On release 1 a student releases a cart from the top of a ramp such that the cart was travel 200 cm before hitting the stopper. The cart increase the speed steadily and reaches the bottom of the ramp in 4.40 seconds. What was the magnitude of the cart acceleration on release 1 ? The student then releases the cart from a lower position such that the cart only travels 70.0 cm. What was the magnitude of the carts acceleration on release 2? How long did the car take to reach the bottom of the ramp on release 2? Please see the picture
Answers
ANSWER
0. a = 20.66 cm/s^2
,1. a = 20.66 cm/s^2
,2. t = 2.60 s
EXPLANATION
1. The initial speed in release 1 is zero. The distance the cart travels is x = 200cm
Which we can write as a function of time:
\(x=x_0+v_0t+\frac{1}{2}at^2\)We know that the cart traveled for 4.40 seconds. Replacing with the information we have:
\(\begin{gathered} 200cm=0+0\cdot4.40s+\frac{1}{2}a\cdot4.40^2s^2 \\ 200cm=\frac{1}{2}\cdot a\cdot4.40^2s^2 \end{gathered}\)Solving for a:
\(a=\frac{2\cdot200cm}{4.40^2s^2}=20.66cm/s^2\)The magnitude of the cart's acceleration is 20.66 cm/s^2, rounded tot he nearest hundredth.
2. Now the students wants the cart to travel x = 70 cm instead. If it is the same cart and the same ramp, with the same slope, then the forces acting upon the cart are the same. This means that it doesn't matter from what position of the ramp the cart is released, it will have the same acceleration every time. However in this case, it will travel for a shorter period of time.
Hence the acceleration in release 2 is the same as in release 2: 20.66 cm/s^2
3. In this case we know that x = 70cm, a = 20.66 cm/s^2 and we have to find t. From this formula:
\(x=x_0+v_0t+\frac{1}{2}at^2\)Replacing with the information we have:
\(70\operatorname{cm}=\frac{1}{2}\cdot20.66\operatorname{cm}/s^2\cdot t^2\)And solving for t:
\(t=\sqrt[]{\frac{70\operatorname{cm}\cdot2}{20.66\operatorname{cm}/s^2}}=2.60s\)We find that the cart traveled for 2.60 seconds, rounded to the nearest hundredth.
Related Questions
Car A is moving at a speed of 45km/h towards car B which is moving at a speed of 55km/h. if the two car were initially separated at a distance of 150km, determine how long it will take the two cars to meet?
Answers
The time taken by the cars to meet is 5.4 x 10³ s.
Speed of car A, v₁ = 45 km/h = 12.5 m/s
Speed of car B, v₂ = 55 km/h = 15.27 m/s
Distance between the cars, d = 150 km = 15 x 10⁴m
The expression for the time taken by the cars to meet can be given as,
Time = Distance/Average speed
t = d/(v₁ + v₂)
Applying the values of d, v₁ and v₂.
t = 15 x 10⁴/(12.5 + 15.27)
t = 15 x 10⁴/27.77
t = 5.4 x 10³ s
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If the hairdryer was put on full power, what would happen? Why?
Answers
Most hair dryers are in the 1800 watt range and lets assume you turned it to the highest speed and heat setting.
PLEASE MARK AS A BRAINLISTIt will heat up the air in the box, failry quickly. Rules of thermodynamics says that the heat will increase the temperature according to the rate of heat input minus the rate of heat loss from the box… due to conduction, radiation and convection (in this case none of the latter because the box is sealed).
HOPE IT WILL HELP YOUThe loss rate depends in the temperature difference and the thermal resistance. THicker and better insulation increases the resistance; larger area decreases the resistance. Eventually the heat (temperature) will rise and the difference between inside and outside will be so large that the temperature will rise no more. But in the case of a 1800 W hairdryer the temperature will exceed the melting point of plastics and wire insulation and if allowed to come to heat equilibrium will probably short out and catch fire or blow an external fuse.
