Answer:
The power of top half of the lens is 0.55 Diopters.
Explanation:
Since, the person can see an object at a distance between 34 cm and 180 cm away from his eyes. Therefore, 180 cm must be the focal length of the upper part of lens, as the top half of the lens is used to see the distant objects.
The general formula for power of a lens is:
Power = 1/Focal Length in meters
Therefore, for the top half of the lens:
Power = 1/1.8 m
Power = 0.55 Diopters
A grinding wheel is a uniform cylinder with a radius of 7.80 cm and a mass of 0.550 kg.
Part A
Calculate its moment of inertia about its center. Express your answer to three significant figures and include the appropriate units.
Part B
Calculate the applied torque needed to accelerate it from rest to 1750 rpm in 7.40 s .
Take into account a frictional torque that has been measured to slow down the wheel from 1500 rpm to rest in 58.0 s .
Answer:
a. I = 167.31 x 10 ⁻³ kg*m²
b. T = 4.59 kg * m² / s²
Explanation:
The moment of inertia of a uniform cylinder:
a.
r = 7.8 cm * 1 m / 100 cm = 0.078 m
I = ½ * m * r²
I = ½ * 0.55 kg * (0.078²m)
I = 167.31 x 10 ⁻³ kg*m²
b.
T = Iα’ + Iα,
α’ = ω’/t = 1750 rpm * (2π/60) / 7.40s = 24.76 rad/s²
α = ω/t = 1500 rpm * (2π/60) / 58 = 2.71 rad/s²
T = (167.31 x 10⁻³ kg*m²)* (24.76 + 2.71 ) rad / s²
T = 4.59 kg * m² / s²
The moment of inertia of the wheel is calculated as 0.00133 kg*m^2. The second part of the question involves determining the net and frictional torques to find the total applied torque.
Explanation:To solve this problem, we need to apply the formulas of moment of inertia and the angular acceleration along with the concept of frictional torque. The moment of inertia for a cylinder rotating about its axis is given by the formula I = 0.5*m*r^2. In this case, where mass (m) is 0.550 kg and radius (r) is 7.80 cm or 0.078 m (since 1cm = 0.01m).
Part A: I = 0.5 * 0.550 kg * (0.078 m)^2 = 0.00133 kg*m^2.
For Part B, we first need to convert the rotational speed from revolutions per minute (rpm) to rad/s. Then we use these values to determine the angular acceleration and calculate the net torque. The frictional torque is then added to this net torque to find the total applied torque.
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A visible disturbance propagates around a crowded soccer stadium when fans, section by section, jump up and then sit back down.
Answer:
Transverse wave formation.
Explanation:
This type of wave formation is called transverse wave formation.
Transverse wave is a wave arising when the medium's pulse is perpendicular to the wave's path of propagation. The crowd was standing vertically, but they were arranged horizontally in the stadium, according to the question. Hence the wave form was transverse.
How much heat energy is required to convert 93.4 g of solid ethanol at − 114.5 ° C to gasesous ethanol at 149.8 ° C ? The molar heat of fusion of ethanol is 4.60 kJ/mol , and its molar heat of vaporization is 38.56 kJ/mol . Ethanol has a normal melting point of − 114.5 ° C and a normal boiling point of 78.4 ° C . The specific heat capacity of liquid ethanol is 2.45 J / g ⋅ ° C , and that of gaseous ethanol is 1.43 J / g ⋅ ° C .
Answer:
Q' = 140.859 kJ
Explanation:
Given that, 93.4 g of solid ethanol at − 114.5 °C is converted to gasesous ethanol at 149.8 ° C.
The molar heat of fusion of ethanol is, ΔH(f) = 4.60 kJ/mol , and its molar heat of vaporization is ΔH(v) = 38.56 kJ/mol .
And also Ethanol has a normal melting point of − 114.5 ° C and a normal boiling point of 78.4 ° C .
The specific heat capacity of liquid ethanol is S(l) = 2.45 J / g ⋅°C , and that of gaseous ethanol is S(g) = 1.43 J / g ⋅°C .
Lets solve this step wise ;
Given 93.4 g of ethanol is taken, but 1 mole of ethanol consists of 46.06 g
⇒ moles of ethanol given = [tex]\frac{93.4}{46.06}[/tex] = 2.02 moles
step 1: solid ethanol to liquid ethanol at melting point of − 114.5 ° C
⇒ 1 mole requires ΔH(f) = 4.60 kJ/mol of heat
⇒ heat required = 4.60 × 2.02 = 9.292 kJ.
step 2: liquid ethanol at -114 °C to liquid ethanol at 78.4 °C
Q = m×S×ΔT ; Q = heat required
m = mass of the substance
S = specific heat of the substance
ΔT = change in temperature
Here S = S(l);
⇒ Q = 93.4×2.45×(78.4-(-114.5))
= 44.141 kJ
step 3: liquid ethanol at 78.4°C to gaseous ethanol at 78.4°C
1 mol of liquid ethanol requires ΔH(v) = 38.56 kJ/mol of heat
⇒ required heat = 38.56×2.02 = 77.89 kJ
step 4: gaseous ethanol at 78.4 °C to gaseous ethanol at 149.8 °C
Q = m×S×ΔT
Here, S = S(g)
Q = 93.4×1.43×(149.8-78.4)
= 9.536 kJ
⇒ Total heat required = 9.292 + 44.141 + 77.89 + 9.536
= 140.859 kJ
⇒ Q' = 140.859 kJ
The total heat energy required to convert 93.4 g of solid ethanol at − 114.5 ° C to gaseous ethanol at 149.8 ° C is 140.859 kJ.
Given data:,
93.4 g of solid ethanol at − 114.5 °C is converted to gaseous ethanol at 149.8 ° C.
The molar heat of fusion of ethanol is, ΔH(f) = 4.60 kJ/mol.
Molar heat of vaporization is ΔH(v) = 38.56 kJ/mol .
And also Ethanol has a normal melting point of − 114.5 ° C and a normal boiling point of 78.4 ° C .
The specific heat capacity of liquid ethanol is S(l) = 2.45 J / g ⋅°C , and that of gaseous ethanol is S(g) = 1.43 J / g ⋅°C .
Since, 93.4 g of ethanol is taken, but 1 mole of ethanol consists of 46.06 g
⇒ moles of ethanol given is
⇒93.04/46.06 = 2.02 moles
Heat required for conversion of solid ethanol to liquid ethanol at melting point of − 114.5 ° C
⇒ 1 mole requires ΔH(f) = 4.60 kJ/mol of heat
⇒ Q = 4.60 × 2.02 = 9.292 kJ.
Heat required to convert liquid ethanol at -114 °C to liquid ethanol at 78.4 °C
Q' = m×s×ΔT
Here,
m = mass of the substance
s = specific heat of the substance
ΔT = change in temperature
Solving as,
Q' = 93.4×2.45×(78.4-(-114.5))
Q' = 44.141 kJ
Heat required to convert liquid ethanol at 78.4°C to gaseous ethanol at 78.4°C
1 mol of liquid ethanol requires ΔH(v) = 38.56 kJ/mol of heat
Q'' = 38.56×2.02 = 77.89 kJ
Heat required to convert gaseous ethanol at 78.4 °C to gaseous ethanol at 149.8 °C
Q''' = m×S×ΔT
Q''' = 93.4×1.43×(149.8-78.4)
= 9.536 kJ
⇒ Total heat required = 9.292 + 44.141 + 77.89 + 9.536
= 140.859 kJ
Thus, we can conclude that the total heat energy required to convert 93.4 g of solid ethanol at − 114.5 ° C to gaseous ethanol at 149.8 ° C is 140.859 kJ.
