why do real gases behave nonideally at very low temperatures and very high pressures?
Answers
Real gases behave non-ideally at very low temperatures and very high pressures because the ideal gas equation doesn't consider the volume of gas molecules and intermolecular forces.
At very high pressures, the volume of the gas molecules becomes significant and, thus, can't be ignored by the ideal gas equation. At very low temperatures, the intermolecular forces are very significant, causing the molecules to stick together and making it difficult for them to move freely. This causes the gas to deviate from the ideal gas law.Real gases are different from ideal gases in many ways. They are not completely empty, and they have some mass. The pressure exerted by real gases is not always constant because the volume of their particles is not negligible. Real gases also have intermolecular forces that can affect their behavior. Real gases' behavior can be predicted using the van der Waals equation. This equation accounts for the volume of the gas molecules and the attractive forces between them. It is given by:
(P + n²a/V²) (V-nb) = nRT,
where P is the pressure, V is the volume, T is the temperature, n is the number of moles of gas, R is the gas constant, a and b are the van der Waals constants, which depend on the gas.
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Related Questions
A ball is thrown straight up with a speed of 30
m/s. What is the maximum height reached by
the ball?
Answers
Answer:
A ball is thrown straight up with a speed of 30
m/s. What is the maximum height reached by
the ball?
An observer recorded the following data for the motion of a car undergoing constant acceleration.
Time (s)
3.0
5.0
6.0
Speed (m/s)
4.0
7.0
8.5
What was the magnitude of the acceleration of the car?
4.5 m/s2
O 2.0 m/s2
3.0 m/s2
1.5 m/s2
Answers
Option D) 1.5 m/s² is the correct answer.
The magnitude of the acceleration of the car is 1.5 m/s².
given the data in the question;
record of the observation;
Time(s) Speed(m/s)
3.0 4.0
5.0 7.0
6.0 8.5
Since its undergoing constant acceleration, we can make use any two from the three data recorded.
So, lets consider motion between t₁ = 3.0s and t₂ = 5.0s
Initial velocity at t₁ ; u = 4.0 m/s
final velocity at t₂; v = 7.0 m/s
Now, Time Elapsed; \(t = t_2 - t_1 = 7.0s - 5.0s = 2.0s\)
To determine the magnitude of acceleration "a", we use the first equation of motion which says;
\(v = u + at\)
we make "a" the subject of the formula
\(a =\frac{v - u}{t}\)
so we substitute in our values;
\(a = \frac{(7.0m/s) - (4.0m/s)}{2.0s}\)
\(a = \frac{3.0m/s}{2.0s}\)
\(a = 1.5 m/s^2\)
Therefore, the magnitude of the acceleration of the car is 1.5 m/s².
Option D) 1.5 m/s² is the correct answer.
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Give two ways humans can reduce the amount of carbon at the location from #3. Give one example that will reduce carbon entering this location and one example that will increase carbon exiting this location. Explain how this reduces the carbon.
Answers
1. Reducing Carbon Entering: Planting more trees at the location is an effective way to reduce the amount of carbon at this location.
What is Carbon?Carbon is an element found in abundance in the Earth's atmosphere and is one of the most abundant elements in the universe. Carbon is a key component of all living organisms and serves as the backbone of life on Earth. Carbon exists in several different forms, including coal, oil, natural gas, and diamonds, and can be found in many other materials, such as plastics, paper, and fabrics.
Trees absorb carbon dioxide from the atmosphere and use it to produce oxygen through photosynthesis. This process helps to reduce the amount of carbon entering the location.
2. Increasing Carbon Exiting: Installing solar panels at the location is an effective way to increase the amount of carbon exiting the location. Solar panels convert solar energy into electricity, releasing carbon dioxide into the atmosphere during the process. This helps to reduce the amount of carbon at the location by releasing it back into the atmosphere.
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What are three other types of new batteries that are being created today? (PLEASE DO NOT GIVE ME ANY LINK, JUST TYPE YOU ANSWER)
Answers
Answer:
Duracell
Amazon Basics 2 Pack CR1616 3 Volt Lithium Coin Cell Battery.
LiCB 20 Pack LR44 AG13 357 303 SR44 Battery 1.5V Button Coin Cell Batteries.
Explanation:
Hope this helps and have a nice day!
