Answers
Answer:
The specific heat of the substance is 393.939 joules per kilogram-degree Celsius.
Explanation:
We notice that the student is mixing a substance with a high temperature and another substance with a low temperature, where the first release heat to the latter one until thermal equilibrium is reached. By the First Law of Thermodynamics and assuming that the entire system has no energy interactions with the surroundings, we get the following model:
\(\Delta U_{x}+\Delta U_{w} = 0\) (1)
Where \(\Delta U_{x}\) and \(\Delta U_{w}\) are the changes in internal energy for the unknown substance and water, measured in joules.
By definition of internal energy, we expand the equation above now:
\(m_{x}\cdot c_{x}\cdot (T_{o,x}-T_{f,x})+m_{w}\cdot c_{w}\cdot (T_{o,w}-T_{f,w}) = 0\) (2)
Where:
\(m_{x}\), \(m_{w}\) - Masses of the unknown substance and water, measured in kilograms.
\(c_{x}\), \(c_{w}\) - Specific heats of the unknown substance and water, measured in joules per kilogram-degree Celsius.
\(T_{o,x}\), \(T_{f,x}\) - Initial and final temperatures of the unknown substance, measured in degrees Celsius.
\(T_{o,w}\), \(T_{f,w}\) - Initial and final temperatures of water, measured in degrees Celsius.
Then, we clear the specific heat of the unknown substance:
\(c_{x} = \frac{m_{w}\cdot c_{w}\cdot (T_{f,w}-T_{o,w})}{m_{x}\cdot (T_{o,x}-T_{f,x})}\)
If we know that \(m_{w} = m_{x} = 0.075\,kg\), \(c_{w} = 4186\,\frac{J}{kg\cdot^{\circ}C}\), \(T_{f,w} = T_{f,x} = 31.15\,^{\circ}C\), \(T_{o,x} = 96.5\,^{\circ}C\) and \(T_{o,w} = 25\,^{\circ}C\), then the heat capacity of the unknown substance is:
\(c_{x} = \frac{(0.075\,kg)\cdot \left(4186\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (31.15\,^{\circ}C-25\,^{\circ}C)}{(0.075\,kg)\cdot (96.5\,^{\circ}C-31.15^{\circ}C)}\)
\(c_{x} = 393.939\,\frac{J}{kg\cdot ^{\circ}C}\)
The specific heat of the substance is 393.939 joules per kilogram-degree Celsius.
Related Questions
03: Hook's law suggests that F is directly proportional to -x, how much true you have found this statement in your experiment? Explain any differences.
Answers
Hooke's Law can be given as follows sometimes:
The restoring force of a spring is equal to the spring constant multiplied by the displacement from its normal position:
F = -kx
Where, F = Restoring force of a spring (Newtons, N)
k = Spring constant (N/m)
x = Displacement of the spring (m)
The negative sign relates to the direction of the applied force and by convention, the minus or negative sign is present in F = -kx. The restoring force F is directly proportional to the displacement (x), according to Hooke's law. When the spring is compressed, the displacement (x) is negative. It is zero when the spring is at its original length and positive when the spring is extended.
Practically, Hooke's Law is applicable only within a limited frame of reference, and through experimenting, this statement proves to be true. Because materials cannot be compressed beyond a certain size or expanded beyond a certain size without some permanent deformation or change of their original state.
The law only applies under some conditions such as a limited amount of force or deformation. Factually, many materials will noticeably deviate from Hooke's law even before those elastic limits are reached.
