Answer:
[tex]s=5.79\ km[/tex]
[tex]\theta=47^{\circ}[/tex] east of south
Explanation:
Given:
distance of the person form the initial position, [tex]d'=8.4\ km[/tex]direction of the person from the initial position, [tex]47^{\circ}[/tex] north of eastdistance supposed to travel form the initial position, [tex]d=5.3\ km[/tex]direction supposed to travel from the initial position, due NorthNow refer the schematic for visualization of situation:
[tex]y=d'.\sin47^{\circ}-d[/tex]
[tex]y=8.4\times \sin47-5.3[/tex] ...............(1)
[tex]x=d'.\cos47^{\circ}[/tex]
[tex]x=8.4\times \cos47^{\circ}[/tex] .................(2)
Now the direction of the desired position with respect to south:
[tex]\tan\theta=\frac{y}{x}[/tex]
[tex]\tan\theta=\frac{8.4\times \sin47}{8.4\times \cos47}[/tex]
[tex]\theta=47^{\circ}[/tex] east of south
Now the distance from the current position to the desired position:
[tex]s=\sqrt{x^2+y^2}[/tex]
[tex]s=\sqrt{(8.4\times \cos47)^2+(8.4\times \sin47-5.3)^2}[/tex]
[tex]s=5.79\ km[/tex]
Suppose you throw a ball vertically upward with a speed of 49 m/s. Neglecting air friction, what would be the height of the ball be 2.0 second later?
Answer:
78.4 m
Explanation:
Using newton's equation of motion,
S = ut + 1/2gt²......................... Equation 1
Where S = Height, t = time, u = initial velocity, g = acceleration due to gravity.
Note: Taking upward to be negative, and down ward positive
Given: u = 49 m/s, t = 2.0 s, g = -9.8 m/s²
Substitute into equation 1
S = 49(2) - 1/2(9.8)(2)²
S = 98 - 19.6
S = 78.4 m
Hence the height of the ball two seconds later = 78.4 m
If two identical conducting spheres are in contact, any excess charge will be evenly distributed between the two. Three identical metal spheres are labeled AA, BB, and CC. Initially, AA has charge qq, BB has charge −q/2−q/2, and CC is uncharged.
Answer:
The solution is in the attached files below
Explanation:
A piece of plastic tape coated with iron oxide is magnet- ized more in some parts than in others. When the tape is moved past a small coil of wire, what happens in the coil? What is a practical application of this?
Answer:
Explanation:when a tape is moved past a small coil of wire due to a non uniform field,the voltage is induced in the coil. A practical example of this is a tape recorder in which a moving tape produces sound energy..
Recall the observed behavior of the compass during the mapping of the magnetic field. Which of the following descriptions best matches the behavior you observed? Two correct answers required for full credit.
A. Where the field is strong the compass needle readily and quickly aligns with rapid oscillations around the equilibrium direction.
B. Where the field is strong the compass needle slowly oscillates around the final equilibrium position.
C. Where the field is weak the compass needle rapidly oscillates around the final equilibrium position.
D. Where the field is weak the compass needle slowly oscillates around the final equilibrium position.
E. Where the field is weak the compass needle readily and quickly aligns with rapid oscillations around the equilibrium direction.
Answer:
Option A & B
Explanation:
This is cause the compass needle is magnetized anointed in such a way that it responds to magnetic field strength.
According to the animation, in the first half of the lunar cycle, the waxing, bright part of the moon appears on the ____ side of the Earth. After full moon when the moon is waning, the illuminated part of the moon appears on the ______ side.
Answer: Blanks in order are: Right, Left
Explanation:
The moon rotates around the Earth. It completes its rotation in roughly 29.5 days which is known as the lunar cycle. The first half of rotation is known as waxing and in this the moon starts to become a full moon so bright side of the moon appears on the left side of Earth. After this ends, the waning part starts which means the decreasing of light of moon so bright side is on the left side of the Earth.
During the first half of the lunar cycle, the bright part of the moon (waxing moon) appears on the right side and after the full moon, the lit part of the moon (waning moon) appears on the left side.
Explanation:According to the animation, in the first half of the lunar cycle, the waxing, or growing, bright part of the moon appears on the right side (from an observer in the northern hemisphere). This phase is known as a waxing moon. After the full moon when the moon starts to shrink, or wane, the illuminated part of the moon appears on the left side. This phase is known as a waning moon.
Learn more about Lunar CycleWhich of the following best summarizes the relationship between dehydration reactions and hydrolysis?
-Dehydration reactions can occur only after hydrolysis.
-Dehydration reactions ionize water molecules and add hydroxyl groups to polymers; hydrolysis reactions release hydroxyl groups from polymers.
- Dehydration reactions assemble polymers, and hydrolysis reactions break down polymers.
-Hydrolysis creates monomers, and dehydration reactions break down polymers.
-Dehydration reactions eliminate water from lipid membranes, and hydrolysis makes lipid membranes water permeable.
Answer:
Dehydration reactions assemble polymers, and hydrolysis reactions break down polymers.
Explanation:
dehydration reaction is a conversion that involves the loss of water from the reacting molecule or ion.
Hydrolysis is defined as any chemical reaction in which a molecule of water ruptures one or more chemical bonds.
Dehydration reactions link monomers together into polymers by releasing water, and hydrolysis breaks polymers into monomers using a water molecule. Monomers are just single unit molecules and polymers are chains of monomers.
Dehydration reactions and hydrolysis have opposite roles in relation to polymers. Dehydration reactions involve the removal of a water molecule to assemble polymers, while hydrolysis involves the addition of a water molecule to break down polymers.
Explanation:The relationship between dehydration reactions and hydrolysis is that they essentially have opposite roles in relation to polymers. Dehydration reactions are processes that involve the removal of a water molecule, which allows for the formation or assembly of polymers. This occurs because a hydroxyl group (OH) is removed from one molecule and a hydrogen (H) from another, enabling the two molecules to bond and form a polymer.
