Answer:
K= 1000 N-m
Explanation:
It is assumed that we asked to find the spring constant k of the spring
We know that under equilibrium condition
weight of the object = force applied by the spring
given m =10 Kg
x= extension in the spring = 9.8 cm
mg=kx
[tex]10\times9.8=k\times9.8\times10^(-2)[/tex]
K= 1000 N-m
The spring constant, k of the vertically suspended spring measured in Newton meter is 1000 N/m
Given the Parameters :
Acceleration due to gravity, g = 9.8 m/s²Mass of object, m = 10 kg Extension, e = 9.8 centimetersUsing the Relation :
F = ke ; k = spring constant Force, F = mass × acceleration due to gravityThe expression can be written thus :
mg = ke
Converting, extension to meters = (9.8/100) = 0.098 m
(10 × 9.8) = 0.098k
98 = 0.098k
k = 98/0.098
k = 1000 N/m
Therefore, the spring constant is 1000 N/m
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In addition to stand-alone discussion boards, all of these sites include discussion boards as part of their features except:
a. LinkedIn.
b. Google.
c. Yahoo.
d. Microsoft Windows Live.
e. All of these sites include discussion boards.
Answer:
A. Linkedln
Explanation:
In addition to stand-alone discussion boards, all of these sites include discussion boards as part of their features except:
A. Linkedln
All other application in addition to stand-alone Google, Yahoo, Microsoft, etc. include discussion board.
While attempting to tune the note C at 523 Hz, a piano tuner hears 2.00 beats/s between a reference oscillator and the string. (a) What are the possible frequencies of the string? (b) When she tightens the string slightly, she hears 3.00 beats/s. What is the frequency of the string now? (c) By what percentage should the piano tuner now change the tension in the string to bring it into tune?
Answer
given,
frequency = 523 Hz
piano tuner hears = 2.00 beats/s
a) possible frequency = 523 ± 2
525 Hz and 521 Hz
b) On tightening piano wire, tension will be increased which will cause an increase of frequency of piano wire. Since no. of beats heard is also increased, the frequency of piano wire must be greater than the frequency of oscillator.
f p' = 5+ f₀
f p' = 3 + 523 Hz = 526 Hz.
c) f = k x √T
Let the tension at 523 Hz be T₀ and at 526 Hz it be T₁
523 = k x √T₀
T₀ = 523²/ k²
T₁ = 526²/ k²
% change in tension
= [tex]\dfrac{T_0- T_1}{T_1}[/tex]
= [tex]\dfrac{523^2- 526^2}{526^2}[/tex]
= 1.13 %
The inner and outer surfaces of a cell membrane carry a negative and a positive charge, respectively. Because of these charges, a potential difference of about 0.076 V exists across the membrane. The thickness of the cell membrane is 7.30 10-9 m.
What is the magnitude of the electric field in the membrane?
Answer:[tex]E=1.041\times 10^7 V/m[/tex]
Explanation:
Given
Potential Difference [tex]\Delta V=0.076 V[/tex]
thickness of cell membrane [tex]d=7.30\times 10^{-9} m[/tex]
Electric Field for this Potential is given by
[tex]E=\frac{\Delta V}{d}[/tex]
[tex]E=\frac{0.076}{7.30\times 10^{-9}}[/tex]
[tex]E=0.01041\times 10^9 V/m[/tex]
[tex]E=1.041\times 10^7 V/m[/tex]
Two train whistles have identical frequencies of 1.54 102 Hz. When one train is at rest in the station and the other is moving nearby, a commuter standing on the station platform hears beats with a frequency of 8.00 beats/s when the whistles operate together. What are the two possible speeds that the moving train can have? moving away from the station m/s moving towards the station m/s
Final answer:
The two possible speeds that the moving train can have are 343.292 m/s when moving away from the station and 355.182 m/s when moving towards the station.
Explanation:
To determine the two possible speeds that the moving train can have, we can use the formula for the beat frequency:
beat frequency = |v - v_s| / v * f_w
v is the speed of sound,
v_s is the speed of the moving train,
f_w is the frequency of the train whistle,
|v - v_s| is the absolute value of the difference between the speed of sound and the speed of the moving train.
Let's solve this equation for two possible speeds:
Case 1 - Moving away from the station:
8.00 beats/s = |331 m/s - v_s| / 331 m/s * 1.54 *[tex]10^2 Hz[/tex]
Solving for v_s, we find v_s = 343.292 m/s (rounded to three decimal places).
