The the acceleration of the sled of mass 100 Kg will be 1.5 m/s².
What is Friction?A drag is force that opposes the motion of an object by acting in the direction opposite to its motion. It is of two types namely - Static friction and kinetic friction. The static friction is given by - F[S] = μ[s] x η and the kinetic friction is given by F[K] = μ[k] x η.
Given is a football blocking sled which has a mass of roughly 100 kg. A football player applies a force of 500 N to the sled, and there is a 350 N frictional force acting on the sled.
Assume that the force applied by the player is - F[P] = 500 N and the force of friction is - F = 350 N.
Now, for the motion of the sled with acceleration [a], we can write -
F[P] - F = ma
a = {F[P] - F}/m
Substituting the values, we get -
a = (500 - 350)/100
a = 150/100
a = 15/10
a = 1.5 m/s²
Therefore, the the acceleration of the sled of mass 100 Kg will be
1.5 m/s².
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. What are the velocities of sound waves in air, helium, and carbon dioxide? Show your data. Which was the fastest? Why do sound waves travel at different speeds in each gas? Explain. Note that the values for helium will be incomplete in these frequency ranges.
Answer:
Taking into account the table we get that, for n = 2, the velocities of sound waves are
Velocity on air: 267.3 m/s
Velocity on helium: 260 m/s
Velocity of carbon dioxide: 829.3 m/s
The velocity of the sound travels at different speeds in each gas because the velocity depends on the properties of the medium like the density. Since the air, helium, and carbon dioxide have different densities, the sound speed will be different.
10. A punter kicks the ball high enough that it spends 5.2 seconds in the air. How fast was it
going when it left his foot?
With the use of first equation of motion, the ball will be going with a velocity of - 51 m/s
What is Velocity ?Velocity is the distance travel in a specific direction per time taken. It is a vector quantity and it is measured in m/s
Given that a punter kicks the ball high enough that it spends 5.2 seconds in the air. To know how fast was it, let us assumed that the initial velocity u of the ball is zero. The ball will be moving under the influence of gravity.
To find the velocity, let us use first equation of motion. That is,
v = u - gt
where
Final velocity v = ?Initial velocity u = 0Acceleration due to gravity g = 9.8 m/s²Time t = 5.2 sSubstitute all the parameters into the equation
v = - 9.8 × 5.2
v = - 50.96 m/s
Therefore, the ball will be going as fast as -51 m/s when it left his foot.
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Help I am so confused on what to do with theses steps to get the answer.
Step 1 convert all the dimensions of the box from cm to m
Step 2 calculate the maximum buoyant force if the box was completely submerged. Fb = pfluid V g
Step 3 calculate the Fg of the box Fg = mg
Step 4 Subtract the Fg from the Fb to get the Force of gravity from extra mass
Step 5. Use the Fg from extra mass =mg equation to solve for mass. Using
The buoyant force if the box was completely submerged is 93.84 N.
The length of the box is 40.8 cm.
The height of the box is 10.4 cm and the depth of the box is 19 cm.
Now we know,
1 m = 100 cm
So,
1 cm = 0.01 m
Therefore,
Length, L = 40.8 cm = 40.8 × 0.01 = 0.408 m
Height, H = 10.4 cm = 10.4 × 0.01 = 0.104 m
Depth, D = 19 cm = 19 × 0.01 = 0.19 m
The density of the fluid is 1187 kg/m³.
ρ = 1187 kg/m³ = 1187 × 10³ g/m³
The mass of the box, m = 2325.4 g
The maximum buoyant force will be:
F = ρgV
Here g = 1 g cc
Therefore,
F = 1187 × 1 × ( 0.408 × 0.104 × 0.19 )
F = 93.846590 N
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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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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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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.
What is thermodynamics Movement of heatMeasure of heatIt’s magic None of them
To find
What is thermodynamics
Explanation
Thermodynamic study about the transfer of energy to and fro from a system.