In contrast, hydrolysis is a reaction that involves the addition of a water molecule, which results in the breakdown of polymers into monomers. During hydrolysis, a water molecule is split into a hydroxyl group and a hydrogen, and these are used to break the bonds within a polymer, resulting in monomers.
So, to summarize, the best choice that describes the relationship between these two processes is: Dehydration reactions assemble polymers, and hydrolysis reactions break down polymers.
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Click on the crate to bring it to a stop, then replace it with the refrigerator. Use the slider to apply a force of about 400 N. After 2 s have elapsed in the simulation, decrease the Applied Force (force exerted) slowly back to zero. Try to do this adjustment in roughly 2 s . While the Applied Force (force exerted) is decreasing, the velocity is:______.
a. constant.
b. increasing.
c. decreasing.
Answer:
While the Applied Force (force exerted) is decreasing, the velocity is decreasing
Explanation:
From Newton's second law of motion, which states that the rate of change of linear momentum is directly proportional to the applied force, and takes place in the direction of the applied force.
Momentum (P) = MV
Thus, F ∝ MV/t
where;
F is the applied force
M is the mass of the object
V is the velocity of the object
From the equation above, force is directly proportional to the velocity of the refrigerator (F∝V). That is, as the applied force is decreasing, the velocity is decreasing and vice versa.
Therefore, while the Applied Force (force exerted) is decreasing, the velocity is decreasing.
Answer:
on mastering physics its b.increasing
Explanation:
Two uncharged metal balls, Y and X, each stand on a glass rod and are touching. A third ball carrying a negative charge, is brought near the first two While the positions of these balls are fixed, ball Y is connected to ground. Then the ground wire is disconnected. While Y and X remain in touch, the ball carring the negative charge is removed. Then ball Y and X are separated.
The phenomenon in question describes charging by induction. The presence of a negative charge induced a redistribution of charges in two uncharged metal balls, and grounding allowed excess charges to move away. The two uncharged balls ended up positively charged after the negatively charged ball was removed and the balls separated.
Explanation:The subject of this question relates to the concept of charging by induction, a common concept in Physics. When the third ball, which carries a negative charge, was brought near Ball Y and X, it caused the redistribution of charges in both balls due to the phenomenon of induction. As Ball Y was grounded, it provided a path for the electric charges to move away, thus becoming neutral again. However, Ball X retained the overall net positive charge due to the negative charge being near earlier.
After the ground wire is disconnected and the negatively charged ball removed, Ball Y and X still have net positive charges. This happens because the negative charge in the vicinity induced a redistribution of charges earlier, and when they were removed, there weren't any free charges nearby to neutralize this charge.
The separation of Ball Y and X made their charges apparent. Both balls are left with a positive charge as the excess electrons moved to the earth during grounding.
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Ball Y ends up with a positive charge, and Ball X with a negative charge due to induction caused by a nearby negatively charged ball. Grounding Ball Y neutralizes any negative charge, leaving it positively charged. The scenario exemplifies charging by induction.
Charging by Induction
This question deals with the concept of charging by induction in physics. Let's break down the scenario step-by-step:
Initially, Balls Y and X are both uncharged and in contact with each other.A third ball carrying a negative charge is brought close to Balls Y and X. Because of this, electrons on Ball Y will be repelled, moving to Ball X, resulting in Ball Y becoming positively charged and Ball X negatively charged.While the negatively charged ball is still near, Ball Y is connected to the ground. This allows the negative charge on Ball Y to be neutralized by the ground, leaving Ball Y positively charged.The ground connection is then removed. Now, Ball Y is positively charged and Ball X might still hold some negative charge due to the residual electrons.The negatively charged ball is then removed, followed by separating Balls Y and X.To address the specific parts of the question:
a. Charges on Each Ball
Ball Y will have a positive charge, and Ball X will have a negative charge. This distribution occurs due to the transfer of electrons from Y to X when the negatively charged third ball is near.
b. How the Balls Get These Charges
The negatively charged third ball induces a redistribution of charges in Balls Y and X. Electrons are repelled from Ball Y towards Ball X, leaving Ball Y positively charged and Ball X negatively charged.
c. Effect of Grounding Ball Y
When Ball Y is grounded while the negatively charged ball is close, the excess electrons on Ball Y will flow into the ground, neutralizing any negative charge and leaving it positively charged. When the ground connection is removed, Ball Y remains positively charged.
This process illustrates induction as the primary method of charging without direct contact.
A woman holds a book by placing it between her hands such that she presses at right angles to the front and back covers. The book has a mass of m = 1.3 kg and the coefficient of static friction between her hand and the book is μs = 0.69. What is the weight of the book in Newton?
The weight of the book is calculated using the formula for weight, which is the mass of the object times the acceleration due to gravity. Given the mass of the book and the standard acceleration due to gravity, the weight of the book is 12.74 Newton.
Explanation:To calculate the weight of the book, we need to remember from our physics knowledge that weight is the force exerted by an object under the influence of gravity. The formula for weight (W) is W = m*g, where m is the mass of the object, and g is the acceleration due to gravity. In most parts of the Earth, the acceleration due to gravity is approximately 9.8 m/s^2.
So in this case, the weight of the book is calculated as follows: W = 1.3 kg * 9.8 m/s^2 = 12.74 N. Therefore, the weight of the book is 12.74 Newton. Note that the coefficient of static friction is not needed to calculate the weight of the book.
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The weight of a book with a mass of 1.3 kg is calculated using the formula W = m x g, resulting in 12.74 N.
The weight of the book in Newtons can be calculated using the formula for weight, which is the mass of the object (m) multiplied by the acceleration due to gravity (g). The acceleration due to gravity is approximately 9.80 m/s² on Earth. Therefore, to find the weight (W) of the book, you use the equation W = m imes g.