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An aluminum rod and a nickel rodare both 5.00 m long at 20.0°C.The temperature of each is raisedto 70.0°C. What is the differencein length between the two rods?AluminumNickela = 23.10-6C+ B = 69.10-6 0-1a = 13.10-6C1 B = 39.10-6-1(Unit = m)Enter
Answers
The difference in length between the two rods = 0.0025m
Explanations:Linear expansivity of a material is given by the formula:
\(\alpha\text{ = }\frac{l_2-l_1}{l_1(\theta_2-\theta_1)}\)For the Aluminium rod:
\(\begin{gathered} l_{A1}\text{ = 5.0m} \\ \theta_{A1}=20^0C \\ \theta_{A2}=70^0C \\ \alpha_A\text{ = }23\times10^{-6}C^{-1} \\ \alpha_A\text{ = }\frac{l_{A2}-l_{A1}}{l_{A1}(\theta_{A2}-\theta_{A1})} \\ \text{ }23\times10^{-6}\text{ = }\frac{l_{A2}-5}{5(70-20)} \\ 5\times50\times\text{ }23\times10^{-6}=\text{ }l_{A2}-5 \\ l_{A2}=\text{ (}5750\text{ }\times10^{-6})\text{ + 5} \\ l_{A2}=\text{ 0.00575+5} \\ l_{A2}=\text{ 5.00575m} \end{gathered}\)For the Nickel rod:
\(\begin{gathered} l_{N1}\text{ = 5.0m} \\ \theta_{N1}=20^0C \\ \theta_{N2}=70^0C \\ \alpha_N=\text{ 13}\times10^{-6}C^{-1} \\ \alpha_N\text{ = }\frac{l_{N2}-l_{N1}}{l_{N1}(\theta_{N2}-\theta_{N1})} \\ \text{ 1}3\times10^{-6}\text{ = }\frac{l_{N2}-5}{5(70-20)} \\ 5\times50\times\text{ 1}3\times10^{-6}=\text{ }l_{A2}-5 \\ l_{N2}=\text{ (32}50\text{ }\times10^{-6})\text{ + 5} \\ l_{N2}=\text{ }0.00325+5 \\ l_{N2}=\text{ 5.00325m} \end{gathered}\)The difference in length between the two rods will be given as:
\(\begin{gathered} l_{A2}-l_{N2}=\text{ 5.00575-5.00325} \\ l_{A2}-l_{N2}=0.0025m \end{gathered}\)The difference in length between the two rods = 0.0025m
"How did your current and voltage measurements differ between the series and parallel circuits you created
Answers
Answer:
Series circuit:
The voltage that is measured across the circuit is different.
The current measured in a series circuit remains the same at all points in the circuit.
Parallel circuit:
The current measured across each resistor varies
The voltage measured across a parallel circuit will remain the same
Explanation:
Series and parallel circuits behave differently when it comes to the circulation of current and the interaction with a potential difference.
In a series circuit, the resistances are connected end to end. As a result, the voltage that is measured across the circuit is different once resistance is encountered. However, the current measured in a series circuit remains the same at all points in the circuit.
A parallel circuit behaves in an exactly opposite manner to the series circuit. In a parallel circuit, the resistances are connected side by side. As a result of this, the current measured across each resistor varies as there are circuit branches through which electric current can flow into. On the other hand, the voltage measured across a parallel circuit will remain the same
A car engine operates at 2000 RPM (revolutions per minute). What is the period? Give your answer in seconds
Answers
Take into account that the period is the inverse of the frequency. In this case, the frequency is 2000 rev/min, then, you have:
\(T=\frac{1}{f}=\frac{1}{2000}=0.0005\min \)The period is 0.0005 minutes.
In seconds the answer results in:
0.0005 min = 0.0005*60 seconds = 0.03 seconds
The period is 0.03 seconds
Secondary succession can happen after primary succession or independently of primary succession. Please select the best answer from the choices provided T F
Answers
Answer: ok
Explanation: “In primary succession, newly exposed or newly formed rock is colonized by living things for the first time. In secondary succession, an area that was previously occupied by living things is disturbed, then re-colonized following the disturbance.”-Khanacademy
I think basically, it is saying that you need primary succession in order to colonize and then you have secondary succession which re-colonizeses the area. It could be like a natural disaster, but secondary succession will rebuild back the ecosystem and may even bring good alterations in some scenarios.
I just took the test.
I put True.
I got it right.
11.) If figure skater spins with a circular acceleration of 275 m/s^2 in a circle 5 poin with a radius of 0.55 meters, calculate the velocity of the skater."
Answers
Answer:
we don't care what ure saying thats absurd
Mosses don't spread by dispersing seeds; they disperse tiny spores. The spores are so small that they will stay aloft and move with the wind, but getting them to be windborne requires the moss to shoot the spores upward. Some species do this by using a spore-containing capsule that dries out and shrinks. The pressure of the air trapped inside the capsule increases. At a certain point, the capsule pops, and a stream of spores is ejected upward at 3.6 m/sm/s, reaching an ultimate height of 20 cm.
Required:
What fraction of the initial kinetic energy is converted to the final potential energy?
Answers
Answer:
U / K = 0.30
Explanation:
For this exercise we must calculate the energy at the two points.
Initial. Where are the spores
K = ½ m v²2
Final. Higher
U = m g h
the fraction of energy is
U / K = 2gh /v²
let's calculate
U / K = 2 9.8 0.20 / 3.6²
U / K = 0.30
therefore 30% of the energy is lost
If you were to come back to our solar system in 6 billion years, what might you expect to find?
