Astronaut on the first trip to mars take along a pendulum that has different period on earth and mars and the free fall acceleration on mars is 3.67m/s².
given:
period on earth T₁ = 1.50 secperiod on mars T₂ = 2.45sec
To calculate free fall acceleration on mars (g₂),
where g₁ free fall acceleration on earth and g₂ free fall acceleration on mars.
The period of the pendulum on the earth is obtained by the equation:
T₁ = 2π√(L/g₁)
The period of the pendulum on the mars is obtained by the equation:
T₂ = 2π√(L/g₂)
On dividing the above equation we get;
T₁/T₂ = √(g₂/g1)
On squaring both the sides,
T₁²/T₂² = g₂/g₁
(1.50)²/(2.45)² = g₂/g₁
g₂ = 2.25/6.0025(9.8)
g₂ = 3.67m/s²
The free fall acceleration g₂ on mars is obtained 3.67m/s².
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15. The diagram shows an electric circuit including a photocell.
The photocell contains a metal plate X that is exposed to
electromagnetic radiation. Photoelectrons emitted from the
surface of the metal are accelerated towards the positive
electrode Y. A sensitive ammeter measures the current in the
circuit due to the photoelectrons emitted by the metal plate
X. The metal plate X has a work function of 2.2 eV. The
maximum kinetic energy of an emitted photoelectron from
this plate is 0.3eV.
G
Electrong
radiation
H
vacuum
a) Calculate the energy of a single photon in eV and in joules.
b) Calculate the frequency of the incident electromagnetic radiation.
c) Deduce the effect on the current if the radiation has the same intensity, but the frequency is
greater than in (b).
16. A negatively charged metal plate is exposed to electromagnetic radiation of frequency (f). The
diagram below shows the variation with (f) of the maximum kinetic energy KEmax of the
photoelectrons emitted from the surface.
The flight of a kicked football follows the quadratic function f(x)=−0. 02x2+2. 2x+2, where f(x) is the vertical distance in feet and x is the horizontal distance the ball travels. How far, in feet, will the ball travel across the field by the time it hits the ground? Round your answer to one decimal place
The turning point's y-value is indeed the largest. As a consequence, we must discover the greatest quadratic function indicating that perhaps the ball would achieve a height limit of 72.5 feet. Which indicates:
[tex]x = 110.90[/tex] or [tex]x = -0.90[/tex]
Given that we're working with time,
[tex]x = 110.90[/tex]
We have 110.9 feet to the nearest decimal place.
Why is it referred to as a quadratic function?
A quadratic issue is a type of challenge in mathematics that deals with something like a variable multiplication by itself, which is defined as squaring. This language stems from the fact that the area of a square is equal to its line segment multiplied by itself. The term "quadratic" is derived from quadratum, which is the Latin word meaning square.
Describe the three different kinds of quadratic functions?
Quadratics are typically utilized in three ways:
1)Standard Form: y = an x 2 + b x + c y=ax2+bx+c y=ax2+bx+c y=ax2+bx+c.
2)Factored Form: y = a (x r 1) (x r 2) y=a(x-r 1)(x-r 2) y=a(xr1)(xr2) y=a(xr1)(xr2).
3)Vertex Form: y = a (x h) 2 + k y=a(x-h)2+k y=a(xh)2+k.
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convert a anthracite coal price of $90/ton to $/mmbtu. heat content of anthracite coal is 15,000 btu/pound.A: $3.00/MMBtu B: $34.50/MMBtu C; $2.80/MMBtu D: $6.42/MMBtu E: $3.46/MMBtu
The anthracite coal price per MMBtu is $3.00/MMBtu. Therefore, the correct option is A.
The heat content of anthracite coal is 15,000 BTU/pound. The question requires the conversion of an anthracite coal price of $90/ton to $/MMBtu. Let us use the given data to solve the problem. 1 ton equals 2000 pounds (lb). Hence,
2000 lb of anthracite coal has a heat content of 2000 lb x 15,000 BTU/lb = 30,000,000 BTU
Thus, the price per MMBtu can be calculated by the given formula;
Price per MMBtu = Price per ton / (BTUs per ton / MMBTUs per ton)
Price per MMBtu = $90 / (30,000,000 / 1,000,000)
Price per MMBtu = $90 / 30
Price per MMBtu = $3.00/MMBtu
Thus, the price per MMBtu of anthracite coal is $3.00/MMBtu. Therefore, the correct answer is option A: $3.00/MMBtu.
