Answer:
0.25
Explanation:
That is the right answer.
The fraction of the initial kinetic energy is lost in the collision is 35.3%.
The given parameters:
Initial velocity of ball A = 8.3 m/sInitial velocity of ball B = 0Final velocity of ball B = 6.4 m/sThe initial kinetic energy of the system collision is calculated as follows;
[tex]K.E_i = \frac{1}{2} mv_1_i^2 + \frac{1}{2} mv_2_i^2\\\\K.E_i = \frac{1}{2} (m)(8.3)^2 + \frac{1}{2} (m) (0)^2\\\\K.E_i = 34.445 m[/tex]
The final velocity of ball A after collision is calculated as follows;
[tex]u_1 + v_1 = u_2 + v_2\\\\8.3 + v_1 = 0 + 6.4\\\\v_1 = 6.4 - 8.3\\\\v_1 = -1.9 \ m/s[/tex]
The final kinetic energy of the system after collision is calculated as follows;
[tex]K.E_f = \frac{1}{2} m(-1.9)^2 + \frac{1}{2} m(6.4)^2\\\\K.E_f = 22.285 \ m \[/tex]
The fraction of the initial kinetic energy is lost in the collision is calculated as follows;
[tex]= \frac{K_i - K_f}{K_i} \\\\= \frac{34.445 - 22.285}{34.445} \\\\= 0.353\\\\= 35.3\%[/tex]
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A person of mass 70 kg stands 0.5 m from one end of a 2 m long uniform plank (of negligible mass) that is
being held in equilibrium by two vertical ropes attached to the ends of the plank. What is the tension in the
rope closer to the person?
Help
Hi there!
We can begin by using a summation of torques, with the fulcrum being at the rope farthest from the person (the other end).
Thus:
F1 = weight of person
R1 = distance of person to fulcrum (farther end)
F2 = Tension of rope closest to person
R2 = distance of rope to fulcrum
∑τ = 0 = R2F2 - R1F1
R2F2 = R1F1
Plug in values:
2F2 = 1.5(700N)
2F2 = 1050 N
F2 = 525 N = Tension of closest rope
The less dense areas created as a sound wave propagates are called
A. rarefactions✅
B. troughs
C. crests
D. compressions
Answer:
C. crests
don't trust me though
Answer:
The less dense areas created as a sound wave propagates are called rarefactions.
Explanation:
Hoped this helped.
What is the volume of 150g of a substance that has a density of 150g of a substance that has a density of 11.3g/cm3
Answer:
25.0 cm3
Explanation:
The volume is 25.0 cm3 .
Some amount of ideal gas with internal energy U was heated from 100^0C to 200^0C. We can predict that internal energy after heating in terms of U is:
The internal energy after heating in terms of U is 100U.
The given parameters;
initial temperature of the gas, T₁ = 100 ⁰Cfinal temperature of the gas, T₂ = 200 ⁰CAssuming a constant pressure, the internal energy of the ideal gas is equal to the change in the enthalpy of the ideal gas.
[tex]\Delta H = U \times \Delta T\\\\\Delta H = U (200 - 100)\\\\\Delta H = 100 U[/tex]
Thus, we can conclude that the internal energy after heating in terms of U is 100U.
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An electromagnet does not attract a piece of iron.Is it true ? Give reason
Answer:
False..
Explanation:
An electoMagnets attract iron due to the influence of their magnetic field upon the iron. ...
I need to choose a theme for my physics assignment My experiment is finding g
How to find g (acceleration due to gravity)
Solution:We know,
Acceleration due to gravity (g)
[tex] = \frac{GM}{ {R}^{2} } [/tex]
where, G = Gravitational constant
[tex] = 6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \\ [/tex]
M = Mass of the earth
[tex] = 6 \times {10}^{24} \: kg[/tex]
R = Radius of the earth
[tex] = 6.4 \times {10}^{6} m[/tex]
Putting these values of G, M and R in the above formula, we get
[tex]g \: = \: \frac{6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \times \: 6 \times {10}^{24} \: kg }{(6.4 \times {10}^{6}m {)}^{2} } \\ = 9.8m/ {s}^{2} [/tex]
So, the value of acceleration due to gravity is
[tex]9.8m/s ^{2} [/tex]
Hope it helps.
Do comment if you have any query.
Help please..