Case 2 - Moving towards the station:
8.00 beats/s = |331 m/s + v_s| / 331 m/s * 1.54 * [tex]10^2 Hz[/tex]
Solving for v_s, we find v_s = -355.182 m/s (rounded to three decimal places). Since the speed cannot be negative, we take the absolute value and find v_s = 355.182 m/s (rounded to three decimal places).
Therefore, the two possible speeds that the moving train can have are:
Moving away from the station: 343.292 m/s (rounded to three decimal places)
Moving towards the station: 355.182 m/s (rounded to three decimal places)
A military base a few miles from a medium- sized city is being decommissioned. The base is a large one, sited along a coastline, and includes some small hills, pristine woodlands and wetlands, streams, a river, and a small lake. There is a railroad spur that connects the base with the nearby city, as well as a four- lane highway. The base contains housing, schools, a hospital, shops, and recreational areas for a population of 10,000. The city's options for using, selling, or leasing the land are numerous, and many groups come forward with proposals. Members of several local environmental groups propose that nearly all of the buildings be removed and the area turned into a regional hiking and camping area, with very limited vehicle access. This will preserve the natural areas while providing recreation for the city dwellers. One issue that is raised by opponents is ________.
Answer:
The lack of jobs and lack of transportation to jobs
Explanation:
As it is seen, the establishment of the camping and hiking areas will dispatch buildings which were hospitals, shops and schools.
These buildings and facilities previously employed a number of workers and their removal hence, will reduce the number of available jobs.
Also, the new constructions will prove difficult in switching from one job to another.
energy expenditure due to physical activity generally accounts for __________ of total energy expenditure.
Answer:
Energy expenditure due to physical activity generally accounts for 20% of total energy expenditure.
Explanation:
The total amount of energy your body uses daily is usually divided as follows:
1. Your Basal Metabolic Rate at rest (50-70%)
Basal Metabolic Rate is the number of calories required to keep your body functioning at rest
2. Consumption by your daily physical activities (20%)
3. Energy to digest food. (10-20%)
This energy has to be deducted from the overall energy content of the food itself.
Then, the answer is:
Energy expenditure due to physical activity generally accounts for 20% of total energy expenditure.
Final answer:
The energy expenditure due to physical activity generally accounts for 10% of total energy expenditure. Physical activity guidelines recommend adults to engage in aerobic and muscle-strengthening activities for substantial health benefits.
Explanation:
The energy expenditure due to physical activity generally accounts for 10% of total energy expenditure.
According to the 2018 Physical Activity Guidelines for Americans issued by the Department of Health and Human Services, adults should do at least 150 minutes to 300 minutes per week of moderate-intensity aerobic activity, or 75 minutes to 150 minutes per week of vigorous-intensity aerobic physical activity, or an equivalent combination of both. Engaging in physical activity beyond 300 minutes per week can result in additional health benefits.
Therefore, the energy expenditure due to physical activity is a crucial component of total energy expenditure and plays a significant role in maintaining overall health.
An object is moving with the speed of light around the Earth. How much time will it take to complete one round trip along the equator on the surface of earth if radius of earth is 6400km
Answer:
0.14 seconds
Explanation:
The speed of light in vacuum is approximately 3.0*10^8. The distance that would be covered by the object would be equivalent to the circumference of the cross-section of the earth on the equator.
Circumference = 2[tex]\pi[/tex]*6400000 =4.02*10^7
Time = distance/speed = 4.2*10^7 / 3.0*10^8 =0.14s
Two cars travel westward along a straight highway, one at a constant velocity of 97 km/h, and the other at a constant velocity of 113 km/h. Assuming that both cars start at the same point, how much sooner does the faster car arrive at a destination 17 km away? Answer in units of h.
Answer:
0.025hr
Explanation:
The full solution is on the image below. The two cars cover the same distance at different time intervals. Since the distance is constant, the velocity is inversely proportional to the time taken to cover the constant distance
Answer:
0.025 h
Explanation:
Let's assume for the first car, the destination is [tex]x_{1}[/tex], the time is [tex]t_{1}[/tex], the velocity is [tex]v_{1}[/tex] and for the second car the destination is [tex]x_{2}[/tex], the time is [tex]t_{2}[/tex], the velocity is [tex]v_{2}[/tex].