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during a ski trip, one of your friends becomes injured on the slope, which makes an angle of 18o with the horizontal. you plan to slide some medical supplies a distance of 15 m up the (frictionless) hill. how fast (in m/s) do you need to push the box to ensure it will reach your friend?
The speed required to push the box is approximately 11.27 m/s.
Given the angle of inclination of the slope, θ = 18°. The distance covered is, d = 15 meters.
The component of gravity parallel to the slope is, g|| = g.sinθ
The force required to push the box up the slope is, F = m.g, The acceleration due to gravity is, g = 9.81 m/s².
The weight of the box is, m.g. The component of gravity parallel to the slope is, g|| = g.sinθ.
The force required to push the box up the slope is, F = m.g||
The force required to push the box up the slope is,F = m.g.sinθ.
The frictionless surface has no frictional force acting upon the object moving upon it. Therefore, no work is done by friction on the object. Hence, the energy of the object in the form of kinetic energy is conserved. This is as given below:
K.E. initial + work done = K.E. finfinalitial kinetic energy is zero. This is because the box is initially at rest. The final kinetic energy is K. This is because we need to push the box up the slope with the required force. The work done is equal to the force exerted multiplied by the distance moved in the direction of the force.
Hence,W = F.dThe final equation becomes,0 + F.d = ½.m.K²The final speed required is,K = √(2.F.d/m)Substituting the values in the above equation, we get the required final speed,K = √(2.m.g.sinθ.d/m)K = √(2.g.sinθ.d)Putting the values of d, g, and θ in the equation, we get,K = √(2.9.81.sin18°.15)≈ 11.27 m/s.
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A wave traveling at 4 m/s is measured to have an amplitude of 0. 5m and a wavelength of 1. 5m. What is its frequency?
The frequency of the wave is 2.66Hz.
It is given that,
Wave velocity = Vw = 4m/s
Wavelength = λ = 1.5m
Frequency = f = ?
The propagation rate The wave travels one wavelength in one period of time, or Vw, the distance it covers in a given amount of time. It is represented by the equation,
Vw = f λ
f = Vw/λ
f = (4m/s)/1.5m
f = 2.66Hz
Therefore, the frequency of the wave would be 2.66Hz.
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Objects in motion tend to stay in motion, while objects at rest, tend to
A start moving
B bounce
C roll
D stay at rest
Data: Magnitude of the charge on the electron = 1.60 × 10-¹9 C How long does it take for a current of 6.0 A to deliver 1.5 x 10¹7 Cu²+ ions in a solution? Assume these ions are the only charged particles moving.
Answer: maybe read your book
Explanation:
tom and his group were making a poster for their classroom to depict the way that electricity and magnetism work together which of the following should the group to their poster
Currents can be created by magnetic fields. A current is capable of creating a magnetic field.
What are the interactions between electricity and magnetism?Magnetic fields are constantly shifting, pushing and pulling electrons. Metals like copper and aluminium have slackly held electrons. As a magnet is moved around a wire or a coil of wire, the electrons in the wire are pushed, creating an electrical current.
Is sound energy potential or kinetic?Sound energy can be made up of both kinetic and potential energy. A musical instrument is one possible illustration. When the instrument is performed, sound waves are produced that have kinetic energy. Nevertheless, potential energy is all that is present when that same musical instrument is at rest.
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the magnitude of the electric field in an em wave is doubled. what happens to the intensity of the wave?
a. Nothing
b. It doubles
c. It quadruples
d. It decreases by a factor of 2
e. It decreases by a factor of 4
When the magnitude of the electric field in an EM wave is doubled, the intensity of the wave increases by a factor of 4. The correct option is c. It quadruples.
The intensity (I) of an electromagnetic wave is given by:
I = (1/2)ε0cE^2
where ε0 is the electric constant, c is the speed of light, and E is the magnitude of the electric field.
If the magnitude of the electric field in an electromagnetic wave is doubled, the intensity of the wave will increase by a factor of four (4), because:
I' = (1/2)ε0c(2E)^2 = 4(1/2)ε0cE^2 = 4I
The correct answer is (c) It quadruples.
Electromagnetic waves are waves that are produced by oscillating electric and magnetic fields. An electromagnetic wave is composed of electric and magnetic fields that oscillate perpendicularly to each other and to the direction of wave propagation. Light, microwaves, X-rays, and radio waves are all examples of electromagnetic waves.