For a book with a mass of 1.3 kg, the calculation is W = 1.3 kg times 9.80 m/s² = 12.74 N.
This is the weight of the book that the woman must counteract with the normal force exerted by her hands, which is equal in magnitude to the weight of the book but in the opposite direction, as explained by Newton's third law of motion.
What is the relationship between the amount of stretch or compression in an elastic object and the amount of force that is applied to that object?
Applied force is one-half the amount of stretch or compression.
They are directly proportional.
They are inversely proportional.
Applied force is double the amount of stretch or compression.
The relationship between the amount of stretch or compression in an elastic object and the amount of force that is applied to that object is directly proportional to each other.
Answer:
Option B
Explanation:
Ideal elastic objects are those which will retain its original form after the removal of external force on it. So this specialty of retaining their original form after deforming due to external force is termed as elasticity. And the objects exhibiting this property is termed as elastic object. The most important law in the field of elasticity which is obeyed by all elastic object is the Hooke's law. As per Hooke's law, the external force applied on any object for deformation will be directly proportional to the changes or deformation exhibited by the object due to force. In other or simple words, the strain produced due to the external force is directly proportional to the stress acting on that object.
Stress ∝ Strain
Stress is the measure of force applied on that object and strain is the measure of amount of stretch or compression due to stress or applied force.
So option B is the correct option.
Thus, the relationship between the amount of stretch or compression in an elastic object and the amount of force that is applied to that object is directly proportional to each other.
The amount of stretch or compression in an elastic object is directly proportional to the force applied, based on the principle known as Hooke's Law. This relationship is linear, meaning that changes in force result in proportional changes in deformation.
Explanation:The amount of stretch or compression in an elastic object is directly proportional to the amount of force that is applied. This principle is captured by Hooke's Law, which states when you apply force to stretch or compress a spring, the resulting spring force is proportional, and in the opposite direction, to the displacement or deformation from its position of equilibrium. Essentially, if you double the force, the amount of stretch or compression doubles, and if you halve the force, the stretch or compression is halved as well. This relationship can be represented by the equation F = -kx, where F is the force, k is the spring's force constant, and x is the displacement from the equilibrium position.
Tension and compression play a role here, where tension forces stretch the object while compression forces push the object together. Hooke's Law applies to ideal springs and other elastic objects provided the deformation does not exceed the material's elastic limit.
Careers in this career cluster also include related professional and technical support activities such as production planning and control, maintenance and manufacturing/process engineering.A. True
B. False
1. Our entire solar system orbits around the center of the Milky Way Galaxyabout once every 230 million years.
2. The Milky Way and Andromeda galaxies are among a few dozen galaxies that make up our Local Group.
3. The Sun appears to rise and set in our sky because Earth orbitsonce each day.
4. You are one year older each time Earth rotatesabout the Sun.
5. On average, galaxies are getting farther apart with time, which is why we say our universeis expanding.
6. Our solar systemis moving toward the star Vega at about 70,000 km/hr.
Answer: Hello! Apparently, your question is incomplete. Those were sentences in which you had to complete with missing information, so here we go:
1- Our entire solar system orbits around the center of the MILKY WAY GALAXY about once every 230 million years.
2- The Milky Way and Andromeda galaxies are among a few dozen galaxies that make up our LOCAL GROUP.
3 - The Sun appears to rise and set in our sky because Earth ROTATES once each day.
4 - You are one year older each time Earth ORBITS about the Sun.
5 - On average, galaxies are getting farther apart with time, which is why we say our UNIVERSE is expanding.
6 - Our SOLAR SYSTEM is moving toward the star Vega about 70,000 km/hr.
A gry is an old English measure for length, defined as 1/10 of a line, where line is another old English measure for length, defined as 1/12 inch. A common measure for length in the publishing business is a point, defined as 1/72 inch. What is an area of 0.68 gry2 in points squared?
Answer:
0.2448 point²
Explanation:
1 gry = 1/10 line
1 line = 1/12 inch
=> 1 gry in inches = 1/10 * 1/12 = 1/120 inch
=> 1 inch = 120 gry
1 point = 1/72 inch
=> 1 inch = 72 points
Therefore,
120 gry = 72 points
=> 1 gry = 3/5 point
Therefore,
1 gry² = (3/5)² point²
1 gry² = 9/25 point²
This means that 0.68 gry² will be:
0.68 gry² = 0.68 * 9/25 point²
=> 0.68 gry² = 0.2448 point²
The area of 0.68 gry2 is 0.2448 point2.
How do you calculate the area of 0.68 gry2 in points?Given that A gry is an old English measure for length, defined as 1/10 of a line. We can write this as given below.
[tex]1 \;\rm gry = \dfrac {1}{10}\;\rm line[/tex]
A line is another old English measure for length, defined as 1/12 inch. We can write this as given below.
[tex]1 \;\rm line = \dfrac {1}{12}\;\rm inch[/tex]
Thus the value of 1 gry in inches will be,
[tex]1 \;\rm gry = \dfrac {1}{10}\times \dfrac {1}{12} \;\rm inch[/tex]
[tex]1 \;\rm gry = \dfrac {1}{120}\;\rm inch[/tex]
[tex]120 \;\rm gry = 1\;\rm inch[/tex]
A common measure for length in the publishing business is a point, defined as 1/72 inch. We can write this as given below.