Most modern hair dryers have a internal thermal fuse that cuts out at temperatures below the melting point and probably this will cut off the dryer before catastrophic meltdown. Its a one time fuse and not readily available, mostly you toss the dryer when the fuse goes
a 1220 kg automobile travels at 75 m/s. what net work would be required to bring it to a stop
Answers
Answer:
W = - 3431250 [N]
Explanation:
In order to solve this problem, we must use the theorem of work and energy conservation. This theorem tells us that the initial mechanical energy of a body plus the work done on this body must be equal to the final mechanical energy of the body. We must remember that the mechanical energy of a body is equal to the sum of kinetic energies plus potential energy plus elastic energy.
In this problem, we only have kinetic energy.
\(E_{1}+W_{1-2}=E_{2}\\where:\\E_{1}=E_{pot}+E_{kine}+E_{elas}\\E_{pot} = 0\\E_{elas}=0\\E_{kine}=\frac{1}{2} *m*v^{2}\)
And we have:
m = mass = 1220 [kg]
v = velocity = 75 [m/s]
As the carriage stops the final kinetic energy is zero.
Now replacing:
\(\frac{1}{2} *1220*(75)^{2} +W_{1-2}=0\\W_{1-2}= - 3431250[N]\)
Note: The negative force means that the force has to be carried out by the carriage. That is, no external force acts on the car to stop it.
Need ASAP! Please Provide the Answer. I do not need an explanation. Thank you.
An iguana runs back and forth along the ground. The horizontal position of the iguana in meters over time is shown below.
Answers
The displacement of the iguana between 0 and 5 seconds is 6 m and the distance is 15 m.
What is displacement?
The displacement of an object is the change in the position of the object.
Displacement = change in position of the iguana
Displacement = position (5 seconds) - Position (0 seconds)
Displacement = 6 m - 0 m = 6 m
Distance traveled by the iguanaThis the total path covered by iguana in the first 5 seconds.
Distance = speed x time
speed = (6 - 0) / (3 - 1) = 3 m/s
Distance traveled by 0 and 5 seconds = 5 x 3 m/s = 15 m
Thus, the displacement of the iguana between 0 and 5 seconds is 6 m and the distance is 15 m.
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the combined effect of all forces acting on an object is the ___ force
Answers
Answer:
net force
Explanation:
A penny of mass 3.10 g rests on a small 20.0 g block supported by a spinning disk with radius of 12.0 cm. The coefficients of friction between block and disk are 0.850 (static) and 0.575 (kinetic) while those for the penny and block are 0.395 (kinetic) and 0.495 (static). What is the maximum rate of rotation in revolutions per minute that the disk can have, without the block or penny sliding on the disk
Answers
Answer:
do this Q yourself because i havent read the chapter
The maximum rate of rotation in revolutions per minute that the disk can have, without the block or penny sliding on the disk is 63 rpm.
How to solveThis is calculated using the coefficient of static friction between the penny and block, which is 0.495.
The maximum angular velocity of the disk is when the force of static friction is just sufficient to prevent the penny from sliding.
This force is equal to the mass of the penny multiplied by the acceleration due to gravity, multiplied by the coefficient of static friction.
The angular velocity of the disk is then calculated from this force and the radius of the disk.
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A 1200 kg car moving +13.7 m/s makes
an elastic collision with a 3200 kg truck,
initially at rest. What is the velocity of the
car after the collision?
(Unit = m/s)
Remember: right is +, left is -
Answers
When a car collides with another object, the total momentum of the system before and after the collision must be conserved. Momentum, on the other hand, is a product of mass and velocity. To find the velocity of a car after a collision, we must first consider the initial momentum of the system before the collision and compare it to the final momentum after the collision.
The total momentum of the system before the collision is calculated as follows:P_initial = m_car x v_carP_initial = 1200 kg x 13.7 m/sP_initial = 16,440 kg*m/s Since the two cars stick together after the collision, their final velocity is the same. Let's suppose the final velocity of the cars after the collision is v_f. Then:P_final = (m_car + m_obstacle) x v_fwhere m_obstacle is the mass of the object the car collided with. Because the car is at rest after the collision, we can assume that the velocity of the object it collided with is zero. Therefore:P_final = m_car x v_fP_final = 1200 kg x v_fThe momentum of the system after the collision must be equal to the momentum of the system before the collision. That means:P_initial = P_final16,440 kg*m/s = 1200 kg x v_fv_f = 13.7 m/s - (16,440 kg*m/s / 1200 kg) v_f = 13.7 m/s - 13.7 m/s v_f = 0 m/sTherefore, the car will come to a stop after the collision.For such more question on velocity
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A 12.0-kg block is pushed across a rough horizontal surface by a force that is angled 30.0◦ below the horizontal. The magnitude of the force is 75.0 N and the acceleration of the block as it is pushed is 3.20 m/s2. What is the magnitude of the contact force exerted on the block by the surface?