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(FLUID MECHANICS)
The apparent weight of an object when a system is not in equilibrium is represented by:
A. Fg (object)
B. Fnet (system)
C. Fb
D. rhoV
Answer:
option B
Explanation:
When a body is immersed in liquid there will be two force is acting on the body.
First one force acting downward due to weight of the body.
And the second force acting on the object will be buoyant force.
If the object is not in equilibrium the apparent weight will be equal to net force acting on the object.
[tex]F_{net} = W - F_b[/tex]
W is the weight of the object acting downward
Fb is the buoyancy force acting upward on the object.
Hence, the correct answer is option B
A spring that is stretched 23 cm from its equilibrium point experiences a force of 103 N.a.) How much energy is stored in the spring?
Answer:
U = 11.85 J
Explanation:
given,
spring is stretched = 23 cm
x = 0.23 m
Force experiences = 103 N
we know,
F = k x
where k is the spring constant
[tex]k =\dfrac{F}{x}[/tex]
[tex]k =\dfrac{103}{0.23}[/tex]
k = 447.83 N/m
energy stored in the spring
[tex]U =\dfrac{1}{2}kx^2[/tex]
[tex]U =\dfrac{1}{2}\times 447.83 \times 0.23^2[/tex]
U = 11.85 J
hence, energy stored in the spring is equal to U = 11.85 J
The elastic potential energy stored in the spring when stretched to a length of 23 cm, given a spring constant of 4 N/cm, is 0.18 J.
Explanation:The question pertains to how much energy is stored in a spring when it is stretched 23 cm from its equilibrium point, where it experiences a force of 103 N. The potential energy stored in the spring can be calculated using the formula U = 1/2kx². Given that the spring constant is 4 N/cm and the displacement of the spring from its unstretched length is 3 cm (since the unstretched length is 20 cm), the potential energy can be calculated as follows: U = 1/2 * 4 N/cm * (3 cm)² = 0.18 J. Thus, the elastic potential energy contributed by the spring when it is stretched to a length of 23 cm is 0.18 J.
A uniform disk, a uniform hoop, and a uniform solid sphere are released at the same time at the top of an inclined ramp. They all roll without slipping. In what order do they reach the bottom of the ramp?A uniform disk, a uniform hoop, and a uniform solid sphere are released at the same time at the top of an inclined ramp. They all roll without slipping. In what order do they reach the bottom of the ramp?hoop, sphere, diskhoop, disk, spheredisk, hoop, spheresphere, disk, hoopsphere, hoop, disk
Answer:
First let's write down the moment of inertia of the objects.
[tex]I_{sphere} = \frac{2}{5}mR^2\\I_{disk} = \frac{1}{2}mR^2\\I_{hoop} = mR^2[/tex]
If they all roll without slipping, then the following relation is applied to all ot them:
[tex]v = \omega R[/tex]
where v is the translational velocity and ω is the rotational velocity.
We will use the conservation of energy, because we know that their initial potential energies are the same. (Here, I will assume that all the objects have the same mass and radius. Otherwise we couldn't determine the difference. )
[tex]K_1 + U_1 = K_2 + U_2\\0 + mgh = \frac{1}{2}I\omega^2 + \frac{1}{2}mv^2[/tex]
For sphere:
[tex]\frac{1}{2}\frac{2}{5}mR^2(\frac{v}{R})^2 + \frac{1}{2}mv^2 = mgh\\\frac{1}{5}mv^2 + \frac{1}{2}mv^2 = mgh\\\frac{7}{10}mv^2 = mgh\\v_{sphere} = \sqrt{\frac{10gh}{7}}[/tex]
For disk:
[tex]\frac{1}{2}\frac{1}{2}mR^2(\frac{v}{R})^2 + \frac{1}{2}mv^2 = mgh\\\frac{1}{4}mv^2 + \frac{1}{2}mv^2 = mgh\\\frac{3}{4}mv^2 = mgh\\v_{disk} = \sqrt{\frac{4gh}{3}}[/tex]
For hoop:
[tex]\frac{1}{2}mR^2(\frac{v}{R})^2 + \frac{1}{2}mv^2 = mgh\\\frac{1}{2}mv^2 + \frac{1}{2}mv^2 = mgh\\mv^2 = mgh\\v_{hoop}= \sqrt{gh}[/tex]
The sphere has the highest velocity, so it arrives the bottom first. Then the disk, and the hoop arrives the last.
Explanation:
The moment of inertia can be defined as the resistance to the rotation. If an object has a high moment of inertia, it resist to rotate more so its angular velocity would be lower. In the case of rolling without slipping, the angular velocity and the linear (translational) velocity are related by the radius, so the object with the highest moment of inertia would arrive the bottom the last.
The order in which a sphere, disk, and hoop reach the bottom of an incline when released from the same height is determined by their moments of inertia. The solid sphere arrives first, followed by the disk, and finally the hoop.
Explanation:In the scenario where a uniform disk, a uniform hoop, and a uniform solid sphere are released from the top of an inclined ramp and roll without slipping, the order in which they reach the bottom depends on their moments of inertia and the distribution of mass. The solid sphere has the smallest moment of inertia relative to its mass (I = 2/5 MR²), which means it will accelerate faster than the other shapes and hence get to the bottom first. The uniform disk, with a moment of inertia of I = 1/2 MR², will follow. The uniform hoop has the largest moment of inertia (I = MR²) for a given mass and radius, so it will accelerate the slowest of the three and reach the bottom last.
Therefore, the objects reach the bottom of the ramp in the following order: sphere, disk, hoop.
In an atom, an electron is confined to a space of roughly 10?10 meters. If we take this to be the uncertainty in the electron's position, what is the minimum uncertainty ?p in its momentum?
Answer:
[tex]5.2728\times 10^{-25}\ kgm/s[/tex]
Explanation:
h = Planck's constant = [tex]6.626\times 10^{-34}\ m^2kg/s[/tex]
[tex]\Delta x[/tex] = Uncertainity in position = [tex]10^{-10}\ m[/tex]
[tex]\Delta p[/tex] = Uncertainty in momentum
According to the Heisenberg uncertainity principle we have
[tex]\Delta x\Delta p=\dfrac{h}{4\pi}\\\Rightarrow \Delta p=\dfrac{h}{4\pi\Delta x}\\\Rightarrow \Delta p=\dfrac{6.626\times 10^{-34}}{4\pi\times 10^{-10}}\\\Rightarrow \Delta p=5.2728\times 10^{-25}\ kgm/s[/tex]
The minimum uncertainty in its momentum is [tex]5.2728\times 10^{-25}\ kgm/s[/tex]
You are baking and run out of an important ingredient. Rather than fire up the car, you decide to use a bicycle. The store is 1/2 mile away. Assume your car gets 30 miles to the gallon. If every driver on the planet made a similar decision once per week for a year, approximately how much gas would be saved for that year? (As this is an estimate, you may make the arithmetic easier by assuming a year has 50 weeks rather than 52 weeks. What number do you use for 'every driver on the planet'?) a. 100 million gallons b. 2 billion gallons c. 20,000 gallons d. 200,000 gallons e. 2 million gallons
Answer: this one is tough. There is a estimated 1.2 billion drivers to help best I could. I did the math the best I could and didn’t get an answer close to the choices. Just guessing I’d guess C. 20,000 because of logic.
