Acceleration of the sled is 0.62m/s^2
Acceleration of the child is 0.13m/s^2
They met at 2.6m from the child's initial position
The question is incomplete, the complete question is given below:
a 40.0-kg child and an 8.40-kg sled are on the surface of an icy lake 15.0 meters apart. the child exerts a 5.20 n force on the sled towards them via a rope they’ve attached.(a) What is the acceleration of the sled? m/s2(b) What is the acceleration of the child? m/s2(c) How far from the child's initial position do they meet? m
Part a
Acceleration of the sled:
Mass of sled=8.40kg
Force=5.20N
Acceleration=F/m
=5.2/8.4
=0.62m/s^2
Part b
Acceleration of the child
Mass of the child=40kg
Force=5.20N
Acceleration=F/m
=5.2/40
=0.13m/s^2
Part c
Distance from the child's initial position they met
They accelerate towards each other at =0.62+0.13
=0.75m/s^2
The time to meet is given by:
We use kinematic equation to find this,
s=ut+1/2at^2
u=0m/s^2
a=0.75m/s^2
s=15m
t=?
15 =0*t+ (1/2) (0.75) t^2
15=(1/2)*0.75t^2
t=6.32 sec
The child moves:
s=ut+1/2at^2, u=0, a=0.13, t=6.32
=(1/2) (0.13) 6.32^2
=2.6m
Hence they met at 2.6m from the child's original position
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two charges have a force of 8 n between them when the distance between them is suddenly increased by a factor of 2. the new force will be _________ n
When the distance between them is suddenly increased by a factor of 2. the new electrostatic force will be 2 N.
We need to know about electrostatic force to solve this problem. The electrostatic force is a force caused by two charges separated at a distance. It can be determined as
F = k . Q1 . Q2 / r²
where F is force, k is coulomb's constant (9 x 10⁹ Nm²/C²), Q is charge and r is radius.
From the question above, we know that
F1 = 8 N
r1 = r m
r2 = 2r1 = 2r
For this question, we can apply the ratio of initial and final force
F1 / F2 = k . Q . Q / r1² / k . Q . Q / r2²
F1 / F2 = r2² / r1²
F1 / F2 = (2r)² / r²
F1 / F2 = 4r² / r²
F1 / F2 = 4
F2 = 1/4 . 8
F2 = 2 N
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WHAT IS THE ANSWER
IS IT A OR C AND PLS EXPLAIN WHY
Answer:
the answer is the options c
in the olympiad of 708 b.c., some athletes competing in the standing long jump used handheld weights called halteres to lengthen their jumps. the weights were swung up in front just before liftoff and then swung down and thrown backward during the flight. suppose a modern 88 kg long jumper similarly uses two 6.00 kg halteres, throwing them horizontally to the rear at his maximum height such that their horizontal velocity is zero relative to the ground. let his liftoff velocity be v with arrow
His range would increase by 0.55m yards if he used the halteres. Therefore, using halteres results in a 0.55 m improvement in range.
A speed velocity of, 9.5, was heard. Before the hotter, I last four J. meters per second. finally at the peak of the puff. Okay. The long jumper went with the bank route. Okay. Helter is also present, so we do have the long jumper. A long jumper is currently in motion at the top. So let's talk about RE X. then assisted her. Moving to the rear roof, indeed. none m/s. Okay, in the frame of the ground.
Okay, So, we're curious to know if the long jumper had the conventional distance added. Okay, so in order to properly answer this topic, we must apply the conservation of momentum at the highest point in addition to the concepts from projectile motion.
Therefore, we have 9.5 I. Without the water, speed remains the same. Thus, we might conclude that this is the V. X. The U. I. is this. We can locate a moment of light, okay. Okay hence, multiplying by two and four. Yes, there were roughly 9.8, which is equivalent to your argument. It's 1/6. The range then became hotter without us having to do anything. Then, this is V. X. multiplied by T. I guess there are 25 aches. 0.816. Let's begin. 7.76. Because. Next, we'll locate a range and figure out how to use it.
Therefore, after a long leap, the time of flights for the for the speed duration of light 0.816 seconds remain. Okay. though 1/2 T. is large and in the. 9.5. He performs it again where T is the other half. So his range would increase by 0.55m.