Thus thermodynamics is the movement of heat
Conclusion
The correct option is movement of heat
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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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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The diagram shows what happens when a polythene rod is rubbed with a dry cloth. Select the statements that are true
When a polythene rod is rubbed with a dry cloth, the dry cloth loses electrons, the polythene rod gains electrons and the dry cloth is left with positive charge.
The electrons on the dry cloth's surface move to the polyethene rod when it is brushed against a polythene rod. It is considered to be negatively charged because more electrons have accumulated within the polythene rod as a result of the rod being electron rich and the electrons moving there.
The amount of electrons inside the cloth has dropped, making the cloth electron-deficient. It is referred to as positively charged because of this.
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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 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).
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
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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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.
Explain how a $10.00 compact fluorescent light bulb (15 W) can be an overall money saver compared to an incandescent light bulb (60 W) that costs $0.50.
Answer:
Explanation:
We would consider the kilowatt hour for both bulbs. It is the energy consumed by an appliance per hour.
Recall, 1 kw = 1000w
15w = 15/1000 = 0.015 kw
60w = 60/1000 = 0.06kw
In 1 hour,
1)
The 15 W bulb would consume 0.015 kw.
For 10 hours per day, the consumption is 0.015 x 10 = 0.15kw
For a month of 30 days, the consumption is 0.15 x 30 = 4.5
Let's assume that the charge per KW is $0.2
Consumption cost per month = 0.2 x 4.5 = $0.9 per month
2)
The 60 W bulb would consume 0.06 kw.
For 10 hours per day, the consumption is 0.06 x 10 = 0.6kw
For a month of 30 days, the consumption is 0.6 x 30 = 18
Let's assume that the charge per KW is $0.2
Consumption cost per month = 0.2 x 18 = $3.6 per month
Thus, the 15W bulb would be an overall money saver as time goes on.
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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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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why do surface waves compared to body waves have a high amplitude?
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
You pull a box 23 m horizontally, If the rope tension is 120 N , and if the rope does 2500 J of work on the box, what angle θ does the rope make with the horizontal?
The angle θ made by the rope with the box is equal to 25.17 degrees.
If the rope is pulled at an angle A and the tension in the rope was T, then the rope will have two components of forces,
One Vertical which is TsinA and one along horizontal which is TcosA,
And we know,
W = F.s.cosα
Where,
F is the force applied,
s is the Displacement along force,
α is the angle between them,
We know, the box is moving forward, so the angle between the vertical component of tension and displacement is 90 degrees, so the vertical component of tension will not do any work, and the angle between displacement and the horizontal component is 0 degrees,
So, the work done by the horizontal component is 0 degrees,
So, we cam write,
W = TcosA.D
where D is the displacement of the box, which is given to be 23m,
T is the tension in the string which is 120N,
W is the work done by the rope on the box which is 25000J,
so, putting all the values,
2500 = 120.cosA.23
cosA = 2500/120x23
cos A = 0.9
So, A is 25.17 degrees.
The angle made by the rope is 25.17 degrees,
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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.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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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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A 1800 kg object moving to the east at 13 m/s, collided with a 200 kg object that was moving to the north at 32 m/s. What was the angle of motion
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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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.
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:
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
A hollow, spherical shell with mass 2.00 kg rolls without slipping down a 33.0 ∘ slope.
Find the acceleration.
Find the friction force.
The acceleration of the spherical shell is determined as 6.37 m/s².
The friction force is 12.74 N.
What is the coefficient of friction of the surface?
The coefficient of friction of the surface is calculated as follows;
Fsinθ = μFncosθ
where;
F is the applied force on the spherical shellFn is the weight of the spherical shellmg(sin θ) = (μmg) cosθ
(sin θ) = μ cosθ
μ = sinθ/cos θ
where;
g is acceleration due to gravityθ is angle of inclination of the slopeμ = tan θ
μ = tan(33)
μ = 0.65
The acceleration of the spherical shell is calculated as;
a = μg
a = 0.65 x 9.8 m/s²
a = 6.37 m/s²
The frictional force is calculated as follows;
Ff = μmg
Ff = 0.65 x 2 x 9.8
Ff = 12.74 N
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