Explanation: I can’t help much but would love to hear how to work this out.
My Notes A container is divided into two equal compartments by a partition. One compartment is initially filled with helium at a temperature of 240 K; the other is filled with nitrogen at a temperature of 315 K. Both gases are at the same pressure. If we remove the partition and allow the gases to mix, what will be their final temperature?
Answer:
[tex]final-temperature = T_{f} = 252.51K[/tex]
Explanation:
we can solve this problem by using the first law of thermodynamics.
[tex]\Delta U= Q-W[/tex]
Q= heat added
U= internal energy
W= work done by system
[tex]E_{final}= E_{initial}[/tex]
[tex]C_{v} (N_{2}) C_{v}(He) T_{f} =C_{v}( N_{2}) T_{N_{2} } C_{v} (He) T_{He}[/tex] (1)
[tex]C_{v}(N_{2})=1.04\frac{KJ}{Kg K}[/tex]
[tex]C_{v}(He)=5.193\frac{KJ}{Kg K}[/tex]
now
From equation 1
[tex]T_{f}=\frac{C_{v}( N_{2}) T_{N_{2} } C_{v} (He) T_{He}}{C_{v} (N_{2}) C_{v}(He)}[/tex]
[tex]T_{f} = \frac{315\times1.04+5.193\times240}{1.04+5.193}[/tex]
[tex]T_{f} = 252.51K[/tex]
We go out to sunbathe on a warm summer day. If we soak up 80 British thermal units per hour [BTU/h] of energy, how much will the temperature of 65 comma 000-gram person increase in 2 hours [h] in units of degrees Celsius [°C]? We assume that since our bodies are mostly water they have the same specific heat as water (4.18 joules per gram degree Celsius [J/(g degrees Upper C)]).
Answer:
0.62127°C
Explanation:
[tex]1\ BTU=1055\ J[/tex]
[tex]80\ BTU/h=80\times 1055=84400\ J/h[/tex]
Heat absorbed by body in 2 hours
[tex]Q=84400\times 2\\\Rightarrow Q=168800\ J[/tex]
m = Mass of person = 65000 g
c = Specific heat of water = 4180 J/kg°C
[tex]\Delta T[/tex] = Change in temperature
Heat is given by
[tex]Q=mc\Delta T\\\Rightarrow 168800=65\times 4180\times \Delta T\\\Rightarrow \Delta T=\dfrac{168800}{65\times 4180}\\\Rightarrow \Delta T=0.62127\ ^{\circ}C[/tex]
The increase in temperature will be 0.62127°C
A cylinder, which is in a horizontal position, contains an unknown noble gas at 4.00 × 10 4 Pa 4.00×104 Pa and is sealed with a massless piston. The piston is slowly, isobarically moved inward 16.3 cm, 16.3 cm, which removes 1.50 × 10 4 J 1.50×104 J of heat from the gas. If the piston has a radius of 30.5 cm, 30.5 cm, calculate the change in the internal energy of the system Δ U ΔU .
Answer:
-13094.55179 J
Explanation:
Q = Heat = [tex]-1.5\times 10^{4}\ J[/tex]
P = Pressure = [tex]4\times 10^4\ Pa[/tex]
[tex]\Delta V[/tex] = Change in volume = [tex]\pi r^2\times -h[/tex](negative because it is decreasing)
h = Height = 16.3 cm
r = Radius = 30.5 cm
Entropy is given by
[tex]\Delta U=Q-W[/tex]
Work done is given by
[tex]W=P\Delta V\\\Rightarrow W=4\times 10^4\times (\pi 0.305^2\times -0.163)[/tex]
[tex]\Delta U=-1.5\times 10^{4}-(4\times 10^4\times (\pi 0.305^2\times -0.163))\\\Rightarrow \Delta U=-13094.55179\ J[/tex]
The change in the internal energy of the system is -13094.55179 J
Which of the following statements are true? A. Earth's gravity has no effect on astronauts inside the International Space Station. B. An astronaut's mass is greater on Earth than on the Moon. C. An astronaut's weight is the same on the Moon as on Earth. D. An astronaut's mass is the same on the International Space Station as it is on Earth. E. None of these statements are true.
The given statement "An astronaut's mass is the same on the International Space Station as it is on Earth" is true.
Answer: Option D
Explanation:
There is usually a slight difference between mass and the weight of an object. The difference is that the mass of any object is independent of its acceleration due to gravity or gravitational influence of the planet where it is present.
Similarly, the weight of any object will be influenced by the gravitational force of that planet as the weight is directly proportional to the acceleration due to gravity of that planet.
So, the other three options are false and those three options states that weight of an object on Earth is equal to the weight of that object on any other planet. This is not true. So, the fourth option related to the mass of an astronaut in and outside Earth is true as it is equal theoretically.
An astronaut's mass being the same on the International Space Station as it is on Earth is a true statement.
What is Mass?Thus is defined as the resistance a matter offers to a change in its speed or position when force is applied.
Gravitational force doesn't determine the mass of objects which is why an astronaut's mass on International Space Station will be the same as on Earth. This therefore makes option D the most appropriate choice.
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A 0.5 kg block of aluminum (caluminum=900J/kg⋅∘C) is heated to 200∘C. The block is then quickly placed in an insulated tub of cold water at 0∘C (cwater=4186J/kg⋅∘C) and sealed. At equilibrium, the temperature of the water and block are measured to be 20∘C.If the original experiment is repeated with a 1.0 kg aluminum block, what is the final temperature of the water and block?If the original experiment is repeated with a 1.0 kg copper (ccopper=387 J/kg⋅∘C) block, what is the final temperature of the water and block?If the original experiment is repeated but 100 g of the 0∘C water is replaced with 100 g of 0∘C ice, what is the final temperature of the water and block? If the original experiment is repeated but 100 g of the 0∘C water is replaced with only 25 g of 0∘C ice, what is the final temperature of the water and block?
These scenarios involve the transfer of heat in a closed system by utilizing calorimetry. The final temperature is affected by the mass of the metal block, the specific heat capacity of the metal, and the introduction of phase changes with the addition of ice into the system.
Explanation:These problems revolve around the principle of conservation of energy, and more specifically the concept of heat transfer, which is often studied using calorimetry. We can solve these problems by using the formula for heat transfer, Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.
For the first scenario, using a 1.0 kg aluminum block instead of a 0.5 kg block, the heat capacity of the system has simply doubled. The final temperature for the aluminum block and the water can be determined by setting the heat gained by the water equal to the heat lost by the aluminum. Since the mass of the block has doubled, the final temperature will be lower than the original experiment.
In the second scenario with 1.0 kg copper block, copper has a lower specific heat capacity than aluminum. This means that when heated to the same temperature, the copper block will contain less energy than the aluminum block. Consequently, when put in contact with the water, the final temperature of the water and block will be lower than in the original experiment.
