When an electron moves between the ends of a 9.00-volt battery, the work done on the electron is 1.44 * 10⁻¹⁸ Jouoles; option C
What is the work done on an electron?The work done in moving an electron between two points is defined as the product of the potential difference between the points and the charge on an electron.
Mathematically;
Work done = ΔV * qwhere;
ΔV is the potential difference between the points
q is the charge on the electron.
The potential difference between the two points is the work done in moving a unit positive charge from one point to the other in the circuit.
The charge, q on an electron is 1.609 * 10⁻¹⁹ Coulombs.
For the given electron that moves between the ends of a 9.00-volt battery;
ΔV = 9.000 Volts
q = 1.609 * 10⁻¹⁹ Coulombs
Work done = 9.000 * 1.609 * 10⁻¹⁹
Work done = 1.44 * 10⁻¹⁸ Jouoles
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Show that entropy change due to heat transfer by conduction is given by
∆S=mC (4 marks)
ln
2
T
T
2
The change in the reservoir, the system or device, and the surroundings are added to determine the overall entropy change. The reservoir's entropy change is. Since entropy is a function of state and we are contemplating a complete cycle (return to initial state), the device's entropy change is zero.
For finite variations at constant T, entropy changes (S) are calculated using the relation G=ΔH - TΔS.
The generation, consumption, conversion, and exchange of thermal energy across physical systems is the focus of the thermal engineering field of study known as heat transfer. Different heat transmission techniques, including thermal conduction, thermal convection, thermal radiation, and energy transfer by phase changes, are categorized.
Conduction is the process through which heat is transported from an object's hotter end to its colder end. Heat naturally transfers from a hotter body to a colder body.
For instance, heat is transferred from the hotplate of an electric stove to the bottom of a saucepan that comes into touch with it.
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What is the relationship of the force on the spring and stretch of the spring.(This is the spring we are looking at weights can be attached to it)
We will have the following:
We know that for a simple spring the following is true:
[tex]\begin{cases}F=-kx \\ \\ F=ma\end{cases}[/tex]So:
[tex]-kx=ma[/tex]So, the stretch of the spring is directly proportional to the force.
What is the average velocity of a car driving down the highway if it’s displacement is 123m west during a time period of 14.0s
The average velocity of a car driving down the highway if it’s displacement is 123 m west during a time period of 14.0 s is 8.79 m/sec.
What is Velocity?The rate at which an object's position changes when observed from a specific point of view and when measured against a specific unit of time is known as its velocity. Unit of velocity is m/sec.
Given in the question a car driving down the highway it’s displacement is 123 m west during a time period of 14.0 s then the average velocity is given as,
Average velocity = displacement / time
= 123/14
= 8.79 m/sec
The average velocity of a car driving down the highway if it’s displacement is 123 m west during a time period of 14.0 s is 8.79 m/sec.
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How do magnification and resolution compare between electron and light microscopes?
magnification and resolution compare between electron and light microscopes by they produce an photograph of a specimen with the aid of using the usage of a beam of electrons instead of a beam of mild.
Electron microscopes range from mild microscopes in that they produce an photograph of a specimen with the aid of using the usage of a beam of electrons instead of a beam of mild. Electrons have a whole lot a shorter wavelength than seen mild, and this lets in electron microscopes to supply higher-decision photos than popular mild microscopes. Electron microscopes may be used to observe now no longer simply complete cells, however additionally the subcellular systems and cubicles inside them.
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A ball is orbits in a circle of radius 10m with a speed 50 m/s. What is its angular velocity?
The angular velocity of the ball orbiting in circle will be 5 rad/s.
What is angular velocity?Angular velocity describes the rate at which an object rotates.
Given is a ball that orbits in a circle of radius 10m with a speed of 50 m/s.