Answers
Answer:
157.36 N
Explanation:
Contact force is the force which is created due to contact and it is applied on the contact point . The force applied by body on the surface is its weight .
If R be the reaction force of the ground
R = mg + F son30
= 12 x 9.8 + 75 sin 30
= 117.6 + 37.5
= 155.10 N .
friction force = f
Net force in forward direction = F cos 30 - f = ma
75cos 30 - f = 12 x 3.2
f = 65 - 38.4
= 26.6 N
Total force on the surface =√( f² + R² )
√ (26.6² + 155.1²)
= √707.56 + 24056²
=√ 24763.57
= 157.36 N.
contact force = 157.36 N .
The buoyant force that acts on a submerged object is equal to
Answers
The buoyant effect on a thing is proportional to the mass of the fluid it is displacing, according to Archimedes' principle.
What is Archimedes' principle?The balance of a body drifting in still water depends on Archimede's Principle, which asserts that a body submerged in a liquid suffers an upthrust proportional to the weight of a fluid expelled .In calm water, a body that really is free-floating experiences the downward force of gravity.
How is the Archimedes Principle applied?It is applied to ship and submarine design. It is employed in lactometers to assess the milk's purity. It's utilized in hydrometers to calculate fluid densities. In hydraulic lifts, it is utilised.
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How does cycling of matter occur in Earth’s mantle?
A. Hot, soft rock rises from the bottom of the mantle toward the top, cools, and sinks back through the mantle.
B. Hot, soft rock sinks to the mantle, cools, and rises to Earth’s crust.
C. Solid rock rises from the bottom of the mantle, cools, and sinks back through the mantle.
D. Solid rock sinks to Earth’s core and then rises to form lava.
i need help fast plz and thank you
11/24/2020
Answers
Answer:
Correct answer to this is: A
Hopefully, this helped out a bit :)
Two students, Alice and Bob, carry out an experiment to measure the speed of
cars. Alice paces out the distance between two lamp posts.
She records:
'Distance between lamp posts = 20 paces'
Bob starts to count when a car passes the first lamp post. He stops
countingwhen he thinks it has passed the second lamp post.
He records:
'My estimate for the time taken for the car to pass between the two lamp posts = 3'
Give three ways the students could improve their experimental procedure.
Answers
Explanation:
1. they could have used a measuring tape to measure the distance between the two lamps.
2. they could use another unit to measure the length.
3. they could use light gates to measure the accurate time.
1. Use a measuring tape to measure the distance instead out pacing it out.
2. Use a watch to check the time instead of counting manually.
3. Alice and Bob could stand at the two lamp posts and mark the time at which the car passes them.
If Alice and Bob try to depend on their own abilities instead of depending on properly calibrated instruments to measure distance and time, it will end up adding considerable error in their calculations which subsequently leads to poor accuracy in the final result.
Hence, by taking the three steps mentioned above, Alice and Bob shall be able to produce more reliable results in their experiment to measure speed of cars.
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I need help with this practice problem solving This practice also asks for your own opinion on it, as stated
Answers
Advantage of solar appliances:
- Reliable (it is important in order not to depen)
derive ideal gas equation for n mole of gas.
Answers
Answer:
Explanation:
Ideal gas equation- The volume (V) occupied by the n moles of any gas has pressure(P) and temperature (T) Kelvin,
the relationship for these variables PV=nRT where R gas constant is called the ideal gas law
Derivation of the Ideal Gas Equation
Let us consider the pressure exerted by the gas to be ‘p,’
The volume of the gas be – ‘v’
Temperature be – T
n – be the number of moles of gas
Universal gas constant – R
According to Boyle’s Law,
it constant n & T, the volume bears an inverse relation with the pressure exerted by a gas.
i.e. v∝1p ………………………………(i)
According to Charles’ Law,
When p & n are constant, the volume of a gas bears a direct relation with the Temperature.