A sample of oxygen gas occupies a volume of 200 mL at a pressure of 720 torr. What volume will the gas occupy at a pressure of 800 torr if temperature is held constant
Answers
We have that the gas will occupy at a pressure of 800 torr if temperature is held constant a Volume of
\(V_2=0.18L\)
From the Question we are told that
Initial Pressure \(P_1=720\)
final Volume \(V_1=200mL\)
Initial Pressure \(P_2=800\)
final Volume \(V_2=?\)
Generally the equation for Boyle's law is mathematically given as
\(P_1V_1= P_2V_2\)
Therefore
\(V_2=\frac{ P_1V_1}{P_2}\\\\V_2=\frac{ 720*200}{800}\\\\V_2=180mL\)
\(V_2=0.18L\)
The gas will occupy at a pressure of 800 torr if temperature is held constant a Volume of
\(V_2=0.18L\)
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times more energy than an earthquake with a magnituide of 2. Ori the Richter scale a magnitude 5 earthquake releases about
A. 30
B. 1000
C. 90
D. 27.000
What is the main caase of the spreading movement of the ocean crust
A. The convection current of the asthenosphere is drasging the ocean crust.
B. The riew magma forming at ocean ridges is pushing the ocean krure
C. There is a stearing motion on translorm plate boundaries
D. All of the above
consribute equally Respr Selestion
A. Increasing temperatures at the top of the asthenosphere
B. There is a greater water content in subducted oceanic crust
c. Decreasing pressure as the overlying crust thins
D. Higher viscosity as the silica content increases E. All of the above Reset Selection
Answers
The main cause of the spreading movement of the ocean crust is the new magma forming at ocean ridges, which pushes the oceanic plates apart. The other options listed do not contribute equally or directly to the spreading movement.
The spreading movement of the ocean crust is primarily driven by the formation of new magma at ocean ridges. As the mantle material beneath the Earth's surface rises and melts, it creates new magma. This molten material then pushes its way upward and fills the gap between the separating oceanic plates. As the new magma solidifies, it forms a new crust, expanding the ocean floor and causing the plates to move apart.
The other options listed do not directly contribute to the spreading movement of the ocean crust. Increasing temperatures at the top of the asthenosphere, the greater water content in subducted oceanic crust, decreasing pressure as the overlying crust thins, and higher viscosity as the silica content increases may have their own geological implications, but they are not the main causes of the spreading movement observed at ocean ridges.
Therefore, the correct answer is B. The new magma forming at ocean ridges is the main cause of the spreading movement of the ocean crust.
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The complete question is:
What is the main cause of the spreading movement of the ocean crust? A. As the ocean crust cools it becomes denser than the asthenosphere and the edge of the slab is pulled downward at subduction zones. B. The convection current of the asthenosphere is dragging the ocean crust C. The new magma forming at ocean ridges is pushing the ocean crust D. There is a shearing motion on transform plate boundaries E. All of the above contribute equally
Can the magnitude of a vector ever (a) be equal to one of its components, or (b) be less than one of its components? 9. Can a particle with constant speed be accelerating? What if it has constant velocity?
Answers
Answer:
a) the other components are zero, in the direction of one of the coordinate axes
b) the magnitude is less than the value of one of its components, it must occur when the vector is in some arbitrary direction
9) constant velocity the acceleration must necessarily be zero,
constant speed can be accelerated since it may be changing the direction of the velocity vector
Explanation:
Vectors are quantities that have modulo (scalar) direction and sense.
a) If in a vector its magnitude is equal to one d its components implies that the other components are zero, therefore the vector must be in the direction of one of the coordinate axes
b) if the magnitude is less than the value of one of its components, it must occur when the vector is in some arbitrary direction, other than the direction of the axes, that is
R² = x² + y²
where R is the magnitude of the vector e x, and are the components
9) When a particle has a constant velocity, the acceleration must necessarily be zero,
v = vo + a t
The bold letters indicate vectors If a = 0 implies that v = vo
If a particle has constant speed it can be accelerated since it may be changing the direction of the velocity vector, this type of acceleration has the name of centripetal acceleration
Determine the force of gravitational attraction between the earth (m = 5.98 x 10^24kg)
and a 70-kg physics student, where the student is at a distance of 6.39 x 10^6 m from earth's
center.