The relation between the linear and angular velocity of a body is given by-
v = rω
where -
v is the linear velocity
r is the radius
ω is the angular velocity
On substituting the values, we get -
50 = 10 x ω
ω = 50/10
ω = 5 rad/s
Therefore, the angular velocity of the ball orbiting in circle will be
5 rad/s.
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Calculate the force between charges 4 x 10^-8 C and 1.8 x 10^-6 C if they are 3.5 m apart.
Explanation
Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects.
[tex]F=k\frac{q_1\cdot q_2}{r^2}[/tex]where k is the coulomb constant
[tex]k=8.9875\cdot10^9\text{ N }\frac{m^2}{C^2}[/tex]Step 1
let
[tex]\begin{gathered} q_1=4\cdot10^{-8}C \\ q_2=1.8\cdot10^{-6} \\ r=3.5\text{ m} \\ k=8.9875\cdot10^9\text{ N }\frac{m^2}{C^2} \end{gathered}[/tex]now, replace.
[tex]\begin{gathered} F=k\frac{q_1\cdot q_2}{r^2} \\ F=8.9875\cdot10^9\text{ N }\frac{m^2}{C^2}(\frac{4\cdot10^{-8}C\cdot1.8\cdot10^{-6}}{(3.5m)^2}) \\ F=5.2824\cdot10^{-5}\text{ N} \end{gathered}[/tex]I hope this helps you
make G the subject of the formula
F = GMM²/1²
HENCE WRITE THE DIMENSION FOR G
The value of G is FR²/M₁M₂. and the dimension of G is [M⁻¹L³T⁻²]..
The provided formula is of gravitational force F between two objects,
F = GM₁M₂/R²
Where M₁ is the mass of first object and M₂ is the mas of the other object while R is the distance between there centers and G is the universal gravitation constant.
To find the dimension of G, making G the subject of formula,
G = FR²/M₁M₂.
As we know, unit of mass is Kilogram (Kg), unit of force is Newton (N) and unit of distance is Meter (M).
Putting all the values, Units in the place of quantities,
G = N.R²/Kg.Kg
Now, using Dimensional analysis, and writing the dimensions of all the other units,
G = [MLT⁻²][L²]/[M][M]
G = [ML³T⁻²]/[M²]
G = [M⁻¹L³T⁻²]
The dimensions of G are [M⁻¹L³T⁻²].
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The process of charging a capacitor consists of transferring charge from the plate at lower ____ to the plate at higher one
The process of charging a capacitor consists of transferring charge from the plate at lower potential to the plate at higher one (higher potential).
What steps are involved in charging a capacitor?When a battery is connected in series with a resistor and capacitor, a large initial current flows when the battery moves charge from one capacitor plate to the other. As the capacitor charges to the battery voltage, the charging current asymptotically decreases until it is zero.
Therefore, a unidirectional flow of electric charge is known as direct current. When the voltage from the power source matches the voltage at the capacitor terminals, the capacitor is fully charged. When the electrical circuit's current stops flowing, the charging phase of the capacitor is complete.
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Students were experimenting with objects in a collision. Ball A moves with a constant acceleration by sliding down a frictionless incline plane before colliding with Ball B. Students used motion detectors to measure the velocity of Ball A at various points. Ball A (mass of 1.00 kg) began from rest, at position 1, on a ramp at a height of 1.25 m. At position 2, ball A was moving at 5.00 m/s. Ball A continues to roll at a constant 5.00 m/s when it collides with Ball B (mass of 1.00 kg) at position 3. Ball A comes to a complete stop. Ball B moves at a constant velocity after the collision.FORMULAS: PE = m•g•h (g=9.8m/s²) KE = 1/2•m•v² momentum = m•v
Given data:
* The mass of ball A is m_1 = 1 kg.
* The mass of ball B is m_2 = 1 kg.
* The initial velocity of ball A is u_1 = 5 m/s.
* The final velocity of ball A is v_1 = 0 m/s.
* The initial velocity of the ball B is u_2 = 0 m/s.
