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.
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.
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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What is the angular velocity of the second hand on a clock? (Hint: It takes the second hand 60 seconds to rotate 2π radians. Divide the number of radians by the number of seconds so that your answer has units of radians/second.)
Explanation:
Time taken to rotate 2π radians = 60 seconds
Angular displacement = 2π radians
Time taken = 60 seconds
Angular displacement = Angular velocity x Time
2π = Angular velocity x 60
Angular velocity = 0.105 rad/s
Angular velocity of the second hand on a clock is 0.105 rad/s
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:
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:
Computers A and B implement the same ISA. Computer A has a clock cycle time of 200 ps and an effective CPI of 1.5 for some program and computer B has a clock cycle time of 250 ps and an effective CPI of 1.7 for the same program. Which computer is faster and by how much?
Answer:
Computer A is 1.41 times faster than the Computer B
Explanation:
Assume that number of instruction in the program is 1
Clock time of computer A is [tex]CT_{A} =200 ps[/tex]
Clock time of computer B is [tex]CT_{B} =250 ps[/tex]
Effective CPI of computer A is [tex]CPI_{A} =1.5[/tex]
Effective CPI of computer B is[tex]CPI_{B} =1.7[/tex]
CPU time of A is
[tex]CPU_{time}=instructions \times CPA_{A} \times CT_{A}\\CPU_{time}=1 \times 1.5 \times 200=300 sec[/tex]
CPU time of B is
[tex]CPU_{time}=instructions \times CPA_{B} \times CT_{B}\\CPU_{time}=1 \times 1.7 \times 250=425 sec[/tex]
Hence Computer A is Faster by [tex]\frac{425}{300} =1.41[/tex]
Computer A is 1.41 times faster than the Computer B
On February 15, 2013, a superbolide meteor (brighter than the Sun) entered Earth’s atmosphere over Chelyabinsk, Russia, and exploded at an altitude of 23.5 km. Eyewitnesses could feel the intense heat from the fireball, and the blast wave from the explosion blew out windows in buildings. The blast wave took approximately 2 minutes 30 seconds to reach ground level. The blast wave traveled at 10° above the horizon. (a) What was the average velocity of the blast wave? (b) Compare this with the speed of sound, which is 343 m/s at sea level.
Answer:
a) Average velocity = 156.7m/s
b) Speed of blast wave is 0.457 times the speed of sound wave
Explanation:
The step by step calculations is as shown in the attached file.
Answer:
Velocity,c = 156.67m/s
Explanation: Altitude, d = 23.5km or 23500m
Time duration, T = 2 minutes + 30 seconds = (120 + 30) seconds = 150 seconds
a) Frequency, F = 1/T= 1/150 Hz
Hence, Velocity, c = d X F = 23500 X 1/150
∴ c = 156.67 m/s
b) The speed of sound increases as the wave moves from a lighter medium to denser medium. While the of the superbolide meteor decreases from a lighter medium to denser medium.
An unbalanced force is defined as a force that __________. allows an object to remain at rest cause an object to accelerate in a direction opposite to the unbalanced force is not canceled out by one or more other forces acting in other directions allows an object to move at constant speed
Answer:
An unbalanced force is defined as a force that is not canceled out by one or more other forces acting in other directions.
Explanation:
According to Newton's first law of motion, “An object in a constant state of motion remains in that constant state of motion unless acted upon by an unbalanced force.” An unbalanced force is a force that changes the speed, position, or direction of an object. It is not canceled out by other forces acting in other directions.
An unbalanced force is defined as a force that is not canceled out by other forces acting in different directions. This results in a net force that causes an object to accelerate, changing its state of motion.
An unbalanced force is defined as a force that is not canceled out by other forces acting in different directions. allows an object to remain at rest cause an object to accelerate in a direction opposite to the unbalanced force is not canceled out by one or more other forces acting in other directions allows an object to move at constant speed.
When forces on an object are unbalanced, it means that the net force acting on the object is not zero, leading to a change in the object's motion. This unbalanced force results in acceleration according to Newton's second law of motion, which states that [tex]\( F = ma \)[/tex] force equals mass times acceleration.
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 componentYou are working out on a rowing machine. Each time you pull the rowing bar (which simulates the "oars") toward you, it moves a distance of 1.5 m in a time of 1.0 s. The readout on the display indicates that the average power you are producing is 82 W. What is the magnitude of the force that you exert on the handle?