Answers
Answer:
F = 683.8 N
Explanation:
The gravitational force of attraction between the Earth and the student is given by Newton's Law of Gravitation as follows:
\(F = \frac{Gm_{1}m_{2}}{r^2}\)
where,
F = Force = ?
G = Universal gravitational constant = 6.67 x 10⁻¹¹ Nm²/kg²
m₁ = mass of Earth = 5.98 x 10²⁴ kg
m₂ = mass of student = 70 kg
r = distance between Earth and student = 6.39 x 10⁶ m
Therefore,
\(F = \frac{(6.67\ x\ 10^{-11}\ Nm^2/kg^1)(5.98\ x\ 10^{24}\ kg)(70\ kg)}{(6.39\ x\ 10^6\ m)^2}\\\)
F = 683.8 N
Plz help me im really confused. The questions are in the second pic and the vehicles are in the first the fetures are in the last 3 the check mars are just to see which data for that vehicle is safer. The check marks are just to see which data is safer for that car.
Answers
Answer:
1.ima say car 2 is safer than car 1.
2.airbags: airbags make it safer because if you get in a reck they will protect you againsed more harm then if you didnt have airbags.
breaking distance: it takes less distance to stop so if there was a reck up the streat you could stop and turn around so you wouldnt get into that wreck.
accrleration: it can make you go faster so incase someoe behind you is speading you could speadup and turn into a parking lot and you wouldent get hurt.
3. car 2 is safer because it has breaking distance and it can stop you way before you get into a wreck than car 1 that doesnt have that ability
Explanation: ima award myself brainliest. Dont ask why i just answerd my own question on here. Yay
Which statement best describes what would happen if the current in the coil of an electromagnet were increased?
A. The electromagnet would stop working until the current became steady
B. The magnetic field would not change
C. The magnetic field would decrease
D. The magnetic field would increase
Answers
Answer:D. The magnetic field would increase.
Explanation:
X The figure below represents a standing wave on a string. Find the frequency if the length L of the string is 5 m and the speed is 20 m/s
Answers
Hence frequency of the standing wave is 2 s⁻¹
What are standing waves ?Standing waves are formed when 2 waves of same wavelength and amplitude travelling in the opposite direction are superposed. Standing waves do not transport energy.
Here,
Length of the string, L = 5 m
Speed of the wave, v = 20 m/s
From the figure we can see that it is a string having two nodes and only one antinode. That means it is the first harmonic.
For first harmonic, L = λ/2
So, wavelength, λ = 2L = 2x 5
λ = 10 m
Therefore frequency, f = v/λ
f = 20/10
f = 2 s⁻¹
Hence frequency of the standing wave is 2 s⁻¹
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Your question was incomplete. Attaching the image here.
what end of the electromagnetic spectrum is more likely to exhibit wave characteristics
Answers
The end of the electromagnetic spectrum more likely to exhibit wave characteristics is the radio wave region.
The electromagnetic spectrum spans from low-energy radio waves to high-energy gamma rays. The wave-like behavior of electromagnetic radiation is determined by its wavelength and frequency. The wavelength (λ) and frequency (ν) of a wave are related by the equation c = λν, where c is the speed of light in a vacuum (approximately 3.00 × 10^8 meters per second).
Radio waves have the longest wavelengths and lowest frequencies in the electromagnetic spectrum, typically ranging from a few millimeters to hundreds of kilometers. Due to their long wavelengths, radio waves are more likely to exhibit wave characteristics such as diffraction and interference. These characteristics allow radio waves to bend around obstacles and interfere constructively or destructively.
In conclusion, the end of the electromagnetic spectrum that is more likely to exhibit wave characteristics is the radio wave region. This is because radio waves have long wavelengths, enabling them to demonstrate wave phenomena like diffraction and interference. Understanding the wave nature of radio waves is essential for various applications, including communication systems, radar, and broadcasting.