Solution:
According to the law of conservation of momentum, the net momentum of the system before the collision is equal to the net momentum of the system after the collision.
Thus,
[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]Substituting the known values,
[tex]\begin{gathered} 1\times5+1\times0=1\times0+p_2 \\ 5+0=0+p_2 \\ p_2=5\text{ kgm/s} \end{gathered}[/tex]where p_2 is the momentum of the ball B,
Thus, the momentum of ball B after the collision is 5 kgm/s.
Hence, the third option is the correct answer.
A projectile is thrown at an angle 30° from horizontal. Which statement about its vertical component of velocity is true?
The true statement is the vertical component of the velocity is half of the magnitude of the velocity.
What is vertical component of velocity?
The vertical component of a velocity is the velocity of the object acting along vertical direction.
The vertical component of the velocity acting along the vertical direction is calculated as follows;
Vy = V sinθ
where;
Vy is the vertical component of the velocityθ is the angle of projectionV is the magnitude of the velocitySubstitute the value of the angle of projection and evaluate the vertical component of the velocity.
Vy = V sin(30)
Vy = V(0.5)
Vy = ¹/₂V.
Thus, the vertical component of the velocity is half of the magnitude of the velocity.
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The complete question is below
A projectile is thrown at an angle 30° from horizontal. Which statement about its vertical component of velocity is true?
the vertical component of the velocity is half of the magnitude of the velocitythe vertical component of the velocity is one-third of the velocitythe vertical component of the velocity is the same as the magnitude of the velocity.A bicycle and rider going 15 m/s approach a hill. Their total mass is 66 kg.
(a) What is their kinetic energy?
7425 J
(b) If the rider coasts up the hill without pedaling, how high above its starting level will the bicycle be when it finally rolls to a stop?
Need help with B Please help!!
a)The kinetic energy is 7425 J.
b) Theight at which the bicycle starts is 1147 m.
For the given problem, we are dealing with kinetic and potential energy, where Kinetic energy is the energy an object has since its movement. On the off chance that we need to accelerate an object, at that point we must apply a force. Applying a force requires us to do work. After work has been done, energy has been transferred to the object, and the object will be moving with a new constant speed,However when its position is modified from its regular equilibrium position, the bow is able to store energy by virtue of its position. This put-away energy of position is alluded to as potential energy. Potential energy is the stored energy of position had by an object.
Since we were given a mass which is 66 kg and a velocity of 15 m/s
Since the formula of kinetic energy is:
K.E= (1/2)mv², where m is the mass of the object and v is the velocity of the object moving .
= 1/2 (66)(15 m/s)²
=7425 J
For the second part the kinetic energy is converted to the potential energy, so the formula for the potential energy is
P.E = mgh, where m is the mass , g is the acceleration due to gravity and h is the height at which the object.
=7425 J =mgh
=>h= 7425/(mg)
=>h =7425/66*15
=>h =11.47 m
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Satellite orbits at a distance from earths center of about 6.6 earth radii and takes 24h to go around once. What distance (in meters) does the satellite travel in one day? What is the orbital velocity (in m/s)?
We know that the radii of the orbit is 6.6 the earth one. To get the distance travel in one day we calculate the circunference of that orbit:
[tex]\begin{gathered} C=(6.6)(6371)(2\pi) \\ C=264,199.14 \end{gathered}[/tex]Therefore the distance travel by the satellite is 264199.14 km per day.
Now, to get the orbital velocity we need to use the equation:
[tex]v=\sqrt[]{\frac{MG}{r}}[/tex]where M is the mass of the object at the center (in this case the earth), G is the gravitational constant and r is the radius of the orbit, then we have:
[tex]\begin{gathered} v=\sqrt[]{\frac{(5.9722\times10^{24})(6.67\times10^{-11})}{(6.6)(6.371\times10^6)}} \\ v=3077.89 \end{gathered}[/tex]Therefore the orbital velocity is 3077.83 m/s.