A) a red giant star
B) a rapidly spinning pulsar
C) a white dwarf
D) a black hole
E) Everything will be essentially the same as it is now
Answers
Answer:
A)a red giant star
a u tube contains a liquid of an unknown density an oil of density is poured into the right arm of the tube until the oil column is 16.0 cm high the oil air interfdace is 4.5 cm above the liquid level in the left arm of the u tube algrebraic expression to deter,ine the density of the unknown fluid
Answers
Answer:
The answer is "\(1155\ \frac{kg}{m^3}\)"
Explanation:
Please find the complete question in the attached file.
\(p = p_0 + ?gh\)
pi = pressure only at two liquids' devices
PA = pressure atmosphere.
1 = oil density
2 = uncertain fluid density
\(h_1 = 11 \ cm\\\\h_2= 3 \ cm\)
The pressures would be proportional to the quantity \(11-3 = 8\) cm from below the surface at the interface between both the oil and the liquid.
\(\to p_A + ?2g(h_1 - h_2) = p_A + ? 1gh_1\\\\\to ?2 = \frac{?1h_1}{(h_1 - h_2)} \\\\\)
\(= \frac{840 \frac{kg}{m^3}}{\frac{11}{8}} \\\\= 1155\ \frac{kg}{m^3}\)
A healthy person has body temperature 37°c,What is temperature on Fahrenheit scale? Show by calculation.
Answers
Answer:
98.6 degrees Fahrenheit.
Explanation:
Human body temperature is 37.0 degrees Celsius or 98.6 degrees Fahrenheit.
A maser is a laser-type device that produces electromagnetic waves with frequencies in the microwave and radio-wave bands of the electromagnetic spectrum. You can use the radio waves generated by a hydrogen maser as a standard of frequency. The frequency of these waves is 1,420,405,751.786 hertz. (A hertz is another name for one cycle per second.) A clock controlled by a hydrogen maser is off by only 1 s in 100,000 years. (The large number of significant figures given for the frequency simply illustrates the remarkable accuracy to which it has been measured.)A. What is the time for one cycle of the radio wave? Express your answer to three significant figures and include the appropriate units.B. How many cycles occur in 4 h ? Express your answer using three significant figures.C. How many cycles would have occurred during the age of the earth, which is estimated to be 4.6×109 years? Express your answer using three significant figures.D. By how many seconds would a hydrogen maser clock be off after a time interval equal to the age of the earth? Express your answer using three significant figures.
Answers
Answer:
a) t₀ = 0.704 10⁻¹² s, b) #cycle = 2.05 10¹⁶ cycles ,
c) #_cycles = 7,418 10³² cycles, d) Δt = 4.6 10⁴ s
Explanation:
For this exercise we must use the relationship between the period and the frequency
f = 1 / T
T = 1 / f
let's apply this equation to our case
a) The time of a cycle is
t = 1 / 1.420405 10¹²
t₀ = 0.704 10⁻¹² s
b) number of cycles in 4 h
t = 4 h (3600 s / 1h) = 14400 s
Let's use a direct rule of proportions (rule of three). If there is 1 cycle in t₀, how much cycle is there
#_cycles = t (1 / t₀)
#_cycle = 14400 s (1 cycle / 0.704 10⁻¹² s)
#_cycle = 2.0454 10¹⁶ cycles
#cycle = 2.05 10¹⁶ cycles
c) cycles since the age of the Earth
t = 4.6 10⁹ years (365 days / year) (24 h / 1 day) (3600 s / 1 h)
t = 5.2224 10²⁰ s
#_cycles = t (1 / to)
# _cycles = 5.2224 10²⁰ (1 / 0.704 10⁻¹²)
#_cycles = 7,418 10³² cycles
d) Let's use the direct proportion rule. If you have a lag of Δt₁ = 1s at t₁ = 1 10⁵ years, what is the lag (Δt) at t = 4.6 10⁹ years
Δt = t (Δt₁ / t₁)
Δt = 4.6 10⁹ years (1 s / 10⁵ years)
Δt = 4.6 10⁴ s
Which statement accurately describes the ultraviolet catastrophe?
O A. A hot lightbulb gave off white visible light instead of ultraviolet
light.
B. An excited gas gave off ultraviolet light rather than an emission
spectrum
O C. A hot lightbulb gave off ultraviolet light instead of white visible
light
D. An excited gas gave off an emission spectrum rather than
ultraviolet light
Answers
Answer:
D
Explanation:
This is a phenomena by a black body. A black body is reckoned to absorb all energy it receives this is because it gives off energy of different frequency ranges hence as the frequency increases the energy rises.