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Please help me, I am following along diligently. V(t) = t^2 -9t+18, with distance, s measured in meters, left or right of 0, and t measured in seconds, with t between 0 and 8 seconds inclusive. The position at time t=0 sec is 1 meter right of 0, that is s(0)= 1Part I: Average velocity over the interval 0 to 8 secondsPart II: The instantaneous velocity and speed at time 5 secsPart III: The time intervals when the particle is moving rightPart IV: The time intervals when the particle is going faster, and slowing downPart V: Total distance the particle has traveled between 0 and 8 seconds
Answers
Given that the velocity at any time t is
\(v(t)=t^2-9t+18\)Also, the time interval is from t = 0 to t = 8 seconds
The position at time t = 0 s is s(0) = 1 m towards right of zero.
The initial time is t = 0 s, so the initial velocity will be
\(\begin{gathered} v_i(t=0)=0^2-9\times0+18\text{ } \\ v_i(0)\text{ = 18 m/s} \end{gathered}\)The final time is t = 8 s, so the final velocity will be
\(\begin{gathered} v_f(t=8)=8^2-9\times8+18 \\ v_f(8)\text{ = 64-72+18} \\ =\text{ 10 m/s} \end{gathered}\)The average velocity will be
\(\begin{gathered} v_{av}=\frac{v_i+v_f}{2} \\ =\frac{18+10}{2} \\ =14\text{ m/s} \end{gathered}\)Thus, the average velocity is 14 m/s.
Part II:
The instantaneous velocity at time t =5 s will be
\(\begin{gathered} v(t=5)=5^2-9\times5+18 \\ =25-45+18 \\ =-2\text{ m/s} \end{gathered}\)The instantaneous speed is the magnitude of instantaneous velocity.
Thus, the instantaneous speed will be 2 m/s.
Part III:
The particle will move towards the right when v(t) > 0
The time intervals will be
\(\begin{gathered} t^2-9t+18>0 \\ t^2-6t-3t+18>0 \\ t(t-6)-3(t-6)>0 \\ (t-6)(t-3)>0 \\ t-6>0\text{ or t>6} \\ t-3>0\text{ ot t>3} \end{gathered}\)Thus, time intervals are t > 3 and t > 6 when the particle is moving towards the right.
Part IV :
The particle will move faster if the acceleration, a(t) > 0
The particle will slow down if the acceleration, a(t) < 0
So, first, we need to find the acceleration, it can be calculated as
\(\begin{gathered} a(t)=\text{ }\frac{d(v(t))}{dt} \\ =\frac{d(t^2-9t+18)}{dt} \\ =2t-9 \end{gathered}\)For the particle moving faster,
\(\begin{gathered} a(t)>0 \\ 2t-9>0 \\ 2t-9+9>9+0 \\ 2t>9 \\ \frac{2t}{2}>\frac{9}{2} \\ t>\frac{9}{2} \\ t>4.5\text{ s} \end{gathered}\)For particle slowing down,
\(\begin{gathered} a(t)<0 \\ 2t-9<0 \\ 2t-9+9<9+0_{} \\ 2t<9 \\ \frac{2t}{2}<\frac{9}{2} \\ t<4.5\text{ s} \end{gathered}\)The total distance can be calculated as
\(\begin{gathered} s(t)=\int ^8_0v(t)dt \\ =\text{ }\int ^8_0(t^2-9t+18)\mathrm{d}t \\ =\lbrack\frac{t^3}{3}\rbrack^8_0-9\lbrack\frac{t^2}{2}\rbrack^8_0+18\lbrack t^{}\rbrack^8_0 \\ =\frac{1}{3}\lbrack512-0\rbrack-9\lbrack64-0\rbrack+18\lbrack8-0\rbrack \\ =\text{ 170.67-576+144} \\ =-261.33\text{ m} \end{gathered}\)Here, the negative symbol indicates it is towards the left from zero.
The length of the life cycle of a star depends on _.
distance from Earth
the nebula’s brightness
mass and the amount of matter
helium and the amount of matter
Answers
Answer:
c.mass and the amount of matter
The great red spot is a gigantic storm located on which planet in our solar system?.
Answers
A golf ball of mass 0.045 kg is hit off the tee at a speed of 34m/s . The golf club was in contact with the ball for 3.50*10^-3s .
Find the impulse imparted to the golf ball.??
Find the average force exerted on the ball by the golf club.??
Express your answer to two significant figures and include the appropriate units.
Answers
The average force exerted on the ball by the golf club is 490 N.
Thus, F = p/t = 0.49 * 10³ N = 490 N.