When is a secondary source more helpful than a primary source?
A. When you want to confirm the conclusion with your own experiment
B. When you are not an expert in the field being studied in the experiment
C. When you do not need to know the conclusion of the experiment
D. When you want to know the exact data values
Option A is the correct answer: A secondary source is more helpful than a primary source when you want to confirm the conclusion with your own experiment.
Primary and secondary sources work well together to support the argument you are trying to make. Although secondary sources demonstrate how your work connects to earlier research, primary sources are more reliable as evidence. When you want to confirm the conclusion of your experiment, secondary sources prove to be more useful.
Original research is built on primary sources. You are able to discover novel information, offer solid justification for your claims, and provide reliable facts on your subject.
Secondary sources are useful for getting a comprehensive perspective of your subject and learning other researchers' methods. They frequently combine numerous primary materials that would take a lot of time and effort to obtain independently.
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The figure shows a standing wave oscillating at 100 Hz on a string. What is the wave speed ?
Answer:
The wavelength is 60 cm.
Explanation:
Speed = frequency x wavelength = 100 x 60 = 6000 cm/s
or if you wanted the answer in m/s
Speed = frequency x wavelength= 100 x 0.60 = 60 m/s
A 58 V AC voltage source is connected across a 3.7 H inductor. What is the current through the inductor if the frequency is 63 Hz?
We will have the following:
First:
[tex]x_l=2\pi(63Hz)(3.7H)\Rightarrow x_l=466.2\pi\Omega[/tex]Then:
[tex]\begin{gathered} I=\frac{58V}{466.2\pi\Omega}\Rightarrow I=0.03960097254...A \\ \\ \Rightarrow I\approx0.04A \end{gathered}[/tex]So, the current is approximately 0.04 A.
Please list two materials that are conductors and two materials that are insulators.
Two materials that are conductors - silver, copper
Two materials that are insulators - rubber, dry wood
If the actual release height is 2 h , calculate the normal force exerted by the track at the bottom of the loop.
The normal force exerted by the track at the bottom of the loop is F = m(v²/h + g).
What is the force exerted at the bottom of the loop?
The normal force at the bottom of a loop, is determined from two opposite forces.
One is reaction to weight and other is from reaction of centripetal force which is due to the curvature of the track.
The normal force exerted at the bottom of a circular loop is calculated as follows;
F = ma + mg
where;
a is the centripetal acceleration of the objectm is mass of the objectg is acceleration due to gravityF = m(v²/r) + mg
where;
v is the velocity of the objectr is the radius of the loopif the release height of the object = 2h
then, the radius of the track = 2h/2 = h
F = m(v²/h) + mg
F = m(v²/h + g)
Thus, the normal force exerted by the track at the bottom of the loop depends on the weight of the object and centripetal force.
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A dog running of constant speed of 3m/s increases it's speed to 7m/s upon seeing a lion. if the mass of the dog is 20kg, the work it does in achieving the new speed is.......
A dog running of constant speed of 3m/s increases it's speed to 7m/s upon seeing a lion. if the mass of the dog is 20kg, the work it does in achieving the new speed is....... 400 J.
What is Speed?
Speed is the time rate at which an object is travelling along a path, whereas velocity is the pace and direction of an object's movement. In other words, velocity is a vector, whereas speed is a scalar valu
Work = Change in Kinetic energy
= ½m(v² - u²)
= ½ × 20 kg × [(7 m/s)² - (3 m/s)²]
= 10 kg × 40 m²/s²
= 400 J
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A football punter accelerates a football from rest to a speed of 13 m/s during the time in which his toe is in contact with the ball (about 0.24 s). If the football has a mass of 0.49 kg, what average force does the punter exert on the ball?_____ N
Given:
The initial speed of the football is: vi = 0 m/s (as initially, the football is at rest)
The final speed of the football is: vf = 13 m/s
The time for which the football accelerates is: t = 0.24 s
The mass of the football is: m = 0.49 kg
To find:
The force exerted on the ball
Explanation:
The acceleration of the ball can be determined by using the following equation.