The power transferred per unit area by an electromagnetic wave is known as the intensity of the wave. The magnitude of the electric field in an EM wave is related to its intensity. When the magnitude of the electric field in an EM wave is doubled, the intensity of the wave quadruples.
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a 0.213 kg baseball is dropped from rest. it has a momentum of 0.85 kg just before it hits the ground. for what amount of time was the ball in the air?
The amount of time the baseball was in air for at a momentum of 0.85kgm/s is 0.399s
How to calculate time using momentum?Momentum is the product of its mass and velocity, or the vector sum of the products of its masses and velocities. It can be calculated as follows:
Momentum = mass (m) × velocity (v)
However, the momentum of an object can be estimated using the following;
ΔM = F × ΔT
Where;
m = massF = forcet = timeF = mass × acceleration
F = 0.213kg × 10m/s² = 2.13N
0.85 = 2.13 × t
t = 0.399s
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one other racer was 6.5 m ahead when the winner started to accelerate, but was too tired to speed up and traveled at 12.2 m/s until the finish line. if the winner continues to cycle at the same speed after crossing the finish line (to celebrate his victory), how far ahead of the loser will the winner be, in meters, when the loser finishes?
If the winner started the acceleration first, he will be 31.24 m ahead of the loser when the loser finishes.
Let the winner's initial position be at x = 0. Assume that the loser is a distance d away from the winner's initial position. Therefore, the loser's initial position is at x = -d.The winner starts to accelerate when the other racer is 6.5 m ahead of him. When the winner reaches a speed of 12.2 m/s, the other racer has traveled a distance of 6.5 m.
The time it takes the winner to catch up to the other racer can be found using the equation:
x = vt + 1/2at^2
where x = 6.5 m, v = 12.2 m/s, and a = 0.
The time t is: 6.5 m = 12.2 m/s × t
Thus, t = 0.533 s.
From that time on, the winner continues to cycle at a constant speed of 12.2 m/s until the finish line. The distance the winner travels between catching up to the other racer and crossing the finish line is: Distance traveled by the winner = 12.2 m/s × (18.3 s – 0.533 s) = 224.43 m. Distance between the winner and the other racer after crossing the finish line is: 12.2 m/s × (18.3 s – 15.0 s) = 40.26 m.
Hence, the distance between the loser and the finish line is (65.0 m – 40.26 m) = 24.74 m. Distance the winner will travel before the loser finishes: 24.74 m + 6.5 m = 31.24 m.
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a spring has a natural length of 10 cm. it takes 8 j to stretch the spring to 15 cm. how much work (in j) would it take to stretch the spring from 15 cm to 20 cm? j
The required work done in stretching the spring from 15 cm to 20 cm is calculated to be 11.2 j.
The original length of the spring is given as 10 cm.
The work done to stretch the spring to 15 cm is 8 j.
Let us find out the value of k spring constant.
W = 1/2 k x²
where,
W is work
k is spring constant
x is elongation of spring
Putting in the values,
8 = 1/2 k (15² - 10²)
8 = 1/2 k (225 - 100)
8 = 1/2 k(125)
k = 8×2/125 = 0.128
Now, let us calculate the work done in stretching the spring from 15 cm to 20 cm.
W = 1/2 k x² = 1/2 (0.128) [(20² - 10²)-(15² - 10²)] = 0.064 [300-125] = 11.2 j
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You have a rope with length 4m. What is the first harmonic
wavelength on the rope?
The wavelength of a wave is the distance between two consecutive points on the wave that are in phase.
The wavelength is the distance between two consecutive nodes or two consecutive antinodes.
The first harmonious wavelength on a rope is given by the equation
λ = 2L/ n
where λ is the wavelength, L is the length of the rope, and n is the harmonious number. For the first harmonious, n = 1.
In this case, the length of the rope is L = 4m. Substituting into the equation over, we get
λ = 2( 4m)/ 1
λ = 8m
thus, the first harmonious wavelength on the rope is 8 measures.
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why is the direction for the force of static friction of the wheel on an incline different from the force of static friction of the wheel on a flat surface.
The direction of the force of static friction on an incline is perpendicular to the surface, while on a flat surface, it is parallel to the surface. This is because the incline introduces a component of the gravitational force that acts perpendicular to the surface.
On a flat surface, the force of static friction is parallel to the surface and opposite in direction to the applied force. However, on an incline, the component of gravitational force perpendicular to the surface creates a normal force that is also perpendicular to the surface. The force of static friction acts perpendicular to both the normal force and the inclined surface since it's always perpendicular to the normal force. This change in direction occurs because the force of static friction is a reactionary force that opposes the motion, and its direction depends on the forces acting on the object. Therefore, the direction of the force of static friction changes based on the orientation of the surface on which the object is placed.
