Astronomers prefer to use infrared telescopes on high-flying airplanes or on satellites in space because the atmosphere of the Earth blocks a significant amount of infrared light. By placing these telescopes in the sky or in space, astronomers can get a better view of infrared light from distant objects in the universe.
Airborne telescopes can offer temporary access to the observable range of the electromagnetic spectrum blocked by water vapor and ozone layer.The atmosphere filters out and absorbs some of the infrared light. The atmosphere consists of many layers. Infrared light does not get through the layer of ozone, water vapor, or carbon dioxide that makes up the lower part of the atmosphere. However, the higher atmosphere is transparent to infrared light. So, flying above this layer can give astronomers a much clearer and less obstructed view of the universe.
Satellites can offer a better view of the universe without interference from the Earth’s atmosphere. In space, the telescope is not hampered by the Earth's atmosphere. For example, the Hubble Space Telescope, which orbits about 570 km above Earth's surface, provides a clearer view of the universe in ultraviolet, visible, and near-infrared wavelengths than ground-based telescopes do. It has discovered galaxies, stars, and phenomena far beyond the reach of Earth-based telescopes.
They offer the possibility of carrying bigger and heavier instruments than a ground-based observatory.The weight of an airborne telescope is significantly reduced in comparison to a ground-based telescope because it is placed above the Earth’s atmosphere, and this has the advantage of allowing for more extensive equipment, including larger and heavier telescopes. Satellites have the same advantage as they are not limited by the size of the payload.
Airborne telescopes and satellites can observe celestial objects in the infrared spectrum.Infrared light is absorbed by the Earth's atmosphere. As a result, infrared telescopes that are placed on the ground will be less successful. As a result, astronomers prefer to place them on airborne telescopes and satellites, which are above the Earth's atmosphere. This makes it possible for infrared telescopes to examine the universe and make discoveries that would otherwise be impossible to observe from the ground.
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A small mass rests on a horizontal platform which vibrates vertically in simple harmonicmotion with period 0.50 s.(a) Find the maximum amplitude of the motion which will allow the mass to stay in contactwith the platform throughout the motion.The maximum acceleration that will allow the object to remain in contact with theplatform at all times is when amax = g = 9:81 m/s.But amax = !222A = (2¼=T )A ) 9:81 = (2¼=0:5)2A = 158A ) A = 0:062 m
The maximum amplitude of the motion which will allow the mass to stay in contact with the platform is 0.062 m.
This can be calculated by using the equation amax = (2π/T)2A, where A is the maximum amplitude, and T is the period of the motion. In this case, T is 0.50 s, and g (the acceleration due to gravity) is 9.81 m/s2, so we can calculate A:
A = (2π/T)2g = (2π/0.50)2 × 9.81 = 158 × 9.81 = 1543.38
Therefore, A = 1543.38/158 = 9.81 m/s2 = 0.062 m.
Alternatively: given,T = 0.50 s,The acceleration due to gravity, g = 9.81 m/s²Maximum acceleration, amax = g = 9.81 m/s². The maximum acceleration that will allow the object to remain in contact with the platform at all times is when amax = !222A = (2π/T )A ) 9.81 ...(1)From the equation (1), we get 158 A = 9.81 (2π/0.50)A = (9.81 (2π/0.50))/158 = 0.062 m. Therefore, the maximum amplitude of the motion which will allow the mass to stay in contact with the platform throughout the motion is 0.062 m.
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Is an object moving with a constany speed around a circular path veloctiy? why? why not?
Answer: The motion of a body with constant speed in a circular path is said to be accelerated, because it is moving with uniform speed, but not with uniform velocity, as velocity is a vector quantity, it can be represented in magnitude as well the direction.
Explanation:
of the three states of matter, which one has the most kinetic energy?
Of the three states of matter (solid, liquid, and gas), gas has the most kinetic energy. This is because the particles in a gas have the highest average speed compared to the particles in solids and liquids.
In a gas, the particles are in constant motion, colliding with each other and the walls of the container. This motion generates kinetic energy, which is proportional to the speed and mass of the particles. In contrast, solids have the lowest kinetic energy because their particles are tightly packed and have limited movement. The particles in a solid vibrate around a fixed position, and only experience small oscillations. Liquids have an intermediate amount of kinetic energy. The particles in a liquid are less tightly packed than in a solid, and can move more freely, resulting in more kinetic energy. However, liquids have more intermolecular forces between the particles compared to gases, which restricts their movement and reduces their average speed. Therefore, of the three states of matter, gases have the most kinetic energy, followed by liquids and then solids.
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suppose the ring rotates once every 4.10 s . if a rider's mass is 51.0 kg , with how much force does the ring push on her at the top of the ride?