i.e. v∝T ………………………………(ii)
According to Avogadro’s Law,
When p & T are constant, then the volume of a gas bears a direct relation with the number of moles of gas.
i.e. v∝n ………………………………(iii)
Combining all the three equations, we have-
v∝nTp
or pv=nRT
where R is the Universal gas constant, which has a value of 8.314 J/mol-K
the earth orbits is oval in shape.explain how the mangnitude of gravitational between the earth and sun changes as the eart moves from position a to b
Answers
As the distance between the Earth and the Sun decreases, the magnitude of gravitational force between the earth and sun increases and vice versa.
What is gravitational force?
Gravitational force is a force that attracts any two objects with mass.
According to Newton's law of universal gravitation, the force of attraction between two objects in the universe is directly proportional to the product of their masses and inversely proportional to the square of the distance between the two objects.
F = Gm₁m₂/R²
where;
m₁ is mass of Earthm₂ is mass of sunR is the distance between the Earth and SunThus, as the distance between the Earth and the Sun decreases, the magnitude of gravitational force between the earth and sun increases and vice versa.
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what the answer???????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????????
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why do the star constellations seem to move across the sky
Answers
Answer: These apparent star tracks are in fact not due to the stars moving, but to the rotational motion of the Earth. As the Earth rotates with an axis that is pointed in the direction of the North Star, stars appear to move from east to west in the sky.
Explanation: why do the star constellations seem to move across the sky
Answer:
because the world moves around and around rotating in a circle and circling around the sun so as we move the stars stay the same but to our view it looks like their moving but in reality we are.
Explanation:
Astronomers observe the motion of four planets that orbit a star similar to the Sun. Each planet follows an elliptical orbit around its star. The astronomers measure each planet's orbital period, as shown in the table.
Planet; Orbital Period (Earth days)
Planet W; 10
Planet X; 640
Planet Y; 80
Planet Z; 270
To determine the distance each planet is from the star, astronomers applied one of Kepler's three laws.
Kepler's first law: The path of each planet around a star is an ellipse, with the star at one focus. Kepler's second law: A planet sweeps out equal areas in equal amounts of time as it revolves around the star.
Kepler's third law: The square of the time for one revolution of a planet is proportional to the cube of the radius of its orbit.
Based on the table, identify the planet that is the farthest distance from the star, and indicate which of Kepler's three laws can be used to justify your answer. Enter your answer in the box provided.
Answers
Answer:
planet that is farthest away is planet X
kepler's third law
Explanation:
For this exercise we can use Kepler's third law which is an application of Newton's second law to the case of the orbits of the planets
T² = (\(\frac{4\pi ^2}{ G M_s}\) a³ = K_s a³
Let's apply this equation to our case
a = \(\sqrt[3]{ \frac{T^2}{K_s} }\)
for this particular exercise it is not necessary to reduce the period to seconds
Plant W
10² = K_s \(a_{w}^3\)
a_w = \(\sqrt[3]{ \frac{100}{ K_s} }\)
a_w = \(\frac{1}{ \sqrt[3]{K_s} }\) 4.64
Planet X
a_x = \(\sqrt[3]{ \frac{640^3}{K_s} }\)
a_x = \frac{1}{ \sqrt[3]{K_s} } 74.3
Planet Y
a_y = \(\sqrt[3]{ \frac{80^2}{K_s} }\)
a_y = \frac{1}{ \sqrt[3]{K_s} } 18.6
Planet z
a_z = \(\sqrt[3]{ \frac{270^2}{K_s} }\)
a_z = \frac{1}{ \sqrt[3]{K_s} } 41.8
From the previous results we see that planet that is farthest away is planet X
where we have used kepler's third law
ONLY ANSWER IF YOU KNOW 100% IF YOU DO A WRONG ANSWER I
What is a controlled procedure that tests the effect of one variable on another?
A. observation
B. experiment
C. model
D. argument
Answers
A controlled procedure that tests the effect of one variable on another is known as experiment.
What is experiment?An experiment is a set of actions and observations, performed in the context of solving a particular problem or question, to support or falsify a hypothesis or research concerning phenomena.
An experiment generally tests how one variable is affected by another.