[tex]a=\frac{v_f-v_i}{t}[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} a=\frac{13\text{ m/s}-0\text{ m/s}}{0.24\text{ s}} \\ \\ a=\frac{13\text{ m/s}}{0.24\text{ s}} \\ \\ a=54.17\text{ m/s}^2 \end{gathered}[/tex]The force exerted by the punter on the ball can be determined by using Newton's second law of motion.
According to Newton's second law of motion,
[tex]F=ma[/tex]Substituting the values in the above equation, we get:
[tex]\begin{gathered} F=0.49\text{ kg}\times54.17\text{ m/s}^2 \\ \\ F=26.54\text{ N} \end{gathered}[/tex]Final answer:
The force exerted by the punter on the ball is 26.54 Newtons (N).
For the following questions, refer to the position versus time graph above.a. What is the object's velocity att = 5 s? (2 sig figs)b. What is the object's velocity at t=40s? (2 sig figs)
We are given a graph of position vs time. We are asked to determine the speed at 5 seconds. To do that we need to determine the slope of the line above the time point of 5 seconds. To determine the slope of the line we will use the following formula:
[tex]m=\frac{y_2-y_1}{x_2-x_1}[/tex]This means that we need to choose two points in the line. We use the graph to determine these points as follows:
The points we have chosen are:
[tex]\begin{gathered} (x_1,y_1)=(0,0)_{} \\ (x_2,y_2)=(10,5) \end{gathered}[/tex]Replacing in the formula for the slope we get:
[tex]m=\frac{5-0}{10-0}[/tex]Solving the operations:
[tex]m=\frac{5}{10}=\frac{1}{2}[/tex]Therefore, the speed at 5 seconds is 0.5 meters per second.
For part B the following points can be taken:
3. A boulder rolls with speed of 3.5 m/s off a cliff. It hits the ground 2.25 m from the base ofthe ledge. A) How high is the ledge? B) How long did it take the boulder to fall to the bottomof the cliff?DrawingVerticalHorizontal
Given data
*The given distance from the base of the ledge is R = 2.25 m
*The given speed is v = 3.5 m/s
The diagram is given below
(a)
Let (h) be the height of the edge
The formula for the distance from the base of the ledge is given as
[tex]\begin{gathered} R=v\times t \\ R=v\times\sqrt[]{\frac{2h}{g}} \\ h=\frac{R^2\times g}{2v^2} \end{gathered}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} h=\frac{(2.25)^2\times9.8}{2\times(3.5)^2} \\ =2.025\text{ m} \end{gathered}[/tex]Hence, the height of the ledge is h = 2.025 m
(b)
The formula for the time taken by the boulder to fall to the bottom of the cliff is given as
[tex]t=\sqrt[]{\frac{2h}{g}}[/tex]Substitute the known values in the above expression as
[tex]\begin{gathered} t=\sqrt[]{\frac{2\times2.025}{9.8}} \\ =0.642\text{ s} \end{gathered}[/tex]Hence, the time taken by the boulder to fall to the bottom of the cliff is t = 0.642 s
Set the cannon to have an initial speed of 20 m/s. For which situation do you think the cannon ball will go father: if it is set at a 60-degree angle, or if it is set at a 70-degree angle?