We are given:
[tex]v_{1}[/tex] = 97 km/h
[tex]v_{2}[/tex] = 113 km/h
If we are asked the time, the destinations must be equal which are also given:
[tex]x_{1}[/tex] = [tex]x_{2}[/tex] = 17
For constant velocity, the equation is x = v * t
Hence,
[tex]x_{1}[/tex] = [tex]v_{1}[/tex] * [tex]t_{1}[/tex] = [tex]x_{1}[/tex] = 97 * [tex]t_{1}[/tex] = 17
⇒ [tex]t_{1}[/tex] = 17/97 = 0.175 h
[tex]x_{2}[/tex] = [tex]v_{2}[/tex] * [tex]t_{2}[/tex] = [tex]x_{2}[/tex] = 113 * [tex]t_{2}[/tex] = 17
⇒ [tex]t_{2}[/tex] = 17/113 = 0.150 h
So,
[tex]t_{1}[/tex] - [tex]t_{2}[/tex] = 0.175 - 0.150 = 0.025 h
The second car arrives 0.025 h sooner.
A particle of charge q is fixed at point P, and a second particle of mass m and the same charge q is initially held a distance r1 from P. The second particle is then released. Determine its speed when it is a distance r2 from P. Let q = 3.1 µC, m = 47 mg, r1 = 0.83 mm, and r2 = 2.5 mm.
_________m/s
Answer:
v=1617.77m/s
Explanation:
A particle of charge q is fixed at point P, and a second particle of mass m and the same charge q is initially held a distance r1 from P. The second particle is then released. Determine its speed when it is a distance r2 from P. Let q = 3.1 µC, m = 47 mg, r1 = 0.83 mm, and r2 = 2.5 mm.
_________m/s
from the law of applied electric force which states that the force of attraction of electric charge particles is directly proportional to the two charge particles and inversely proportional to the square of their distances apart
dk= -dU
kinetic energy equals potential energy
[tex]\frac{1}{2} mv^{2} -\frac{1}{2} mu^{2} =-(\frac{kq^2}{r2} -\frac{kq^2}{r1} )[/tex]
[tex]\frac{1}{2} 47*10^-6v^{2} -\frac{1}{2} 47*10^-6*0^{2} =-(\frac{9*10^9*(3.1*10^-6)^2}{(2.5*10^-3)} -\frac{9*10^-9*(3.1*10-6)^2}{0.83*10^-3} )[/tex]
23.5*10^-6v^2=96.1-34.596)
v^2=61.504/23.5*10^-6)
v^2=2617191.48
v=1617.77m/s
v=1.617km/s
A particle of charge q is fixed at point P, and a second particle of mass m and the same charge q is initially held a distance r1 from P. The second particle is then released. Determine its speed when it is a distance r2 from P. Let q = 3.1 µC, m = 47 mg, r1 = 0.83 mm, and r2 = 2.5 mm.
v=1617.77m/s.
Charge refers to a fundamental property of matter that determines how it interacts with electric and magnetic fields. It is one of the basic building blocks of nature and comes in two types: positive and negative.
The concept of charge is closely related to the phenomenon of electricity. Electric charge is responsible for the creation and interaction of electric fields, which can exert forces on other charged objects. Like charges repel each other, while opposite charges attract each other.
Protons, which carry a positive charge, have a charge of +e, while electrons, which carry a negative charge, have a charge of -e.
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A steel, rigid container is filled with an ideal gas. The gas inside is heated such that its temperature, as measured in kelvin, doubles. Choose the following:A. The pressure inside will rise by a factor of 4.A. The pressure inside will rise by a factor of 2. B. The pressure inside will drop by a factor of 4.C. The pressure inside will drop by a factor of 2.D. Nothing will happen to the pressure because this is an isobaric process.
Answer:
If the gas inside a rigid steel vessel (constant volume) is heated in such a way that its temperature, measured in Kelvin degrees, doubles then the internal pressure will increase by a factor of 2.
Explanation:
Gay-Lussac's law can be expressed mathematically as follows:
[tex]\frac{P}{T} =k[/tex]
where V = volume, T = temperature, K = Constant
This law indicates that the ratio between pressure and temperature is constant.
This law indicates that, as long as the volume of the container containing the gas is constant, as the temperature increases, the gas molecules move faster. Then the number of shocks against the walls increases, that is, the pressure increases. That is, the gas pressure is directly proportional to its temperature. This must be fulfilled since the relationship must remain constant
In short, when there is a constant volume, as the temperature increases, the gas pressure increases. And when the temperature decreases, gas pressure decreases.