Momentum is a metric for power and how challenging it is to stop an object. Zero momentum applies to any object that is not moving. tremendous, slow-moving objects have tremendous amounts of momentum.
A small, swiftly moving object also possesses a significant momentum. A bowling ball, for instance, has more momentum than a ping-pong ball if their velocities are equal.
This is because bowling balls are larger in mass than ping-pong balls.
Thus, The average force exerted on the ball by the golf club is 490 N.
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Which of the following is the minimum amount of work done by a hydraulic lift
to raise a 150-kg aluminum block 2.0 m vertically?
Answers
Answer: 300 J
Explanation:
Force = 150 KJ
Distance = 2.0 m
The minimum amount of work done by a hydraulic lift to raise a 150-kg aluminum block 2.0 meters vertically would be 2940 Joules, as the work done is the product of the force applied to the displacement of the object, therefore the correct answer is option B.
What is work done?The total amount of energy transferred when a force is applied to move an object through some distance
The work done is the multiplication of applied force with displacement.
Work Done = Force × Displacement
As given in the problem we have to find out the minimum amount of work done by a hydraulic lift to raise a 150-kg aluminum block 2.0 meters vertically.
The minimum work done = 150 × 9.8 × 2
= 2940 Joules
Thus, The minimum amount of work done by a hydraulic lift to raise a 150-kg aluminum block 2.0 meters vertically would be 2940 Joules.
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12. A screwdriver is which type of a simple machine?
O wedge
O screw
O wheel and axle
O lever
Answers
Answer:
wheel and axle
Explanation:
hope this helps
You are outside on a sunny afternoon and observe a thunderstorm far to the west. An hour later, the storm passes over you. Is it more likely to have been an air mass thunderstorm or some sort of mesoscale convective system?
Answers
It is more likely that the storm you observed, which passed over you after an hour, was a mesoscale convective system (MCS) rather than an air mass thunderstorm.
Air mass thunderstorms typically form and dissipate within the boundaries of a single air mass. They are generally short-lived and do not typically travel large distances. Therefore, it is less likely for an air mass thunderstorm to travel from a location far to the west to your current location within just an hour.
On the other hand, mesoscale convective systems are larger-scale weather systems that can cover hundreds of miles and persist for several hours or even longer. They often have organized structures, such as squall lines or clusters of thunderstorms, and can move over significant distances. It is more plausible for a mesoscale convective system to travel from a distant location to your area within the span of an hour.
Considering the time and distance covered, the fact that the storm passed over you after an hour suggests that it was likely a mesoscale convective system.
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What physical property of salt allowed you to separate it from the sand?
Answers
Which statement about a turtle's energy source is true?
Responses
A. A turtle performs photosynthesis to gain its energy.
B. Nutrients in the water are the ultimate source of energy for a turtle
C. A turtle relies indirectly on the sun as an energy source.
Answers
Answer:
C. A turtle relies indirectly on the Sun as an energy source.
Explanation:
Turtles are animals, meaning they are unable to perform photosynthesis, thus option A is inapplicable. Turtles consume food such as algae and seagrass. These aquatic plants gain their energy directly from the Sun through photosynthesis, and as a result when the turtle consumes these plants, that energy from the Sun is indirectly feeding the turtle.
what is the work done when a 400.n force is use to lift a 400.n object 3.5 meters strait up
Answers
Answer:
i think it's 1400
Explanation:
w = Force x displacement
w = 400 x 3.5
How long would it take a 7.5 KW motor to raise a 500 kg piano to an apartment window 10
meters above the ground?
Answers
The time taken for the motor to raise the piano is 6.67 s
Data obtained from the question:Power (P) = 7.5 KW = 7.5 × 1000 = 7500 WMass (m) = 500 KgHeight (h) = 10 mAcceleration due to gravity (g) = 10 m/s²Time (t) =? How to determine the timeThe time taken for the motor to raise the piano can be obtained as follow:
P = mgh / t
7500 = (500 × 10 × 10) / t
7500 = 50000 / t
Cross multiply
7500 × t = 50000
Divide both side by 7500
t = 50000 / 7500
t = 6.67 s
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What potential difference, acting over a distance of 2.5 cm, would be needed to balance the downward force of gravity so that an electron would remain stationary
Answers
The potential difference, acting over a distance of 2.5 cm is 1.4*10^-12V.