[tex]1 \;\rm point = \dfrac {1}{72}\;\rm inch[/tex]
[tex]72 \;\rm point = 1 \;\rm inch[/tex]
Now we can equate the values of 1 inch, we get the expression as,
[tex]120 \;\rm gry = 72 \;\rm point[/tex]
[tex]1 \;\rm gry = \dfrac {72}{120}\;\rm point[/tex]
[tex]1 \;\rm gry = 0.6 \;\rm point[/tex]
[tex](1\;\rm gry )^2 = (0.6 \;\rm point)^2[/tex]
[tex]1\;\rm gry ^2 = 0.36 \;\rm point^2[/tex]
We have the value of 1 gry2 in terms of point2. Thus,
[tex]0.68 \;\rm gry^2 = 0.68 \times 0.36 \;\rm point^2[/tex]
[tex]0.68\;\rm gry^2 = 0.2448 \;\rm point^2[/tex]
Hence we can conclude that the area of 0.68 gry2 in points squared is 0.2448.
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How high would you need to lift a 2 kg bottle of soda to increase its potential energy by 50 Joules? (use G=10 N/KG.)
Answer:
2.5m
Explanation:
PE = MGH
50 = 2*10*H
50 = 20H
H = 2.5m
"1.0 kg mass is attached to the end of a spring. The mass has an amplitude of 0.10 m and vibrates 2.0 times per second. Find its speed when it passes the equilibrium position."
Answer:
1.3m/s
Explanation:
Data given,
Mass,m=1.0kg,
Amplitude,A=0.10m,
Frequency,f=2.0Hz.
From the equation of a simple harmonic motion, the displacement of the object at a given time is define as
[tex]x=Acos\alpha \\[/tex]
we can express the velocity by the derivative of the displacement,
Hence
[tex]V=-Awsin\alpha \\[/tex]
at equilibrium, the velocity becomes
[tex]V=wA\\w=2\pi f[/tex]
Hence if we substitute values we arrive at
[tex]V=2\pi fA\\V=2\pi *2*0.1\\V=1.3m/s[/tex]
Answer:
haha lol
Explanation:
sorry but i dont know this
Which of Galileo's observations directly disproved Ptolemy's epicycle model of the Solar System, showing that the Sun is at the center and not Earth?
Answer:
Venus observation.
Explanation:
Galileo had learned regarding the heliocentric (Sun-centered) idea of Copernicus, and acknowledged it. However, the theory was proven by Galileo's observations of Venus. Galileo concluded that Venus should travel round the Sun, sometimes passing behind and then beyond, instead of directly rotating around the Earth.
The amplitude of a sound wave is measured in terms of its maximum gauge pressure. By what factor does the amplitude of a sound wave increase if the sound level goes up by 26.0 dB?
Answer:
The amplitude of wave increases by a factor of 20
Explanation:
The Sound level in decibel β is described as
[tex]\beta =(10dB)log(\frac{I}{I_{o} })[/tex]
Where I₀=10⁻¹²W/m² is reference intensity
Suppose β₁ is initial sound intensity and β₂ is final sound intensity.Therefore the change in sound intensity is given as
β₁-β₂=26.0 dB
[tex](10dB)log(\frac{I_{2} }{I_{o}} )-(10dB)log(\frac{I_{1} }{I_{o}} )=26.0dB\\(10dB)[log(\frac{I_{2} }{I_{o}} )-log(\frac{I_{1} }{I_{o}} )]=26.0dB\\log(\frac{\frac{I_{2}}{I_{o}} }{\frac{I_{1}}{I_{o}} })=\frac{26dB}{10dB}\\ log\frac{I_{2}}{I_{1}}=2.6\\ \frac{I_{2}}{I_{1}}=10^{2.6}[/tex]
The intensity of sound is directly proportional to the amplitude of wave squared
So
I∝A²
[tex]\frac{I_{1}}{I_{2}}=(\frac{A_{1}}{A_{2}})^{2}\\ (\frac{A_{1}}{A_{2}})=\sqrt{(\frac{I_{1}}{I_{2}})}\\ (\frac{A_{2}}{A_{1}})=\sqrt{(\frac{I_{2}}{I_{1}})}\\ (\frac{A_{2}}{A_{1}})=\sqrt{10^{2.6} }\\ (\frac{A_{2}}{A_{1}})=20[/tex]
Therefore the amplitude of wave increases by a factor of 20
Krista is playing tennis at the park. When the tennis ball flies toward her, Krista hits the ball with her racket, which causes the ball to fly in the opposite direction. According to Newton's third law of motion, which of the following is true?A. When the racket hits the tennis ball with a force, the ball does not apply any reaction force to the racket.
B. When the racket hits the tennis ball with a force, the tennis ball applies an equal but opposite force to the racket.
C. When the racket hits the tennis ball with a force, the ball applies a much weaker force in the opposite direction of the racket's force.
D. When the racket hits the tennis ball with a force, the ball applies an equal force in the same direction as the racket's force.
Answer:
B. When the racket hits the tennis ball with a force, the tennis ball applies an equal but opposite force to the racket
Explanation:
Newton's third law of motion states that for every action there is an equal and opposite reaction. Therefore, when Krista's racket hits the tennis ball with a force, the tennis ball applies an equal but opposite force to the racket.
When the racket hits the tennis ball with a force, the tennis ball applies an equal but opposite force to the racket. Therefore option B is correct.
According to Newton's third law of motion, for every action, there is an equal and opposite reaction.
In this scenario, when Krista hits the tennis ball with her racket (the action), the tennis ball exerts an equal but opposite force back onto the racket (the reaction).
This is depicted by option B, which correctly states that the tennis ball applies an equal but opposite force to the racket. This law underscores the idea that forces always come in pairs, and the forces are exerted on two different objects.
The force Krista applies to the ball is met with an equal force from the ball back onto her racket, causing the ball to fly in the opposite direction.
Therefore option B is correct.
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A mass attached to a spring oscillates and completes 51 full cycles in 35 s. What is the time period (in s) and frequency (in Hz) of this system?
The time period of the mass attached to a spring that oscillates and completes 51 full cycles in 35 seconds is approximately 0.686 seconds, and the frequency is about 1.458 Hz.
