000\frac{2}{500}\times 10 = 40\ N[/tex]
From the fig.1:
N = Mg + T
40 = 2(10) + T
T = 40 - 20 = 20 N
[tex]N = \rho_{w}Vg[/tex]
7. A stick of length L and mass M is hanging at rest from its top edge from a ceiling hinged at that point so that it is free to rotate. Find the distance from the top of the stick where an impulse, FΔt, is applied such that there is no horizontal component to the force of the hinge on the stick. This point is called the center of percussion.
Answer:
The distance from the top of the stick would be 2l/3
Explanation:
Let the impulse 'FΔt' acts as a distance 'x' from the hinge 'H'. Assume no impulsive reaction is generated at 'H'. Let the angular velocity of the rod about 'H' just after the applied impulse be 'W'. Also consider that the center of percussion is the point on a bean attached to a pivot where a perpendicular impact will produce no reactive shock at the pivot.
Applying impulse momentum theorem for linear momentum.
FΔt = m(Wl/2), since velocity of center of mass of rod = Wl/2
Similarly applying impulse momentum theorem per angular momentum about H
FΔt * x = I * W
Where FΔt * x represents the impulsive torque and I is the moment of inertia
F Δt.x = (ml² . W)/3
Substituting FΔt
M(Wl/2) * x = (ml². W)/3
1/x = 3/2l
x = 2l/3
A wing with an elliptical planform is flying through sea-level air at a speed of 55 m/s. The wing is untwisted, has the same section from root to tip and its loading is W/S = 1000 N/m2 . The 2D lift curve slope is a0 = 5.7. The span of the wing is 15 m and the aspect ratio if 5. Find, a) The sectional lift coefficient. b) The sectional induced-drag coefficient. c) The effective, induced and geometric angles of attack if L=0 = -2 degrees.
Answer:
Explanation:
a.) To find Secrional liftSectional lift Coefficient [tex](c_{I})=\frac{W}{(0.5*Rho*V^{2})}[/tex]
where Rho = 1.225Kg/m³ (density of air at sea level)
[tex](c_{I})=\frac{1000}{(0.5*1.225*55^{2})}=0.5397[/tex]
b.) To obtain the sectional induced coefficient:
[tex]C_{D} = \frac{(C_{L})^2}{π(AR)}[/tex]
[tex]C_{D} = \frac{(0.5397)^2}{π(5)}=0.0185[/tex]
C.) For the effective angle of attack ∝0
∝0 = [tex]\frac{C_{l}}{m_{0}}\\=\frac{0.5397}{5.7}\\=0.0947rad[/tex]
For induced angle of attack ∝i:
∝i= [tex]\frac{-C_{D}}{C_{L}}\\=\frac{-0.0185}{0.5397}\\=-0.0343rad[/tex]
For absolute angle of attack ∝a:
∝a = ∝0 - ∝i = 0.0947 - (-0.0343) = 0.1290 rad
For geometric angle of attack ∝g:
∝g = ∝a + ∝L given ∝L = -2°≅-0.0349 rad
= 0.1290-0.0349 = 0.0941rad
Final answer:
The sectional lift coefficient is 0.3102, the sectional induced-drag coefficient is 0.0185, and the effective, induced, and geometric angles of attack are -2 degrees.
Explanation:
The sectional lift coefficient can be found using the formula: CL = W / (0.5 * density * V^2 * S), where CL is the sectional lift coefficient, W is the wing loading, density is the air density, V is the airspeed, and S is the wing area.
Substituting the given values, we get: CL = 1000 / (0.5 * 1.29 * (55^2) * (15/5)), which simplifies to CL = 0.3102
The sectional induced-drag coefficient can be found using the formula: CDi = CL^2 / (pi * AR * e), where CDi is the sectional induced-drag coefficient, CL is the sectional lift coefficient, AR is the aspect ratio, and e is the span efficiency factor. Substituting the given values, we get: CDi = 0.3102^2 / (pi * 5 * 0.8), which simplifies to CDi = 0.0185
The effective angle of attack can be found using the formula: alpha_eff = alpha_i + alpha_0, where alpha_eff is the effective angle of attack, alpha_i is the induced angle of attack, and alpha_0 is the geometric angle of attack.