What are the effects of distance on potential difference?The potential actually grows as you get farther away from the charge, becoming less negative as you get closer to it and eventually reaching zero. The potential for both positive and negative charges is zero at infinite distances from the charge.
Exactly what is potential energy?The energy that a body has because of its location or deformation is known as potential energy. Both the body's bulk and the height to which it is lifted play a role.
Given:
Distance (d) = 2.5cm = 0.025 m
For an electron,
\(m= 9.109*10^-31kg\)
\(q=1.602*10^-19C\)
We isolate the electric field:
\(E=\frac{mg}{q}\)
\(g=9.8 m/s^2\)
\(E=\frac{(9.109*10^-31kg)(9.8m/s^2)}{1.602*`10^-19C}\)
\(E=5.5723*10^-11V/m\)
Electric potential,
\(E=\frac{V}{d}\)
\(V=Ed\)
\(V= (5.5723*10^-11V/m)(0.025m)\)
\(V=1.4*10^-12V\)
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With a mass of 109 kg, Baby Bird is the smallest monoplane ever
flown. Suppose the Baby Bird and pilot are coasting along the
runway when the pilot jumps horizontally to the runway behind
the plane. The pilot's velocity upon leaving the plane is 5.11 m/s
backward. After the pilot jumps from the plane, the plane coasts
forward with a speed of 3.27 m/s. If the pilot's mass equals 61.4
kg, what is the velocity of the plane and pilot before the pilot
jumps? Round to the nearest hundredth.
Answers
Total resultant velocity=5.11-3.27=1.84m/s
m_1=61.4kgm_2=109kgv_1=1.84m/sv_2=?\(\\ \sf\longmapsto ∆P=P\)
\(\\ \sf\longmapsto m_1v_1=m_2v_2\)
\(\\ \sf\longmapsto v_2=\dfrac{m_1v_1}{m_2}\)
\(\\ \sf\longmapsto v_2=\dfrac{61.4(1.84)}{109}\)
\(\\ \sf\longmapsto v_2=112.976/109\)
\(\\ \sf\longmapsto v_2\approx 1.3m/s\)
The velocity of the plane and the pilot before the pilot jumps is 0.25 m/s.
The given parameters;
mass of the pilot, m₁ = 61.4 kgvelocity of the pilot, u₁ = 5.11 m/s backwardsvelocity of the plane, u₂ = 3.27 m/s forwardmass of the plane, m₂ = 109 kgThe velocity of the plane and the pilot before the pilot jumps is calculated by applying the principle of conservation of linear momentum for inelastic collision as follows;
\(m_1u_1 + m_2u_2 = v(m_1 + m_2)\\\\61.4(-5.11) \ + \ 109(3.27) = v(61.4 + 109)\\\\42.68 = v(170.4)\\\\v = \frac{42.68}{170.4} \\\\v = 0.25 \ m/s\)
Thus, the velocity of the plane and the pilot before the pilot jumps is 0.25 m/s.
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pls help me with it..fast as possible...pls .....
Answers
Answer:
Because the light only spears to part of the water so it would appear less deep
Pick the word which best completes the sentence. The dark ____ cloud was a sure sign that it was doing to storm.
a. cumulus
b. cumulonimbus
c. stratus
d.cirrus
Answers
Answer:
the dark cumulus cloud was a sure sign that it was going to storm.
Explanation:
hope this helps :)
Answer:
a. Cumulonimbus
Explanation:
A 16-year-old employee working for Southern Virginia College's (SVC) bookstore during the summer months is helping prepare for Fall sales. It's a good way to make extra money, and the teen is saving for a car.
Books from one supplier are shipped to the SVC bookstore in large crates equipped with rope handles on all sides. On one occasion, the teen momentarily pulled with a force of 713 N at an angle of 35.8° above the horizontal to accelerate a 114-kg crate of books. The coefficient of friction between the crates and the vinyl floor is 0.541.
Determine the acceleration experienced by the crate in m/s2. Use the approximation g ≈ 10 m/s2.