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13.
a) Einstein's photoelectric equation may be written as: hf = 0 + mvmax²
Identify the terms hf, and mvmax².
2
b) The surface of sodium metal is exposed to electromagnetic radiation of wavelength
6.5 x 10-7 m. This wavelength is the maximum for which photoelectrons are released.
i. Calculate the threshold frequency.
ii. Show that the work function energy of the metal is 1.9 eV.
c) For a particular wavelength of incident light, sodium releases photoelectrons. State how the
rate of releases of photoelectrons changes with the intensity of light is doubled. Explain
your answer.
Answer:
13. a) In Einstein's photoelectric equation, hf represents the energy of a single photon of the electromagnetic radiation, and mvmax² represents the maximum kinetic energy of the emitted photoelectrons.
b) i. To calculate the threshold frequency, we can use the equation c = fλ, where c is the speed of light, f is the frequency of the electromagnetic radiation, and λ is the wavelength:
c = fλ
f = c/λ
f = (3.00 x 10^8 m/s)/(6.5 x 10^-7 m)
f = 4.62 x 10^14 Hz
Therefore, the threshold frequency is 4.62 x 10^14 Hz.
ii. We can use the equation hf = Φ + KEmax, where hf is the energy of a single photon, Φ is the work function energy of the metal, and KEmax is the maximum kinetic energy of the emitted photoelectrons. We know that the wavelength of the incident electromagnetic radiation is 6.5 x 10^-7 m, which corresponds to a frequency of f = 4.62 x 10^14 Hz (as calculated in part i). We also know that this wavelength is the maximum for which photoelectrons are released, which means that the energy of the photons is equal to the work function energy:
hf = Φ
Substituting the values for h and f, we get:
(6.63 x 10^-34 J s)(4.62 x 10^14 Hz) = Φ
Φ = 1.93 x 10^-19 J
Converting this to electronvolts (eV), we get:
Φ = (1.93 x 10^-19 J)/(1.60 x 10^-19 J/eV)
Φ = 1.21 eV
Therefore, the work function energy of the metal is 1.9 eV.
c) If the intensity of the incident light is doubled, the rate of release of photoelectrons will also double. This is because the intensity of light is directly proportional to the number of photons incident on the metal surface per unit time. Each photon can cause the emission of one photoelectron, so doubling the number of photons will double the number of emitted photoelectrons per unit time. However, the kinetic energy of the emitted photoelectrons will not change, since this is determined only by the frequency (and therefore the energy) of the incident photons, and not by their intensity.
Imagine you live in a Martian colony and are tasked with using geometric parallax to determine a specific star’s (a) distance away and (b) intrinsic luminosity. You’re familiar with such methods from Earth. Briefly explain the steps you will take to complete this task, including the measurements you must make. Then identify the way the use of one of these methods would differ on Mars compared with Earth.
Select a white dwarf or other star that is known to be relativity close and steady in its brightness as your reference point. When Mars is on separate sides of the sun, see the target star twice, six months apart.
How can we calculate an astronomical object's distance using the parallax method?The parallax angle can be used to calculate distance. It is occupied at the two sites on Earth that are separated by b by the far-off celestial object.
What does the parallax method—which measures the separation between Earth and planet—entail?When measuring the distance between the earth and other planets and stars using the parallax method, we estimate the position of a first from one position, then from another. Calculating the distance from a star or planet involves measuring the amount of shift that occurs in its location.
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what is the difference in momentum between a 50.0 kg runner moving at a speed of 3.00 m/s and a 3000 kg truck moving at a speed of only 1.00 m/s?
Answer:
It would be 2850 kg m/s
Explanation:
two objects are sliding on ice and moving in the same direction. the first object has a mass of 4.71 kg and is moving at 27.2 m/s when it collides with the second object that has a mass of 7.86 kg and is moving at 13.7 m/s. after the collisions the objects stick together. what is the final speed of the second object after the collision?
The final speed of the second object after the collision is 27.91 m/s.