Explanation:To determine the time period (T) and frequency (f) of the oscillating mass attached to a spring, we can use the relationship between the number of cycles (n), total time (t), and the time period (T = t/n). If the mass completes 51 full cycles in 35 seconds, the time period (T) is calculated as:
Calculate the total time for one cycle (the time period T) by dividing the total time by the number of cycles: T = t/n.Using the given values, T = 35 s / 51 cycles.Therefore, the time period (T) is approximately 0.686 seconds.The frequency (f), which is the number of cycles per second, is the inverse of the time period: f = 1/T.So, f is approximately 1/0.686 Hz, which gives us a frequency of about 1.458 Hz.In conclusion, the time period of the system is approximately 0.686 seconds, and the frequency is about 1.458 Hz.
If you were to drill a hole into the wooden sphere, and fill it with lead, how much of the volume of the wood would have to be replaced to make the whole sphere neutrally buoyant (i.e. have the same density as water)?
Answer:
[tex]\displaystyle V_l=V\ \frac{\rho_o -\rho_w }{\rho_l-\rho_w}[/tex]
Explanation:
Density
It's a physical magnitude that relates the mass of an object with the volume it occupies. The formula is
[tex]\displaystyle \rho = \frac{m}{V}[/tex]
Equivalently
[tex]m=\rho V[/tex]
The density of water is 1 gr/ml, 1 Kg/lt, 1000 kg/m^3 or any equivalent unit.
Assume the wood sphere has a volume V and a density [tex]\rho_w[/tex], thus
[tex]m_w=\rho_w V[/tex]
A hole is to be drilled inside the wood so it means part of its volume will be filled with lead, which mass will be
[tex]m_l=\rho_l V_l[/tex]
The volume of wood is the total volume V minus the (unknown) volume of lead, thus
[tex]V_w=V-V_l[/tex]
The total mass of the modified sphere is
[tex]m_s=m_w+m_l=\rho_w V_w+\rho_l V_l[/tex]
Substituting Vw
[tex]m_s=m_w+m_l=\rho_w (V-V_l)+\rho_l V_l[/tex]
Operating
[tex]m_s=m_w+m_l=\rho_w V-\rho_w V_l+\rho_l V_l[/tex]
[tex]m_s=m_w+m_l=\rho_w V+ V_l(\rho_l -\rho_w)[/tex]
The total volume of the sphere doesn't change, which means
[tex]V_s=V[/tex]
The new density of the modified sphere is
[tex]\displaystyle \rho_s = \frac{m_s}{V}[/tex]
[tex]\displaystyle \rho_s = \frac{\rho_w V+ V_l(\rho_l -\rho_w)}{V}[/tex]
This density must be equal to the density of water [tex]\rho_o[/tex]
[tex]\displaystyle \rho_o = \frac{\rho_w V+ V_l(\rho_l -\rho_w)}{V}[/tex]
Operating and solving for vl
[tex]\displaystyle V_l=\frac{\rho_o V-\rho_w V}{\rho_l-\rho_w}[/tex]
Or equivalently
[tex]\displaystyle \boxed{V_l=V\ \frac{\rho_o -\rho_w }{\rho_l-\rho_w}}[/tex]
The equation would be enough for the volume of wood to keep the sphere neutrally buoyant would be [tex]Vt= Vp0-Pw/Pt-Pw[/tex]
Calculations and Parameters:The formula for the physical magnitude is p= m/v
Given that the density of water is 1 gr/ml, 1 Kg/lt, 1000 kg/m^3 or any equivalent unit.
We would assume the wood sphere has a volume V and a density Pw,
[tex]Mw= PwV[/tex]
Since a hole is to be drilled inside the wood so it means part of its volume will be filled with lead, the mass will be [tex]Mt= PtVt[/tex]
The volume of wood is the total volume V minus the (unknown) volume of lead, thus
[tex]Vw= V- Vt[/tex]
Hence, after further evaluation, the density would be [tex]Vt= V Po- Pw/Pt- Pw[/tex]
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A Honda Civic travels in a straight line along a road. Its distance x from a stop sign is given as a function of time t by the equation x(t)= α t2− β t3, where α = 1.53 m/s2 and β = 4.80×10−2 m/s3. Calculate the average velocity of the car for each time interval:_______.
a) t = 0 to t = 2.00 s.
b) t = 0 to t = 4.00 s.
c) t = 2.00 s to t = 4.00 s.
Answer:
(a) [tex]V_{avg}=2.868m/s[/tex]
(b) [tex]V_{avg}=5.352m/s[/tex]
(c) [tex]V_{avg}=7.836m/s[/tex]
Explanation:
Given data
x(t)=αt²-βt³
α=1.53m/s²
β=0.0480m/s³
First we need to find distance x at these time so
x(t)=1.53t²-0.0480t³
at t=0
x(0)=1.53(0)²-0.0480(0)³=0m
at t=2
x(2)=1.53(2)²-0.0480(2)³=5.736m
at t=4s
x(4)=1.53(4)²-0.0480(4)³=21.408 m
For(a) Average velocity at t=0s to t=2s
The average velocity is given as
Vavg=Δx/Δt
[tex]V_{avg}=\frac{x_{f}-x_{i} }{t_{f}-t_{i}}\\ As\\x(0)=0m\\x(2)=5.736m\\V_{avg}=\frac{5.736m-0m }{2s-0s}\\V_{avg}=2.868m/s[/tex]
For(b) Average velocity at t=0s to t=4s
The average velocity is given as
Vavg=Δx/Δt
[tex]V_{avg}=\frac{x_{f}-x_{i} }{t_{f}-t_{i}}\\ As\\x(0)=0m\\x(4)=21.408m\\V_{avg}=\frac{21.408m-0m }{4s-0s}\\V_{avg}=5.352m/s[/tex]
For(c) Average velocity at t=2s to t=4s
The average velocity is given as
Vavg=Δx/Δt
[tex]V_{avg}=\frac{x_{f}-x_{i} }{t_{f}-t_{i}}\\ As\\x(2)=5.736m\\x(4)=21.408m\\V_{avg}=\frac{21.408m-5.736m }{4s-2s}\\V_{avg}=7.836m/s[/tex]
Final answer:
The average velocity of the Honda Civic over the time intervals from t=0 to t=2.00 s is 2.868 m/s, from t=0 to t=4.00 s is 5.352 m/s, and from t=2.00 s to t=4.00 s is 7.836 m/s, calculated using the given equation and the variables α and β.