The ring rotates once every 4.10 s. If a rider's mass is 51.0 kg, how much force does the ring push on her at the top of the ride is 500 N.
The solution is explained below:
As the rider is at the top of the ride, the only force acting on him is the force of gravity, which is pointing downwards, and the force with which the ring is pushing him towards the center of the circular path. By equating both forces, we can determine the required force to maintain the rider at the top of the ride.
Hence, the answer to the question is that the force with which the ring pushes the rider at the top of the ride is equal to the force of gravity, which is given as F = mgF = (51.0 kg)(9.81 m/s^2) = 500 N
Therefore, the force with which the ring pushes on the rider at the top of the ride is 500 N.
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the car passes over the top of a vertical curve at a with a speed of 50 km/hr and then passes through the bottom of a dip at b. the radii of curvature of the road at a and b are both 70 m. find the speed of the car at b if the normal force between the road and the tires at b is twice that at a. the mass center of the car is 1.2 meter from the road.
The speed of the car at b if the normal force between the road and the tires at b is twice that at a is about 44.1 km/h.
What is Speed?Speed of the car at A = 50 km/h
Radius of curvature at A = 70 m
Radius of curvature at B = 70 m
Normal force between the road and the tires at B = 2 × Normal force between the road and the tires at A= 2N
Mass center of the car = 1.2 m
The speed of car at B be v km/h
From the conservation of energy at the point A and B, we get:
1/2 mv² + mgh = 1/2 m(50)² + mg(70 - r)
1/2 mv² + mg(70 + r) = 1/2 m(50²)
1/2 mv² = 1/2 m50² - mg(70 + r) …… equation (1)
From the conservation of energy at point B, we get:
1/2 mv² + mg(2r + 1.2) = 1/2 m(50)² + mg(70 - r)
2× Normal force between the road and the tires at A = Normal force between the road and the tires at B
Normal force between the road and the tires at B = 2 × Normal force between the road and the tires at A
Therefore, mg - 2 × N = mv²/rmg - N = mv²/2r
2mg - 4N = mv²/rmg - 2N = mv²/2r
Subtracting, we get:
N = mg/3
Normal force between the road and the tires at A = mg/3
Normal force between the road and the tires at B = 2mg/3
Normal force between the road and the tires at B = 2(mg/3) = mg/3
From the above equations, we get the value of v. Putting the values, we get:
1/2 mv² = 1/2 m(50)² - mg(70 + r) - mg(2r + 1.2) + mg(70 - r)1/2 v² = 1/2(50)² - g(70 + r) - g(2r + 1.2) + g(70 - r)v = 44.1 km/h
Therefore, the speed of the car at B is 44.1 km/h.
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why do nuclear reactors have three separate water loops instead of just a single one that runs from the water source, through the reactor, then back to the cooling tower?
Nuclear reactors have three separate water loops instead of just a single one that runs from the water source, through the reactor, then back to the cooling tower because the water running through the reactor is highly radioactive.
What are nuclear reactors?A nuclear reactor is a device that controls and maintains a sustained nuclear chain reaction for the purpose of generating heat or power, as well as the materials that make up a nuclear reactor.
The water running through the reactor is highly radioactive, which means that it cannot be released into the atmosphere or allowed to come into touch with humans or the environment. As a result, nuclear reactors are designed with three separate water loops.
The first loop circulates ordinary water that passes through the reactor and generates heat. The second loop, which is a separate circuit, brings this water to a steam turbine. The third loop, which is also a closed circuit, recovers the cooling water after it has passed through the turbine and transports it back to the reactor's inlet.In summary, nuclear reactors have three separate water loops instead of a single one that runs from the water source, through the reactor, and back to the cooling tower because the water running through the reactor is highly radioactive.
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discuss whether the values found in parts (a) and (b) seem consistent with the fact that tidal effects with earth have caused the moon to rotate with one side always facing earth.
Yes, the values found in parts (a) and (b) are consistent with the fact that tidal effects with earth have caused the moon to rotate with one side always facing earth.
This is because part (a) states that the moon rotates on its axis in the same amount of time it takes to complete one orbit around the Earth, which is a phenomenon known as tidal locking. Part (b) further indicates that the same side of the moon always faces the Earth, further supporting the notion that tidal effects have caused the moon to rotate with one side always facing Earth.
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In outer space will a liquid in a beaker exert a pressure on the bottom or on the sides of a beaker?
Answer:
Yo dude, if you had a beaker of liquid in outer space, it wouldn't push down on the bottom or the sides of the beaker like it would on Earth. In space, there's no gravity to make the liquid settle down, so it forms a round shape because of surface tension. So basically, the liquid would just float around in a ball inside the beaker. If you moved the beaker around, the liquid would just roll around with it like a bouncy ball.
the end result of a theory that is not verified is
Unproven theories ultimately cannot be regarded as scientific facts or principles and are not generally recognised by the scientific community.