It is desired to study two different states, an initial state and an final state. You have a gas that is at a pressure P1 and at a temperature T1 at the beginning of the experiment. When the temperature varies to a new T2 value, then the pressure will change to P2, and the following will be true:
[tex]\frac{P1}{T1} =\frac{P2}{T2}[/tex]
Given the above, it is possible to say that if the gas inside a rigid steel vessel (constant volume) is heated in such a way that its temperature, measured in Kelvin degrees, doubles then the internal pressure will increase by a factor of 2 (doubles too)
A 2.5-kg book slides horizontally and falls from a shelf 3.0 m above the floor. How much work does the force of gravity do on the book as it falls 3.0 m?
Answer:
73.5 N
Explanation:
W = PE = mgh = 2.5 * 9.8 * 3.0 = 73.5 N
Final answer:
The work done by gravity on a 2.5-kg book falling from a 3.0 m height is calculated to be 73.5 joules, using the formula for work in the presence of gravitational force.
Explanation:
To calculate the work done by gravity on the 2.5-kg book as it falls 3.0 m, you can use the formula work (W) = force (F) × distance (d). Since we're dealing with gravity, the force is the weight of the book, which is its mass multiplied by the acceleration due to gravity (9.8 m/s²). Therefore, the force is 2.5 kg × 9.8 m/s². The distance the book falls is given as 3.0 m.
The formula simplifies to W = m × g × d, which gives us W = 2.5 kg × 9.8 m/s² × 3.0 m, resulting in 73.5 J. The work done by gravity on the book as it falls to the ground is 73.5 joules.
A large storage tank open at the top and filled with water develops a small hole in its side at a point 16.0 m below the water level. The rate of flow from the leak is found to be 2.50 × 10^(-3) m^3/min.
Determine; (a) the speed (m/s) at which the water leaves the hole and (b) the diameter of the hole (mm).
Final answer:
The speed at which the water leaves the hole is 17.7 m/s, and the diameter of the hole is approximately 4.8 mm using Torricelli's theorem and the flow rate equation.
Explanation:
The scenario described involves principles of fluid dynamics within the field of physics, specifically Bernoulli's equation and the equation of continuity. To calculate the speed at which the water leaves the hole (a) and the diameter of the hole (b), we employ the Torricelli's theorem and the flow rate equation.
Using Torricelli's theorem, the speed (v) of water exiting the hole can be determined by the formula v = √(2gh), where g is the acceleration due to gravity (9.8 m/s²) and h is the height of the water column above the hole (16.0 m).
v = √(2 * 9.8 m/s² * 16.0 m) = √(313.6 m²/s²) = 17.7 m/s
To find the diameter of the hole, we use the flow rate Q = A * v, where Q is the flow rate (2.50 × 10³ m³/min), A is the area of the hole, and v is the speed of water leaving the hole. First, we convert the flow rate to m³/s by dividing by 60 (since there are 60 seconds in a minute):
Q (in m³/s) = (2.50 × 10³ m³/min) / 60 = 4.17 × 10µ m³/s
To find the area: A = Q / v, then to find the diameter (d), we use the area of a circle A = π * (d/2)². Rearranging and solving for d gives us:
d = 2 * √(Q / (π * v))
d = 2 * √((4.17 × 10µ m³/s) / (π * 17.7 m/s)) = 4.8 mm
(a) The speed at which the water leaves the hole is approximately 17.7m/s.
(b) The diameter of the hole is approximately 1.74mm.