Answer:
A hot lightbulb gave off white visible light instead of ultraviolet light.
Explanation:
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2
Your professor asks you to come up with an example to demonstrate organic solidarity. What is your BEST example?
OA I rely on my grocer because I am not able to produce my own food.
OB. I feel connected to my community because we all worship the same god.
OC. I can evaluate a relationship by pretending to be on the outside of it
OD. I am one of the wealthiest people, so I am benefitting from the poorest people.
Answers
The best example of organic solidarity is; I rely on my grocer because I am not able to produce my own food. Option A
What is organic solidarity?Organic solidarity is a concept developed by French sociologist Émile Durkheim to describe the unity and cohesion that arises from the interdependence of individuals in a society.
In societies with organic solidarity, individuals are dependent on one another for the fulfillment of their needs and the functioning of the society as a whole.
This dependence arises from the division of labor and specialization of tasks, which leads to a complex network of interrelated roles and functions. Organic solidarity is seen as a characteristic of modern, industrialized societies.
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I need help with us history
Answers
Answer:
English is the language
Answer: im most likely wrong but i think its A
Explanation:
similarities and differences between a lunar eclipse and the disappearing sun. HELP FAST
Answers
Answer: A solar eclipse results when the moon passes in between the earth and the sun hiding the sun fully or partly for some time. A lunar eclipse occurs when the earth passes in between the moon and the sun casting its shadow on the moon and thus hiding it fully or partly for some time.
How does the earth orbit the sun?
Answers
Answer:
The Sun's gravity pulls on the planets, just as Earth's gravity pulls down anything that is not held up by some other force and keeps you and me on the ground.
Explanation:
Hope that helps
A frying pan is connected to a 1500 volt circuit. If the resistance of the frying pan is 25 ohms, how many amperes does the frying pan draw?
Answers
The current (in amperes) the frying pan draws from the 500 volt circuit, given that it has a resistance of 25 ohms is 60 amperes
How do i determine the current drawn by the frying pan?From the question given above, the following data were obtained:
Voltage of circuit (V) = 1500 V Resistance of frying pan (R) = 5 Ω Current (I) =?The current drawn by the frying pan can be obtained as follow:
Voltage (V) = Current (I) × resistance (R)
Inputting the given parameters, we have:
1500 = Current × 25
Divide both sides by 25
Current = 1500 / 25
Current = 60 amperes
Thus, from the above calculation we can conclude that the current drawn by the frying pan is 60 amperes
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A force acts on a body of mass 13 kg initially at rest. The force
acts for 10 seconds, and once it quits, the body covers 60 min
the next 6 seconds. Find the force that acted on the body.
Answers
Answer:
Force that acted on the body was F = 13 N
Explanation:
If once accelerated, the body covers 60 meters in 6 seconds, then its velocity is 60/6 m/s = 10 m/s
When the force was acting (for 10 seconds) the object accelerated from rest (initial velocity vi = 0) to 10 m/s (its final velocity). therefore we can use the kinematic equation for the velocity in an accelerated motion given by:
\(v_f=v_i+a*t\)
which in our case becomes;
\(10\,m/s=0+a*(10\,s)\)
and we can solve for the acceleration as:
a = 10/10 m/s^2 = 1 m/s^2
Therefore the force acting on the body, based on Newton's 2nd Law expression: F = m * a is:
F = 13 kg * 1 m/s^2 = 13 N
What is one of the causes of mechanical weathering?
acid rain
oxidation
animal actions
carbon dioxide
Answers
The cause of mechanical weathering among the given options is animal actions. The correct answer is option C
WEATHERINGThis is the breaking down of rocks and minerals into particles of matters
There are three types of weathering. They are:
Physical weatheringchemical weatheringbiological weatheringThe one which is the cause of mechanical weathering among the given options is animal actions.
Therefore, the correct answer is option C
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Answer:
the answer is C animal actions
Explanation:
EDG 2022
g=9.8m/s2A block of mass m = 18 kg accelerates across a rough table when a force F = 100 N acts upon it at an angle = 35° with the horizontal. The force of friction on the block is f = 33 N.(a) What is the acceleration of the block? m/s2(b) What is the coefficient of friction of the block?