A well-supported explanation of a natural occurrence in science that has passed rigorous examination and is backed by empirical data is referred to as a theory. A hypothesis, however, cannot be regarded as a scientific fact or principle if it is not backed up by empirical data or if it has not undergone extensive testing and verification. The scientific community frequently rejects unproven notions with scant empirical backing and may even label them as pseudoscientific or non-scientific. This is so that scientific theories and findings may be evaluated and verified frequently. Science does this by using evidence-based reasoning and critical thinking. Unproven theories are therefore eventually not regarded as being a part of the corpus of scientific knowledge.
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a particle's velocity is described by the function vx=kt2 , where vx is in m/s , t is in s , and k is a constant. the particle's position at t0=0s is x0 = -5.40 m . at t1 = 2.00 s , the particle is at x1 = 5.80 m .
A particle's velocity is described by the function vx=kt2 , where vx is in m/s , t is in s , and k is a constant. The particle's position at t0=0s is x0 = -5.40 m. At t1 = 2.00 s , the particle is at x1 = 5.80 m. The value of k is 2.80 m/s2.
The given equation describes the velocity of a particle in terms of a constant, k, and time, t. The velocity, vx, is given in m/s. The initial position of the particle at t0=0s is x0=-5.40 m, and at t1=2.00 s the particle is at x1=5.80 m. To find the value of the constant k, we can solve the equation for the change in velocity Δvx.
Δvx = vx1 – vx0 = k(t12 – t02)
Δvx = 5.80 – (-5.40) = 11.20 m/s
k = (11.20 m/s) / (2.002 s2) = 2.80 m/s2
Now that we have found the value of the constant k, we can use it to find the velocity of the particle at any time t. For example, at t2=4.00 s the velocity of the particle is vx2=11.20 m/s. This can be calculated using the equation vx2 = k(t22) = 2.80(4.002) = 11.20 m/s.
From the velocity equation, we can also calculate the position of the particle at any time t. The position of the particle at t2=4.00 s is x2= 11.20(4.00) = 44.80 m. We can also calculate the position of the particle at any other time t, by simply substituting in the corresponding value of t into the equation.
In conclusion, the equation vx = kt2 describes the velocity of a particle in terms of a constant, k, and time, t. Using this equation, we can calculate the velocity and position of the particle at any given time.
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Complete Question:
A particle’s velocity is described by the function vx = [tex]kt^2m/s[/tex], where k is a constant and t is in s. The particle’s position at [tex]t_0[/tex] = 0s is [tex]x_0[/tex] = -5.40 m. At [tex]t_1[/tex] = 2.00 s, the particle is at [tex]x_1[/tex] = 5.80 m. Determine the value of the constant k. Be sure to include the proper units
A 2. 00-kg object is attached to an ideal massless horizontal spring of spring constant 100. 0 N/m and is at rest on a frictionless horizontal table. The spring is aligned along the x-axis and is fixed to a peg in the table. Suddenly this mass is struck by another 2. 00-kg object traveling along the x-axis at 3. 00 m/s, and the two masses stick together. What are the amplitude and period of the oscillations that result from this collision? 0. 300 m, 1. 26 s 0. 424 m, 5. 00 s 0. 424 m, 0. 889 s 0. 300 m, 0. 889 s 0. 424 m, 1. 26 s
The correct option is A, the amplitude and period of the oscillations that result from this collision are 0.300 m in 1.26s.
The expression for Period of spring is,
[tex]T = 2\pi\sqrt{\frac{2m}{k} }[/tex]
Here, m is the mass of the spring and k is the spring constant
Substitute 2 kg
for m
and 100N/m
for k
in equation [tex]T = 2\pi\sqrt{\frac{2m}{k} }[/tex]
and solve for T .
[tex]T = 2\pi\sqrt{\frac{(2)2 kg}{100 N/m} }[/tex]
T = 1.26s
In physics, amplitude refers to the maximum displacement or distance moved by a wave from its equilibrium or mean position. It is a measure of the intensity or strength of a wave, and it is usually represented as the height of the crest or depth of the trough of the wave.
The amplitude of a wave can be measured in various units, depending on the type of wave and the context in which it is being studied. For example, the amplitude of a sound wave is measured in decibels (dB), while the amplitude of an electromagnetic wave is measured in volts per meter (V/m). Amplitude plays an important role in the behavior of waves. It determines the energy carried by the wave and affects other properties such as frequency, wavelength, and phase.