Substituting the given values, we get: alpha_eff = -2 + 0, which simplifies to alpha_eff = -2 degrees.
Kara Less was applying her makeup when she drove into South's busy parking lot last Friday morning. Unaware that Lisa Ford was stopped in her lane 30 feet ahead, Kara rear-ended Lisa's rented Taurus. Kara's 1300-kg car was moving at 11 m/s and stopped in 0.14 seconds.a. Determine the momentum change of Kara's car.b. Determine the impulse experienced by Kara's car.c. Determine the magnitude of the force experienced by Kara's car.
Answer
given,
Mass of Kara's car = 1300 Kg
moving with speed = 11 m/s
time taken to stop = 0.14 s
final velocity = 0 m/s
distance between Lisa ford and Kara's car = 30 m
a) change in momentum of Kara's car
Δ P = m Δ v
[tex]\Delta P = m (v_f-v_i)[/tex]
[tex]\Delta P = 1300 (0 - 11)[/tex]
Δ P = - 1.43 x 10⁴ kg.m/s
b) impulse is equal to change in momentum of the car
I = - 1.43 x 10⁴ kg.m/s
c) magnitude of force experienced by Kara
I = F x t
I is impulse acting on the car
t is time
- 1.43 x 10⁴= F x 0.14
F = -1.021 x 10⁵ N
negative sign represents the direction of force
Answer:
Explanation:
Given that,
Mass of kara car m =1300kg
Velocity at which kara car was moving Vi =11m/s
The car stopped after t= 0.14sec
Therefore the final velocity is Vf = 0m/s
a. What is the change in momentum?
Change is momentum can be determine using
∆p = ∆MV
∆p = M∆V
∆p = m(Vf-Vi)
∆p = 1300(0-11)
∆p = 1300×-11
∆p = —14,300 kgm/s
The change in momentum of kara's car is —14,300kgm/s.
b. Impulse felt be kara car?
Impulse can be determine by using Newton second law of motion
Ft = ∆p
Impulse is Ft
I = ∆p = -14,300kgm/s.
Then, the impulse felt by Kara's car is -14,300kgm/s
c. Magnitude of force experienced by Kara's car?
From the impulse formula,
Ft = ∆p
Therefore,
F = ∆p/t
F = -14,300/0.14
F = -102,142.9N.
F ≈ —102,143N
The force experienced by kara's is -102,143N.
We can also determined the deceleration of Kara's car
From Newton second law
F=ma
Then, a = F/m
a = -102,143/1300
a = -78.57m/s²
The negative sign show deceleration
We can also calculate the distance the car moved before coming to halt.
Using equation of motion
X =ut+½at²
X = 11×0.14 + ½(-78.57) × 0.14²
X = 1.54 —0.77
X = 0.77m
A factory siren indicating the end of a shift has a frequency of 80 Hz where the speed of sound in air is 343 m/s. What frequency is perceived by the occupant of a car that is traveling at 30 m/s (a) towards the factory and (b) away from the factory
To develop this problem it is necessary to apply the concepts related to the Dopler Effect.
The apparent frequency can be calculated through the expression
[tex]f' = \frac{v\pm v_0}{v}*f[/tex]
Where,
v = Velocity of the sound or light
[tex]v_0 =[/tex]Velocity of the observer
f=Real frequency
Our values are given as,
[tex]v = 343m/s[/tex]
[tex]v_0 = 30m/s[/tex]
[tex]f = 80Hz[/tex]
PART A ) Towards the factory we need to apply the plus sign
[tex]f' = \frac{v+v_0}{v}*f[/tex]
[tex]f' = \frac{343+30}{343}*80[/tex]
[tex]f' = 86.99Hz[/tex]
PART B ) Away from the factory we need to apply the minus sign
[tex]f' = \frac{v-v_0}{v}*f[/tex]
[tex]f' = \frac{343-30}{343}*80[/tex]
[tex]f' = 73 Hz[/tex]
A large uniform chain is hanging from the ceiling, supporting a block of mass 46 kg. The mass of the chain itself is 19 kg, and the length of the chain is 1.9 m. The acceleration of gravity is 9.81 m/s 2 . Find the tension in the chain at the point where the chain is supporting the block. Answer in units of N.