Thus, a controlled procedure that tests the effect of one variable on another is known as experiment.
The correct answer is option B. "experiment"
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Answer:
The experiment.
The small spherical planet called "Glob" has a mass of 7.88×10^18 kg and a radius of 6.32×10^4 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×10^3 m, above the surface of the planet, before it falls back down.
1. What was the initial speed of the rock as it left the astronaut's hand? (Glob has no atmosphere, so no energy is lost to air friction. G = 6.67×10^-11 Nm2/kg2.)
2. A 36.0 kg satellite is in a circular orbit with a radius of 1.45×10^5 m around the planet Glob. Calculate the speed of the satellite.
Answers
Answer: The small spherical planet called "Glob" has a mass of 7.88×1018 kg and a radius of 6.32×104 m. An astronaut on the surface of Glob throws a rock straight up. The rock reaches a maximum height of 1.44×103 m, above the surface of the planet, before it falls back down.
1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.
2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.
Explanation: To find the answer, we need to know about the different equations of planetary motion.
How to find the initial speed of the rock as it left the astronaut's hand?We have the expression for the initial velocity as,\(v=\sqrt{2gh}\)
Thus, to find v, we have to find the acceleration due to gravity of glob. For this, we have,\(g_g=\frac{GM}{r^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2}= 0.132\)
Now, the velocity will become,\(v=\sqrt{2*0.132*1.44*10^3} =19.46 m/s\)
How to find the speed of the satellite?As we know that, by equating both centripetal force and the gravitational force, we get the equation of speed of a satellite as,\(v=\sqrt{\frac{GM}{r} } =\sqrt{\frac{6.67*10^{-11}*7.88*10^{18}}{1.45*10^5} } =3.624km/s\)
Thus, we can conclude that,
1) the initial speed of the rock as it left the astronaut's hand is 19.46 m/s.
2) A 36.0 kg satellite is in a circular orbit with a radius of 1.45×105 m around the planet Glob. Then the speed of the satellite is 3.624km/s.
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The tiny planet known as "Glob" has a radius of 6.32× 10^4 meters and a mass of 7.88× 10^18 kg. On Glob's surface, an astronaut launches a rock straight upward. Before falling back down, the rock rises to a maximum height of 1.44×10^3 m above the planet's surface.
1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.
2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.
Understanding the planetary motion equations is necessary in order to determine the solution.
How to determine the rock's original speed when it left the astronaut's hand?The starting velocity's expression is as follows:\(V=\sqrt{2gh}\)
So, in order to determine v, we must determine the acceleration of glob caused by gravity. We already have,\(a=\frac{GM}{r^2} =\frac{6.67*10^{-11}*7.88*10^{18}}{(6.32*10^4)^2} \\a=0.132m/s^2\)
The velocity will now change to,\(V=\sqrt{2*0.132*1.44*10^3} =19.46m/s\)
How can I determine the satellite's speed?As we are aware, the centripetal force and gravitational force are equivalent, and thus leads to the following satellite speed equation:\(v=\sqrt{\frac{GM}{r} } =3,624km/s\\where,\\M=7.88*10^{18}kg\)
Consequently, we can say that
1) The rock was moving at 19.46 m/s when it first left the astronaut's palm.
2) A 36.0 kg spacecraft is orbiting the planet Glob in a sphere with a radius of 1.45 105 meters. The satellite is moving at 3.624 km/s at that point.
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what is the physics behind why electric parallel plates move from positive to negative
Answers
The physics behind the movement of electric charges between parallel plates is based on the principles of electrostatics. Electric charges are either positive or negative, and they are affected by electric fields.
Electric fields are created by a difference in electric potential, which is measured in volts. When a voltage is applied to a set of parallel plates, the charges within the plates will be affected by the electric field, and will move in response to it.
What are electric parallel plates?When a voltage is applied to a set of parallel plates, the positive charges in the plate connected to the positive voltage will be attracted to the negative voltage, while the negative charges in the plate connected to the negative voltage will be attracted to the positive voltage.
The movement of charges between the plates is also affected by the presence of any obstacles or resistances in the electric field, such as resistance in the wire. This can slow down the movement of charges and result in a decrease in the current flowing through the circuit.