In the final two scenarios, adding ice to the system adds an additional phase change. This means, it’s necessary to add in the energy used to convert the ice at 0 degrees Celsius into water at 0 degrees Celsius before heating the water to the final temperature. The amount of ice being added in these instances changes the energy dynamics of the system and thus resulting in different final temperatures.
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A carpenter builds an exterior house wall with a layer of wood 3.0 cm thick on the outside and a layer of Styrofoam insulation 2.4 cm thick on the inside wall surface. The wood has k=0.080W/(m?K), and the Styrofoam has k= 0.010 W/(m?K). The interior surface temperature is 20.0 ?C , and the exterior surface temperature is -13.0 ?C
A.)What is the temperature at the plane where the wood meets the Styrofoam? _______ Celsius
B.)What is the rate of heat flow per square meter through this wall? ______W/m^2
Answer:
Explanation:
Given
thermal conductivity of wood [tex] K_w=0.08 W/m^2-K[/tex]
thermal conductivity of insulation [tex]K_i=0.01 W/m^2-K[/tex]
thickness of wood [tex]t_2=3 cm[/tex]
thickness of insulation [tex]t_1=2.4 cm[/tex]
[tex]T_i=20^{\circ}C[/tex]
[tex]T_o=-13^{\circ}C[/tex]
we know heat Flow is given by
[tex]Q=kA\frac{dT}{dx}[/tex]
[tex]dT=[/tex] change in temperature
[tex]dx=[/tex] thickness
K=thermal conductivity
A=Area of cross-section
A is same
Suppose T is the temperature of Junction
as heat Flow is same thus
[tex]\frac{k_w(20-T)}{3}=\frac{k_i(T-(-13))}{2.4} [/tex]
[tex]\frac{0.08(20-T)}{3}=\frac{0.01(T+13)}{2.4}[/tex]
[tex]T=19.36 ^{\circ}C[/tex]
(b)Rate of heat flow
[tex]Q=\frac{k_w(T+13)}{3\times 10^{-2}}[/tex]
[tex]Q=\frac{0.08\times 32.36}{0.03}[/tex]
[tex]Q=86.303 W/m^2[/tex]
The temperature at which the wood meets the Styrofoam is determined using the proportional temperature drops across each material, considering their respective thermal resistances. The rate of heat flow per square meter through the wall is calculated using Fourier's law, considering the combined thermal resistance of both layers.
Explanation:Temperature at the Wood-Styrofoam Plane and Heat Flow Rate
To find the temperature at the plane where the wood meets the Styrofoam, we need to use the concept of thermal resistance and the fact that the heat flow through both materials is the same. With the given thermal conductivity (k values) and thicknesses of the wood and Styrofoam, we can calculate their respective thermal resistances (R = thickness/k). Then, by setting up a proportion, we can find the temperature drop across the wood (ΔTw) and the Styrofoam (ΔTs) using the formula ΔT = (k*A* ΔT)/(thickness), where A is the area (which cancels out as it's the same for both layers). Knowing the temperature drop across each and the outer and inner surface temperatures, we can find the temperature at the plane where the layers meet.
To calculate the rate of heat flow per square meter through this wall, we'll use Fourier's law of thermal conduction, which is given by Q = k*A* ΔT/d, where Q is the heat flow rate, A is the area, ΔT is the temperature difference, and d is the thickness. Since the wall consists of two layers with different thermal conductivities and thicknesses, we need to calculate the equivalent thermal resistance for the combined wall and then use the overall temperature difference to find the heat flow rate.
A 10-kg block on a rough horizontal surface is attached to a light spring (force constant = 1400 N/m). The block is pulled 10.0 cm to the right from its equilibrium position and released from rest. The frictional force between the block and surface has a magnitude of 30 N. What is the kinetic energy of the block as it passes through its equilibrium position?
To develop this problem it is necessary to apply the concepts related to the conservation of Energy. In this case the definition concerning kinetic energy from the simple harmonic movement.
From the conservation of energy we know that the kinetic energy would be conserved through the work done by the frictional force and the simple harmonic potential energy, in other words:
[tex]KE = PE_s +W_f[/tex]
[tex]KE = \frac{1}{2}kA^2 + W_f[/tex]
Where,
[tex]KE =[/tex] Kinetic Energy
[tex]PE_s =[/tex]Potential Harmonic Simple Energy
[tex]W_f =[/tex] Work made by friction.
Our values are given as,
[tex]m = 10Kg \rightarrow[/tex] mass
[tex]k = 1400N/m \rightarrow[/tex] Spring constant
[tex]A = 0.1m \rightarrow[/tex]Amplitude
[tex]f_f = 30N \rightarrow[/tex] Frictional Force
Replacing we have,
[tex]KE = \frac{1}{2}kA^2 + W_f[/tex]
[tex]KE =\frac{1}{2} 1400 * 0.1^2 + ( - 30 * 0.1)[/tex]
[tex]KE = 4 J[/tex]
Therefore the Kinetic Energy of the block as it passes through its equlibrium position is 4J.
The kinetic energy of the 10-kg block as it passes through its equilibrium position is 4 Joules. This is calculated by converting the potential energy stored in the spring to kinetic energy and then subtracted the energy lost due to friction.
Explanation:The first step in solving this problem is to understand the two forces acting on the block in this question: the spring force and the frictional force. The spring potential energy when the block is pulled 10 cm to the right is given by the formula U = 1/2kx^2, where k is the force constant and x is the displacement. Substituting the given values, we have U = 1/2(1400 N/m)(0.1 m)^2 = 7 Joules. This is the initial potential energy stored in the spring. As the block passes through its equilibrium position, this potential energy is fully converted to kinetic energy.
We also need to take into account the work done against frictional force which is equal to the frictional force times the displacement, i.e., W_friction = Friction * displacement = 30N * 0.1m = 3 Joules. This is the energy lost due to friction.
Finally, we subtract the work done by the frictional force from the potential energy to achieve the kinetic energy. Therefore, the kinetic energy of the block as it passes through its equilibrium position is K = U - W_friction = 7J - 3J = 4 Joules.
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In lab, your instructor generates a standing wave using a thin string of length L = 1.65 m fixed at both ends. You are told that the standing wave is produced by the superposition of traveling and reflected waves, where the incident traveling waves propagate in the +x direction with an amplitude A = 2.45 mm and a speed vx = 10.5 m/s . The first antinode of the standing wave is a distance of x = 27.5 cm from the left end of the string, while a light bead is placed a distance of 13.8 cm to the right of the first antinode. What is the maximum transverse speed vy of the bead? Make sure to use consistent distance units in your calculations.
Given that the distance from the left end of the string to the first antinode is 27.5 cm , calculate the wavelength of the standing wave on the string. Remember to convert all measurements into units of meters before performing this calculation.
Answer:
On the standing waves on a string, the first antinode is one-fourth of a wavelength away from the end. This means
[tex]\frac{\lambda}{4} = 0.275~m\\\lambda = 1.1~m[/tex]
This means that the relation between the wavelength and the length of the string is
[tex]3\lambda/2 = L[/tex]
By definition, this standing wave is at the third harmonic, n = 3.