Answer:
T = 451.26 N
Explanation:
It is given that,
The mass of block, m = 46 kg
Mass of the chain, m' = 19 kg
Length of the chain, l = 1.9 m
Let T is the the tension in the chain at the point where the chain is supporting the block. It is clearly equal to the product of mass and acceleration.
[tex]T=mg[/tex]
[tex]T=46\ kg\times 9.81\ m/s^2[/tex]
T = 451.26 N
So, the tension in the chain at the point where the chain is supporting the block is 451.26 N. Hence, this is the required solution.
You place three nonflammable objects in a fire. They are identical in shape and size, but one object is black, the second is white, and the third is shiny silver. After a few minutes, all three objects are at the same temperature: 1800° C. They remain solid and are now glowing with thermal radiation. Which one is glowing most brightly? The first object (black). They are all glowing with equal brightness. The third object (silver). The second object (white).
Answer:
The first object (black)
Explanation:
The thermal radiation of a body is governed by the Stefan Boltzmann Law, which is mathematically given as:
[tex]E=\epsilon. \sigma.T^4[/tex]
where:
[tex]\sigma =5.67\times 10^{-8}\ W.m^{-2}.K^{-4}[/tex] (Stefan Boltzmann constant)
Also the absorptivity and emissivity of a perfectly black body is maximum ( =1 ) therefore it absorbs all the heat energy incident on it and glows the brightest due to the perfect emission.So here also the black object will glow brighter than the others.
Answer:
The guy above me is definitely correct just read it.
Explanation:
it is what it is.
Sarah attempts to do the popular physics trick of swinging a pail of water around in a vertical circle fast enough so the water doesn't spill out when the pail is upside down. If her arm is 0.417 meters long, what is the minimum speed with which she can swing the pail so the water does not spill out at the top of the path? (Assume that you may neglect the length of the pail).
Answer:
The minimum speed will be 1.997 m /sec at which water does not spill out
Explanation:
We have given that length of arm r = 0.417 m
Acceleration due to gravity [tex]g=9.8m/sec^2[/tex]
We have to find the minimum speed so that water does not spill out
We know that minimum speed is given by
[tex]v_{min}=\sqrt{rg}=\sqrt{0.417\times 9.8}=1.997m/sec[/tex]
So the minimum speed will be 1.997 m /sec at which water does not spill out
A ball is thrown straight up. At what point does the ball have the most energy? Ignore air resistance.
a. At the highest point of its path.
b. When it is first thrown.
c. Just before it hits the ground.
d. Everywhere; the energy of the ball is the same at all of these points.
e. When the ball is halfway to the highest point of its path.
Answer:
Everywhere; the energy of the ball is the same at all of these points.
Explanation:
A ball is thrown straight up. The total energy of an object remains constant as per the law of conservation of energy. At the highest point the object have only potential energy and at lowest point it have only kinetic energy. As a result, the total energy remains constant.
So, at every point the energy of the ball is the same at all of these points. Hence, the correct option is (d).
The ball has the most kinetic energy when it is first thrown. As it rises, its kinetic energy decreases while its potential energy increases, but the total energy is conserved and remains constant unless air resistance is factored in. So the correct option is b.
Explanation:The answer is b. When it is first thrown. The total energy of the ball in this context is its kinetic energy (energy of motion) and potential energy (energy of position). At the moment the ball is thrown, it has the maximum kinetic energy and zero potential energy. As it rises, its kinetic energy decreases while its potential energy increases, but the total energy (kinetic plus potential) is conserved and remains constant if air resistance is ignored. So, the ball doesn't really have 'the most energy' at any one point, but it has the maximum kinetic energy when it is first thrown, which is probably what the question is asking.