To solve this problem, we need to use the conservation of momentum principle, Before the collision, the total momentum of the system is:
[tex]P = m1v1 + m2v2\\\P = (4.71 kg)(27.2 m/s) + (7.86 kg)(13.7 m/s)\\\P = 236.56 kgm/s + 107.82 kgm/s\\\P = 344.38 kg*m/s[/tex]
After the collision, the two objects stick together and move with a common velocity, which we will call v. Therefore, the total momentum of the system after the collision is:
[tex]P' = (m1 + m2)*v\\\\\P' = (4.71 kg + 7.86 kg)v\\\P' = 12.57 kgv[/tex]
Since the total momentum of the system is conserved, we can equate the two expressions for the momentum:
[tex]P = P'\\\236.56 kgm/s + 107.82 kgm/s = 12.57 kgv\\\\\v = (236.56 kgm/s + 107.82 kg*m/s) / 12.57 kg\\\\\v = 27.91 m/s[/tex]
Therefore, the final speed of the second object after the collision is 27.91 m/s.
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A student connects a coil of insulated wire to a galvanometer. What would be seen if a bar magnet with its north pole towards one face of the coil is:(a) moved quickly towards it?(b) moved quickly away from the coil?(c) placed near the face of the coil?Name the phenomenon involved.
The galvanometer briefly swung to one side. The galvanometer is in the other way. There is no galvanometer deflection. Electromagnetic induction is the relevant phenomenon. The galvanometer is in the opposite direction from that in the previous case. There is no galvanometer deflection.
The observations that will occur for each choice when a coil of insulated copper wire is coupled to a galvanometer are as follows:
(I) Electromagnetic induction will cause the coil to create an electric current if a bar magnet is put into it.
(ii) If a bar magnet is removed from the inside, electromagnetic induction will again induce current in the copper wire, but this time the direction of the current in the galvanometer will be reversed.
(iii) If a bar magnet is kept stationary inside the coil, no current is produced, and as a result, the galvanometer does not deflect.
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why should the ammeter and voltmeter be read simultaneously to determine the resistance of a circuit component?
The ammeter and voltmeter should be read simultaneously to determine the resistance of a circuit component because the resistance depends on the current passing through the component and the voltage drop across it. By measuring both the current and voltage, Ohm's law (V=IR) can be applied to calculate the resistance of the component.
The resistance of a circuit component represents how much it impedes the flow of current through it. According to Ohm's law, the voltage drop across a component is directly proportional to the current flowing through it, and the constant of proportionality is the resistance of the component. Therefore, to determine the resistance of a component, we need to know both the current passing through it and the voltage drop across it.
An ammeter is used to measure the current flowing through a circuit, while a voltmeter is used to measure the voltage difference between two points in the circuit. By reading both instruments simultaneously, we can use Ohm's law to calculate the resistance of the component being measured.
It is important to note that the readings must be taken simultaneously because the current and voltage can change over time, and any delay between taking the readings can result in inaccurate measurements. Therefore, it is necessary to read the ammeter and voltmeter at the same time to obtain an accurate measurement of the resistance of the component.
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when you place a thumbtack 54.0 cm c m in front of a lens, the resulting image is real, inverted, and the same size as the thumbtack. is the lens converging or diverging?
When you place a thumbtack 54.0 cm in front of a lens, the resulting image is real, inverted, and the same size as the thumbtack, then the lens in this case is a converging lens.
Converging lenses are thick in the middle and thin at the edges. They are convex in shape, which means they bulge outwards. Converging lenses can converge light rays to a point on the other side of the lens. They are also known as convex lenses, and they have a positive focal length.
The image produced by a converging lens can be real or virtual depending on the position of the object relative to the lens. The image of the object is real when the object is placed beyond the focal point of the lens, as in this situation.
Therefore, when you place a thumbtack in front of a lens, the resulting image is real, inverted, and the same size as the thumbtack. that means the lens is a converging lens.
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What are the causes of an earthquake?
100- 150 worded geographic terms and concept paragraph.
Answer:
Explanation:
An earthquake, with its sudden and cataclysmic rupturing of the earth's crust, is a sublime exhibition of the extraordinary might of nature. This geological upheaval arises from the movements of tectonic plates - vast and ponderous slabs of rock comprising the earth's outermost layer. These plates glide past each other, collide, or diverge, accumulating seismic energy that they discharge as seismic waves, shaking the earth to its core.
The factors that underlie an earthquake are complex and multifarious, spanning the entire spectrum of geological events, from volcanic activity to anthropogenic interventions such as drilling and mining. Nevertheless, the dominant causal factor is the intricate interplay of tectonic plates, ceaselessly engaged in a dynamic dance of motion, friction, and release, culminating in an earthquake's fearsome display of raw power.