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the acceleration due to gravity on the surface of a planet is four times as large as it is on the surface of earth. the mass density of the planet is known to be eight times that of earth. what is the radius of this planet in terms of earth's radius?
The radius of the planet is √2 times radius of earth or 1.414 times the radius of Earth.
What is gravitational force?
According to Newton's law of universal gravitation, every object in the universe attracts each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
F = GmM/R²
where;
G is universal gravitation constantm, M are the massesR is the distance between the objectsThe acceleration due to gravity on surface of Earth is given as;
g = GM/R²
G = gR²/M
For a planet whose acceleration due to gravity is 4 times of Earth and mass is 8 times of earth, the radius of the planet is calculated as;
(4gR_p²)/(8M) = (gR²)/(M)
where;'
R_p is the radius of the planet(4R_p²)/8 = R²
4R_p² = 8R²
R_p² = 8R²/4
R_p² = 2R²
R_p = √(2R²)
R_p = R√2
R_p = 1.414 (R)
Thus, the radius of the planet is root two times radius of earth or 1.414 times the radius of Earth.
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Consider a toy made out of a ball attached to a 7 inch string attached to a pole. the pole is motorized so that the ball at the end of the string moves in a circle clockwise around the pole. the stand is located on origin of an x and y axis and the speed of the ball is constant (22 in/s). find the magnitude and direction.
The ball swings around the pole while maintaining its angular momentum because its angular velocity rises with decreasing radius and mechanical energy falls.
Think about the board at the top of the circle and the speed the bull is traveling at. We converse, and the velocity is the same when the board is at the base of the circle. According to Newton's Second Lawyer Motion, we are aware of this. the overall force applied to an item that is equal to the acceleration of a mustang underneath the object.
Therefore, a minimum velocity is necessary. Keep the board rotating in a circle. If not, the ball will drop to the ground. Therefore, the ground national wait that is acting on the circular's top board.
Additionally, for the minimal velocity, the centripetal force pressing on the kid must be equal to the gravitational weight acting under effort. Additionally, centripetal force can be expressed as fc = M. The bull is at the top of the circle with the lowest velocity, so according to Newton's Second Law of Motion, M. G. must be equal to him.
From the beginning, we debated in a minimum square. Find the least velocity at the circle's top using that expression. Therefore, by rearranging, we can cancel em from both sides of the minimal square G. R., making it equal. taking the highway. The group of G. R. will have the lowest velocity.
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A pitcher throws a fastball horizontally toward home plate 18.39 meters away. If the ball is traveling at a speed of 40.0 m/s when it leaves his hand, what is the ball's acceleration if it takes 0.500s to reach home plate?
A -6.44 m/s2
B -12.9 m/s2
C 6.44 m/s2
D 12.9 m/s2
The ball's acceleration if it takes 0.500s to reach home -12.9 ms⁻².
The problem we are dealing with is related to acceleration, which is the rate of change of the velocity, Since speed could be a speed and a direction, there are as it were two ways for you to accelerate: alter your direction or change your speed—or alter both. We are given the distance of the home plate 18.39 meters, traveling with the initial velocity of 40.0 ms⁻¹ and the time taken to reach the home plate 0.500s.
Since referring to the equation of motion:
s=ut+1/2at^2
=>a=2(s-ut)/t^2
= 2(18.39-(40*0.500)/(0.500^2)
=-12.88
= -12.9 ms⁻²
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What is the frequency of light when the energy for a mole of photons is 1. 55 × 1013 j?.
The frequency of light is 3.88 x 10^22Hz.
What is electromagnetic radiation?
The electromagnetic spectrum is the area where the electromagnetic radiation that we see as light, or visible light, is found. Visible light usually has wavelengths between 400 and 700 nanometers (nm) or frequencies between 750 and 420 terahertz, which are between the longer wavelengths of infrared and ultraviolet (with shorter wavelengths).
E = 1.55 x10^13 J/ mol
E = hv
v = 1.55 x 10^13/1mol x 1mol/6.023 x )/6.626 x
velocity = 3.88x10^22 s-1
since 1 Hz = 1/ s = 1
velocity =3.88x 10^22 Hz
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ASAP:You live on a part of Earth where the angle of the sun's rays are very high, and the latitude is low. Which temperatures do you experience quite often?'
the angular area of the entire sky is about 41,250 square degrees. how many pictures would the hst have to take with its camera to obtain a complete picture of the entire sky?