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A solid steel cylinder is standing (on one of its ends) vertically on the floor. The length of the cylinder is 3.2 m and its radius is 59 cm. When an object is placed on top of the cylinder, the cylinder compresses by an amount of 5.6 10-7 m. What is the weight of the object?
To solve this problem it is necessary to apply the concepts related to Young's Module and its respective mathematical and modular definitions. In other words, Young's Module can be expressed as
[tex]\Upsilon = \frac{F/A}{\Delta L/L_0}[/tex]
Where,
F = Force/Weight
A = Area
[tex]\Delta L[/tex]= Compression
[tex]L_0[/tex]= Original Length
According to the values given we have to
[tex]\Upsilon_{steel} = 200*10^9Pa[/tex]
[tex]\Delta L = 5.6*10^{-7}m[/tex]
[tex]L_0 = 3.2m[/tex]
[tex]r= 0.59m \rightarrow A = \pi r^2 = \pi *0.59^2 = 1.0935m^2[/tex]
Replacing this values at our previous equation we have,
[tex]\Upsilon = \frac{F/A}{\Delta L/L_0}[/tex]
[tex]200*10^9 = \frac{F/1.0935}{5.6*10^{-7}/3.2}[/tex]
[tex]F = 38272.5N[/tex]
Therefore the Weight of the object is 3.82kN
To calculate the weight of the object placed on top of the cylinder, we need to use the concept of pressure and the formulas for pressure and force. The weight of the object is equal to the force, which can be calculated using the given compression and pressure values, along with the area of the cylinder. The weight of the object is approximately 1.22 x 10^5 N.
Explanation:In order to calculate the weight of the object placed on top of the cylinder, we need to use the concept of pressure and the formula for pressure, which is force divided by area. First, we calculate the area of the cylinder using the formula for the area of a circle, πr2. The radius of the cylinder is 59 cm, so the area is calculated as A = π(0.59 m)2.
Then, we need to calculate the force using the given compression and the formula for force, which is pressure times area. The compression of the cylinder is given as 5.6 x 10-7 m and the pressure is given as 6.9 x 106 N/m2. Now we can calculate the force with the formula F = (6.9 x 106 N/m2) (0.59 m)2. The weight of the object is equal to the force, so the weight of the object is approximately 1.22 x 105 N.
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An astronaut is being tested in a centrifuge. The centrifuge has a radius of 9 m and, in starting, rotates according to ? = 0.30t2, where t is in seconds and ? is in radians.
(a) When t = 9.0 s, what is the magnitude of the astronaut's angular velocity?
(b) When t = 9.0 s, what is the magnitude of the astronaut's linear speed?
(c) When t = 9.0 s, what is the magnitude of the astronaut's tangential acceleration?
(d) When t = 9.0 s, what is the magnitude of the astronaut's radial acceleration?
Answer:
5.4 rad/s
48.6 m/s
5.4 m/s²
262.44 m/s²
Explanation:
r = Radius of centrifuge = 9 m
[tex]\theta=0.3t^2[/tex]
Differentiating with respect to time
[tex]\frac{d\theta}{dt}=0.6t=\omega[/tex]
At t = 9 s
[tex]\omega=0.6\times 9=5.4\ rad/s[/tex]
Angular velocity is 5.4 rad/s
Linear speed
[tex]v=r\omega\\\Rightarrow v=9\times 5.4=48.6\ m/s[/tex]
The linear speed of the astronaut is 48.6 m/s
Differentiating [tex]\omega[/tex] with respect to time
[tex]\frac{d\omega}{dt}=0.6=\alpha[/tex]
Tangential acceleration is given by
[tex]a_t=\alpha r\\\Rightarrow a_t=0.6\times 9=5.4\ m/s^2[/tex]
Tangential acceleration of the astronaut is 5.4 m/s²
Radial acceleration is given by
[tex]a_r=\frac{v^2}{r}\\\Rightarrow a_r=\frac{48.6^2}{9}\\\Rightarrow a_r=262.44\ m/s^2[/tex]
The radial acceleration of the astronaut is 262.44 m/s²
A tightened string vibrates with a standing wave. Which of the following statements is correct? Group of answer choices Points on the string undergo the same displacement. Points on the string vibrates with different amplitude. Points on the string vibrates with different frequencies. Points on the string undergo the same speed. Points on the string vibrate with the same energy..