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Complete Question: -
A 2.00-kg object is attached to an ideal massless horizontal spring of spring constant 100.0 N/m and is at rest on a frictionless horizontal table. The spring is aligned along the x-axis and is fixed to a peg in the table. Suddenly this mass is struck by another 2.00-kg object traveling along the x-axis at 3.00 m/s, and the two masses stick together. What are the amplitude and period of the oscillations that result from this collision
A) 0.300 m, 1.26 s
B) 0.300 m, 0.889 s
C) 0.424 m, 0.889 s
D) 0.424 m, 1.26 s
E) 0.424 m, 5.00 s
if the 2 currents are same direction and forces are attractive, what is the direction of force wire 1 on wire 2
The direction of the force from wire 1 on wire 2 is attractive, as the two currents are in the same direction.
If two currents are flowing in the same direction and the forces between the wires are attractive, then the direction of the force on wire 2 due to wire 1 will be towards wire 1. This is because the magnetic field created by the current in wire 1 will induce a magnetic field in wire 2, and the interaction between these two magnetic fields will result in an attractive force between the wires.
In summary, if two currents are flowing in the same direction and the forces are attractive, the direction of the force on wire 2 due to wire 1 will be towards wire 1.
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In which of the following cases does a car have a negative velocity and a positive acceleration? A car that is traveling in the ................. (A) -x direction at a constant 10 m/s. (B) - direction increasing in speed. (C) +x direction increasing in speed. (D) - direction decreasing in speed. (E) +x direction decreasing in speed.
In the case where the car is traveling in the -x direction and decreasing in speed, it has a negative velocity and a positive acceleration. Therefore, option D is the correct answer. In this case, the car is traveling in the - direction and decreasing in speed. Therefore, it has a negative velocity and a positive acceleration.
Let's discuss the given options one by one:
(A) In this case, the car is traveling in the -x direction at a constant speed. Therefore, it has a negative velocity and zero acceleration. This option is incorrect.
(B) In this case, the car is traveling in the - direction and increasing its speed. Therefore, it has a negative velocity and a positive acceleration. However, the given direction is not specified, and thus this option is not accurate.
(C) In this case, the car is traveling in the +x direction and increasing in speed. Therefore, it has a positive velocity and a positive acceleration. This option is incorrect.
(D) In this case, the car is traveling in the - direction and decreasing in speed. Therefore, it has a negative velocity and a positive acceleration. This option is correct.
(E) In this case, the car is traveling in the +x direction and decreasing in speed. Therefore, it has a positive velocity and a negative acceleration. This option is incorrect.
Therefore, Option D ( - direction decreasing in speed) is correct.
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Review your answer to part c. In addition, reread the portion of your physics text that discusses Newton's third law. Then consider a book on a level table: e. Which force completes the Newton's third law (or action-reaction) force pair with the normal force exerted on the book by the table?
In this case, the normal force exerted by the table on the book is the action force and the reaction force is the force that the book exerts on the table. This force is equal in magnitude to the normal force and acts in the opposite direction.
Newton's third law states that for every action, there is an equal and opposite reaction. This means that when one object exerts a force on another object, the second object exerts a force back on the first object that is equal in magnitude and opposite in direction.
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you live on an island in the pacific. an earthquake of magnitude 8.5 off the coast of japan, 8000 km away, generates a tsunami with a wavelength of 200 km. the average water depth between your island and japan is 4900 m. if a tsunami warning is issued for your island, how many hours will you have before the waves arrive?
If a tsunami warning is issued for the island, they will have approximately 11.7 hours before the waves arrive.
What is Magnitude?
Magnitude is a measure of the strength or intensity of a physical quantity or phenomenon, such as an earthquake or a sound wave. It is often expressed using a numerical scale, with higher values indicating greater strength or intensity. In the case of earthquakes, magnitude is typically measured using the Richter scale or the moment magnitude scale, which take into account the amplitude of seismic waves and the energy released by the earthquake.
To calculate the time it takes for a tsunami to travel from Japan to the island, we can use the following formula:
t = (2 * pi * d) / g * ln(1 + sqrt(h/d))
where t is the time it takes for the tsunami to travel, d is the average water depth, h is the wave height, and g is the acceleration due to gravity (9.8 m/s^2).
Magnitude of the earthquake: 8.5
Wavelength of the tsunami: 200 km = 200,000 m
Average water depth: 4,900 m
To calculate the wave height, we can use the following formula:
h = (M / 5) * (D / 10)^1/2
where M is the magnitude of the earthquake and D is the distance between the earthquake epicenter and the observation point (in this case, the island). Note that this formula is an approximation and may not be accurate for all cases.