It will takes years of time to obtain a complete picture of the entire sky of angular area is about 41,250 square degrees.
Divide 41250 by 0.0036 , we will get 11458333 and 1/3 image. but to take image for all of the sky it will takes years of time.
An angular distance used to describe how big a sphere or circle seems from a certain point of view is known as the angular diameter, angular size, apparent diameter, or apparent size. It is known as the visual angle in the vision sciences and as the angular aperture in optics (of a lens).
The angular diameter may also be thought of as the angle that a camera or eye must spin around in order to view from one side of an apparent circle to the other. Humans are able to distinguish sizes of up to around 1 arcminute (about 0.017° or 0.0003 radians) with their unaided eyes.
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at midday, solar energy strikes the earth with an intensity of about 1000. w/m2. what is the area of a solar collector that could collect 150 mj of energy in 1 h? this is roughly the energy content of 1 gallon of gasoline.
41.67 m² is the area of a solar collector that could collect 150 mj of energy in 1 h.
Power received per unit surface is the definition of intensity.
[tex]I=P/A=1kW/m2(1)[/tex]
Power is simply defined as the rate at which energy changes over time.
[tex]P=\Delta E/\Delta t(2)[/tex]
ΔE = 150 MJ
Δt = 1 hour = 3600 sec.
[tex]P=150 MJ/3600 s=41.67 kW(3)[/tex]
Replacing (3) in (1), and solving for Area, we get:
Area = 41.67 m²
The measurement that indicates the size of a region on a plane or curved surface is called area. Surface area refers to the area of an open surface or the border of a three-dimensional object, whereas the area of a plane region or plane area refers to the area of a form or planar lamina.
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why is chlorophyll green? why is chlorophyll green? it absorbs wavelengths in only the blue and red parts of the visible spectrum. it absorbs wavelengths only in the red portions of the spectrum (680-700 nm). it absorbs wavelengths only in the blue part of the visible spectrum (450-480 nm). it absorbs all wavelengths in the visible spectrum.
chlorophyll is green because It absorbs wavelength in only the blue and red parts of the visible spectrum
The pigment chlorophyll is what gives green plants their green color. A portion of the visible light spectrum is absorbed by chlorophyll. Chlorophyll is a pigment that absorbs visible light in the red (long wavelength) and blue (short wavelength) portions of the spectrum. The plant appears green because it reflects rather than absorbs green light.
The chloroplasts of plants contain chlorophyll. Plants have chlorophylls a and b, though there are other forms as well. Only one side chain's makeup distinguishes these two forms of chlorophyll from one another.
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electricity iii goals define the concept of obstacle determine how obstacles combine when in series and in parallel. expand model of electricity to include effects of obstacle size
A circuit is necessary for electricity to flow. The same materials can be used in multiple ways to create a complete circuit.
Diffraction is the interference or bending of waves via an aperture into the area that is geometrically in the shadow of the obstruction or aperture. Effectively, the wave's secondary source is the diffracting element or aperture. The term "diffraction" was created by Italian scientist Francesco Maria Grimaldi, who also made the first precise observations of the phenomena in 1660.
The Huygens-Fresnel principle, which views each point in a propagating wavefront as a collection of unique spherical wavelets, describes the diffraction phenomenon in classical physics. [3] As seen in the enclosed figure, the distinctive bending pattern is most obvious when a wave from a coherent source (like a laser) comes into contact with a slit or aperture that is roughly the same size as its wavelength.
A circuit is necessary for electricity to flow. The same materials can be used in multiple ways to create a complete circuit. Electric circuits of various types exhibit varying behavior (parallel and simple circuits) Light, heat, and music can all be produced with the use of electricity electrical conductors and insulators.
This results from the addition or interference of various wavefront points (or, alternatively, each wavelet) that follow pathways to the registering surface of varying lengths. Multiple, closely spaced apertures (like those in a diffraction grating) can produce a complicated pattern with different intensities.