Answers
The free body diagram of the block can be shown as,
Part (a)
According to free body diagram, the net force which acts along x-axis is given as,
\(F_x=F\cos \theta-f\)According to Newton's law, the net force acting along x-axis is,
\(F_x=ma\)Plug in the known value,
\(\begin{gathered} F\cos \theta-f=ma \\ a=\frac{F\cos \theta-f}{m} \end{gathered}\)Substitute the known values,
\(\begin{gathered} a=\frac{(100\text{ N)cos35-33 N}}{18\text{ kg}} \\ =\frac{(100N)(0.819)-33\text{ N}}{18\text{ kg}} \\ =\frac{48.9\text{ N}}{18\text{ kg}}(\frac{1kgm/s^2}{1\text{ N}}) \\ =2.72m/s^2 \end{gathered}\)Thus, the acceleration of the block is 2.72 m/s2.
Part (b)
According to free body diagram, the net force acting along y-axis is,
\(F_y=N+F\sin \theta-mg\)Since the block accelerates along x-axis therefore, the net force along y-axis is zero which can be expressed as,
\(\begin{gathered} 0=N+F\sin \theta-mg \\ N=mg-F\sin \theta \end{gathered}\)Substitute the known values,
\(\begin{gathered} N=(18\text{ kg)}(9.8ms^{-2})(\frac{1\text{ N}}{1kgm/s^2})-(100\text{ N)sin35} \\ =176.4\text{ N-}(100\text{ N)(0.574)} \\ =176.4\text{ N-}57.4\text{ N} \\ =119\text{ N} \end{gathered}\)The formula to calculate the frictional force is,
\(f=\mu N\)Substitute the known values,
\(\begin{gathered} 33\text{ N=}\mu(119\text{ N)} \\ \mu=\frac{33\text{ N}}{119\text{ N}} \\ \approx0.277 \end{gathered}\)Thus, the coefficient of friction of block is 0.277.
PLEASE HELP ME ANSWER THIS QUESTION.
Answers
When the segment is shrunk to one-third of its original length, the ratio of the final linear charge density (?f) to the initial linear charge density (?i) is f/?i = 3
This is because the charge is the same, but the length of the segment has decreased to one-third of its original length. Therefore, the charge density has increased by a factor of 3.
The ratio of the electric force on the proton after the segment is shrunk (Ff) to the force before the segment was shrunk (Fi) is also 3. This is because the electric force is directly proportional to the charge density, so if the charge density increases by a factor of 3, the electric force will also increase by a factor of 3.
If the original segment of wire is stretched to 15 times its original length, the charge density will decrease by a factor of 15. To keep the linear charge density unchanged, we need to add 14 times the original charge to the wire. This is because the original charge will be spread out over 15 times the original length, so we need to add 14 times the original charge to make up for the decrease in charge density.
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Kindly please tell me the answer to this question...
Answers
Following are the answers:
Pressure = \(force/area = 1.92 N/0.196 m^2 = 9.79 N/m^2\) or 9.79 PaDensity = Pressure/ (acceleration due to gravity x height) =\(9.79 N/m^2 / (9.8 m/s^2 x 0.92 m)\) = 1060 kg/m^3.What is the pressure and density?1. To calculate the pressure exerted by the water column on the surface of the mercury, we can use the formula:
Pressure = force/area
The force is the weight of the water column and the area is the cross-sectional area of the container.
The weight of the water column is given by the mass of the water times the acceleration due to gravity:
mass = density x volume
volume = area x height
So, mass = density x area x height = 1000 kg/m^3 x pi x (0.025 m)^2 x 0.25 m = 0.196 kg
Weight = mass x acceleration due to gravity =\(0.196 kg * 9.8 m/s^2\) = 1.92 N
The cross-sectional area of the container is pi x (0.025 m)^2 =\(0.196 m^2.\)
So, Pressure = force/area = \(1.92 N/0.196 m^2 = 9.79 N/m^2\) or 9.79 Pa
2. To calculate the density of the oil, we can use the formula:
density = mass/volume
Since the height of the oil column is 0.92 m and the cross-sectional area of the container is 0.196 m^2, the volume of the oil column is 0.196 m^2 x 0.92 m = 0.18012 m^3.
We do not know the mass of the oil, but we can calculate it using the pressure exerted by the oil column on the surface of the mercury:
Pressure = force/area = density x acceleration due to gravity x height
So, density = Pressure/ (acceleration due to gravity x height) = 9.79 N/m^2 / (9.8 m/s^2 x 0.92 m) = 1060 kg/m^3.
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Communication cannot be reversed
Answers
Communication cannot be reversed. This statement is clearly true. Because, we cannot take the words back and undo all the initial effects during communication.
What is reversibility ?An action that can be reversed back to its initial state is called reversible otherwise it is called irreversibility. Communication is irreversible and can be defined as a principle of interpersonal interaction in which we cannot take or retrieve information we say or pass to another party, whether it is what we intended to say or not.