Furthermore, the standing wave equation is as follows:
[tex]y(x,t) = (A\sin(kx))\sin(\omega t) = A\sin(\frac{\omega}{v}x)\sin(\omega t) = A\sin(\frac{2\pi f}{v}x)\sin(2\pi ft) = A\sin(\frac{2\pi}{\lambda}x)\sin(\frac{2\pi v}{\lambda}t) = (2.45\times 10^{-3})\sin(5.7x)\sin(59.94t)[/tex]
The bead is placed on x = 0.138 m. The maximum velocity is where the derivative of the velocity function equals to zero.
[tex]v_y(x,t) = \frac{dy(x,t)}{dt} = \omega A\sin(kx)\cos(\omega t)\\a_y(x,t) = \frac{dv(x,t)}{dt} = -\omega^2A\sin(kx)\sin(\omega t)[/tex]
[tex]a_y(x,t) = -(59.94)^2(2.45\times 10^{-3})\sin((5.7)(0.138))\sin(59.94t) = 0[/tex]
For this equation to be equal to zero, sin(59.94t) = 0. So,
[tex]59.94t = \pi\\t = \pi/59.94 = 0.0524~s[/tex]
This is the time when the velocity is maximum. So, the maximum velocity can be found by plugging this time into the velocity function:
[tex]v_y(x=0.138,t=0.0524) = (59.94)(2.45\times 10^{-3})\sin((5.7)(0.138))\cos((59.94)(0.0524)) = 0.002~m/s[/tex]
In this exercise we have to use the knowledge of mechanics to be able to calculate the wavelength, so we can say that it will correspond to:
[tex]v_y=0.002m/s[/tex]
On the standing waves well-behaved, the first antinode happen one of four equal parts of a intuitiveness out completely. This means:
[tex]\lambda=1.1 m[/tex]
This way that the connection middle from two points the wavelength and the extent of object of the strand happen:
[tex]L=3\lambda /2[/tex]
By definition, this standing wave exist at the after second harmonious, n = 3.
Furthermore, the standing wave equating happen in this manner:
[tex]y(x,t)=(Asin(kx))sin(wt)=Asin(w/vx)sin(wt)=((2.45*10^{-3})sin(5.7X)sin(59.94t)[/tex]
The droplet exist established on x = 0.138 m. The maximum speed exist place the derivative of the speed function equals to nothing:
[tex]v_y=wAsin(kx)cos(wt)\\a_y=-w^2Asin(kx)sin(wt)\\a_y=0[/tex]
Find the time, we have:
[tex]59.94t=\pi\\t=0.0524s[/tex]
This exist moment of truth when the speed is maximum. So, the maximum speed maybe raise by stop up existing time into the velocity function:
[tex]v_y(x=0.138,t=0.0524)=(59.94)(2.45*10^{-3})sin((5.7)(0.138))cos((59.94)(0.0524))\\=0.002 m/s[/tex]
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A particle moves according to the law of motion s(t) = t^{3}-8t^{2}+2t, t\ge 0, where t is measured in seconds and s in feet. a.) Find the velocity at time t. Answer: b.) What is the velocity after 3 seconds? Answer: c.) When is the particle at rest? Enter your answer as a comma separated list. Enter None if the particle is never at rest. At t_1= and t_2= with t_1
Answer:
Explanation:
Given
displacement is given by
[tex]s(t)=t^3-8t^2+2t[/tex]
so velocity is given by
[tex]v(t)=\frac{\mathrm{d} s(t)}{\mathrm{d} t}[/tex]
[tex]v(t)=3t^2-16t+2[/tex]
(b)velocity after [tex]t=3 s[/tex]
[tex]v(3)=3(3)^2-8\cdot 3+2[/tex]
[tex]v(3)=19 m/s[/tex]
(c)Particle is at rest
when its velocity will become zero
[tex]v(t)=0[/tex]
i.e. [tex]3t^2-16t+2=0[/tex]
[tex]t=\frac{16\pm \sqrt{16^2-4\cdot 3\cdot 2}}{2\cdot 3}[/tex]
[tex]t=\frac{16\pm 15.23}{6}[/tex]
[tex]t=5.20 s[/tex]
At a steam power plant, steam engines work in pairs, the heat output of the first one being the approximate heat input of the second. The operating temperatures of the first are 750◦C and 440◦C respectively, and of the second 415◦C and 270◦C.
Answer:
The rate at which coal burned is 111.12 kg/s.
Explanation:
Given that,
First initial temperature =750°C
First final temperature =440°C
Second initial temperature =415°C
Second final temperature =270°C
Suppose If the heat of combustion of coal is [tex]2.8×10^{7}\ J/kg[/tex], at what rate must coal be burned if the plant is to put out 950 MW of power? Assume the efficiency of the engines is 65% of the Carnot efficiency.
The work done by first engine is
[tex]W=eQ[/tex]
The work done by second engine is
[tex]W'=e'Q'[/tex]
[tex]W'=e'Q(1-e)[/tex]
Total out put of the plant is given by
[tex]W+W'=950\ MW[/tex]
Put the value into the formula
[tex]eQ+e'Q(1-e)=950\times10^{6}[/tex]....(I)
We need to calculate the efficiency of first engine
Using formula of efficiency
[tex]e=65%(1-\dfrac{T_{L}}{T_{H}})[/tex]
[tex]e=0.65(1-\dfrac{440+273}{750+273})[/tex]
[tex]e=0.196[/tex]
We need to calculate the efficiency of second engine
Using formula of efficiency
[tex]e'=65%(1-\dfrac{T_{L}}{T_{H}})[/tex]
[tex]e'=0.65(1-\dfrac{270+273}{415+273})[/tex]
[tex]e'=0.136[/tex]
Put the value of efficiency for first and second engine in the equation (I)
[tex]Q(0.196+0.136(1-0.196))=950\times10^{6}[/tex]
[tex]Q=\dfrac{950\times10^{6}}{(0.196+0.136(1-0.196))}[/tex]
[tex]Q=3111.24\times10^{6}\ W[/tex]
We need to calculate the rate at which coal burned
Using formula of rate
[tex]R=\dfrac{Q}{H_{coal}}[/tex]
[tex]R=\dfrac{3111.24\times10^{6}}{2.8×10^{7}}[/tex]
[tex]R=111.12\ kg/s[/tex]
Hence, The rate at which coal burned is 111.12 kg/s.
A car of mass 900 kg is traveling at 20 m/s when the brakes are applied. The car then comes to a complete stop in 5 s. What is the average power that the brakes produce in stopping the car?
A. 36,000 W
B. 7200 W
C. 3600 W
D. 1800 W
Answer:
A. 36,000 W
Explanation:
[tex]m[/tex] = mass of the car = 900 kg
[tex]v_{o}[/tex] = Initial speed of the car = 20 m/s
[tex]v_{f}[/tex] = Final speed of the car = 0 m/s
[tex]W[/tex] = Work done by the brakes on the car
Magnitude of work done on the car by the brakes is same as the change in kinetic energy of the car.hence
[tex]W = (0.5) m (v_{o}^{2} - v_{f}^{2})\\W = (0.5) (900) ((20)^{2} - (0)^{2})\\W = 180000 J[/tex]
[tex]t[/tex] = time taken by the car to come to stop = 5 s
[tex]P[/tex] = Average power produced by the car
Average power produced by the car is given as
[tex]P = \frac{W}{t} =\frac{180000}{5} \\P = 36000 W[/tex]
Cold water (cp = 4180 J/kg·K) leading to a shower enters a thinwalled double-pipe counterflow heat exchanger at 15°C at a rate of 1.25 kg/s and is heated to 60°C by hot water (cp = 4190 J/kg·K) that enters at 100°C at a rate of 4 kg/s. If the overall heat transfer coefficient is 880 W/m2 ·K, determine the rate of heat transfer and the heat transfer surface area of the heat exchanger.