Despite their potential for devastation, comprehending the causes of earthquakes can inform and enhance our preparedness for these events, and equip us with the knowledge to mitigate their impact on human society. It may also allow us to deepen our appreciation of the natural world's enigmatic processes, and our place within it, engendering a richer understanding of our planet and its precarious balance.
describe the behavior of the beam in the uniform field, particularly how it differs from the bar magnet induced behavior.
In a uniform field, a beam has no magnetic poles and therefore experiences a force equal to the product of its length and the field intensity. whereas bar magnet is influenced by the interaction between its magnetic poles and the field intensity.
When placed in a uniform field, the bar magnet experiences a torque, which causes it to align itself with the field direction. In contrast, a beam placed in a uniform field experiences a uniform force, which acts parallel to the field direction and perpendicular to the beam length. This force tends to cause the beam to move in the direction of the field intensity. The magnitude of the force on the beam is proportional to the length and field intensity, as well as the product of its magnetic permeability and the field intensity.
The force on the beam also tends to cause it to rotate about its center, in the direction of the field intensity. This rotation will result in the beam becoming oriented along the field direction, but it will not cause the beam to align itself with the field direction as with a bar magnet. Instead, the beam will experience a net force, which tends to move it in the direction of the field intensity.
In conclusion, the behavior of a beam in a uniform magnetic field differs from the behavior of a bar magnet in several ways. A beam placed in a uniform field experiences a uniform force, which acts parallel to the field direction and perpendicular to the beam length. This force tends to cause the beam to move in the direction of the field intensity and rotate about its center, in the direction of the field intensity.
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ppositely
Which best describes the result of moving the charge to
the point marked X?
OIts electric potential energy increases because it has
the same electric field.
OIts electric potential energy increases because the
electric field increases.
Its electric potential energy stays the same because
the electric field increases.
OIts electric potential energy stays the same because
it has the same electric potential.
The statement that best describes the result of moving the charge to the point marked X is: B. Its electric potential energy increases because the electric field increases.
Which best describes the result of moving the charge to the point marked X?Moving the charge to the point marked X will change its electric potential energy because the electric field at that point is different from the electric field at its initial position.
At the initial position of the charge, the electric potential energy is determined by the electric potential at that point and the charge of the object. When the charge is moved to point X, the electric potential at that point changes, which means that the electric potential energy of the charge also changes. This is because the electric potential at a point is directly proportional to the electric field at that point.
Therefore, since the electric field is different at point X than at the initial position, the electric potential energy of the charge increases as it moves to point X.
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The height of a man is 5 feet. He observes the depth of swimming pool 5 feet. Is it wise to jump in the swimming pool to swim if he is not perfect on swimming ?
Answer:
no
Explanation:
because when he jumps into the swimming pool he will not be able to swim comfortably
I'm having a really hard time trying to solve this
A block with mass mb = 1.3 kg is connected by a rope across a 50-cm-diameter, 2.0 kg pulley, as shown in (Figure 1). There is no friction in the axle, but there is friction between the rope and the pulley; the rope doesn't slip and the pulley can be modeled as a solid cylinder. The weight is accelerating upward at 1.2 m/s^2.
What is the tension in the rope on the right side of the pulley?
Answer:
Approximately [tex]15.5\; {\rm N}[/tex], assuming that [tex]g = 9.81\; {\rm N \cdot kg^{-1}}[/tex].
Explanation:
To find the tension in the rope on the right side of the pulley, apply the following steps:
Find the tension that the rope exerts on the block, which is equal to the tension [tex]T_{\text{left}}[/tex] on the left side of the pulley.Find the torque [tex]\tau_{\text{left}}[/tex] resulting from the tension [tex]T_{\text{left}}[/tex] on the left side of the pulley.Find the moment of inertia [tex]I[/tex] of the pulley and the net torque [tex]\tau_{\text{net}}[/tex].Add the torque on the left [tex]\tau_{\text{left}}[/tex] to the net torque [tex]\tau_{\text{net}}[/tex] to find [tex]\tau_{\text{right}}[/tex], the torque on the right side of the pulley. Divide [tex]\tau_{\text{right}}[/tex] by radius of the pulley [tex]r[/tex] to find the tension on the right side, [tex]T_{\text{right}}[/tex].