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If an object has a mass of 85g and a volume 92cm^3
The density of the object mentioned in the question when mass and volume are given is 0.924 g[tex]/cm^{3}[/tex]
According to the problem we have the following data given,
The mass of object m= 85g
The volume of object V= 92[tex]cm^{3}[/tex]
To find the density of the object we use the formula [tex]density=\frac{mass}{volume}=\frac{m}{V}[/tex]
∴density ρ [tex]=\frac{m}{V}=\frac{85}{92}=0.924g/cm^{3}[/tex]
Thus the density of the object = 0.924 g/[tex]cm^{3}[/tex]
Your question was not complete but most probably the question was if an object of mass 85g and volume 92cm^3 is given, what is its density?
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the bottom profile normal to the shore off a section of the coast has the following average depths (in m) at the given segment distance lengths (in nautical miles): 2.7 m (3 n.m.), 3.8 m (7 n.m.), 4 m (10 n.m.), 11 m (20 n.m.), 12.2 m (30 n.m.), 15.5 m (40 n.m.), 27 m (50 n.m.), 48 m (60 n.m.), 73 m (70 n.m.), and 200 m (80 n.m.). there is a sustained wind blowing directly onshore at an average speed of 72 knots. calculate the wind-induced setup at the shore. note: to solve this problem you can assume a flat bottom in each segment (we covered this solution in class) and march your solution from offshore to onshore through each segment just like a simple computer storm surge model would. keep in mind that as you march your solution shoreward, the water level in each segment is influenced by the water level from the previous one.
Entering 20 35 4 is not a valid value. This is the same as 27.6 at 50 frames per second multiplied by nine.16 Two plus one thus. Why I visited the buying power. 10 60 x 10 to the power is equal to the required speed of 20.5 mph.
He had utilized equal to allow the power of 0.16 to take over. U. Cuban is equivalent to two. What 4.16 is used for. First late invitation. U. Cuban is equivalent to one. wherefore in light of part.16.
As a result, Cubans find it difficult to offer you what you desire. Why 2 30? Power 300.16. Why? I continued to solve it up to a power of 0.16. Why do I purchase one at a power of 10.16 30?
The third major issue. Do not enter the value 20 35 4. This is equal to 27.6 at 50 frames per second nine to the power.16
Part B. As a result, two equals one. Why I went to the power to buy.4. The needed speed of 20.5 mph is equivalent to 10 60 x 10 to the power.4.
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consider a mach 4 airflow at a pressure of 4 atm. we wish to slow this flow to subsonic speed through a system of shock waves with as small a loss in total pressure as possible.
The density of the gas remains constant and the flow of gas can be characterized by momentum and energy conservation if the speed of the object is substantially slower than the gas's speed of sound.
We must examine the impact of the gas's compressibility when the object's speed approaches that of sound. As the gas is compressed by the object, its local density changes.
Entropy is constant for compressible flows with little to no flow turning, and the flow process is reversible. The isentropic relations then provide the change in flow attributes. Constant entropy is the definition of entropy. However, the flow process becomes irreversible and the entropy rises when an object travels faster than the speed of sound and the flow area suddenly decreases. There are shock waves produced. Shock waves are incredibly small areas in a gas where the gas's properties change significantly.
The static pressure, temperature, and gas density all rise almost instantly across a shock wave. The total enthalpy and the total temperature remain constant because a shock wave doesn't produce any work and there isn't any heat addition.
The total pressure downstream of the shock is, however, always lower than the total pressure upstream of the shock since the flow is non-isentropic. A shock wave is associated with a drop of overall pressure. The slide displays the total pressure ratio. We cannot apply the standard (incompressible) form of Bernoulli's equation across the shock due to variations in total pressure. Over a shock wave, the flow's speed and Mach number both fall.
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f the coefficient of kinetic friction between tires and dry pavement is 0.80, what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 31.3 m/s ?
The shortest distance in which you can stop an automobile by locking the brakes when traveling at 31.3 m/s is 62.48m
Force of friction f=N=mg,
where m is the mass of the car.
Acceleration,
a=f/m=mg/m=g
=0.8 x9.8
=7.84m/s2
The distance travelled, S before coming to rest (v = 0)
S=(v2- u2)/2
=(0 - 31.3x31.3)/2x7.84
=62.48 m
The shortest distance in which you can stop an automobile by locking the brakes when traveling at 31.3 m/s is 62.48m
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At which part a of a river would you have the best luck
growing food??
Answer:
Delta !
Explanation:
What is the mass of an object a of 77.5 N?