Question 2 options:
60 degrees
70 degrees
The cannon ball will go farther if the the angle of projection is set at 60 degrees
How to determine which angle will result in farther distance
Case 1:
Initial velocity (u) = 20 m/sAngle of projection (θ) = 60 ° Acceleration due to gravity (g) = 9.8 m/s²Horizontal distance (R) =?R = u²Sine(2θ) / g
R = 20² × Sine (2×60) / 9.8
R = 346.41 / 9.8
R = 35.35 m
Case 2:
Initial velocity (u) = 20 m/sAngle of projection (θ) = 70 ° Acceleration due to gravity (g) = 9.8 m/s²Horizontal distance (R) =?R = u²Sine(2θ) / g
R = 20² × Sine (2×70) / 9.8
R = 257.12 / 9.8
R = 26.24 m
From the above calculations, we can conclude that the ball will go farther, if the angle is 60 °
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What is the magnification of a curved mirror if a 10.0 cm tall object is placed 8.00 cmfrom the mirror and produces an image 12.00 cm in front of the mirror?-0.6671.50-1.500.667
We will have that the magnification will be given by:
[tex]m=-\frac{v}{u}[/tex]That is:
[tex]\begin{gathered} m=-\frac{8cm}{12cm}\Rightarrow m=-\frac{2}{3} \\ \\ \Rightarrow m\approx-0.667 \end{gathered}[/tex]So, the magnification is approximately -0.667.
Each set of protons and electrons represents a different atom. Place the atoms in order of their overall charge. Order them from most positive to most negative.swap_vert22 protons, 18 electronsswap_vert12 protons, 10 electronsswap_vert17 protons, 10 electrons
Protons have positive charge
Electrons have negative charge
• 22 protons (+22) + 18 electrons (-18 )
22 - 18 = 4 (positive)
• 12 -10 = 2
• 17 - 10 = 7
• 2 - 1 = 1
From most positive to most negative overall charges.
7 - 4 - 2 - 1
29. [-/10 Points]DETAILSSERCPWA11 4.WA.010.An object of mass 0.77 kg is initially at rest. When a force acts on It for 2.9 ms It acquires a speed of 15.5 m/s. Find the magnitude (In N) of the average force acting on the object during the2.9 ms time interval.
We have:
m = mass = 0.77 kg
t = time = 2.9 ms = 0.0029 s
vf = final speed = 15.5 m/s
vi= initial speed = 0 m/s
Apply:
• F = m (vf-vi) / t
Replacing:
F = 0.77 (15.5 - 0 )/0.0029
F= 4,155.5 N
A rope pulls a 72.5 kg skier upa 21.7°slope with /k = 0.120.The rope is parallel to the slope,and exerts a force of 383 N. What isthe acceleration of the skier?(Unit = m/s?)Enter
A rope pulls a 72.5 kg skier upa 21.7°slope with /k = 0.120.The rope is parallel to the slope,and exerts a force of 383 N. The acceleration of the skier is approximately 2.97 m/s².
To find the acceleration of the skier, we need to use Newton's second law of motion and consider the forces acting on the skier.
Identify the forces acting on the skier:
The forces acting on the skier are the force of gravity (mg) and the force exerted by the rope (T), where m is the mass of the skier (72.5 kg) and g is the acceleration due to gravity (9.8 m/s²). The force exerted by the rope is parallel to the slope and can be calculated using the given value of 383 N.
Resolve the forces:
Since the rope is parallel to the slope, we need to resolve the force of gravity into components parallel and perpendicular to the slope. The component parallel to the slope is m * g * sin(21.7°).
Apply Newton's second law of motion:
Newton's second law states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, Fnet = ma.
Determine the net force:
The net force acting on the skier is the difference between the force exerted by the rope and the component of the force of gravity parallel to the slope. Fnet = T - m * g * sin(21.7°).
Calculate the acceleration:
Using Newton's second law, we can rearrange the equation Fnet = ma to solve for the acceleration (a). a = Fnet / m.
Substitute the values and solve:
Substitute the known values into the equation to find the acceleration.
Therefore, the acceleration of the skier is approximately 2.97 m/s².
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Answer:
about 0.567 m/s²
Explanation:
You want the acceleration of a 72.5 kg skier up a 21.7° slope with µk = 0.120, towed by a rope exerting a force of 383 N.
ForcesThe force up the slope is 383 N.