Using the given values, we get:
D = 8,000 km = 8,000,000 m
h = (8.5 / 5) * ((8,000,000 / 10)^1/2) = 2,738.6 m
Substituting these values into the formula for t, we get:
t = (2 * pi * 4,900) / 9.8 * ln(1 + sqrt(2,738.6/4,900)) = 11.7 hours
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A student produces severa standing waves on string by adjusting the (requency vibration at ona end olthe string: The student measures the wavelength and frequency for each standing wave produced Which of the following procedures and calculations will allow the student I0 determine Ihe wave speed on the string? a.Graph function of 1\f The slope of the Iine equal t0 the wave speed;
b. Graph a5 a function of f The slope of the Ilne equal to he wave speed:
c. Graph A a5 function of 1\f The area under Ihe Iine I5 equal to Ihe wave speed d. Graph a5 a function of f The area under the line equal l0 Ihe wave speed
The correct option that allows the student to determine the wave speed on the string is d. Graph a5 a function of f The area under the line equal l0 Ihe wave speed.
Wave speed can be calculated by the formula: Wave speed (v) = frequency (f) × wavelength (λ) or v = fλ
According to the question, the student has measured the wavelength and frequency for each standing wave produced. Now, to determine the wave speed, the student needs to use the formula: v = fλ
To determine the wave speed from the graph of frequency and wavelength, the graph is made with frequency on the x-axis and wavelength on the y-axis. The slope of the line gives the speed of the wave. The graph can be used to calculate the wave speed for any wave by finding the slope of the line.
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A small grinding wheel has a moment of inertia of 4. 0×10−5 kg⋅m2
k
g
⋅
m
2. What net torque must be applied to the wheel for its angular acceleration to be 150 rad/s2
r
a
d
/
s
2
?
A net torque of [tex]6.0×10^−3 N⋅m[/tex] is sufficient to produce the desired angular acceleration of [tex]150 rad/s^2[/tex].
The net torque required to produce an angular acceleration in a rotating object can be calculated using the formula: net torque = moment of inertia × angular acceleration In this case, the moment of inertia of the grinding wheel is given as 4.0×10^−5 kg⋅m^2 and the angular acceleration required is 150 rad/s^2.
Therefore, the net torque required can be calculated as: net torque = [tex](4.0×10^−5 kg⋅m^2) × (150 rad/s^2) = 6.0×10^−3 N⋅m[/tex]To explain this result, we need to understand the relationship between torque and angular acceleration. Torque is the rotational equivalent of force and it is defined as the product of force and the perpendicular distance between the line of action of the force and the axis of rotation.
When a torque is applied to a rotating object, it produces an angular acceleration in the object, which is a measure of how quickly the object's rotational speed changes.
The moment of inertia of an object is a measure of its resistance to changes in its rotational motion. It depends on the object's mass distribution and the distance of each element of mass from the axis of rotation. Objects with larger moments of inertia require more torque to produce a given angular acceleration than objects with smaller moments of inertia.
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amanda weighs about 600 n on earth, but would only weigh about 100 n on the moon. which best explains why amanda would weigh less on the moon than on earth? A. the mass of the moon is less than that of earth, therefore it has a weaker gravitational force. B. the circumference of the moon is smaller than earth, therefore it has less gravity. C. the pull from the gravity from earth decreases the pull of gravity from the moon. D. the lack of air pressure on the moon weakens the gravitational force of the moon.
Option A is the correct answer. The mass of the moon is less than that of earth, therefore it has a weaker gravitational force.
The correct option that explains why Amanda would weigh less on the moon than on earth is "A. the mass of the moon is less than that of the earth, therefore it has a weaker gravitational force." This is because weight is the result of the gravitational force that acts on an object, which is determined by both the mass of the object and the gravitational force acting on it. Therefore, the weight of an object varies depending on the mass and gravity.
The gravity of an object is the force that attracts it towards the center of the earth or the celestial object. The amount of gravity an object has depends on its mass and the mass of the object that it is attracting. The moon has a smaller mass than the earth, which means that it has a weaker gravitational force.
Consequently, the pull of gravity on the moon is weaker than on earth. The weight of Amanda is less because pull of gravity on the moon is weaker than on earth. Therefore, option A is the correct answer.
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A spring attached to a mass is at rest in the initial position (not shown). The spring is compressed in position A and is then released, as shown in position B. Which equation describes the conservation of energy in position A?