Given:
- The depth of the hole below the water level, [tex]\( h = 16.0 \)[/tex] m
- The rate of flow, [tex]\( Q = 2.50 \times 10^{-3} \) m\(^3\)/[/tex]min
Part (a): Speed of Water Leaving the Hole
We can use Torricelli's Law to determine the speed at which the water leaves the hole:
[tex]\[ v = \sqrt{2gh} \][/tex]
where:
- [tex]\( g \)[/tex] is the acceleration due to gravity[tex](\( 9.8 \) m/s\(^2\))[/tex]
- [tex]\( h \)[/tex] is the height of the water above the hole
Plugging in the values:
[tex]\[ v = \sqrt{2 \times 9.8 \times 16.0} \][/tex]
[tex]\[ v = \sqrt{2 \times 9.8 \times 16.0} \][/tex]
[tex]\[ v = \sqrt{313.6} \][/tex]
[tex]\[ v \approx 17.7 \text{ m/s} \][/tex]
Part (b): Diameter of the Hole
First, convert the flow rate to cubic meters per second:
[tex]\[ Q = 2.50 \times 10^{-3} \text{ m}^3/\text{min} \][/tex]
[tex]\[ Q = \frac{2.50 \times 10^{-3}}{60} \text{ m}^3/\text{s} \][/tex]
[tex]\[ Q = 4.17 \times 10^{-5} \text{ m}^3/\text{s} \][/tex]
The flow rate ( Q ) can also be expressed as:
[tex]\[ Q = A \cdot v \][/tex]
where:
- ( A ) is the cross-sectional area of the hole
- ( v ) is the speed of the water leaving the hole
Rearranging to solve for ( A ):
[tex]\[ A = \frac{Q}{v} \][/tex]
[tex]\[ A = \frac{4.17 \times 10^{-5}}{17.7} \][/tex]
[tex]\[ A \approx 2.36 \times 10^{-6} \text{ m}^2 \][/tex]
The area ( A) of a circular hole is given by:
[tex]\[ A = \pi \left( \frac{d}{2} \right)^2 \][/tex]
Solving for the diameter[tex]\( d \)[/tex]:
[tex]\[ d = 2 \sqrt{\frac{A}{\pi}} \][/tex]
[tex]\[ d = 2 \sqrt{\frac{2.36 \times 10^{-6}}{\pi}} \][/tex]
[tex]\[ d = 2 \sqrt{7.51 \times 10^{-7}} \][/tex]
[tex]\[ d \approx 1.74 \times 10^{-3} \text{ m} \][/tex]
[tex]\[ d \approx 1.74 \text{ mm} \][/tex]
A person with which of the following BMI values has the highest increased risk for premature mortality?
a.29 kg/m2
b.35 kg/m2
c.25 kg/m2
d.18.5 kg/m2
Answer:
b.
Explanation:
BMI is body mass index, it is calculated by dividing the weight of a person in kilogram by square of height in meters.
A high BMI is an indicator of obesity. Obese people are prone to death causing diseases like High Blood pressure, cardiac arrest and diabatese. So, a high value of BMI values has the highest increased risk for premature mortality.
So the correct answer is 35 kg/m^2.
Because Earth is rotating, horizontal winds everywhere except at the equator are deflected to the right or left relative to Earth's surface. This so-called Coriolis Effect arises because the air is moving over a surface which itself is continually turning because of Earth's rotation. In the Northern Hemisphere, the Coriolis Effect causes wind blowing toward the west to turn toward the___________.
Answer:
Because Earth is rotating, horizontal winds everywhere except at the equator are deflected to the right or left relative to Earth's surface. This so-called Coriolis Effect arises because the air is moving over a surface which itself is continually turning because of Earth's rotation. In the Northern Hemisphere, the Coriolis Effect causes wind blowing toward the west to turn toward the North Western.
What is the farthest distance at which a typical "nearsighted" frog can see clearly in air?
A. 12m
B. 80 m
C. 6.0m
D.17m
Answer: the correct option is D (17m).
Explanation: The farthest distance at which a typical "nearsighted" frog can see clearly in air is 17m.
A pumpkin with a mass of 3.3 kg is launched from a catapult at an initial height of 3.8 m off the ground, with an initial speed of 13.9 m/s. What height is the pumpkin at when the speed is half its initial speed? Round your answer to 2 decimal places.
Answer:
11.2m
Explanation:
Suppose the pumpkin is launched vertically and the speed of 13.9 m/s is absolutely vertical. As the pumpkin rises up, its kinetic energy is converted to potential energy, let the reference point be the round, we can create the following equation from the law of energy conservation:
[tex] E_1 + P_1 = E_2 + P_2[/tex]
[tex] 0.5mv_1^2 + mgh_1 = 0.5mv_2^2 + mgh_2[/tex]
we can substitute [tex]g = 9.8m/s^2, v_1 = 13.9m/s, v_2 = 0.5v_1 = 0.5*13.9 = 6.95 m/s, h_1 = 3.8[/tex]
We can also divide both sides by m and 0.5
[tex]13.9^2 + 2*9.8*3.8 = 6.95^2 + 2*9.8*h_2[/tex]
[tex]19.6h_2 = 219.3875[/tex]
[tex]h_2 = 11.2m[/tex]
Strange as it may seem, it is as hard to accelerate a car on a level surface on the moon as it is here on the Earth. This is because?1.the weight of the car is independent of gravity.2.the mass of the car is independent of gravity.3.Nonsense! A car is much more easily accelerated on the moon than on the Earth.