Answer:
the rate of heat transfer Q is Q =235.125 kJ/s
the heat transfer surface area A of the heat exchanger is A= 15.30 m²
Explanation:
Assuming negligible loss to the environment, then the heat flow of the hot water goes entirely to the cold water
Denoting a as cold water and b as hot water , then
Q= Fᵃ* cpᵃ * ( T₂ᵃ - T₁ᵃ)
where
F= mass flow
cp = specific heat capacity at constant pressure
T₂= final temperature
T₁ = initial temperature
replacing values
Q = Fᵃ* cᵃ * ( T₂ᵃ - T₁ᵃ) = 1.25 kg/s* 4180 J/kg·K* ( 60°C-15°C) * 1 kJ/1000J= 235.125 kJ/s
if all there is no loss to the surroundings
Qᵃ + Qᵇ = Q surroundings = 0
Fᵃ* cpᵃ * ( T₂ᵃ - T₁ᵃ) + Fᵇ* cpᵇ * ( T₂ᵇ - T₁ᵇ) = 0
T₂ᵇ = T₁ᵇ - [Fᵃ* cpᵃ / (Fᵇ* cpᵇ) ]* ( T₂ᵃ - T₁ᵃ)
replacing values
T₂ᵇ =100°C - [1.25 kg/s* 4180 J/kg·K/ (4 kg/s* 4190 J/kg·K)]* ( 60°C-15°C)
T₂ᵇ = 85.97 °C
the heat transfer surface of the heat exchanger is calculated through
Q = U*A* ΔTlm
where
U= overall heat transfer coefficient
A = heat transfer area of the heat exchanger
ΔTlm = (ΔTend - ΔTbeg)/ ln ( ΔTend - ΔTbeg)
ΔTbeg = temperature difference between the 2 streams at the inlet of the heat exchanger ( hot out - cold in) = 85.97 °C - 15°C = 70.97 °C
ΔTbeg = temperature difference between the 2 streams at the end of the heat exchanger ( hot in - cold out ) = 100°C - 60 °C = 40°C
then
ΔTlm = (ΔTend - ΔTbeg)/ ln ( ΔTend - ΔTbeg) =( 70.97 °C- 40°C)/ ln( 70.97°C/40°C) = 17.455 °C
ΔTlm = 17.455 °C
then
Q = U*A* ΔTlm
A = Q/(U*ΔTlm) = 235.125 kJ/s/(17.455 °C *880 W/m²*K) *1000 J/kJ = 15.30 m²
A= 15.30 m²
Final answer:
Calculate the heat transfer rate and heat transfer surface area in a double-pipe counterflow heat exchanger using given water properties and overall heat transfer coefficient.
Explanation:
Cold Water: mw = 1.25 kg/s, cp = 4180 J/kg·K, Tin = 15°C, Tout = 60°C
Hot Water: mh = 4 kg/s, cp = 4190 J/kg·K, Tin = 100°C, Tout = ?
Overall Heat Transfer: U = 880 W/m²·K
Calculate Heat Transfer Rate:
Calculate Q using Q = mcΔT for each water type.Calculate ∆T using Tin and Tout values.Use the overall heat transfer coefficient equation: Q = U × A × ∆Tlm.Solve for A, the heat transfer surface area.A ≈ 1m²
So, the rate of heat transfer is approximately 55341.28 W and the heat transfer surface area of the heat exchanger is approximately 1 m².
A simple piping system consisting of a small pipe (diameter 6 cm) connected to a larger pipe (diameter 30 cm) is used to transfer and magnify force from one end to the other. The system is full of water and completely enclosed by two pistons, one on each end. If a force 11.2 N is exerted (additional to the forces already present) on the small piston, then how much additional force is exerted at the larger piston
Answer:
280N
Explanation:
Pascal's law states that the pressure in a fluid is transmitted across every point in the fluid system.
Hence, the pressure in both tubes must remain same;
So, pressure = F1/A1 = F2/A2
where F1 = initial force on small pipe, A1 = area of small pipe
F2 = force on larger piston, A2 = area of larger piston
Given:
F1 = 11.2N
D1 = diameter of smaller pipe = 6 cm
D2 = diameter of larger piston = 15 cm
A1 = π*(r1)² = π*(6/2)² =π*9 (As radius, r = diameter/2)
A2 = π*(r2)² = π*(30/2)² =π*225
Hence 11.2/π*9 = F2/π*225
solving, we have
F2 = 280N
The capacity of a storage battery, such as those used in automobile electrical systems, is rated in ampere-hours (A?h). A 50 A?h battery can supply a current of 50 Afor 1.0 h, or 25 A for 2.0 h, and so on.A) What total energy can be supplied by a 13V , 60A?h battery if its internal resistance is negligible?Answer= ...... JB) What volume (in liters) of gasoline has a total heat of combustion equal to the energy obtained in part (a)? (See Section 17.6; the density of gasoline is 90 kg/m 3.)Answer= ........ LC) If a generator with an average electrical power output of 0.45 kW is connected to the battery, how much time will be required for it to charge the battery fully?Answer= ........ h
Answer: (A) 780J
(B) 1.89×10^-11L
(C)1.67×10^-4 h
Explanation:
Energy of the battery = IVt
=13×60 = 780J
Heat combustion of
1g of gasoline relax 46000J
Therefore 780J will release 780/46000
= 0.017g
Density = mass/volume
Volume = mass/density
Volume =0.017× 10^-3 / 900
= 1.89× 10^-8 m3
= 1.89×10^-11 litres
P=IVt
t=P/IV
= 450/60×13
1.67×10^-4 hours
The energy of a battery, volume of gasoline and time required to charge the battery is required.
The energy is 2808000 J
The volume is 0.683 L.
The time required is 1.733 h
It = Current-time = 60 Ah = [tex]60\times 3600\ \text{As}[/tex]
t = Time
V = Voltage = 13 V
Energy is given by
[tex]E=IVt=ItV\\\Rightarrow E=60\times 3600\times 13\\\Rightarrow E=2808000\ \text{J}[/tex]
[tex]\rho[/tex] = Density = [tex]90\ \text{kg/m}^3=\dfrac{90}{1000}=0.09\ \text{kg/L}[/tex]
C = Thermal heat capacity = [tex]4.57\times 10^7\ \text{J/kg}[/tex]
m = Mass
Power is given by
[tex]P=mC\\\Rightarrow m=\dfrac{P}{C}\\\Rightarrow m=\dfrac{2808000}{4.57\times 10^7}=\dfrac{702}{11425}\ \text{kg}[/tex]
Volume is given by
[tex]V=\dfrac{m}{\rho}\\\Rightarrow V=\dfrac{\dfrac{702}{11425}}{0.09}\\\Rightarrow V=0.683\ \text{L}[/tex]
P = Power = 0.45 kW
Time is given by
[tex]t=\dfrac{E}{P}\\\Rightarrow t=\dfrac{2808000}{450}=6240\ \text{s}\\\Rightarrow t=\dfrac{6240}{3600}=1.733\ \text{h}[/tex]
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The oscillating current in an electrical circuit is as follows, where I is measured in amperes and t is measured in seconds. I = 7 sin(60πt) + cos(120πt). Find the average current for each time interval. (Round your answers to three decimal places.) (a) 0 ≤ t ≤ 1 60 amps
The average current in the given oscillating current function, I = 7 sin(60πt) + cos(120πt), can be found by integrating each component of the function over the given time interval and taking the average. Due to the complexity of the function, it is recommended to use mathematical software or a calculator with integral computation capability.