Kepler’s third law states that:
A. the orbits of the planets are elliptical.
B. the planets move slower when they are closer to the Sun and faster when they are farther from the Sun.
C. the square of the ratio of the periods of any two planets revolving around the Sun is equal to the cube of the ratio of their average distance from the Sun.
D. objects attract other objects with a force that is directly proportional to the product of their masses, and inversely proportional to the square of the distance between them.
Answer:
c
Explanation:
6. A 15.53 kg bag of soil falls 5.50 m at a construction site. If all the energy is retained by the soil in the bag, how much will its temperature increase
The increase in the temperature of the soil when bag fall from the given height is 0.065 ⁰C.
The given parameters:
mass of the bag, m = 15.53 kgheight of fall, h = 5.5 mspecific heat capacity of soil, C = 0.200 kcal/kg ⁰CApply the principle of conservation of energy as follows;
[tex]mgh = mC \Delta T\\\\gh = C \Delta T\\\\\Delta T= \frac{gh}{C}[/tex]
Convert the value of C in kcal/kg ⁰C to J/kg ⁰C
1 kcal = 4180 J
[tex]0.200 \ kcal/kg ^0 C= 0.2 \times 4180 (J/kg ^0C) = 836 \ J/kg^0 C[/tex]
The increase in the temperature of the soil is calculated as follows;
[tex]\Delta T= \frac{gh}{C}\\\\\Delta T= \frac{9.8 \times 5.5}{836} \\\\\Delta T= 0.065 \ ^0C[/tex]
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During a picnic, you and two of your friends decide to have a three-way-tug-of-war, with three ropes in the middle tied into a knot. Roberta pulls to the west with 277 N of force; Michael pulls to the south with 603 N. a) With what force should you pull to keep the knot from moving
Answer:
Explanation:
I'm not 10% sure but i think that if you were to do:
603 + 277 = 880
1000 - 880 = 120
So to keep it from moving, i think you would need to pull with a force of 120N
The force with which it should be pulled to keep the knot from moving is 663.58 N.
What is Newton's third law ?Newton's third law states that, for every action, there is an equal and opposite reaction.
Here,
Force exerted by Roberta, F₁ = 277 N
Force exerted by Michael, F₂ = 603 N
Since, Roberta is pulling to the west and Michael is pulling towards the south, the angle between the forces applied by them is 90°.
The force needed to keep the knot from moving is the resultant force of these two forces.
Therefore, the resultant force,
Fₙ = √F₁² + F₂²
Fₙ = √(277)² + (603)²
Fₙ = 663.58 N
Hence,
The force with which it should be pulled to keep the knot from moving is 663.58 N.
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)Suppose you have two insulated buckets containing the same amount of water at room temperature. You also happen to have two blocks of metal of the same mass, both at the same temperature, warmer than the water in the buckets. One block is made of aluminum and one is made of copper. You put the aluminum block into one bucket of water, and the copper block into the other. After waiting a while you measure the temperature of the water in both buckets. Which is warmer
The water in the bucket containing the aluminium block is warmer than the bucket containing the copper block.
The specific heat is the amount of heat needed or required to elevate the temperature of 1 gram of a substance by 1° C.
At standard conditions;
the specific heat of aluminium = 0.215 cal/gm/° Cthe specific heat of copper = 0.0923 cal/gm/° CWe know that:
Heat loss by the metal block = Heat gained by the water bucketTherefore, since the specific heat of aluminium is higher than that of copper, the water bucket containing aluminium block will be warmer than the bucket containing the copper block.
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HELPP
When two forces are in opposite directions, and they are the exact same magnitude, the forces will _______.
a. subtract from each other
b. cancel out
c. go on infinitely
d. eventually reach equilibrium
Answer:
i think the correct answer is B. cancel out
Consider 5.00 mol of liquid water. (a) What volume is occupied by this amount of water? The molar mass of water is 18.0 g/mol. (b) Imagine the molecules to be, on average, uniformly spaced, with each molecule at the center of a sma11 cube. What is the length of an edge of each sma11 cube if adjacent cubes touch but don't overlap? (c) How does this distance compare with the diameter of a molecule?
This question involves the concepts of density, volume, and mass.
(a) The volume occupied by this amount of water is "90 m³".
(b) The length of an edge of each small cube is "83 μm".
(c) The length of the edge of each small cube is "equal" to the diameter of a molecule.