Answer:
Explanation:
Given:
F = 77.5 N
g = 9.8 m/s²
__________
m- ?
m = F /g = 77.5 / 9.8 ≈ 7,9 kg
Question 2 of 6
Read the problem below and find the solution. Draw a diagram on your own
paper to help solve it.
If your teacher tells you to do questions 28 through 41 in your math book for
homework, how many questions is that?
(Do not include units in your answer.)
Answer here 14
If your teacher tells you to do questions 28 through 41 in your math book for homework, 14 questions are required, and this can be found out through general mathematical concepts such as addition and subtraction.
What are the general mathematical concepts?The general mathematical concepts can be said as the basic concepts or formulas that are taught to everybody. these include the concepts such as Addition, Subtraction, Multiplication and Division.
These are the base formulated concepts basing upon which the whole lot of other generalized equations and formulas are set up.
As per the question, we see that your teacher tells you to do questions a total of 28 through 41 in your math book for homework. Thus, we are then asked to find out the total number of questions.
Since we need to solve a total of questions 28 through 41, this means we have to solve question number 28 and 41 as well.
We already know that 28-37 would be 10 questions and then further 38-41 would be about 4 questions.
So, the total questions required = 10 +4
Total questions = 14
Thus, you just need to solve 14 questions in total to complete the homework.
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There are several methods of extrasolar planet detection currently in use. Which properties of a planet does each method discover? sort the following methods into the appropriate categories.
The complete box as per methods of extrasolar planet detection
Diameter of the
Transit methodMass of the planet
Microlensing methodRadial velocity methodAstrometry methodThis is further explained below.
What are the methods of extrasolar planet detection?Generally, The techniques for finding planets beyond the solar system
Microlensing caused by gravitational forces occurs when a star that contains a planet moves in front of another star. This causes the light to bend in a peculiar manner.
The transit method takes into account the varying luminosities of the stars. When a planet passes in front of a star, the star's apparent brightness decreases, which provides information about the planet's radius or diameter.
The radial velocity method analyzes the Doppler shift of a star to determine whether the star is moving closer or farther away from us.
Method of astronomy that is based on variations in the positions of stars in the sky.
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A mechanic uses a hydraulic car jack to lift the front end of a car to change the oil. The jack used exerts 8000 N of force from the larger piston. To pump the jack, 4000 N of force is exerted on the small piston, which has an area of 2 m2 . What is the area of the large piston?
Answer:
Hope the picture will help you
What makes a method accurate or inaccurate?
Answer:
Accuracy is how far off a measurement is from its true value. A measured value that's far from a true value is inaccurate, while a measure that is close to a true value is accurate.
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a worker is standing on a wet concrete floor, holding an electric drill having a metallic case. the metallic case is connected through the ground wire of a three-terminal power outlet to power-system ground. the resistance of the ground wire is rg. the resistance of the worker's body is rw
The resistance of the ground wire is Rg is mathematically given as Rg=0.025Ф
This is further explained below.
What is resistance?Take into account that the amount of electricity flowing through the worker's body does not surpass 0.1 mA.
The value of the earth's resistors should be extremely low in order to ensure that the greatest possible current is carried by the earth wire and, as a result, that the worker is not shocked.
Determine the highest possible value for R g s o.
[tex]$\begin{aligned}&0.1 \mathrm{~mA}=(2)\left(\frac{R_g}{R_g+500}\right) \\&\frac{R_g+500}{R_g}=20000 \\&R_g+500=20000 R_g \\&\begin{aligned}R_g &=\frac{500}{20000-1} \\&=0.025 \Omega\end{aligned}\end{aligned} \\\\ Rg=0.025Ф[/tex]
The current that flows through the worker is less than or equal to 0.1mA.
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CQ
A worker is standing on a wet concrete floor, holding an electric drill having a metallic case. The metallic case is connected through the ground wire of a three-terminal power outlet to the power-system ground. The resistance of the ground wire is Rg. The resistance of the worker's body is Rw = 500 ?. Due to faulty insulation in the drill, a current of 2 A flows into its metallic case. The circuit diagram for this situation is shown in the figure.Find the maximum value of Rgso that the current through the worker does not exceed 0.1 mA.
a current density of 4.50 10-13 a/m2 exists in the atmosphere at a location where the electric field is 133 v/m. calculate the electrical conductivity of the earth's atmosphere in this region.