[tex]E=\frac{1}{2} mv^{2} \\E=mgh\\E=\frac{1}{2} kx^{2} \\E=\frac{1}{2} k2kx^{2}[/tex]
Answer:
Explanation:
The energy conservation is equal to half of the product of the spring constant and the square of displacement of the spring, so option C is correct.
two blocks with masses 4m and 7m are on a collision course with the same initial speeds vi. the block with mass 4m is traveling to the left, and the 7m block is traveling to the right. they undergo a head-on elastic collision and each bounces back, retracing its original path. find the final speeds of the particles. (enter your answers in terms of
The final speeds of the particles expressed in terms of the initial velocity are |v1'| = |v1| = 27/8|vi| and |v2'| = |v2| = 27/14|vi|
The conservation of momentum can be applied. The total momentum of the system before the collision is:
P before = m1v1 + m2v2
where m1 and v1 are the mass and velocity of the 4m block and m2 and v2 are the mass and velocity of the 7m block. Since the two blocks have the same initial speed, the momentum before the collision is:
P before = (4m)(-vi) + (7m)(vi)
P before = 3mvi
After the collision, the two blocks bounce back, so their final velocities are:
v1' = -v1
v2' = -v2
where v1 and v2 are the velocities of the blocks after the collision. Using the conservation of momentum again, the total momentum of the system after the collision is:
Pafter = m1v1' + m2v2'
Pafter = -4mv1 - 7mv2
Pafter = -4m(-v1) - 7m(-v2)
Pafter = 4mv1 + 7mv2
Since the collision is elastic, the total kinetic energy of the system is conserved. Therefore, the kinetic energy before the collision is equal to the kinetic energy after the collision:
Kbefore = Kafter
where Kbefore is the kinetic energy of the system before the collision and Kafter is the kinetic energy of the system after the collision. The kinetic energy can be expressed as:
K = 1/2mv²
Therefore, the total kinetic energy of the system before the collision is:
Kbefore = 1/2(4m)(vi)² + 1/2(7m)(vi)²
Kbefore = 27/2m(vi)²
The total kinetic energy of the system after the collision is:
Kafter = 1/2(4m)(-v1)² + 1/2(7m)(-v2)²
Kafter = 1/2(4m)(v1)² + 1/2(7m)(v2)²
Using the conservation of kinetic energy, Kbefore = Kafter:
27/2m(vi)² = 1/2(4m)(v1)² + 1/2(7m)(v2)²
Simplifying, the final velocities can be expressed in terms of the initial velocity:
v1 = 27/8vi
v2 = 27/14vi
Therefore, the final speeds of the particles are: |v1'| = |v1| = 27/8|vi| and |v2'| = |v2| = 27/14|vi|
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2.2 VECTORS IN TWO 120 N bearing 70° and 160 N bearing 40°
Answer:
Explanation:
Assuming you want to find the resultant vector of the two given vectors:
We can use the graphical method or the component method to find the resultant vector. Here, I will demonstrate the component method:
Step 1: Convert the given vectors into their component form (i.e., horizontal and vertical components).
Vector 1: 120 N bearing 70°
Horizontal component = 120 cos(70°) ≈ 38.23 N
Vertical component = 120 sin(70°) ≈ 113.41 N
Vector 2: 160 N bearing 40°
Horizontal component = 160 cos(40°) ≈ 122.15 N
Vertical component = 160 sin(40°) ≈ 103.08 N
Step 2: Add the horizontal components and vertical components separately to get the components of the resultant vector.
Horizontal component of resultant vector = 38.23 N + 122.15 N ≈ 160.38 N
Vertical component of resultant vector = 113.41 N + 103.08 N ≈ 216.49 N
Step 3: Use the Pythagorean theorem to find the magnitude of the resultant vector.
Magnitude of resultant vector = √(160.38 N)^2 + (216.49 N)^2 ≈ 268.15 N
Step 4: Find the direction of the resultant vector.
Direction of resultant vector = tan^-1(216.49 N / 160.38 N) ≈ 53.12°
Therefore, the resultant vector of the two given vectors is approximately 268.15 N at a bearing of 53.12°.
when a 2.75-kg fan, having blades 18.5 cm long, is turned off, its angular speed decreases uniformly from 10.0 rad/s to 6.30 rad/s in 5.00 s. (a) what is the magnitude of the angular acceleration of the fan?
The angular acceleration of the fan is 0.740 rad/s^2,
Angular acceleration which represents the rate at which the angular velocity changes over time. The unit used to measure angular acceleration is radians per square second (rad/s2), according to the International System of Units. The Greek alphabet symbol alpha (α) is used to denote angular acceleration.
To calculate the angular acceleration of the fan, the formula α = Δω/Δt is used. Here, α represents angular acceleration, Δω represents the change in angular speed, and Δt represents the change in time.
In this scenario, Δω is equal to 10.0 - 6.30 = 3.70 rad/s, and Δt is equal to 5.00 s. By substituting these values into the formula, we obtain α = 3.70/5.00 = 0.740 rad/s^2.