Nodes and antinodes on a vibrating string standing wave has different amplitudes.
Option B
Explanation:
For a standing wave produced from a tightened string vibration, points of maximum and minimum displacement are formed termed as nodes and antinodes. Hence (A) is false.The amplitudes at nodes are zero and increases gradually to the maximum when it approaches the antinode. Statement (B) is true. The string vibrates with a single natural frequency and a number of resonant frequencies. The nodes and antinodes have same frequencies. (C) is false.Velocity at the nodes is zero and it increases gradually at the antinodes. Statement (D) is false. Since nodes have a displacement equal to zero, no energy is present. All energy is confined in between two nodes. Energy at any point on the string is always constant and is not transferred.(E) is false.Final answer:
Points on a tightened string vibrating with a standing wave have different amplitudes, with nodes having no displacement and antinodes having the maximum. The frequency and speed of the wave are constant along the string, but energy varies with amplitude.
Explanation:
When a tightened string vibrates with a standing wave, points on the string vibrate with different amplitudes. This variation in amplitude can be seen by observing the points of maximum displacement (antinodes) and the points of no displacement (nodes). While points at the antinodes experience the greatest amplitude, the points at the nodes do not move at all. The frequency of oscillation is the same for all points on the string because a standing wave is established by the interference of waves traveling in opposite directions with the same frequency. Consequently, the speed of the wave along the string is also the same because it is determined by the tension and the linear mass density of the string which we assume are constant. However, the energy of the points is not the same due to the varying amplitudes; points at the antinodes have more energy as compared to those at the nodes.
A small ball of mass 62 g is suspended from a string of length 72 cm and whirled in a circle lying in the horizontal plane. r v 9.8 m/s 2 72 cm 62 g 32◦ If the string makes an angle of 32◦ with the vertical, find the centripetal force experienced by the ball. The acceleration of gravity is ____.
Answer:
Centripetal force, [tex]F_x=0.379\ N[/tex]
Explanation:
It is given that,
Mass of the ball, m = 62 g = 0.062 kg
Length of the string, l = 72 cm = 0.72 m
Angle, [tex]\theta=32^{\circ}[/tex]
To find,
The centripetal force experienced by the ball.
Solution,
According to the attached free body diagram,
[tex]T\ cos\theta-mg=0[/tex]
[tex]T=\dfrac{mg}{cos\theta}[/tex]
[tex]T=\dfrac{0.062\times 9.8}{cos(32)}[/tex]
T = 0.716 N
The centripetal force will be experienced by the ball in horizontal direction. It is equal to :
[tex]F_x=T\ sin\theta[/tex]
[tex]F_x=0.716\ N\ sin(32)[/tex]
[tex]F_x=0.379\ N[/tex]
So, the centripetal force experienced by the ball is 0.379 N.
The centripetal force experienced by the small ball if mass 62 g and attached to the string is 0.379 N.
What is centripetal force?Centripetal force is the force which is required to keep rotate a body in a circular path. The direction of the centripetal force is inward of the circle towards the center of rotational path.
Given information-
The mass of the small bag is 62 grams.
The length of the string is 72 cm.
Suppose T is the tension in the string, m is the mass of ball and g is the gravitational force.