The net force on the block is:
[tex]F_{\text{net}} = m_{\text{b}} \, a[/tex], where
[tex]m_{\text{b}} = 1.3\; {\rm kg}[/tex] is the mass of the block, and[tex]a = 1.2\; {\rm m\cdot s^{-2}}[/tex] is the linear acceleration of the block.At the same time, the net force on the block can also be expressed as:
[tex]\begin{aligned}F_{\text{net}} &= T_{\text{left}} - (\text{weight}) \\ &= T_{\text{left}} - m_{\text{b}}\, g \end{aligned}[/tex], where
[tex]g = 9.81\; {\rm N\cdot kg^{-1}}[/tex] by assumption, and[tex]T_{\text{left}}[/tex] is the tension the rope exerted on the block. This tension is equal to the tension on the left side of the pulley.Rearrange and solve for [tex]T_{\text{left}}[/tex]:
[tex]T_{\text{left}} - m_{\text{b}}\, g = F_{\text{net}} = m_{\text{b}}\, a[/tex].
[tex]\begin{aligned}T_{\text{left}} &= m_{\text{b}}\, a + m_{\text{b}}\, g \\ &= m_{\text{b}}\, (a + g) \\ &= 1.3\, (1.2 + 9.81)\; {\rm N} \\ &= 14.313\; {\rm N}\end{aligned}[/tex].
Let [tex]r[/tex] denote the radius of the pulley. It is given that the diameter of the pulley is [tex]50\; {\rm cm}[/tex]. In standard units, the radius of the pulley would be [tex]r = 25\; {\rm cm} = 0.25\; {\rm m}[/tex].
On the left side of the pulley, tension in the rope exerts a torque of [tex]\tau_{\text{left}} = T_{\text{left}}\, r[/tex] on the pulley:
[tex]\begin{aligned}\tau_{\text{left}} &= T_{\text{left}}\, r \\ &= (14.313)\, (0.25)\; {\rm N\cdot m} \\ &= 3.57825\; {\rm N\cdot m} \end{aligned}[/tex].
Under the assumptions, the moment of inertia [tex]I[/tex] of this cylindrical pulley would be:
[tex]\begin{aligned} I &= \frac{1}{2}\, m\, r^{2} \end{aligned}[/tex], where
[tex]m = 2.0\; {\rm kg}[/tex] is the mass of the pulley, and[tex]r = 0.25\; {\rm m}[/tex] is the radius of the pulley.[tex]\begin{aligned} I &= \frac{1}{2}\, m\, r^{2} \\ &= \frac{1}{2}\, (2.0)\, (0.25)^{2}\; {\rm kg \cdot m^{2}} \\ &= 0.0625\; {\rm kg\cdot m^{2}} \end{aligned}[/tex].
Since the rope doesn't slip on the pulley, linear acceleration of the pulley would be equal to that of the rope, [tex]a = 1.2\; {\rm m\cdot s^{-2}}[/tex]. Divide this linear acceleration by the radius of the pulley to find the angular acceleration [tex]\alpha[/tex] of the pulley:
[tex]\begin{aligned}\alpha &= \frac{a}{r} \\ &= \frac{1.2}{0.25}\; {\rm s^{-2}} \\ &= 4.8\; {\rm s^{-2}}\end{aligned}[/tex].
Multiply angular acceleration by the moment of inertia to find the net torque [tex]\tau_{\text{net}}[/tex] on the pulley cylinder:
[tex]\begin{aligned}\tau_{\text{net}} &= I\, \alpha \\ &= (0.0625)\, (4.8)\; {\rm kg \cdot m^{2}\cdot s^{-2}}\\ &= 0.3\; {\rm kg \cdot m^{2} \cdot s^{-2}} \end{aligned}[/tex].
Note that the net torque of the pulley [tex]\tau_{\text{net}}[/tex] is in the same direction as [tex]\tau_{\text{right}}[/tex], but the opposite of [tex]\tau_{\text{left}}[/tex]. Hence:
[tex]\begin{aligned}\tau_{\text{right}} &= \tau_{\text{net}} + \tau_{\text{left}} \\ &= 0.3\; {\rm N\cdot m} + 3.57825\; {\rm N\cdot m} \\ &= 3.87825\; {\rm N\cdot m}\end{aligned}[/tex].