Answer:
54.67 N
Explanation:
The total energy produced is the product of power and time duration:
E = Pt = 82 * 1 = 82 J
Which is converted from work, product of forced extended over a displacement
W = E = Fs = F*1.5 = 82
F = 82 / 1.5 = 54.67 N
So the magnitude of the force exerting on the handle is 54.67 N
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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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.
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Which 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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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.
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
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.
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
Nathan drops marbles down two ramps that have different lengths. It takes the marbles 10 seconds to reach the bottom of both ramps. Which statement is TRUE?
Nathan drops marbles down two ramps that have different lengths. It takes the marbles 10 seconds to reach the bottom of both ramps.Which statement is TRUE?
answer choices
Marble 1 has a faster speed than Marble 2.
Marble 2 has a faster speed than Marble 1.
Both the marbles travel at the same speed.
There is not enough data to compare the speeds of marbles.
Answer:Marble 2 has a faster speed than Marble 1.
Option B.
Explanation:The speed is defined as the distance covered per unit time. Here in the question, 2 balls cover equal distances in same time.
Time taken by the ball = 10 seconds.
Distance covered by 1st ball = 20 cm.
Distance covered by 2nd ball = 3cm.
So speed of the 1st ball = 2cm/sec.
Speed of the 2nd ball = 3 cm /sec.
So,it's very much evident that speed of 2nd Marble is much higher than the speed of the 1st marble.
A brass lid screws tightly onto a glass jar at 20 degrees C. To help open the jar, it can be placed into a bath of hot water. After this treatment, the temperature of the lid and the jar are both 60 degrees C. The inside diameter of the lid is 8.0 cm at 20 degrees C. Find the size of the gap (difference in radius) that develops by this procedure.
Answer:
0.0016 cm
Explanation:
[tex]\alpha_b[/tex] = Thermal coefficient of expansion of brass = [tex]19\times 10^{-6}\ /^{\circ}C[/tex]
[tex]\alpha_g[/tex] = Thermal coefficient of expansion of glass = [tex]9\times 10^{-6}\ /^{\circ}C[/tex]
[tex]\Delta T[/tex] = Change in temperature = [tex](60-20)^{\circ}C[/tex]
[tex]R_0[/tex] = Initial radius = 4 cm
Change in radius of material is given by
[tex]R=R_0(1+\alpha\Delta T)[/tex]
Difference in radii of the lid and jar
[tex]\Delta R=R_b-R_g\\\Rightarrow \Delta R=R_0(1+\alpha_b\Delta T)-R_0(1+\alpha_g\Delta T)\\\Rightarrow \Delta R=R_0(\alpha_b-\alpha_g)\Delta T\\\Rightarrow \Delta R=4\times (19\times 10^{-6}-9\times 10^{-6})\times (60-20)\\\Rightarrow \Delta R=0.0016\ cm[/tex]
The size of the gap is 0.0016 cm or 0.000016 m
Final answer:
Heating the brass lid and glass jar causes the lid to expand more than the jar, creating a gap between them. To find the size of the gap, we can calculate the change in radius of the lid. Using the given information, we can find that the size of the gap is approximately 19.2 x [tex]10^-^3[/tex] cm.
Explanation:
When the brass lid and glass jar are heated in hot water, their temperatures increase. The lid, being made of metal, expands more than the glass jar due to the higher coefficient of expansion for metals. This causes the lid to loosen its grip on the jar, creating a gap between them.
To find the size of the gap, we need to calculate the difference in radius of the lid before and after heating. The inside diameter of the lid is given as 8.0 cm at 20 degrees C. We can calculate the radius using the formula r = ½ d, where d is the diameter. The radius at 20 degrees C is therefore 4.0 cm.
Now, let's calculate the radius at 60 degrees C. To do this, we need to know the coefficient of linear expansion for brass. Assuming it is 19x10^-6 /°C, we can use the formula: ∇r = αr0∇T, where ∇r is the change in radius, α is the coefficient of linear expansion, r0 is the initial radius at a reference temperature, and ∇T is the change in temperature. Plugging in the values, we get: ∇r = (19x[tex]10^-^6[/tex] /°C) x (4.0 cm) x (60 °C - 20 °C) = 19.2 x [tex]10^-^3[/tex] cm.
Therefore, the size of the gap that develops by heating the lid and jar is approximately 19.2 x [tex]10^-^3[/tex] cm.
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
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.
To know more about the area, follow the link given below.
https://brainly.com/question/16151549.
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.
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.
Learn more about Charging by Induction here:https://brainly.com/question/10254645
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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.
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.
"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
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
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
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
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 CycleA 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..