In all, the movement of charges between electric parallel plates is the result of the electric field created by a difference in electric potential, and the movement of charges is called drift velocity. The movement is also affected by the presence of resistance.
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Which correctly describes a different evolutionary stage of a star like the sun?
A) it’s forms from a cold, dusty molecular cloud
B) During a yellow giant stage, it burns carbon in its core and helium in the shell surrounding the core.
C) After leaving the main sequence, its core is stable due to electron degeneracy
D) It becomes a white dwarf after exploding as a supernova
E)During a red giant stage, its core contracts and cools
Answers
Answer:
Explanation:
The correct option that describes a different evolutionary stage of a star like the sun is:
D) It becomes a white dwarf after exploding as a supernova
This is because a star like the sun does not have enough mass to undergo a supernova explosion. After it has exhausted all the fuel in its core, it will evolve into a red giant and then a planetary nebula, leaving behind a small, hot, dense remnant known as a white dwarf. Supernovae occur in much more massive stars that have cores that can collapse to form a neutron star or black hole.
E._____________________
Answers
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Two atoms of the same element only differ because one of the atoms has more electrons, making it an ion. Which statement is true? They have the same A-number and the same Z-number. They have the same A-number but different Z-number. They have a different A-number but the same Z-number. They have different A-numbers and different Z-numbers.
Answers
The correct answer is Option B. The statement "they have the same A-number but different Z-number" is true .
Atoms of the same element only differ because one of the atoms has more electrons, making it an ion.
This difference does not affect the mass of the atom, which is determined by the sum of its protons and neutrons, represented by the atomic mass or A-number.
The number of protons in an atom is called the atomic number or Z-number.
The Z-number of an element is unique to it. All the atoms of a given element have the same number of protons.
Thus, for example, all carbon atoms have six protons, making the Z-number of carbon 6.
However, different isotopes of an element can have different numbers of neutrons.
This means that they have a different atomic mass or A-number.
Therefore, they have the same A-number but different Z-number.
Therefore the correct Option is B.
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24. A body A rests on a smooth horizontal table. Two bodies of mass 2 kg and 10 kg hanging freely, are attached to A by strings which pass over smooth pulleys at the edges of the table. The two strings are taut. When the system is released from rest, it accelerates at 2 m/s2 . Find the mass of A.
Answers
The two strings are taut. When the system is released from rest, it accelerates at 2 m/s2 then, Mass of A = 8m/5 kg.
Let the mass of the body A be ‘m’.
The two strings are taut so they exert a tension ‘T’ on body A.
Let ‘a’ be the acceleration produced in the system.
The free body diagram of body A is given below: mA + 2T = mA + ma = mA + m(2)mA + 10T = mA + ma = mA + m(2)
As the two strings are taut, we can say that tension in both strings is equal.
Therefore 2T = 10T or T = 5T As the body A is resting on a smooth horizontal table, there is no friction force acting on the body A.
The net force acting on body A is the force due to tension in the strings. ma = 2T – mg …(1)
As per the given problem, the system is released from rest.
Hence the initial velocity is zero.
Also, we are given that the system accelerates at 2 m/s2.
Therefore a = 2 m/s2 …(2)
From the equations (1) and (2), we get, m(2) = 2T – mg …(3)⇒ m(2) = 2×5m – mg⇒ 2m = 10m – g⇒ g = 8m/5
Thus, the mass of A is 8m/5 kg.
Answer: Mass of A = 8m/5 kg.
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which of the following changes will increase the frequency of an oscillating pendulum?
a. an increase in the mass of the pendulum.
b. an increase in the initial height of release.
c. an increase in the length of the rope.
d. more than one of the above
e. none of the above
explain your answer in your own words.
NO LINKS.
Answers
Answer:
b (i think)
Explanation:
The refractive indices of materials A and B have a ratio of nA/nab=1,33. The speed of light in material A is 1.25 times 10^8m/s. What is the speed of light in material B?
Answers
The speed of light in material B is 1.6625 × 108 m/s.
The refractive index of a material is its optical density relative to that of a vacuum.
Material B has a refractive index of nB, and its speed of light is vB.
The speed of light in material A is given as 1.25 x 108 m/s.
The refractive indices of materials A and B have a ratio of nA/nB = 1.33.