The forces down the slope will be the sum of the force due to gravity and the friction force.
GravityThe force down the slope due to gravity is ...
F = m·g·sin(θ) = (72.5 kg)(9.8 m/s²)(sin(21.7°) ≈ 262.7 N
FrictionThe force due to friction will be the product of µk and the force normal to the slope:
F = m·g·cos(θ)·µk = (72.5 kg)(9.8 m/s²)cos(21.7°)·0.120 ≈ 79.22 N
Net ForceThe net force up the slope is ...
383 N -262.7 N -79.22 N ≈ 41.08 N
This will accelerate a mass of 72.5 kg in the amount of ...
A = F/m = 41.08 N/(72.5 kg) ≈ 0.567 m/s²
The acceleration of the skier is about 0.567 m/s² up the slope.
__
Additional comment
We don't have to figure the forces. Rather we can figure the gross acceleration due to the tow rope, then subtract the accelerations due to gravity and friction. This saves a few math operations as we don't have to multiply, then divide, by 72.5 kg.
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. A circular loop of wire of area 10 cm2 carries a current of 25 A. At a particular instant, the loop lies in the xy-plane and is subjected to a magnetic field B=(2.0iˆ+6.0jˆ+8.0kˆ)×10−3T. As viewed from above the xy-plane, the current is circulating clockwise. (a) What is the magnetic dipole moment of the current loop? (b) At this instant, what is the magnetic torque on the loop?
The magnetic dipole moment of the current loop is 0.025 Am².
The magnetic torque on the loop is 2.5 x 10⁻⁴ Nm.
What is magnetic dipole moment?The magnetic dipole moment of an object, is the measure of the object's tendency to align with a magnetic field.
Mathematically, magnetic dipole moment is given as;
μ = NIA
where;
N is number of turns of the loopA is the area of the loopI is the current flowing in the loopμ = (1) x (25 A) x (0.001 m²)
μ = 0.025 Am²
The magnetic torque on the loop is calculated as follows;
τ = μB
where;
B is magnetic field strengthB = √(0.002² + 0.006² + 0.008²)
B = 0.01 T
τ = μB
τ = 0.025 Am² x 0.01 T
τ = 2.5 x 10⁻⁴ Nm
Thus, the magnetic dipole moment of the current loop is determined from the current and area of the loop while the magnetic torque on the loop is determined from the magnetic dipole moment.
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009 (part 1 of 2) 10.0 points When a water gun is fired while being held horizontally at a height of 1.19 m above ground level, the water travels a horizontal distance of 1.98 m. Find the initial velocity of the water. The acceleration of gravity is 9.81 m/s^2
Answer in units of m/s.
010 (part 2 of 2) 10.0 points A child, holding the same gun in a horizontal
position, slides down a 33.0◦incline at a constant speed of 1.40 m/s. The child fires the gun when it is 4.36 m above the ground and the water takes 0.868 s to reach the ground. How far will the water travel horizontally?
Answer in units of m.
Answer:
Speed = 4 m/s
Explanation:
Given:
009 (part 1 of 2)
H =1 .19 m
L = 1.98 m
g = 9.81 m/c²
____________
V₀ - ?
Equation of motion horizontally:
L = V₀*t (1)
Equation of vertical motion:
H = g*t² / 2 (2)
From equation (2):
Time:
t = √ (2*H / g) = √ ( 2*1.19 / 9.81) ≈ 0,49 s
From equation (1):
Horizontal speed:
V₀ = L / t = 1.98 / 0.49 ≈ 4 m/s
why do surface waves compared to body waves have a high amplitude?
If your 1000 kg car ran out of gas and you had to push it into the gas station with a
force of 50 N, what would the acceleration of the car be?
Answer:
a=0.05m/s²
Explanation:
force=mass × acceleration
f=ma
f=50N
m=1000kg
a=?
50=1000×a
50=1000a
a=50/1000
a=0.05m/s²
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