Answer: ___________ m/s2 (rounded to the hundredths or thousandths place)
Answers
The acceleration experienced by the crate is approximately 0.844 m/s
How to solve for the accelerationWeight of the crate:
Weight = mass × acceleration due to gravity
Weight = 114 kg × 10 m/s^2
Weight = 1140 N
Force of friction:
Force of Friction = coefficient of friction × normal force
Force of Friction = 0.541 × 1140 N
Force of Friction ≈ 616.74 N
Net force:
Net Force = Applied Force - Force of Friction
Net Force = 713 N - 616.74 N
Net Force ≈ 96.26 N
Acceleration:
Acceleration = Net Force / mass
Acceleration = 96.26 N / 114 kg
Acceleration ≈ 0.844 m/s
Therefore, the acceleration experienced by the crate is approximately 0.844 m/s
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A 0.25 kg ball rolling at a 1m/s and overtakes a 0.3kg ball rolling in the same direction at 0.5m/s. The ball sticks together on impact. What is the velocity of the two balls after collision?
Answers
The velocity of the two balls after the collision is 0.73 m/s.
The velocity of the two balls after the collision can be calculated using the formula below.
Formula:mu+m'u' = V(m+m')............... Equation 1Where:m = mass of the first ballm' = mass of the second ballu = initial velocity of the first ballu' = initial velocity of the second ballV = velocity of the two balls after the collision.make V the subject of the equation
V = (mu+m'u')/(m+m')................ Equation 2From the question,
Given:m = 0.25 kgm' = 0.3 kgu = 1 m/su' = 0.5 m/sSubstitute these values into equation 2
V = [(0.25×1)+(0.3×0.5)](0.25+0.3)V = 0.4/0.55V = 0.73 m/s.Hence, the velocity of the two balls after the collision is 0.73 m/s
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A metal rectangular loop (heighth and width w) with resistance R is fixed in place with one third of its length located inside a region of space where there is a time-varying magnetic field B = Bo - bl pointing out of the page. h w B0 B=0 A. Determine the magnitude and direction of the current I(t) induced in the loop. B. If the loop were not fixed in place, it would move due to the magnetic force exerted on it by the external magnetic field. What is the magnitude of the magnetic force felt by the loop? What direction would the loop move towards, if it were not fixed in place?
Answers
The magnitude is I(t) = (blA)/R, and its direction is clockwise. The magnetic force exerted on the loop would move towards the left with maximum acceleration, the direction of the motion is perpendicular to the magnetic field and the current flowing in the loop.
A metal rectangular loop (height h and width w) with resistance R is fixed in place with one-third of its length located inside a region of space where there is a time-varying magnetic field B = Bo - bl pointing out of the page.
Based on this, we have to determine the magnitude and direction of the current I(t) induced in the loop and the magnitude of the magnetic force felt by the loop, and in what direction the loop would move if it were not fixed in place.
When a time-varying magnetic field is applied to a loop, it induces a current in it. The magnitude of the induced current is given by the equation emf = -N(dФ/dt), where N is the number of turns, Ф is the magnetic flux, and emf is the electromotive force.
Here, we have one loop with N=1, and the magnetic flux through the loop is Ф = B.A = (Bo - bl)A, where A is the area of the loop. Since the loop is stationary, the induced emf must balance the resistive force, which is equal to I(t)R, where I(t) is the current flowing in the loop at time t.
Thus, we have emf = I(t)R = -d/dt(BA) = -d/dt((Bo - bl)A) = blA. Therefore, the magnitude of the induced current is I(t) = (blA)/R, and its direction is clockwise.
The magnetic force on the loop is given by F = I(t)l x B, where l is the length of the loop in the direction of the current flow. Here, l = w/3, and the current flows clockwise. Thus, the direction of l is out of the page. Also, the magnetic field is pointing out of the page. Therefore, the direction of the force is to the left, as per Fleming's left-hand rule.
The magnitude of the force is given by F = I(t)lB = (blA/3)(Bo - bl), where A = hw is the area of the loop. The force is maximum when bl = Bo/2, and its magnitude is Fmax = (Bo²hw)/(12R). If the loop were not fixed in place, it would move towards the left due to the magnetic force acting on it.
The force is maximum when bl = Bo/2, and its magnitude is Fmax = (Bo²hw)/(12R). Thus, the loop would move towards the left with a maximum acceleration given by a = Fmax/m, where m is the mass of the loop. The direction of the motion is perpendicular to the magnetic field and the current flowing in the loop.
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