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Calculate the translational speed of a cylinder when it reaches the foot of an incline 7.20 m high. Assume it starts from rest and rolls without slipping.
Express your answer using three significant figures and include the appropriate units. Thank you!!
The translational speed of the cylinder when it reaches the foot of the incline is approximately 9.43 m/s.
We can use conservation of energy to solve this problem. The initial energy of the cylinder is all potential energy, and the final energy is all kinetic energy. The potential energy at the bottom of the incline is zero.
The potential energy of the cylinder at the top of the incline is given by:
PE = mgh
where m is the mass of the cylinder, g is the acceleration due to gravity, and h is the height of the incline. Substituting the given values, we get:
PE = (mass of cylinder) x (acceleration due to gravity) x (height of incline) = mgh
The kinetic energy of the cylinder at the bottom of the incline is given by:
KE = (1/2)mv^2
where v is the translational speed of the cylinder at the bottom of the incline.
According to the conservation of energy, the initial potential energy is equal to the final kinetic energy, so we can set these two expressions equal to each other:
mgh = (1/2)mv^2
We can cancel the mass of the cylinder from both sides, and solve for v:
v = sqrt(2gh)
Substituting the given values, we get:
v = sqrt(2 x 9.81 m/s^2 x 7.20 m) ≈ 9.43 m/s
Therefore, the translational speed of the cylinder when it reaches the foot of the incline is approximately 9.43 m/s.
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a cliff diver drops from rest to the water below. how many seconds does it take for the driver to go from 0 mi/h to 60 mi/h? (for comparison, it takes about 3.5 s to 4.0 s for a powerful car to go from 0 to 60 mi/h.)
Assuming that the only force acting on the diver is gravity and neglecting air resistance, we can use the kinematic equations of motion to determine that it takes 2.7 s for the diver to reach a speed of 60 mi/h (or 88 ft/s).
Since the diver starts from rest, we can use the kinematic equation:
[tex]$$v_f = v_i + at$$[/tex]
where [tex]$v_i$[/tex] is the initial velocity (0 mi/h), [tex]$v_f$[/tex] is the final velocity (60 mi/h or 88 ft/s), [tex]$a$[/tex] is the acceleration due to gravity [tex](32.2 ft/s$^2$)[/tex], and [tex]$t$[/tex] is the time it takes to reach the final velocity.
Converting the final velocity to feet per second, we get:
[tex]$$v_f = 60\ \text{mi/h} \times \frac{5280\ \text{ft/mi}}{3600\ \text{s/h}} = 88\ \text{ft/s}$$[/tex]
Substituting the given values, we get:
[tex]$$88\ \text{ft/s} = 0\ \text{ft/s} + (32.2\ \text{ft/s}^2)t$$[/tex]
Solving for [tex]$t$[/tex], we get:
[tex]t = \frac{88\ \text{ft/s}}{32.2\ \text{ft/s}^2}[/tex]
Therefore, it takes approximately 2.73 seconds for the diver to go from 0 mi/h to 60 mi/h.
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This is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Pin A, which is attached to link AB, is constrained to move in the circular slot CD. At t=0, the pin starts from rest and moves so that its speed increases at a constant rate of 1.2 in/s2 D 3.5 in. А B Determine the magnitude of its total acceleration when t= 0. The magnitude of its total acceleration is in/s2
The magnitude of the total acceleration of the pin when t=0 is 1.2 in/s^2.
To explain further, the acceleration of the pin is the sum of two components: tangential acceleration and centripetal acceleration. The tangential acceleration is responsible for increasing the speed of the pin, and its magnitude is constant at 1.2 in/s^2.
The centripetal acceleration is due to the circular motion of the pin in the slot CD and is directed towards the center of the circle.
To find the magnitude of the total acceleration at t=0, we need to first find the magnitude of the tangential acceleration and the centripetal acceleration separately. We know that the tangential acceleration is 1.2 in/s^2, and we can use the formula for centripetal acceleration, a_c = v^2/r, where v is the velocity of the pin and r is the radius of the circle. At t=0, the velocity of the pin is zero, and the radius of the circle is 3.5 inches.
Therefore, the centripetal acceleration is also zero.
Since the centripetal acceleration is zero, the magnitude of the total acceleration is equal to the magnitude of the tangential acceleration, which is 1.2 in/s^2.
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a particle moving along the x axis in simple harmonic motion starts from its equilibrium position, the ori- gin, at t 5 0 and moves to the right. the amplitude of its motion is 2.00 cm, and the frequency is 1.50 hz. (a) find an expression for the position of the particle as a function of time. determine (b) the maximum speed of the particle and (c) the earliest time (t . 0) at which the particle has this speed. find (d) the maxi- mum positive acceleration of the particle and (e) the earliest time (t . 0) at which the particle has this accel- eration. (f) find the total distance traveled by the par- ticle between t 5 0 and t 5 1.00 s.