Divide the torque on the right [tex]\tau_{\text{right}}[/tex] by radius [tex]r[/tex] to find the tension in the string on the right [tex]T_{\text{right}}[/tex]:
[tex]\begin{aligned}T_{\text{right}} &= \frac{\tau_{\text{right}}}{r} \\ &= \frac{3.87825}{0.25}\; {\rm N} \\ &= 15.513\; {\rm N}\end{aligned}[/tex].
a car moving at constant speed around a circular track at a speed of v. the force of friction provides the necessary centripetal force to keep the car on the track. if the speed of the car is doubled, what will be the frictional force that is needed to hold the car on the road? g
The frictional force required to keep the car on the circular track will increase by a factor of four when the speed of the car is doubled.
The centripetal force required to keep a car moving in a circular path is given by the formula F = mv²/r, where m is the mass of the car, v is its speed, and r is the radius of the circular path. In this case, the force of friction provides the necessary centripetal force to keep the car on the track.
When the speed of the car is doubled, the centripetal force required to keep it on the circular path increases by a factor of four, because the velocity appears squared in the formula for centripetal force. Therefore, the frictional force needed to keep the car on the road must also increase by a factor of four.
This means that if the original frictional force was F, then the new frictional force needed will be 4F.
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a trucker sees the image of a car passing her truck in her diverging rear view mirror whose focal length is-60 cm. If the car is 1.5 m high and 6.0 m away, what is the size and location of the image?
The size of the image is 20 times smaller than the size of the car, and the image is located 120 cm behind the mirror.
What is length?Length is a measure of the magnitude of a line segment, straight or curved, between two points in space. It is a fundamental physical quantity in many fields of mathematics, physics, and engineering. Length is usually measured in units of distance such as inches, feet, meters, and kilometers. Length is an important concept in geometry, which is the study of shapes and their properties. In geometry, length is used to describe the size of lines, angles, and other shapes. Length can also be used to measure the size and distance between two points in space.
The size and location of the image of the car in the trucker’s diverging rear view mirror is determined by the focal length of the mirror. The focal length of the mirror is -60 cm. Using the thin lens equation, the image distance can be calculated as:
1/f = 1/di + 1/do
1/-60 = 1/di + 1/6
di = -120 cm
The height of the image can be calculated using the magnification formula:
m = -di/do
m = -120/6
m = -20
Therefore, the size of the image is 20 times smaller than the size of the car, and the image is located 120 cm behind the mirror.
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the bolts on the cylinder head of certain engines require tightening to a torque of 90 nm. if a wrench is 20 cm long, what force perpendicular to the wrench must the mechanic exert at its end?
The bolts on the cylinder head of certain engines require tightening to a torque of 90 nm. The force perpendicular to the wrench that the mechanic must exert at its end is 450 N.
What is a torque?Torque is a measure of how much a force acting on an object causes the object to rotate. One component of torque is force, the other is distance. If the distance between the point where the force is applied and the point where the object rotates is shorter, a higher torque can be produced using less force.The torque formula is given by τ = r × F, where τ is the torque, r is the lever arm, and F is the force.
What is the torque formula?The formula for torque is as follows:
Torque = Force x Distance perpendicular to the applied force
Where:
Force is the amount of force applied to the lever.Distance perpendicular to the applied force is the distance between the applied force and the pivot point.
What is the torque calculation for the given question?Given that:
torque = 90 Nm r = 20 cm.To convert cm to m, divide by 100. So, the distance perpendicular to the applied force is 0.2 m.
torque = force x distance perpendicular to the applied force90 Nm = force x 0.2 mforce = 90 Nm/ 0.2 m= 450 NThus, the force perpendicular to the wrench that the mechanic must exert at its end is 450 N.
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a point located on the second hand of a large clock has a radial acceleration of 0.14 cm/s2. how far is the point (in cm) from the axis of rotation of the second hand?
The radial acceleration of a point on the second hand of a large clock is 0.14 cm/s². Using the formula for radial acceleration, the distance of the point from the axis of rotation is approximately 11.3 cm.
The radial acceleration of a point on a rotating object is the acceleration that points towards the center of rotation. This acceleration arises due to the object's circular motion and is given by the formula a = v²/r, where a is the radial acceleration, v is the tangential velocity of the object, and r is the radius of the circular path. In this case, we know that the radial acceleration of the point on the second hand of the clock is 0.14 cm/s². Since the second hand completes one full revolution in 60 seconds, we can find its tangential velocity using the formula v = 2πr/T, where T is the time taken for one revolution. Substituting T = 60 seconds, we get v = 0.1047 cm/s. We can now use the formula for radial acceleration to find the radius of the circular path followed by the point on the second hand. Rearranging the formula, we get r = v²/a, which gives us a value of approximately 11.3 cm for the distance of the point from the axis of rotation.
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