Explanation:The given function describes the oscillating current in an electrical circuit: I = 7 sin(60πt) + cos(120πt). The average current, Iave, is conceptually the net charge, ΔQ, that passes through a given cross-sectional area per unit time, Δt. Due to the complexity of this function, we must split it into two integrals to find the average currents.
For the time interval 0 ≤ t ≤ 1/60 seconds, we have two integrals for each component of the current: ∫01/607 sin(60πt) dt and ∫01/60cos(120πt) dt. Calculating these will give us the average current for this time interval. Due to the complexity of the function and the requirement of calculus to solve it, it's recommended to use mathematical software or a calculator with integral computation capability to get the numerical values, always rounding your answers to three decimal places.
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A heat engine takes thermal energy QH from a hot reservoir and uses part of this energy to perform work W. Assuming that QH cannot be changed, how can the efficiency of the engine be improved?
a) Increase the work W, the thermal energy QC rejected to the cold reservoir increasing as a result.
b) Decrease the work W, the rejected QC decreasing as a result.
c) Increase the work W, the rejected QC remaining unchanged.
d) Increase the work W, the rejected QC decreasing as a result.
e) Decrease the work W, the rejected QC remaining unchanged
Answer:
d) Increase the work W, the rejected QC decreasing as a result.
Explanation:
By the second law of thermodynamics the efficiency of a heat engine working between two reservoirs is:
[tex]\eta=\frac{W}{Q_{H}} [/tex] (1)
With W the work and [tex] Q_{H} [/tex] the heat of the hot reservoir, note in (1) that efficiency is directly proportional to the work and inversely proportional to the heat of the hot reservoir, so if we remain [tex]Q_{H} [/tex] constant we should increase the work to increase the efficiency.
Also, efficiency is:
[tex] \eta=1-\frac{Q_{C}}{Q_{H}}[/tex] (2)
With [tex]Q_{C} [/tex] the heat released to the cold reservoir, it is important to note that because second law of thermodynamics the efficiency of a heat engine should be between 0 and 1 ([tex]0\leq\eta\leq1 [/tex]), so the ratio [tex]\frac{Q_{C}}{Q_{H}} [/tex] always is positive and its maximum value is 1, that implies if [tex]Q_{H} [/tex] remains constant and efficiency increases, [tex]Q_{C} [/tex] will decrease and the ratio [tex]\frac{Q_{C}}{Q_{H}} [/tex] too.
So, the correct answer is d)
Final answer:
The efficiency of a heat engine is increased by increasing the work output while decreasing the rejected heat to the cold reservoir, in accordance with the first and second laws of thermodynamics.
Explanation:
To improve the efficiency of a heat engine, one must increase the work output, W, while simultaneously decreasing the heat rejected to the cold reservoir, Qc. The correct choice is:
d) Increase the work W, the rejected Qc decreasing as a result.
The efficiency, η, of a heat engine is defined as the ratio of the work done, W, to the heat absorbed from the hot reservoir, QH. By the first law of thermodynamics, QH = W + Qc, meaning that the efficiency can be increased either by increasing W or decreasing Qc. According to the second law of thermodynamics, there is a minimum amount of QH that cannot be used for work and must be rejected as Qc. Therefore, the aim is to minimize this rejected heat without altering QH, which cannot be changed in this scenario. The ideal is to approach the efficiency of a Carnot engine, which has the maximum possible efficiency between two given temperatures by operating in a reversible manner and reducing entropy generation.
An ideal heat engine operates of fixed difference of temperatures between hot and cold reservoirs, say 100K. What will provide greater efficiency: operation with the hot reservoir as hot as possible, or operation with a cold reservoir as cold as possible?
To solve this problem we will apply the concepts given for the efficiency of an engine which is given as
[tex]\eta = 1-\frac{T_C}{T_H}[/tex]
[tex]\eta = \frac{T_H-T_C}{T_H}[/tex]
Where
[tex]T_C[/tex] = Temperature of the cold reservoir
[tex]T_H[/tex] = Temperature of the hot reservoir
The efficiency maximum would be given only if [tex]T_C = 0[/tex]
Replacing this value we have
[tex]\eta = \frac{T_H-0}{T_H}[/tex]
[tex]\eta = 1[/tex]
Therefore: Cold reservoir as cold as possible provide the greater efficiency.
Final answer:
The greater efficiency in an ideal heat engine is obtained when the hot reservoir is as hot as possible and the cold reservoir is as cold as possible.
Explanation:
The greater efficiency in an ideal heat engine operating with a fixed temperature difference between hot and cold reservoirs is obtained when the hot reservoir is as hot as possible and the cold reservoir is as cold as possible.
Efficiency is determined by the ratio of the temperature of the cold reservoir (Tc) to the temperature of the hot reservoir (Th). The greater the temperature difference, the easier it is to convert heat transfer to work.
Therefore, maximizing the temperature difference by making the hot reservoir as hot as possible and the cold reservoir as cold as possible will result in greater efficiency.
A 44.5 mA current is carried by a uniformly wound air-core solenoid with 500 turns, a 18.5 mm diameter, and 14.0 cm length. (a) Compute the magnetic field inside the solenoid. µT (b) Compute the magnetic flux through each turn. T·m2 (c) Compute the inductance of the solenoid. mH (d) Which of these quantities depends on the current? (Select all that apply.)
Answer:
a. Magnetic Field =1.997×[tex]10^{-4}[/tex] T
b. Area= 2.68×[tex]10^{-4}[/tex][tex]m^{2}[/tex]
Magnetic Flux= 5.367×[tex]10^{-8}[/tex]T[tex]m^{2}[/tex]
c. Inductance= 6.013×[tex]10^{-4}[/tex]H
Explanation:
Parameters from the question
I= 44.5×[tex]10^{-3}[/tex]A
N=500 turns
Diameter=18.5mm
Radius = (diameter/2) = 9.25mm =9.25×[tex]10^{-3}[/tex]m
L= 14cm = 0.14m
Permitivity [tex]U_{o}[/tex]=4π×[tex]10^{-7}[/tex]H/m
The Formulars Used are
B(Magnetic Field) =[tex]\frac{U_{o}. N. I }{l}[/tex]
Mag Flux= B.A
Inductance= [tex]\frac{U_{o}.N^{2} .A }{l}[/tex]
Answer:
a) 199.716 μT
b) [tex]5.368 * 10^{-8}[/tex] T·m^2
c) 0.603 mH
d) B and Ф
Explanation:
I am giving the explanation with my handwritten solution in the paper.
Check the attachment please.
The free-fall acceleration on Mars is 3.7 m/s2.
(a) What length of pendulum has a period of 1.2 s on Earth? cm
(b) What length of pendulum would have a 1.2-s period on Mars? cm An object is suspended from a spring with force constant 10 N/m.