Answer:
2.the mass of the car is independent of gravity.
Explanation:
When mass and acceleration is multiplied we get the force.
[tex]F=ma\\\Rightarrow a=\dfrac{F}{m}[/tex]
Mass of an object is always constant anywhere in the universe. Acceleration of the car would be the ratio of the force and mass. As the mass is constant force the same force applied the acceleration is the same on Earth or Moon.
Hence, the mass of the car is independent of gravity.
An 8.31-m, 267-N uniform ladder rests against a smooth (frictionless) wall. The coefficient of static friction between the ladder and the ground is 0.582, and the ladder makes a 52.6? angle with the ground. A 928 N person is standing on the ladder a distance d from the bottom end of the ladder.
(a) Draw a free-body diagram of the ladder. (Turn in part (a) on the sheet provided in your packet.)
(b) How far up the ladder (distance d) can the person climb before the ladder begins to slip? Hint: Among other things, you may want to look at the sum of the torques about the point where the ladder touches the wall. Be careful with angles! ___________ m ( ± 0.02 m)
To determine how far up the ladder a person can climb before the ladder begins to slip, we need to consider the forces and torques acting on the ladder. The condition for the ladder to begin slipping is when the torque due to the person's weight is greater than the torque due to the static friction force.
Explanation:To determine how far up the ladder a person can climb before the ladder begins to slip, we need to consider the forces and torques acting on the ladder. The forces include the normal reaction force from the floor, the static friction force between the ladder and the floor, the weight of the ladder, and the normal reaction force from the wall. The torques are calculated by multiplying the force by the lever arm, which in this case is the distance between the center of mass of the ladder and the point of contact with the floor.
The condition for the ladder to begin slipping is when the torque due to the person's weight is greater than the torque due to the static friction force. We can calculate the torque due to the person's weight by multiplying the person's weight by the distance between their position on the ladder and the point of contact with the floor. Next, we calculate the torque due to the static friction force by multiplying the static friction force by the distance between the point of contact with the floor and the point of contact with the wall.
Setting these two torques equal to each other and solving for the distance d yields the distance up the ladder that the person can climb before it begins to slip.
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One light-hour is the distance that light travels in an hour. How far is this, in kilometers? (Recall that the speed of light is 300,000 km/s.) _______A) 18 million kmB) 1.08 billion kmC) 300,000 kmD) 9.46 trillion kmE) 100 million km
Answer:
B 1.08 BILLION
Explanation:
SEE ATTACHMENT
600 J of work is done on a system in a process that decreases the thermal energy of the system by 300 J.
How much heat energy is transferred to or from the system as heat?
Write positive value if the energy is transferred to the system and negative if the energy is transferred from the system.
Answer:
900 J
Explanation:
[tex]\Delta U[/tex] = Change in entropy = -300 J (Decrease)
[tex]W[/tex] = Work done = 600 J
[tex]Q[/tex] = Heat transferred
Change in internal energy is given by
[tex]\Delta U=W+Q\\\Rightarrow Q=\Delta U-W\\\Rightarrow Q=-300-600\\\Rightarrow Q=-900\ J[/tex]
The heat transferred to or from the system as heat is -900 J.
900 J is transferred. As heat is transferred from the system the sign is negative.
When 600 J of work is done on a system and the thermal energy of the system decreases by 300 J, the heat energy transferred to or from the system can be calculated using the First Law of Thermodynamics. In this case, -900 J of heat is transferred from the system.
Explanation:When 600 J of work is done on a system and the thermal energy of the system decreases by 300 J, the change in internal energy of the system can be calculated using the First Law of Thermodynamics.
The First Law of Thermodynamics states that the change in internal energy of a system is equal to the heat added to or removed from the system plus the work done on or by the system. Mathematically, it can be expressed as:
ΔU = Q + W
Where ΔU represents the change in internal energy, Q represents the heat added to or removed from the system, and W represents the work done on or by the system.