Explanation:
To find the average velocity of the Honda Civic over specific time intervals using the equation x(t) = α t^2 - β t^3, where α = 1.53 m/s^2 and β = 4.80×10^-2 m/s^3, we must calculate the change in position (Δx) over the change in time (Δt) for each interval.
a) t = 0 to t = 2.00 s
Initial position, x(0) = 0 (since any term with t = 0 results in 0).
Final position, x(2) = 1.53×2^2 - 4.80×10^-2×2^3 = 6.12 - 0.384 = 5.736 m.
Δx = 5.736 - 0 = 5.736 m, Δt = 2 s.
Average velocity = Δx/Δt = 5.736 m / 2 s = 2.868 m/s.
b) t = 0 to t = 4.00 s
Final position, x(4) = 1.53×4^2 - 4.80×10^-2×4^3 = 24.48 - 3.072 = 21.408 m.
Δx = 21.408 - 0 = 21.408 m, Δt = 4 s.
Average velocity = Δx/Δt = 21.408 m / 4 s = 5.352 m/s.
c) t = 2.00 s to t = 4.00 s
Using the previously calculated final positions for t = 2 and t = 4:
Δx = 21.408 - 5.736 = 15.672 m, Δt = 4 s - 2 s = 2 s.
Average velocity = Δx/Δt = 15.672 m / 2 s = 7.836 m/s.
A certain five cent coin contains 5.00 g of nickel. What fraction of the nickel atoms' electrons, removed and placed 1.00 m above it, would support the weight of this coin? The atomic mass of nickel is 58.7, and each nickel atom contains 28 electrons and 28 protons. (answer in ×10^{-11})
Answer:
Fraction of the nickel atoms removed = 1.02 x 10 raised to power -11
Explanation:
The detailed and step by step derivation with appropriate substitution is as shown in the attached files.
To calculate the fraction of nickel atoms' electrons needed to support the weight of a nickel coin when placed 1 meter above it, we use the weight of the coin and Coulomb's law to determine the necessary charge. This charge divided by the elementary charge gives us the number of electrons needed, providing the fraction when compared to the total number of electrons in the coin.
Explanation:To calculate what fraction of a nickel coin's electrons would need to be removed and placed 1.00 m above it to support its weight, we should use the concept of electrostatic force. We'll begin by determining the weight of the coin in newtons (since weight is a force). Next, we will determine the number of moles of nickel in the coin using its atomic mass, and from that, calculate the number of nickel atoms. With the atomic structure of nickel, we find the total number of electrons in the coin. Using Coulomb's Law, we can equate the electrostatic force due to the separated charges to the weight of the coin to solve for the fraction of electrons needed.
First, we need to convert the mass of the coin into weight using the formula weight (W) = mass (m) * gravity (g). Since the coin has a mass of 5.00 g (which is 0.005 kg) and the gravitational constant is approximately 9.8 m/s², the weight W is 0.049 N. Next, we use the atomic mass of nickel (58.7 u) to find the number of moles in 5.00 g of nickel. This number is given by the mass of the coin divided by the atomic mass of nickel. The number of atoms is the number of moles times Avogadro's number (approximately 6.022 × 10²³ atoms/mole).
Therefore, the total number of electrons in the 5.00 g of nickel is the number of atoms times the number of electrons per atom, which for nickel is 28. To support the coin at a distance of 1.00 m, the electrostatic force (Felectrostatic) should be equal to the weight of the coin, or 0.049 N. Using Coulomb's law, Felectrostatic = k * |q₁ * q₂| / r², where k is Coulomb's constant (approximately 8.988 × 10⁹ N·m²/C²), q₁ and q₂ represent the charges, and r is 1.00 m in our case. By solving this equation for q (assuming both q₁ and q₂ would be the same since it's the same electrons that have been separated), we can find the necessary charge to create a force equal to the weight of the coin.
After finding this charge, we divide by the elementary charge (approximately 1.602 × 10⁻¹⁹ C) to get the number of electrons needed. Dividing this number by the total number of electrons in the coin gives us the fraction of the nickel atoms' electrons needed to support the weight of the coin.
Xenon fluoride can be prepared by heating a mixture of Xe and F2 gases to a high temperature in a pressure-proof container. Assume that xenon gas was added to a 0.25 liter container until its pressure reached 0.12 atm at 0.0°C. Fluorine gas was then added until the total pressure reached 0.72 atm at 0.0°C. After the reaction was complete, the xenon was consumed completely, and the pressure of the F2 remaining in the container was 0.36 atm at 0.0°C. What is the empirical formula of the xenon fluoride?
Answer:
XeF₄
Explanation:
Dalton law's of partial pressure states that the total pressure equal to the sum of the partial pressure of the individual gases that make up the mixture.