The horizontal force applied on the body is,
[tex]\sum F_x=T\cos \theta-mg=0\\T=\dfrac{mg}{\cos \theta}[/tex]
The centripetal force applied on the body is,
[tex]F_c=T\sin \theta\\T=\dfrac{F_c}{\sin \theta}[/tex]
Compare both the values of the T, we get,
[tex]\dfrac{F_c}{\sin\theta}=\dfrac{mg}{\cos \theta}\\{F_c}=\dfrac{mg\times{\sin\theta}}{\cos \theta}\\{F_c}={mg\times{\tan\theta}}[/tex]
Put the values as,
[tex]F_c=0.062\times9.8\times\tan(32)\\F_c=0.379\rm N[/tex]
Thus, the centripetal force experienced by the small ball if mass 62 g and attached to the string is 0.379 N.
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A car travels 10 m/s east. Another car travels 10 m/s north. The relative speed of the first car with respect to the second is:
a. more than 20 m/s.
b. equal 20 m/s
c. 0 m/s.
d. less than 20 m/s.
Answer:
d. less than 20m/s
Explanation:
To the 2nd car, the first car is travelling 10m/s east and 10m/s south. So the total velocity of the first car with respect to the 2nd car is
[tex]\sqrt{10^2 + 10^2} =10\sqrt{2}=14.14m/s
As 14.14m/s is less than 20m/s. d is the correct selection for this question.
On a cold winter day when the temperature is −20∘C, what amount of heat is needed to warm to body temperature (37 ∘C) the 0.50 L of air exchanged with each breath? Assume that the specific heat of air is 1020 J/kg⋅K and that 1.0 L of air has mass 1.3×10−3kg.
To calculate the heat required to warm the air from -20°C to body temperature with each breath, first find the mass of the air, then apply the heat transfer formula, which gives around 38 Joules of heat.
Explanation:The question is asking for the amount of heat needed to warm up air from a very cold outside temperature (-20°C) to body temperature (37°C). This is a typical physics problem dealing with heat transfer and specific heat.
First, find the temperature difference: ΔT = T_final - T_initial = 37°C - (-20°C) = 57°C. Then, we convert volume to mass, using given density of air (1.3×10^-3 kg/L), so mass m = 0.50 L * 1.3×10^-3 kg/L = 0.00065 kg.
Next, apply the heat transfer formula: Q = mcΔT, where m is the mass of the air, c is the specific heat of air, and ΔT is the change in temperature. With m = 0.00065 kg, c = 1020 J/Kg°C, and ΔT = 57°C, we find Q = 0.00065 kg * 1020 J/Kg°C * 57°C = 37.887 J.
So, approximately 38 Joules of heat is needed to warm the air with each breath from -20°C to 37°C.
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The amount of heat needed to warm the 0.50 L of air to body temperature is approximately 37.79 J.
We can use the formula for heat transfer (Q) in a substance:
[tex]\[ Q = mc\Delta T \][/tex]
where:
[tex]\( Q \)[/tex] is the heat energy required (in joules),
[tex]\( m \)[/tex] is the mass of the substance (in kilograms),
[tex]\( c \)[/tex] is the specific heat capacity of the substance (in joules per kilogram per kelvin),
[tex]\( \Delta T \)[/tex] is the change in temperature (in kelvins or degrees Celsius).