In this case, we know that the work done on the system is 600 J, and the change in thermal energy is -300 J (since it decreases).Substituting these values into the equation, we get:ΔU = Q + 600 J
-300 J = Q + 600 J
Now we can solve for Q, which represents the heat transferred to or from the system.Rearranging the equation:Q = -300 J - 600 J
Q = -900 J
The negative sign indicates that the heat is transferred from the system.
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How does using a ramp make work easier when loading a piano onto a truck three feet above the ground?A. It allows a larger force to be applied over a shorter distance.B. It allows a larger force to be applied over a longer distance.C. It allows a smaller force to be applied over a shorter distance.D. It allows a smaller force to be applied over a longer distance.
Answer:
D) True. The value of the work is constant and the force is less, therefore it is easier to raise the piano
Explanation:
In this exercise the work to raise the piano is given by
W = F d cos θ
The distance and the displacement have an angle of zero degrees
W = F d
If we use energy conservation
W = ΔEm
W = [tex]Em_{f}[/tex] - Em₀
Suppose the piano starts from rest and reaches the top also at rest
W = mh y₂ - mg y₁
W = mg h
h= 3 ft
We see that the work to raise the body is always the same, regardless of the path, so, in the work relationship or is the product of force for distance if one goes up the other must decrease so that the product is the same.
Let's examine the answers
A) False. The distance on a ramp is greater
B) False. Change the value of work
C) False. Change the value of work
D) True. The value of the work is constant and the force is less, therefore it is easier to raise the piano
A ball is thrown straight up at 20 m. What is the balls velocity as it hits the ground?
Answer:
The velocity of the ball as it hit the ground = 19.799 m/s
Explanation:
Velocity: Velocity of a body can be defined as the rate of change of displacement of the body. The S.I unit of velocity is m/s. velocity is expressed in one of newtons equation of motion, and is given below.
v² = u² + 2gs.......................... Equation 1
Where v = the final velocity of the ball, g = acceleration due to gravity, s = the height of the ball
Given: s = 20 m, u = 0 m/s
Constant: g = 9.8 m/s²
Substituting these values into equation 1,
v² = 0 + 2×9.8×20
v² = 392
v = √392
v = 19.799 m/s.
Therefore the velocity of the ball as it hit the ground = 19.799 m/s
You and a friend are riding on a ferris wheel, which is moving with constant velocity. You, as a seeker of deep understanding, are sitting on a bathroom scale, wondering what will happen. Your friend observes 3 bathroom scale readings, one at the top, Rt , one at the middle, Rm, and one at the bottom, Rb .
Which of the following is the correct relationship between Rb , Rt , Rm? top middle bottom
1. Rt = Rb < Rm
2. Rt > Rm > Rb
3. Rt = Rb = Rm
4. Not enough information
The correct relationship between the bathroom scale readings at the top, middle, and bottom of a ferris wheel ride is that they are all the same.
Explanation:When you and your friend are riding on a ferris wheel moving with constant velocity, the bathroom scale readings at the top, middle, and bottom of the ride can be compared. Since the ferris wheel is moving at a constant velocity, there is no acceleration and thus no net force acting on you. This means that your apparent weight will be the same at all three points on the ride. Therefore, the correct relationship is option 3: Rt = Rb = Rm.
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To test the resiliency of its bumper during low-speed collisions, a 3 010-kg automobile is driven into a brick wall. The car's bumper behaves like a spring with a force constant 6.00 106 N/m and compresses 3.26 cm as the car is brought to rest. What was the speed of the car before impact, assuming no mechanical energy is transformed or transferred away during impact with the wall? m/s
Answer:
1.45549 m/s
Explanation:
m = Mass of car = 3010 kg
v = Velocity of car
k = Spring constant = [tex]6\times 10^6\ N/m[/tex]
x = Displacement of spring = 3.26 cm
As the energy of the system is conserved
[tex]\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2\\\Rightarrow v=\sqrt{\dfrac{kx^2}{m}}\\\Rightarrow v=\sqrt{\dfrac{6\times 10^6\times 0.0326^2}{3010}}\\\Rightarrow v=1.45549\ m/s[/tex]
The speed of the car before impact is 1.45549 m/s
In a client with lower crossed syndrome, which of the following muscles is lengthened?
Answer:
Internal Oblique.
Explanation:
Lower crossed syndrome is a condition in which there are strong and weak muscles. So there is an imbalance of muscle strengths. It occurs when some muscles constanly get shortened or lengthened just like in this case internal oblique muscle got lengthened.