We can wish we hadn't said something, regret it, and apologies later, but we can't take it back because once it's out, it's out.
We're only talking to make up for the unintended consequences of previous communication errors. We believe that taking more care with what we say in the first place can break this seemingly never-ending cycle.
Once a word, phrase, or comment is spoken, or an impulsive text message or e-mail is sent, it cannot be erased from the memory of others. Because communication is irreversible, one should always be mindful of what they communicate to others.
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Your question is incomplete. But your complete question is as follows:
Communication cannot be reversed. True/false ?
A 75.0 kg astronaut is training for accelerations that he will experience upon reentry. He is placed in a centrifuge (r = 20.0 m) and spun at a constant angular velocity of 15.0 rpm (revolutions per minute). He is then slowed and brought to a stop in 2.0 minutes.
Find the magnitude and direction of the centripetal acceleration and force when he is spinning at constant angular velocity.
How many g’s is the astronaut experiencing when moving at constant angular velocity?
Find the torque that is needed to bring the centrifuge to a stop knowing the centrifuge has a mass of 5500.0 kg (ignore all other forces) and the force is applied at the edge of the centrifuge (20.0 m radius). Hint: torque is based on the change of linear velocity.
Answers
a. The magnitude and direction of the centripetal acceleration and force when he is spinning at constant angular velocity is 8.72 m/s^2 and 654.0 N respectively.
b. The astronaut is experiencing 0.89 g when moving at constant angular velocity.
c. The torque that is needed to bring the centrifuge to a stop 6875 Nm.
What is angular velocity?
Angular velocity is described as a pseudovector representation of how fast the angular position or orientation of an object changes with time.
The magnitude of the centripetal acceleration and force, we will use the formula: a = v^2 / r, where v is the tangential velocity and r is the radius of the centrifuge.
a = (2pi20m15.02pi/60)^2 / 20m = 8.72 m/s^2
To calculate the force, we will use the formula
F_ = ma, where m is the mass of the astronaut, 75.0 k
F_ = 75.0 kg * 8.72 m/s^2 = 654.0 N
b. To calculate the number of g's the astronaut is experiencing when moving at constant angular velocity, we will divide the centripetal acceleration by the acceleration due to gravity, 9.8 m/s^2
8.72 m/s^2 / 9.8 m/s^2 = 0.89 g
c.
Torque = I * alpha, where I is the moment of inertia and alpha is the angular acceleration.
I = (1/2) * 5500.0 kg * 20.0m^2 = 55000 kgm^2
The angular acceleration can be found using the formula
Alpha = (change in angular velocity) / (change in time)
The change in angular velocity is 15.0 rpm - 0 rpm = 15.0 rpm and the change in time is 2.0 minutes = 120 seconds
alpha = 15.0 rpm / 120 s = 0.125 rad/s^2
Torque = 55000 kgm^2 * 0.125 rad/s^2 = 6875 Nm
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A 5 kg block is released from rest at the top of a quarter- circle type curved frictionless surface. The radius of the curvature is 3.8 m. When the block reaches the bottom o the curvature it then slides on a rough horizontal surface until it comes to rest. The coefficient of kinetic friction on the horizontal surface is 0.02.
a. What is the kinetic energy of the block at the bottom of the curved surface?
b. What is the speed of the block at the bottom of the curved surface?
c. Find the stopping distance of the block?
d. Find the elapsed time of the block while it is moving on the horizontal part of the track.
e. How much work is done by the friction force on the block on the horizontal part of the track?
Answers
Answer:
a. 186.2 J b. 8.63 m/s c. 190 m d. 43.2 s e. 186.2 J
Explanation:
a. From conservation of energy, the potential energy loss of block = kinetic energy gain of the block.
So, U + K = U' + K' where U = initial potential energy of block = mgh, K = initial kinetic energy of block = 0, U' = final potential energy of block at bottom of curve = 0 and K' = final kinetic energy of block at bottom of curve.
So, mgh + 0 = 0 + K'
K' = mgh where m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s², h = initial height above the ground of block = radius of curve = 3.8 m
So, K' = 5 kg × 9.8 m/s² × 3.8 m = 186.2 J
b. Since the kinetic energy of the block K = 1/2mv² where m = mass of block = 5 kg, v = velocity of block at bottom of curve
So, v = √(2K/m)
= √(2 × 186.2 J/5 kg)
= √(372.4 J/5 kg)
= √(74.48 J/kg)
= 8.63 m/s
c. To find the stopping distance, from work-kinetic energy principles,
work done by friction = kinetic energy change of block.