(a)
The volume can be found using the following formula:
[tex]V = \frac{nM}{\rho}[/tex]
where,
V = volume occoupied = ?
n = no. of moles = 5 mol
M = molar mass = 18 g/mol
[tex]\rho[/tex] = density of water = 1 g/m³
Therefore,
[tex]V=\frac{(5\ mol)(18\ g/mol)}{1\ g/m^3}\\\\[/tex]
V = 90 m³
(b)
First, we will find the volume of an individual molecule:
[tex]V_i =\frac{V}{nN_A}[/tex]
where,
[tex]N_A[/tex] = Avogadro's number = 6.02 x 10²³ molecules/mol
Therefore,
[tex]V_i=\frac{90\ m^3}{5\ mol(6.02\ x\ 10^{23}\ molecules/mol)}[/tex]
Vi = 3 x 10⁻²³ m³
This volume can be given as a volume of the sphere:
[tex]V_i=\frac{4}{3}\pi r^3\\\\r=\sqrt[3]{\frac{3(3\ x\ 10^{-23}\ m^3)}{4\pi}}[/tex]
r = 4.15 x 10⁻⁷ m
Diameter = d = 2r = 2(4.15 x 10⁻⁷ m)
d = 8.3 x 10⁻⁷ m = 83 μm
since the cubes are adjacent to each other. Therefore, the diameter will be equal to the edge length.
Edge Length = L = d
L = 8.3 x 10⁻⁷ m = 83 μm
(c)
The edge length is equal to the diameter of the molecule.
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i mark u brainliest answer please answer me
Answer:
bathing and wear clean clothes 2 pollute 3throwing waste in land air water pollution
As a bicycle pump inflates a tyre, it pressure rises from 30 kPa to 40 kPa at constant temperature of 30 °C. By assuming the air acts as an ideal gas, calculate the work done per mol of the air.
A. -80.35 J
B. 80.35 J
C. -811.93 J
D. 811.93 J
(please show calculation)
can use this formula W=nRT ln(p1/p2)
Answer:
B.-80.35 J
i dont know the calculation
The coefficient of kinetic friction for a 22 kg bobsled on a track is 0.10. What force is required to push it down a 5.0 degree incline and achieve a speed of 62 km/h at the end of 75 m
Hi there!
We must begin by converting km/h to m/s using dimensional analysis:
[tex]\frac{62km}{1hr} * \frac{1hr}{3600sec}*\frac{1000m}{1km} = 17.22 m/s[/tex]
Now, we can use the kinematic equation below to find the required acceleration:
vf² = vi² + 2ad
We can assume the object starts from rest, so:
vf² = 2ad
(17.22)²/(2 · 75) = a
a = 1.978 m/s²
Now, we can begin looking at forces.
For an object moving down a ramp experiencing friction and an applied force, we have the forces:
Fκ = μMgcosθ = Force due to kinetic friction
Mgsinθ = Force due to gravity
A = Applied Force
We can write out the summation. Let down the incline be positive.
ΣF = A + Mgsinθ - μMgcosθ
Or:
ma = A + Mgsinθ - μMgcosθ
We can plug in the given values:
22(1.978) = A + 22(9.8sin(5)) - 0.10(22 · 9.8cos(5))
A = 46.203 N
. If the frequency of a system undergoing simple harmonic motion doubles, by what factor does the maximum value of acceleration change
Answer:
a1 = -w^2 A sin w t where w is the angular frequency
a2 = - (2 w)^2 A sin 2 w t where w2 = 2 w1
a2 / a1 = 4 sin 2 w t / sin w t
Since sin 2 w t / sin w t = 2 sin w t will change twice as fast
a2 / a1 = 8 the max acceleration must change 8 tims as fast
Another box of samples is hoisted up by the same rope. If the rope is shaken with the same frequency as before, and the wavelength is found to be 7.9 m , what is the mass of this box of samples
The motion of the rope which is perpendicular to the direction of the
propagation of the wave is a transverse wave motion.