Pt = Pxe + Pf₂
0.72 = 0.12 + Pf₂
0.72 - 0.12 = Pf₂
0.60 atm = Pf₂
after the the reaction was complete, the pressure of F₂ remaining = 0.36 atm
pressure of the consumed F₂ = 0.60 - 0.36 = 0.24 atm
Pxe / total pressure = number of mole / total number of mole of gas present
0.24 / 0.72 = nf / nt
0.12 / 0.72 = nxe / nt
comparing the two
(1/ 3) / ( 1/6) = (nf/ nt) / ( nxe/ nt)
nf / nxe = 2 / 1
the emperical formula = XeF₄
The empirical formula of the xenon fluoride is :
-XeF₄
"Dalton law's of partial pressure"It states that the Total pressure break even with to the whole of the fractional weight of the person gasses that make up the blend.
Pt = Pxe + Pf₂0.72 = 0.12 + Pf₂0.72 - 0.12 = Pf₂0.60 atm = Pf₂Pressure of F₂ remaining = 0.36 atm
Pressure of the consumed F₂ = 0.60 - 0.36 = 0.24 atm
Pxe / total pressure = number of mole / total number of mole of gas present
0.24 / 0.72 = nf / nt
0.12 / 0.72 = nxe / nt
comparing the two:
(1/ 3) / ( 1/6) = (nf/ nt) / ( nxe/ nt)
nf / nxe = 2 / 1
The emperical formula = XeF₄
Learn more about "Emperical Formula" :
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A hammer taps on the end of a 4.0-m-long metal bar at room temperature. A microphone at the other end of the bar picks up two pulses of sound, one that travels through the metal and one that travels through the air. The pulses are separated in time by 11.0 ms.
1. What is the speed of sound in this metal? v = 6060 m/s
2.If the bar was twice as long, what would be the time difference?
Answer: c*m on now
Explanation:
u dumb b***h
You are driving to the grocery store at 18 m/s. You are 130 m from an intersection when the traffic light turns red. Assume that your reaction time is 0.50 s and that your car brakes with constant acceleration.a. How far are you from the intersection when you begin to apply the brakes? b. What acceleration will bring you to rest right at the intersection? c. How long does it take you to stop?
Answer
given,
initial speed = 18 m/s
distance, d = 130 m
reaction time, t_r = 0.50 s
a) distance traveled in the reaction time
d= s x t_r
d=18 x 0.5
d = 9 m
distance of the traffic light, d = 130 - 9 = 121 m
b) deceleration
[tex]a = \dfrac{v^2-u^2}{2s}[/tex]
[tex]a = \dfrac{0^2-18^2}{2\times 121}[/tex]
a = -1.34 m/s²
c) Using equation of motion
v = u + a t
0 = 18 - 1.34 t
t = 13.43 s
Total time, T = 13.43 + 0.5 = 13.93 s
Part A. The distance from the intersection is 121 m.
Part B. The acceleration required is 1.34 m/s2
Part C. The total time to stop at the signal is 13.93 s.
How do you calculate the distance, acceleration and time?Given that the initial speed is 18 m/s. The distance is 130 m from an intersection when the traffic light turns red. The rejection time is 0.50 s.
Part A
The distance traveled in rejection time is given below.
distance traveled in rejection time = speed [tex]\times[/tex] rejection time
[tex]d_r = s\times t_r[/tex]
[tex]d_r = 18\times 0.50[/tex]
[tex]d_r = 9\;\rm m[/tex]
Hence the distance from the traffic light is given below.
[tex]d = 130 -d_r[/tex]
[tex]d = 130 -9[/tex]
[tex]d = 121\;\rm m[/tex]
The distance from the intersection is 121 m.
Part B
The acceleration can be calculated as given below.
[tex]a = \dfrac {v^2 - u^2}{2s}[/tex]
Where u is the initial speed, v is the final speed and s is the distance from the intersection.
[tex]a = \dfrac {0^2 - (18)^2}{2\times 121}[/tex]
[tex]a = - 1.34 \;\rm m/s^2[/tex]
The acceleration required is 1.34 m/s2. Here negative sign shows that this is deacceleration.
Part C
The time to stop at the signal can be calculated as given below.
[tex]v = u +at[/tex]
[tex]0 = 18 - 1.34 \times t[/tex]
[tex]t = \dfrac {18}{1.34}[/tex]
[tex]t = 13.43 \;\rm s[/tex]
Total Time = [tex]13.43 + 0.5[/tex]
Total Time = 13.93 s
Hence the total time to stop at the signal is 13.93 s.
To know more about speed and acceleration, follow the link given below.
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Two cars collide at an icy intersection and stick together afterward. The first car has a mass of 1050 kg and was approaching at 6.00 m/s due south. The second car has a mass of 750 kg and was approaching at 25.0 m/s due west.
Answer:
v = 11.0 m/s at 198.6° (18.6° south of west)
ΔKE = -145 kJ
Explanation:
I assume you want to find the final velocity and the change in kinetic energy.
Take east to be +x and north to be +y.
Momentum is conserved in the x direction:
(1050 kg) (0 m/s) + (750 kg) (-25.0 m/s) = (1050 kg + 750 kg) vₓ
vₓ = -10.4 m/s
Momentum is conserved in the y direction:
(1050 kg) (-6.00 m/s) + (750 kg) (0 m/s) = (1050 kg + 750 kg) vᵧ
vᵧ = -3.50 m/s
The magnitude of the final velocity is:
v² = (-10.4 m/s)² + (-3.50 m/s)²
v = 11.0 m/s
The direction of the final velocity is:
θ = atan(-3.50 m/s / -10.4 m/s)
θ = 198.6°
The initial kinetic energy is:
KE₀ = ½ (1050 kg) (6.00 m/s)² + ½ (750 kg) (25.0 m/s)²
KE₀ = 253,275 J
The final kinetic energy is:
KE = ½ (1800 kg) (11.0 m/s)²
KE = 108,682 J
The change in kinetic energy is:
ΔKE = 108,682 J − 253,275 J
ΔKE ≈ -145,000 J
The energy released by the exploding gunpowder in a cannon propels the cannonball forward. Simultaneously, the cannon recoils. The mass of the cannonball is less than that of the cannon. Which has the greater kinetic energy, the launched cannonball or the recoiled cannon? Assume that momentum conservation applies.