So ΔK = -fd where ΔK = K" - K' where K" = final kinetic energy = 0 J (since the block stops)and K' = initial kinetic energy = 186.2 J, f = frictional force = μmg where μ = coefficient of kinetic friction = 0.02, m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s² and d = stopping distance
ΔK = -fd
K" - K' = - μmgd
d = -(K" - K')/μmg
Substituting the values of the variables, we have
d = -(0 J - 186.2 J)/(0.02 × 5 kg × 9.8 m/s²)
d = -(- 186.2 J)/(0.98 kg m/s²)
d = 190 m
d. Using v² = u² + 2ad where u =initial speed of block = 8.63 m/s, v = final speed of block = 0 m/s (since it stops), a = acceleration of block and d = stopping distance = 190 m
So, a = (v² - u²)/2d
substituting the values of the variables, we have
a = (0² - (8.63 m/s)²)/(2 × 190 m)
a = -74.4769 m²/s²/380 m
a = -0.2 m/s²
Using v = u + at, we find the time t that elapsed while the block is moving on the horizontal track.
t = (v - u)/a
t =(0 m/s - 8.63 m/s)/-0.2 m/s²
t = - 8.63 m/s/-0.2 m/s²
t = 43.2 s
e. The work done by friction W = fd where
= μmgd where f = frictional force = μmg where μ = coefficient of kinetic friction = 0.02, m = mass of block = 5 kg, g = acceleration due to gravity = 9.8 m/s² and d = stopping distance = 190 m
W = 0.02 × 5 kg × 9.8 m/s² × 190 m
W = 186.2 J
The potential energy of the loss of the block will be equal to the kinetic energy gain. The kinetic energy of the block is 186.2 J at the bottom of the curved surface.
From the conservation of energy:
The potential energy of the loss of the block will be equal to the kinetic energy gain.
So,
\(U = mgh\)
Where,
\(U\) - potential energy
\(m\) - mass of block = 5 kg
\(g\) - gravitational acceleration = 9.8 m/s²
\(h\) = height = radius of curve = 3.8 m
Put the values in the formula,
\(U = 5 \times 9.8 \times 3.8 \\\\ U = 186.2 \rm \ J\)
Therefore, the kinetic energy of the block is 186.2 J at the bottom of the curved surface.
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HELP ME the mean free path λ and the mean collision time τ of the molecules of a diatomic gas of molecular mass 6.00 × 10⁻²⁵ kg and radius r = 1.0 x 10⁻¹⁰ m are measured. From these microscopic data can we obtain macroscopic properties such as temperature T and pressure P? If so, consider λ = 4.32 x 10⁻⁸ m and τ = 3.00 x 10⁻¹⁰ s and calculate T and P.
Answers
The temperature of the diatomic gas is 300.5 K and the pressure is 5.33 atm.
The given parameters;
Mass of the gas molecules, m = 6 x 10⁻²⁵ kgRadius of the gas, r = 1 x 10⁻¹⁰ mMean free path, \(\lambda_{rms}\) = 4.32 x 10⁻⁸ mMean collision time, \(\tau = 3 \times 10^{-10} \ s\)The mean velocity of the gas molecules is calculated as follows;
\(\tau = \frac{\lambda _{rms}}{V_{rms}} \\\\V_{rms} = \frac{\lambda _{rms}}{\tau} \\\\V_{rms} = \frac{4.32 \times 10^{-8} }{3 \times 10^{-10}} \\\\V_{rms} = 144 \ m/s\)
The temperature of the gas molecules is calculated as follows;
\(V_{rms} = \sqrt{\frac{3kT}{M} } \\\\V_{rms}^2 = \frac{3kT}{M} \\\\T = \frac{V_{rms} ^2 M}{3k}\)
where;
k is Boltzmann constant\(T = \frac{V_{rms} ^2 M}{3k} \\\\T = \frac{(144)^2 \times (6.0 \times 10^{-25})}{3 \times 1.38 \times 10^{-23}} \\\\T = 300.5 \ K\)
The number of gas molecules per unit volume is calculated as follows;
\(\lambda = \frac{1}{4\pi \sqrt{2} \ r^2 n} \\\\n = \frac{1}{\lambda 4\pi \sqrt{2} \ r^2} \\\\n = \frac{1}{(4.32 \times 10^{-8}) \times 4 \pi \times \sqrt{2} \ \times (1\times 10^{-10})^2} \\\\n = 1.303 \times 10^{26} \ molcules/m^3\)
The pressure of the gas molecule is calculated as follows;
\(n = \frac{P}{kT} \\\\P = nkT\\\\P = (1.303 \times 10^{26} ) \times (1.38 \times 10^{-23}) \times (300.5)\\\\P = 540,341.07 \ Pa\\\\P = 5.33 \ atm\)
Thus, the temperature of the diatomic gas is 300.5 K and the pressure is 5.33 atm.