The mass of the box is approximately 9.93 kgReasons:
The given function for the wave speed is presented as follows;
[tex]\displaystyle v = \sqrt{\frac{T}{\mu} } \\[/tex]
Where;
[tex]\displaystyle \mu = \frac{Mass \ of \ rope }{Length \ of \ rope}[/tex]
Taking the mass of the rope as, m = 2.00 kg
The length of the rope, L = 80.0 m
The mass hanging on the rope, M = 20.0 kg
We have;
T = 20.0 kg × 9.81 m/s² = 196.2 N
[tex]\displaystyle \mu = \frac{2.0 }{80.0} = \frac{2.0 }{80.0} = 0.025[/tex]
Therefore;
Taking the wavelength as, λ = 7.9 m, and the frequency as 20 Hz, we have;
v = f × λ
Therefore;
v = 7.9 Hz × 7.9 m = 62.41 m/s
Which gives;
[tex]\displaystyle 62.41 = \sqrt{\frac{T}{0.025} }[/tex]
T = 62.41² × 0.025 = 97.3752025
[tex]\displaystyle Mass, \ m = \mathbf{\frac{T}{g}}[/tex]
Where;
g = The acceleration due to gravity which is approximately 9.81 m/s²
[tex]\displaystyle Mass, \ m = \frac{97.3752025}{9.81} \approx 9.93[/tex]
Therefore;
The mass of the box, m ≈ 9.93 kg
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The parameters obtained from a similar question online are;
[tex]\displaystyle The \ equation \ applied, \, v = \sqrt{\frac{T}{\mu} } \\[/tex]
Length of the rope, L = 80.0 m
Mass of the rope, m = 2.0 kg
Frequency of a point on the rope, f = 20 Hz
2) A ray of light in air is approaching the boundary with water at an
angle of 52 degrees. Determine the angle of refraction of the light
ray. (Refractive index of air = 1, water = 1.33)
=
Answer:
Explanation:
ASSUMING the 52° is the angle of incidence measured from the perpendicular to the surface
n₁sinθ₁ = n₂sinθ₂
1 sin52 = 1.33sinθ₂
θ₂ = arcsin(sin52 / 1.33)
θ₂ = 36°
as measured from the perpendicular to the surface
What will be the speed of these waves (in terms of V) if we increase M by a factor of 18.0, which stretches the wire to double its original length
The speed of the chord wave allows finding a new speed when the mass is increased and the chord length is:
The velocity is: v = [tex]\frac{v_o}{3}[/tex]
A wave is a periodic movement of the particles that carries energy, but not matter, the speed of the waves is related to the properties of the medium by the relationship.
[tex]v = \sqrt{\frac{T}{\mu } }[/tex]
Where v is the speed of the wave, T the force and μ is the linear density.
Indicates that the applied mass increases by a factor of 18.0 and the length of the head is increased to twice the original.
The linear density of the cable is
[tex]\mu = \frac{m}{l}[/tex]
Where m is the mass of the cable and l is the length.
Let's use the subscript "o" for the initial conditions.
M = 18.0 m₀
l = 2 l₀
We look for the density.
[tex]\mu = \frac{18.0 m_o}{2.0 l_o}[/tex]mu = 18.0 mo / 2 lo
[tex]\mu = 9.0 \mu_o[/tex]
We substitute in the expression for the velocity assuming that the tension is kept constant.
[tex]v= \sqrt{\frac{T}{9.0 \mu} }[/tex]
[tex]v= \frac{v_o}{3}[/tex]
In conclusion, using the speed of the chord wave we can find a new speed when the mass is increased and the chord length is:
The velocity is: v = [tex]\frac{v_o}{3}[/tex]
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If it takes 50.0 seconds to lift 10.0 Newtons of books to a height of 7.0
meters, calculate the power required to lift it?
name two examples where the cohesive force dominates over the adhesive force and vice versa
Cohesive forces are the forces that draw molecules of the same type together. Adhesive forces are those that draw molecules of various types together.
What is the difference between cohesive force and adhesive force?Cohesive forces are the forces that draw molecules of the same type together. Adhesive forces are those that draw molecules of various types together.
The force that draws molecules of the same substance together is called the cohesive force. The force that holds molecules of various substances together is known as the adhesive force.
Between molecules of the same substance, there are cohesive forces. There is a natural tendency to resist separation due to these intermolecular forces between like elements. Conversely, adhesive forces draw disparate molecules together.
In physics, cohesion refers to the intermolecular attraction that exists between two adjacent parts of a substance, especially one that is solid or liquid. A piece of matter is held together by this force. Adhesion is a term for the intermolecular forces that act when two dissimilar substances come into contact.