First, we need to find the mass of 0.50 L of air using the given density:
[tex]\[ m = \text{density} \times \text{volume} \][/tex]
[tex]\[ m = 1.3 \times 10^{-3} \, \text{kg/L} \times 0.50 \, \text{L} \][/tex]
[tex]\[ m = 6.5 \times 10^{-4} \, \text{kg} \][/tex]
Next, we need to calculate the change in temperature [tex](\( \Delta T \))[/tex]:
[tex]\[ \Delta T = T_{\text{final}} - T_{\text{initial}} \][/tex]
[tex]\[ \Delta T = 37 \, ^\circ\text{C} - (-20 \, ^\circ\text{C}) \][/tex]
[tex]\[ \Delta T = 37 + 20 \, ^\circ\text{C} \][/tex]
[tex]\[ \Delta T = 57 \, ^\circ\text{C} \][/tex]
Now, we convert the change in temperature from degrees Celsius to kelvins:
[tex]\[ \Delta T = 57 \, \text{K} \][/tex]
Finally, we can calculate the heat required:
[tex]\[ Q = mc\Delta T \][/tex]
[tex]\[ Q = (6.5 \times 10^{-4} \, \text{kg}) \times (1020 \, \text{J/kg} \cdot \text{K}) \times (57 \, \text{K}) \][/tex]
[tex]\[ Q = 37.79 \, \text{J} \][/tex]
An electric motor has an eccentric mass of 10 kg (10% of the total mass) and is set on two identical springs (k=3200 N/m). The motor runs at 1750 rpm, and the mass eccentricity is 100 mm from the center. The springs are mounted 250 mm apart with the motor shaft in the center. Neglect damping and determine the amplitude of vertical vibration.
Answer:
The amplitude is 0.010 m
Solution:
As per the question:
Eccentric Mass, [tex]m_{e} = 10\ kg[/tex]
[tex]m_{e}[/tex] = 10%M kg
0.1M = 10
Total mass, M = 100 kg
Spring constant, k = 3200 N/m
Eccentric center, e = 100 mm = 0.1 m
Speed of motor, N = 1750 rpm
Distance between two springs, d = 250 mm = 0.25 m
Now,
Angular velocity, [tex]\omega = \frac{2\pi N}{60} = \frac{2\pi \times 1750}{60} = 183.26\ rad/s[/tex]
For the vertical vibrations:
[tex]\omega_{n} = \sqrt{\frac{2k}{M}} = \sqrt{\frac{6400}{100}} = 8\ rad/s[/tex]
Now, the frequency ratio is given by:
r = [tex]\frac{\omega}{\omega_{n}} = \frac{183.26}{8} = 22.91[/tex]
Thus the amplitude is given by:
A = [tex]e^{\frac{m_{e}}{M}}.\frac{r^{2}}{|1 - r^{2}|}[/tex]
A = [tex]e^{\frac{10}{100}}.\frac{22.91^{2}}{|1 - 22.91^{2}|} = 0.010 m[/tex]
Final answer:
The amplitude of vertical vibration for the electric motor mounted on springs is 100 mm or 0.1 meters, considering the eccentric mass and neglecting damping.
Explanation:
The subject of this question is Physics, and it pertains to a college-level study of mechanics in the context of oscillations and waves, specifically relating to a system experiencing simple harmonic motion (SHM). The motor with an eccentric mass which causes vertical vibrations can be modeled as a mass-spring system, characteristic of simple harmonic oscillators.
Given the mass and the spring constant, we can calculate the natural angular frequency (not represented in this problem due to lack of damping) of this system. The driving frequency can also be converted to an angular frequency to check against the natural frequency. However, the question directly asks for the amplitude of vertical vibration, which in the undamped case of SHM, depends on the initial displacement of the mass from its equilibrium position. Since the question neglects damping, the amplitude can be considered equal to the eccentricity of the mass in the static case. Therefore, the amplitude of vertical vibration is 100 mm, or 0.1 meters.
You have just moved into a new apartment and are trying to arrange your bedroom. You would like to move your dresser of weight 3,500 N across the carpet to a spot 5 m away on the opposite wall. Hoping to just slide your dresser easily across the floor, you do not empty your clothes out of the drawers before trying to move it. You push with all your might but cannot move the dresser before becoming completely exhausted. How much work do you do on the dresser?
Answer:
0 Joules
Explanation:
The work done is given by
[tex]W=F\times s\times cos\theta[/tex]
where,
F = Force applied
s = Displacement of the object = 0 m
[tex]\theta[/tex] = Angle between the force applied and the horizontal = 0
[tex]W=F\times 0\times cos0\\\Rightarrow W=0\ J[/tex]
Work is only observed when there is a displacement.
The work done by me is 0 Joules as I was unable to move it.