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Điện tích Q = 8. 10-6C đặt cố định trong
không khí , điện tích q = - 10. 10-6C di
chuyển trên đường thẳng xuyên qua Q,
từ M cách Q một khoảng 100cm, lại
gần Q thêm 50cm. Tính công của lực
điện trường trong dịch chuyển đó?
Answers
Answer:
0.72J
Explanation:
A person pushes horizontally on a heavy box and slides it across the level floor at constant velocity. The person pushes with a 60.0 N force for the first 16.4 m at which time he begins to tire. The force he exerts then starts to decrease linearly from 60.0 N to 0.00 N across the remaining 6.88 m. How much total work did the person do on the box
Answers
Over the first 16.4 m, the person performs
W = (60.0 N) (16.4 m) = 984 J
of work.
Over the remaining 6.88 m, they perform a varying amount of work according to
F(x) ≈ 60.0 N + (-8.72 N/m) x
where x is in meters. (-8.72 is the slope of the line segment connecting the points (0, 60.0) and (6.88, 0).) The work done over this interval can be obtained by integrating F(x) over the interval [0, 6.88 m] :
W = ∫₀⁶˙⁸⁸ F(x) dx ≈ 206.4 J
(Alternatively, you can plot F(x) and see that it's a triangle with base 6.88 m and height 60.0 N, so the work done is the same, 1/2 (6.88 m) (60.0 N) = 206.4 J.)
So the total work performed by the person on the box is
984 J + 206.4 J = 1190.4 J ≈ 1190 J
Two loudsspeakers emit identical sound waves along the x axis. THe osund at a point on the axis has maximum intensity when the speakers are 40 cm apart. The sound intensity decreases as the distance between the speakers is increased, reaching zero at a separation of 50 cm. If the distance between the speakers continuse to increase, at what separation will the sound intensity again be a maximum?
Answers
Answer: 30 cm.
Solving
The situation described is that of two sources of sound waves that are separated by some distance. The two waves interfere with each other constructively at some points and destructively at others. When they interfere constructively, the amplitude (and intensity) of the sound wave is greater than when they interfere destructively.
When the speakers are 40 cm apart, the waves that they produce are in phase at some points on the axis, leading to constructive interference and a maximum in the intensity of the sound. As the distance between the speakers is increased beyond 40 cm, the points of constructive interference move farther apart, and the intensity of the sound decreases. When the speakers are 50 cm apart, the waves that they produce are exactly out of phase at some points on the axis, leading to complete destructive interference and a minimum in the intensity of the sound.
If the separation between the speakers continues to increase, the points of constructive interference will move closer together again, and the intensity of the sound will increase. The separation between the speakers at which the intensity of the sound will again be a maximum can be found using the following equation:
d = λ/2 + nλ
where d is the separation between the speakers, λ is the wavelength of the sound wave, and n is an integer that represents the number of half-wavelengths between the speakers.
At the maximum, the separation is an even multiple of half the wavelength, so we can use the formula above with n = 1. The wavelength can be found from the distance between the speakers at the minimum, which is 50 cm, and the distance at the maximum, which is 40 cm:
λ = 2(d_max - d_min) = 20 cm
Substituting λ and n into the formula gives:
d = λ/2 + nλ = 10 cm + 20 cm = 30 cm
Therefore, the sound intensity will be a maximum again when the separation between the speakers is 30 cm.
A metal bar has a volume of 32 cm3. The mass of the bar is 256 g. What is the density of the metal? A. 290 g/cm3 6 B. 8,200 g/cm C. 8.0 g/cm3 O D. 220 g/cm
Answers
The density of the metal is ρ = 8.0g/cm³.
Why is density important?The measure of material how densely it is packed together is called density. As the mass per unit volume, it has that definition. Symbol for density: D or Formula for Density: When is the density, m is the object's mass, and V is its volume, the equation is: = m/V.
Because it enables us to predict which compounds will float and which will sink in a liquid, density is a crucial notion. As long as an object's density is lower than the liquid's density, it will often float.
Equation :To the given equation we have :
mass of the bar = 256g
volume of metal bar = 32cm³
So according to the formula of density
ρ = m/V
So, putting values
ρ = 256g /32cm³
ρ = 8.0g/cm³
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