We will use the formula:
nA/nB = vB/vA = nA/nB.
Therefore, nA/nB = vB/1.25 x 108 m/s.
This equation can be rearranged to give the speed of light in material B:
vB = nA/nB × 1.25 x 108 m/s.
Therefore, vB = 1.33 × 1.25 × 108 m/s.
We will perform this calculation:
vB = 1.6625 × 108 m/s.
Therefore, the speed of light in material B is 1.6625 × 108 m/s.
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When a skater pulls her arms in, it
reduces her moment of inertia from
2.12 kg m² to 0.699 kg-m². If she was
initially spinning 3.25 rad/s, what is
her final angular velocity?
Answers
The skater's final angular velocity is approximately 9.86 rad/s.
The skater's final angular velocity can be calculated using the principle of conservation of angular momentum. The equation for angular momentum is given by:
L = Iω
where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Initially, the skater has an angular momentum of:
L_initial = I_initial * ω_initial
Substituting the given values:
L_initial = 2.12 kg m² * 3.25 rad/s
The skater's final angular momentum remains the same, as angular momentum is conserved:
L_final = L_initial
The final moment of inertia is given as 0.699 kg m². Therefore, the final angular velocity can be calculated as:
L_final = I_final * ω_final
0.699 kg m² * ω_final = 2.12 kg m² * 3.25 rad/s
Solving for ω_final:
ω_final = (2.12 kg m² * 3.25 rad/s) / 0.699 kg m²
Hence, the skater's final angular velocity is approximately 9.86 rad/s.
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Problem 05.086 - Water pumped from a lower reservoir to a higher reservoir Water is pumped from a lower reservoir to a higher reservoir by a pump that provides 21 kW of useful mechanical power to the water. The free surface of the upper reservoir is 45 m higher than the surface of the lower reservoir. If the flow rate of water is measured to be 0.03 m3/s, determine the irreversible head loss of the system and the lost mechanical power during this process. Take the density of water to be 1000 kg/m3. The irreversible head loss of the system is 26.35 m. The lost mechanical power in this process is kW.
Answers
Answer:
Lost Mechanical Power = 7.7565 KW
Head Loss = 26.35 m
Explanation:
First, we will find the useful mechanical power used to transport water to the higher reservoir:
\(P_{useful} = \rho ghV\)
where,
P_useful = Useful mechanical Power = ?
ρ = density of water = 1000 kg/m³
g = acceleration due to gravity = 9.81 m/s²
h = height = 45 m
V = Volume flow rate = 0.03 m³/s
Therefore,
\(P_{useful} = (1000\ kg/m^3)(9.81\ m/s^2)(45\ m)(0.03\ m^3/s)\\P_{useful} = 13243.5\ W = 13.2435\ KW\)
Now, the lost mechanical power will be:
\(Lost\ Mechanical\ Power = Total\ Mechanical\ Power - Useful\ power\\Lost\ Mechanical\ Power = 21\ KW - 13.2435\ KW\\\)
Lost Mechanical Power = 7.7565 KW
Now, for the head loss:
\(Lost\ Mechanical\ Power = \rho g(Head\ Loss)V\\Head\ Loss = \frac{Lost\ Mechanical\ Power}{\rho gV} \\\\Head\ Loss = \frac{7756.5\ W}{(1000\ kg/m^3)(9.81\ m/s^2)(0.03\ m^3/s)} \\\)
Head Loss = 26.35 m
How do very small nuclei release energy?
A. Increase size
B. Not enough info
C. Stay the same size
D. Decrease size
Answers
Have you donated blood ? If so describe your experience. If not, explain whether you would or would not donate blood and why.
Answers
Many of the rocks of the canyon wall appear as horizontal bands. What do the bands indicate about the way the rocks formed?
Answers
Given what we know, we can confirm that the bands indicate that the canyon was carved due to erosion caused by moving water.
What do these bands tell us about the canyon?The bands are the result of sediments being deposited as the canyon was being formed. These sediments are deposited by the running water that carves the canyon. This results in horizontal bands that can tell us much about the history of the canyon as well as the river that formed it.
Therefore, we can confirm that the bands indicate that the canyon was carved due to erosion caused by moving water.
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