Answer:
the launched cannonball
Explanation:
In a cannon firing a cannonball, while the total momentum is conserved, the cannonball has greater kinetic energy than the recoiling cannon due to its higher velocity and the relationship between kinetic energy and velocity.
The subject of this question is Physics, specifically dealing with momentum conservation and kinetic energy in the context of a cannon firing a cannonball. The law of conservation of momentum states that the total momentum before and after an event is constant if no external forces are acting on the system. When a cannon fires a cannonball, the momentum is conserved, meaning the cannon and the cannonball together possess the same total momentum post-explosion as the stationary cannon had before firing (which is zero). Though both the cannon and the cannonball have momentum of equal magnitude but opposite direction, the distribution of kinetic energy between them is not equal.
Kinetic energy, which depends on both mass and the square of velocity, will not be conserved in the same way as momentum. The kinetic energy is given by the relation KE = (1/2)mv2. Considering the masses of the two objects, a lighter cannonball and a heavier cannon, if both the cannon and cannonball have the same magnitude of momentum, the cannonball will have a much higher velocity due to its smaller mass. Since kinetic energy depends on the square of velocity, the cannonball, with a higher velocity, will have greater kinetic energy than the heavier, slower-moving cannon. Thus, even though momentum is conserved, the kinetic energies will differ, with the cannonball gaining more kinetic energy than the recoiling cannon.
In summary, the law of conservation of energy applies, but kinetic energy is not conserved like momentum. The chemical energy stored in the gunpowder is converted into heat and the kinetic energy of both the cannon and the cannonball. Although the cannon has a larger mass, it will recoil with less velocity than the fired cannonball, which means the cannonball will have a greater kinetic energy.
do you think toxicity is a qualitative or quantitative property? explain
Answer:
Toxicity is a quantitative property
Explanation:
Qualitative property of a object cannot be measured it can just be observed Quantitative property of a substance can be measured and be assigned a numerical value .The toxicity level of a substance can be measured and be assigned a numeral value.
You are standing in an elevator that is accelerating upward at 2 m/s2. What is the normal force acting on you by the elevator if your mass is 70 kg?
Answer:
784 N.
Explanation:
The expression for the Normal reaction acting on a body inside an elevator that is moving upward is given as
R = m(g+a)................. Equation 1
Where R = normal reaction, g = acceleration due to gravity, a = acceleration, m = mass .
Given: m = 70 kg, a = 2 m/s²
Constant: g = 9.8 m/s²
Substitute into equation 1
R = 70(2+9.2)
R = 70(11.2)
R = 784 N.
Hence the normal force = 784 N.
A small pumpkin is launched from a very versatile catapult at a variety of speeds and angles. The angles are measured relative to the horizontal. Rank each of the following trials by the maximum height reached by the pumpkin. (1=highest) Consider air resistance to be negligible.
1 2 3 4 Speed: 30.7m/s, Angle: 35.7°
1 2 3 4 Speed: 47.9m/s, Angle: 33.2°
1 2 3 4 Speed: 21.8m/s, Angle: 67.7°
1 2 3 4 Speed: 7.7m/s, Angle: 62.1°
Answer:
Explanation:
Maximum height of a projectile
[tex]H_{max}=\frac{u^2\sin ^2\theta }{2g}[/tex]
(1)[tex]u=30.7\ m/s[/tex]
[tex]\theta =35.7^{\circ}[/tex]
[tex]H_1=\frac{30.7^2\times \sin ^2(35.7)}{2\times 9.8}[/tex]
[tex]H_1=16.37\ m[/tex]
(2)[tex]u=47.9\ m/s[/tex]
[tex]\theta =33.2^{\circ}[/tex]
[tex]H_2=\frac{47.9^2\times \sin ^2(33.2)}{2\times 9.8}[/tex]
[tex]H_2=35.09\ m[/tex]
(3)[tex]u=21.8\ m/s[/tex]
[tex]\theta =67.7^{\circ}[/tex]
[tex]H_3=\frac{21.8^2\times \sin ^2(67.7)}{2\times 9.8}[/tex]
[tex]H_3=20.75\ m[/tex]
(4)[tex]u=7.7\ m/s[/tex]
[tex]\theta =62.1^{\circ}[/tex]
[tex]H_4=\frac{7.7^2\times \sin ^2(62.1)}{2\times 9.8}[/tex]
[tex]H_4=2.36\ m[/tex]
Therefore height reached by pumpkin can be arranged as
[tex]H_4<H_1<H_3<H_2[/tex]
Final answer:
The maximum height reached by the pumpkin depends on the vertical component of the initial velocity, which is highest for the trial with a speed of 21.8m/s and angle of 67.7 degrees and lowest for the speed of 7.7m/s at an angle of 62.1 degrees.
Explanation:
To rank the maximum height reached by the pumpkin in each trial, we can rely on the principle of projectile motion. The maximum height depends primarily on the vertical component of the initial velocity and can be compared by isolating this component using the initial speeds and angles given.
Using the formula for the vertical component of velocity: [tex]Vy = V * sin(\(\theta\)),[/tex] where V is the initial speed and [tex]\(\theta\)[/tex] is the launch angle. Then, the maximum height H can be roughly compared using [tex]H \(\propto Vy^2[/tex], assuming gravitational acceleration is constant and air resistance is negligible.
Speed: 21.8m/s, Angle: [tex]67.7\(\deg\)[/tex] - Highest vertical componentSpeed: 47.9m/s, Angle: [tex]33.2\(\deg\)[/tex]Speed: 30.7m/s, Angle: [tex]35.7\(\deg\)[/tex]Speed: 7.7m/s, Angle: [tex]62.1\(\deg\)[/tex] - Lowest vertical component