(a) The position of the particle as a function of time is given by:
x(t) = A cos(2πft)
where A is the amplitude (2.00 cm), f is the frequency (1.50 Hz), and cos is the cosine function.
Substituting the given values, we get:
x(t) = 2.00 cos(3πt)
(b) The maximum speed of the particle occurs at the equilibrium position, where the displacement is zero. At this point, the velocity is maximum and is given by:
vmax = Aω
where ω is the angular frequency and is equal to 2πf. Substituting the given values, we get:
vmax = 2.00 × 2π × 1.50 = 18.85 cm/s
(c) The earliest time at which the particle has this speed is when it passes through the equilibrium position. This happens at t = 0, so the earliest time is t = 0.
(d) The maximum positive acceleration of the particle occurs at the ends of its motion, where the displacement is maximum. At these points, the acceleration is given by:
amax = Aω^2
Substituting the given values, we get:
amax = 2.00 × (2π × 1.50)^2 = 282.74 cm/s^2
(e) The earliest time at which the particle has this acceleration is when it reaches the maximum displacement. This happens at t = 1/4T, where T is the period of the motion. The period is given by:
T = 1/f = 2/3 s
So, t = 1/4T = 1/4 × 2/3 = 0.33 s
(f) The total distance traveled by the particle between t = 0 and t = 1.00 s is equal to one complete cycle of its motion. The distance traveled in one complete cycle is equal to four times the amplitude, or:
4A = 8.00 cm
Therefore, the total distance traveled is:
8.00
on a sunny day, a rooftop solar panel delivers 55 w of power to the house at an emf of 17 v . part a how much current flows through the panel? express your answer with the appropriate units.
On a sunny day, a rooftop solar panel delivers 55 w of power to the house at an emf of 17 v . the current flows through the panel is 3.235 A.
The amount of current flowing through the panel can be calculated using Ohm's Law (I = p/v). When working with the formula, Power = Voltage × Current (P = V × I), one can determine the current that flows through the panel by rearranging the formula to:
Current = Power/Voltage (I = P/V).
The calculation of the current (I) is given as follows: I = P/V = 55 W / 17 V = 3.235 A.
Therefore, 3.235 A of current flows through the solar panel on a sunny day.
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A car has an intial velocity of 50 km hr after 5 h, its final velocity is 70 km hr. solve for the car acceleration
Answer:
4 km/hr^2
Explanation:
We can use the formula for acceleration:
a = (v_f - v_i) / t
where:
a = acceleration
v_f = final velocity
v_i = initial velocity
t = time taken
Substituting the given values, we get:
a = (70 km/hr - 50 km/hr) / 5 hr
a = 20 km/hr / 5 hr
a = 4 km/hr^2
Assume that a drop of mercury is an isolated sphere. What is the capacitance in picofarads of a drop that results when two drops each of radius R = 5.61 mm merge?
The formula C=4R, where is the permittivity of open space, may be used to determine the capacitance of a merged mercury drop, assuming it is an isolated sphere. The capacitance is around 1.68 pF with R = 5.61 mm.
The formula C=4R, where R is the drop's radius and is the permittivity of free space, may be used to determine the capacitance of a merged mercury drop. As the capacitance of an isolated sphere is exactly proportional to its radius, the capacitance produced by the merger of two drops with similar radii is equal to the total of the capacitances of the individual drops. Given that the radius of the combined drop in this instance is R = 5.61 mm, the capacitance can be estimated using the formula C = 4(8.85 x 10-12 F/m) (5.61 x 10-3 m)2, yielding a capacitance of around 1.68 pF.
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the maximum energy of photoelectrons from aluminium is 2.3 ev for radiation of 2000 a and 0.90 ev for radiation of 3130 a. use this data to calculate plancks constant and the work function of aluminium
The maximum energy of photoelectrons from aluminium is 2.3 eV for radiation of 2000 Å and 0.90 eV for radiation of 3130 Å.
To calculate Planck's constant and the work function of aluminium, we need to use the equation:
h = E2 - E1/ λ2 - λ1
Where h is Planck's constant, E1 and E2 are the maximum energy of photoelectrons for each wavelength, and λ1 and λ2 are the wavelengths.
Using the given data, we have:
h = (2.3 - 0.90) / (2000 - 3130)
Therefore, h = -1.4 eV / -930 Å, which simplifies to h = 0.0015 eVÅ.
The work function of aluminium is equal to the maximum energy of the photoelectrons for the longest wavelength, in this case, 0.90 eV. Therefore, the work function of aluminium is 0.90 eV.
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