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the wheel of bicycle has a radius of 25cm. what will be the magnitudes of the angular displacement in radian and revolution respectively, when the wheel has rolled a distance of 350cm on straight level road?
Answer:
Explanation:
350 cm / 25 cm = 14 radians
14 rad / 2π rad/rev = 2.23 revolutions
How does climate change?
Answer:
Con el cielo del agua
Explanation:
espero que te ayude
I need the answer to this question
Answer:
I don't see anything so you should repost
A racecar can be slowed with a constant acceleration of -14 m/s2. If the car is going 75 m/s, how many meters will take to stop?
Answer:
Explanation:
v² = u² + 2as
s = (v² - u²)/2a
s = (0² - 75²) / (2(-14))
s = 200.8928
s = 200 m
Answer:
Identification
Explanation:
:-;..........
A camera takes a properly exposed photo at f/5.6 and 1/250 s. What shutter speed should be used if the lens is changed to f/4.0?
a.1/65 s
b.1/125 s
c.1/250 s
d.1/500 s
e.1/1000 s
The shutter speed that should be used if the lens is changed to f/4.0 is; Choice D: 1/500 s.
The relationship between the shutter speed and the focal ratio is an inverse relationship.
Ratio of Areas = 5.6²/4²
A1/A2 = 1.96 = 2Since; T1A1 = T2A2
T2 = T1(A1/A2)T2 = 250 × 2 = 500 secondsIn essence; when the focal ratio is reduced as in this case; from f/5.6 to f/4.0; the shutter speed is increased.
Ultimately, the shutter speed of the camera in discuss increases to; 1/500 s.
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Kinesha and her friend were watching a solar eclipse. Kinesha explains to her friend that a solar eclipse means that Earth is located between the Sun and the Moon. Her friend tells Kinesha that her explanation is incorrect. Why?
this is where the sun and moon line up where you asleep only a tiny bit of the sun it's pretty cool to see
Explanation:
a solar eclipse means when the moon goes infront of the sun and the earth turns dark
A 5 kg bowling ball travelling at 2 m/s hits a motionless 10 kg bowling ball. If the smaller ball bounces back at a speed of -1 m/s, what will be the speed of larger ball after the collision? Hint: Use the conservation of momentum equation to solve this problem.
Answer:
1.5 m/s
Explanation:
Conservation of momentum means the momentum of the system before the collision is the same as after.
The before, after momentum of each ball is ...
5 kg ball: (5 kg)(2 m/s), (5 kg)(-1 m/s)
10 kg ball: (10 kg)(0 m/s), (10 kg)(v)
The sum of the "before" products is the same as the sum of the "after" products:
(5 kg)(2 m/s) +0 = (5 kg)(-1 m/s) +(10 kg)v
(10 +5) kg·m/s = (10 kg)·v . . . . . add (5 kg)(1 m/s) to both sides
v = (15 kg·m/s)/(10 kg) = 1.5 m/s
The speed of the larger ball will be 1.5 m/s. Its direction of motion will be the opposite of that of the 5 kg ball after the collision.
Explain different layers of atmosphere and the pressure in each layer. Draw diagram
Answer:
Our atmosphere has five different layers. They are:
1. Troposphere: This is the most important layer of the atmosphere with an average height of 13 km from the earth. It is in this layer that we find the air that we breathe. Almost all the weather phenomena such as rainfall, fog and hailstorm occur here.
2. Stratosphere: This layer extends up to a height of 50 km. It presents the most ideal condition for flying airplanes. It contains a layer of ozone gas which protects us from the harmful effect of the sun rays.
3. Mesosphere: This layer extends up to a height of 80 km. Meteorites bum up in this layer on entering from the space.
4. Thermosphere: In this layer, the temperature rises very rapidly with increasing height. The ionosphere is a part of this layer. It extends between 80-400 km. This layer helps in radio transmission. Radio waves transmitted from the earth the reflected back to the earth by this layer.
5. Exosphere: It is the uppermost layer where there is very thin air. Light gases such as helium and hydrogen float into space from here.
At a distance of 1 cm from the source, the flux from the isotropic source is one hundred times brighter. Which source is brighter fifty centimeters away
Answer:
op
Explanation:
At a distance of 1 cm from the source, the flux from the isotropic source is one hundred times brighter. Which source is brighter fifty centimeters away