All of the orbitals in a given subshell have the same value as the ________ quantum number.
A) principal
B) spin
C) magnetic
D) A and B
E) B and C
Answer:
The given subshell have the same value as the spin quantum number and principal quantum number. Option D is correct.
Explanation:
Every shell has some orbitals. For example, 1st shell has only s-orbital, 2nd shell has s and p-orbitals, 3rd shell has s, p and d-orbitals, 4th shell has s, p, d and f-orbitals.
Now, every orbital has a fix number of subshells. s-orbital has 1 subshell, p-orbital has 3 subshells, d-orbital has 5 subshells, and f-orbital has 7 subshells.
Every subshell of the orbital has same principal quantum number because it is associated with the same shell. Magnetic quantum of every subshell of any orbital is different as it specifies the subshells. And the spin quantum number can be [tex]+1/2 \texttt{ or }-1/2[/tex] based on the spin of the electrons. Every subshell contains two electrons with both the opposite spins.
Let's take an example of 4f-orbital.
Here, the principal quantum number for all the subshells will be 4. The magnetic quantum number will vary from -3 to +3. Every subshell can has two electrons of spin quantum number [tex]+1/2 \texttt{ or }-1/2[/tex].
Therefore, a given subshell have the same value as the spin quantum number and principal quantum number. Option D is correct.
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Experimental Inquiry: What factors influence the loss of nutrients from a forest ecosystem?
A glass window is coated with a transparent film of refractive index n2 =1.25. This film will cause destructive interference in the reflected light when light of wavelength λ = 550nm is incident normally on the window. What is the minimum thickness of film required to produce this anti-reflection coating?
Answer:
0.00000011 m
Explanation:
n = 0 for minimum thickness
t = Thickness
[tex]n_2[/tex] = Refractive index of the film = 1.25
[tex]\lambda[/tex] = Wavelength = [tex]550\times 10^{-9}\ m[/tex]
We have the formula
[tex]2n_2t=(2n+1)\dfrac{\lambda}{2}[/tex]
[tex]\\\Rightarrow t=(2n+1)\dfrac{\lambda}{4n_2}\\\Rightarrow t=(2\times 0+1)\dfrac{550\times 10^{-9}}{4\times 1.25}\\\Rightarrow t=0.00000011\ m[/tex]
The minimum thickness of the film is 0.00000011 m
For all turning vehicles, the rear wheels follow a ______ than the front wheels.
a. slower pathb. faster pathc. longer pathd. shorter path
Answer:
Shorter path
Explanation:
For all turning vehicles, the rear wheels follow Shorter path than the front wheels.
Any turning vehicle, the rear(the back part of something, especially a vehicle.) wheels follow a shorter path than the front wheels. The longer the vehicle is, the greater the difference will be in path. Trucks initially swing out before making a turn
An electron is accelerated from rest by a potential differ- ence of 350 V. It then enters a uniform magnetic field of magni- tude 200 mT with its velocity perpendicular to the field. Calculate (a) the speed of the electron and (b) the radius of its path in the magnetic field.
Answer:
a) the speed of the electron is 1.11 × 10⁷ m/s
b) the radius of electron's path in the magnetic field is 3.16 × 10⁻⁴ m
Explanation:
a) Let's assume that we have an electron accelerated using a potential difference of V = 350, which gives the ion a speed of v. In order to find this speed we set the potential energy of the electron equal to its kinetic energy. Thus,
eV = 1/2 m v²
where
e is the charge of the electronm is the mass of the electronv is the speed of the electronThus,
v = √[2eV / m]
v = √[2(1.6 × 10⁻¹⁹ C)(350 V) / 9.11 × 10⁻³¹ kg]
v = 1.11 × 10⁷ m/s
Therefore, the speed of the electron is 1.11 × 10⁷ m/s
b) Then the electron enters a region of uniform magnetic field, it moves in a circular path with a radius of:
r = mv / eB
where
m is the mass of the electronv is the speed of the electrone is the charge of the electronB is the magnetic fieldThus,
r = (9.11 × 10⁻³¹ kg)(1.11 × 10⁷ m/s) / (1.6 × 10⁻¹⁹ C)(200 × 10⁻³ T)
r = 3.16 × 10⁻⁴ m
Therefore, the radius of electron's path in the magnetic field is 3.16 × 10⁻⁴ m