This measurement demonstrates the conductivity of a certain material. When compared to rubber, which has poor electrical conductivity and high electrical resistance, metal, for instance, has high electrical conductivity and low electrical resistivity.
Electrical resistivity's physical opposite is electrical conductivity. This measurement demonstrates how well a certain material can conduct an electric current.
Rubber has a poor electrical conductivity and a high electrical resistivity, whereas metal has a high electrical conductivity and a low electrical resistance.
We are aware of the following thanks to the values that have been provided:
J = 6.00 10 13 A / m 2 and E = 100 V / m
The following equation provides the current density in terms of the electric field:
J = σ E
Here, E is the electric field, J is the current density, and is the electrical conductivity.
σ = J / E
The known values are then substituted.
σ = ( 6.00 × 10 − 13 A / m 2 ) / ( 100 V / m )
σ = ( 6.00 × 10 − 15 / m ) / ( 100 V )
The corresponding unit for V / A is.
Condense the units.
σ = 6.00 × 10 − 15 Ω − 1 m − 1.
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Two airplanes leave an airport at the same time. The velocity of the first airplane is 740 m/h at a heading of 42.8°. The velocity of the second is 580 m/h at a heading of 114°.
How far apart are they after 1.6 h? Answer in units of m.
You’re driving a car East on the 91 fwy for a total distance of 230 km. If it took you 2.3 hours to reach your destination, then what was your speed?
What was GIVEN?
Answer:
Explanation:
Given:
S = 230 km
t = 2.3 h
_________
V - ?
Speed:
V = S / t
V = 230 / 2.3 = 100 km/h
how much power is required to do 200 joules of work in 4 seconds?
what's Given:
asked:
formula:
substitute:
final answer:
Why is using a car not a good reason to use as a reference point
A car not a good reason to use as a reference point because it is not a fixed object relative to earth.
What is a Reference point?This is referred to as a point which is used to describe or discover the location of an object. The types of objects which are used as a good reference point are those which are fixed relative to earth.
On the other hand, cars are not fixed relative to earth such as trees, land which is therefore the reason why car is a good reason not to be used as a reference point.
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A student drops a ball and it falls downward. What types of energy are involved in this energy transformation? Two answers are correct
A student drops a ball and it falls downward. Types of energy are involved in this energy transformation are kinetic and gravity potential.
What is energy?Energy is defined as the capacity of a physical system to perform work.
Types of energy are Kinetic energy and potential energy.
The energy an object has as a result of motion is known as kinetic energy. Applying force to an object will cause it to accelerate.
Potential energy is the energy that is stored in any object or system as a result of its position or component arrangement.
A student drops a ball and it falls downward then there is kinetic energy due to motion and potential energy due to height.
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Users in motion, wireless connectivity, and a cloud-based resource are the elements in a ________ system.
Users in motion, wireless connectivity, and a cloud-based resource are the elements in a mobile system.
Smartphones, tablets, 2-in-1 PCs, smart speakers, smart watches, and smart eyewear all use mobile operating systems. Because they were originally developed for desktop computers, which historically did not have or require specific mobile capability, the operating systems that are used on "mobile" machines, like ordinary laptops, are often not referred to as mobile operating systems.
The distinction between mobile and other forms has become more hazy recently as a result of current hardware becoming more compact and portable than earlier equipment. This distinction is being muddled by the introduction of tablet computers and lightweight laptops, two major advancements.
Mobile operating systems combine desktop operating system elements with additional functionality that is useful for handheld or mobile use. For telephony and data connectivity, these systems typically have a wireless integrated modem and SIM tray.
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Users in motion, wireless connectivity, and a cloud-based resource are the elements of a mobile system which is the operating system used to run mobile devices like smartphones, tablets, etc.
A mobile system is a computing system with mobile entities. These can be hardware or software. The system runs on a mobile operating system such as android and iOS which are the two major mobile operating systems among others. A mobile operating system (OS) allows smartphones, tablet PCs, and other devices to run applications and programs.
Because mobile systems help in the real-time execution of several applications these must use wireless forms of connectivity and a cloud-based resource and storage system. Also, the fact that users can be in motion should be kept in mind and hence these are important elements in a mobile system
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