(c) Find the mass suspended from this spring that would result in a period of 1.2 s on Earth. kg
(d) Find the mass suspended from this spring that would result in a period of 1.2 s on Mars.
Answer:
(a) The length of the pendulum on Earth is 36.8cm
(b) The length of the pendulum on Mars is 13.5cm
(c) Mass suspended from the spring on Earth is 0.37kg
(d) Mass suspended from the spring on Mars is 0.36kg
Explanation:
Period = 1.2s, free fall acceleration on Earth = 9.8m/s^2, free fall acceleration on Mars = 3.7m/s^2
( a) Length of pendulum on Earth = [( period ÷ 2π)^2] × acceleration = (1.2 ÷ 2×3.142)^2 × 9.8 = 0.0365×9.8 = 0.358m = 35.8cm
(b) Length of the pendulum on Mars = (1.2÷2×3.142)^2 × 3.7 = 0.0365×3.7 = 0.135cm = 13.5m
(c) Mass suspended from the spring on Earth = (force constant×length in meter) ÷ acceleration = (10×0.358) ÷ 9.8 = 0.37kg
(d) Mass suspended from the spring on Mars = (10×0.135)÷3.7 = 0.36kg
The length of a pendulum with a period of 1.2 s on Earth is approximately 36.95 cm, while on Mars it is around 16.99 cm. The mass suspended from a spring that would result in a period of 1.2 s on Earth is approximately 0.722 kg, and on Mars it is approximately 0.329 kg.
Explanation:(a) What length of pendulum has a period of 1.2 s on Earth? cm
Using the equation for the period of a pendulum, T = 2π √(L/g), where T is the period, L is the length of the pendulum, and g is the acceleration due to gravity, we can solve for L. Rearranging the equation, we have L = (T/2π)² * g.
Given that the free-fall acceleration on Earth is approximately 9.8 m/s², substituting the values into the equation, we have:
L = (1.2/2π)² * 9.8 = 0.3695 m = 36.95 cm
(b) What length of pendulum would have a 1.2-s period on Mars? cm
Using the same equation, L = (T/2π)² * g, we can substitute the values for the period and acceleration due to gravity on Mars:
L = (1.2/2π)² * 3.7 = 0.1699 m = 16.99 cm.
(c) Find the mass suspended from this spring that would result in a period of 1.2 s on Earth. kg
For a spring-mass system, the period is given by T = 2π √(m/k), where T is the period, m is the mass, and k is the spring constant. Rearranging the equation, we have m = (T/2π)² * k.
Given that the spring constant is 10 N/m, substituting the values into the equation, we have:
m = (1.2/2π)² * 10 = 0.722 kg.
(d) Find the mass suspended from this spring that would result in a period of 1.2 s on Mars. kg
Using the same equation, m = (T/2π)² * k, we can substitute the values for the period and spring constant:
m = (1.2/2π)² * 10 = 0.329 kg.
Flywheels are large, massive wheels used to store energy. They can be spun up slowly, then the wheels energy can be released quickly to accomplish a task that demands high power. An industrial flywheel has a 1.5 m diameter and a mass of 250 kg.
1. A motor spins up the flywheel with a constant torque of 50 N*m. How long does it take the flywheel to reach top angular speed of 1200 rpm
2. How much energy is stored in the flywheel?
Final answer:
To find the time it takes for the flywheel to reach its top angular speed, we can use the formula: ω = Δθ / Δt. In this case, the flywheel has a diameter of 1.5 m and spins up to an angular speed of 1200 rpm. It takes approximately 0.0375 seconds for the flywheel to reach its top angular speed. The energy stored in the flywheel is approximately 554,414.06 joules.
Explanation:
To find the time it takes for the flywheel to reach its top angular speed, we can use the formula:
ω = Δθ / Δt
Where ω is the angular velocity, Δθ is the change in angle, and Δt is the change in time.
In this case, the flywheel has a diameter of 1.5 m, so the radius is 0.75 m. The flywheel spins up from rest to an angular speed of 1200 rpm, which is equivalent to 125.66 rad/s.
Using the formula, we can rearrange to solve for Δt:
Δt = Δθ / ω
Δθ is equal to the circumference of the flywheel, which is 2π times the radius:
Δθ = 2π × 0.75 m = 4.7124 rad
Plugging in the values:
Δt = 4.7124 rad / 125.66 rad/s = 0.0375 s
So it takes approximately 0.0375 seconds for the flywheel to reach its top angular speed.
Now, to calculate the energy stored in the flywheel, we can use the formula:
KE = 0.5 × I × ω^2
Where KE is the kinetic energy, I is the moment of inertia, and ω is the angular velocity.
The moment of inertia for a solid disk is given by:
I = 0.5 × m × r^2
Where m is the mass of the flywheel and r is the radius.
Plugging in the values:
I = 0.5 × 250 kg × (0.75 m)^2 = 70.3125 kg×m^2
Now we can calculate the energy:
KE = 0.5 × 70.3125 kg×m^2 × (125.66 rad/s)^2 = 554,414.06 J
So, the flywheel stores approximately 554,414.06 joules of energy.
A student in an undergraduate physics lab is studying Archimede's principle of bouyancy. The student is given a brass cylinder and, using a triple beam balance, finds the mass to be 3.21 kg. The density of this particular alloy of brass is 8.62 g/cm 3 . The student ties a massless string to one end of the cylinder and submerges it into a tank of water where there is an apparent reduction in the weight of the cylinder. With this information, calculate the volume, V , of the cylinder and the tension, T , in the string when it is submerged in the tank of water. The density of water is 1.00 g/cm 3 , and the acceleration due to gravity is gVolume of the cylinder: _____ cm3Tension in the string: ______ N
Answer:
V = 0.3724 m³
T = 27.836 N
Explanation:
Given :
m = 3.21 kg , W= 3.21 * 9.81 m / s² = 31.4901 N
ρ = 8.62 g / cm ³ = 8620 kg / m³
V = m / ρ = 3.21 kg / 8620 kg / m³
V = 0.3724 m³
when submerged the weight of brass cylinder is equal to the tension in string:
F = (0.3724m³) * (1000 kg / m³) * (9.81 m/s²²) = 3.653 ≈ 3.65 N
T = 31.4901 N - 3.65 N
T = 27.836 N
Snorkeling by humans and elephants. When a person snorkels, the lungs are connected directly to the atmosphere through the snorkel tube and thus are at atmospheric pressure. In atmospheres, what is the difference Δp between this internal air pressure and the water pressure against the body if the length of the snorkel tube is
(a)24 cm (standard situation) and
(b)4.1 m (probably lethal situation)?
Answer:
2354.4 Pa
40221 Pa
Explanation:
[tex]\rho[/tex] = Density = 1000 kg/m³
g = Acceleration due to gravity = 9.81 m/s²
h = Depth
The pressure difference would be
[tex]\Delta P=\rho gh\\\Rightarrow \Delta P=1000\times 0.24\times 9.81\\\Rightarrow \Delta P=2354.4\ Pa[/tex]
The pressure difference in the first case is 2354.4 Pa
[tex]\Delta P=\rho gh\\\Rightarrow \Delta P=1000\times 4.1\times 9.81\\\Rightarrow \Delta P=40221\ Pa[/tex]
The pressure difference in the second case is 40221 Pa