Hi there!
Since the collision is elastic, we must also satisfy the following condition:
Ei = Ef, or:
KEi = KEf
Begin by writing an expression for momentum. (p = mv) Remember that one ball's direction is negative; in this instance, we can let the second ball be moving LEFT.
mv1 + mv2 = mvf1 + mvf2
0.220(1.84) + 0.220(-.530) = 0.220(vf1 + vf2)
0.2882/0.220 = vf1 + vf2
1.31 = vf1 + vf2
Now, we can express this as a conservation of energy:
1/2mv1² + 1/2mv2² = 1/2mvf1² + 1/2mvf2²
Plug in values and simplify:
0.403315 = 1/2m(vf1² + vf2²)
Simplify further:
3.6665 = vf1² + vf2²
Use the equation derived from momentum above and solve for one variable:
vf2 = 1.31 - vf1
Plug in this expression for vf2:
3.6665 = vf1² + (1.31 - vf1)²
Expand:
3.6665 = vf1² + 1.7161 - 2.62vf1 + vf1²
Simplify:
1.9504 = -2.62vf1 + 2vf1²
Solve for vf1 using a graphing calculator:
vf1 = -0.53 m/s or 1.84 m/s; we must figure out which one is correct.
Since v1 is heading to the right initially with a velocity of 1.84 m/s, we know that the ball's velocity could not have stayed the same in both magnitude and direction, so the final velocity must be -0.53 m/s.
Now, we can solve for the velocity of the other ball (initial of 0.53 m/s):
vf2 = 1.31 - (-0.53) = 1.84 m/s.
Now, you could have also made the connection that when two balls of the SAME MASS experience an ELASTIC collision, the velocities are simply "exchanged" from one to another. I just used this more "extensive" method to prove this.
calculate the surface area of a box whose mass is 200 kg and exerts a pressure of 100 Pascal on the floor.
Answer:
Explanation:
If 2×2 is 4 so 1 kg can be 1 gram if it belive on it self some people change
A constant horizontal force of 30.0 N is exerted by a string attached to a 5.0 kg block being pulled across a tabletop. The block also experiences a frictional force of 5.0 N due to contact with the table. What is the horizontal acceleration of the block?
Answer:
A 5.00- kg block is placed on top of a 10.0 -kg block (Fig. P5.68). A horizontal force of 45.0 N is applied to the 10.0-kg block, and the 5.00- kg block is tied to the wall. The coefficient of kinetic friction between all surfaces is 0.200. (a) Draw a free-body diagram for each block and identify the action-reaction forces between the blocks.
(b) Determine the tension in the string and the magnitude of the acceleration of the 10.0-kg block.
The horizontal acceleration of the block is 5 m/s².
To calculate the horizontal acceleration on the block, we use the formula below.
Formula:
ma = (F-F')............... Equation 1Where:
m = mass of the blocka = Horizontal acceleration of the blockF = Horizontal force exerted on the stringF' = Frictional forceMake "a" the subject of the equation.
a = (F-F')/m............... Equation 2Substitute these values into equation 2
F = 30 NF' = 5 Nm = 5.0 kgSubstitute these values into equation 2
a = (30-5)/5a = 25/5a = 5 m/s²Hence the horizontal acceleration of the block is 5 m/s².
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A merry-go-round of radius R, shown in the figure, is rotating at constant angular speed. The friction in its bearings is so small that it can be ignored. A sandbag of mass m is dropped onto the merry-go-round, at a position designated by r. The sandbag does not slip or roll upon contact with the merry-go-round.
The figure shows a top view of a merry-go-round of radius capital R rotating counterclockwise. A sandbag is located on the merry-go-round at a distance lowercase r from the center.
Rank the following different combinations of m and r on the basis of the angular speed of the merry-go-round after the sandbag "sticks" to the merry-go-round.
The angular speed of the merry-go-round reduced more as the sandbag is
placed further from the axis than increasing the mass of the sandbag.
The rank from largest to smallest angular speed is presented as follows;
[m = 10 kg, r = 0.25·R]
[tex]{}[/tex] ⇩
[m = 20 kg, r = 0.25·R]
[tex]{}[/tex] ⇩
[m = 10 kg, r = 0.5·R]
[tex]{}[/tex] ⇩
[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]
[tex]{}[/tex] ⇩
[m = 10 kg, r = 1.0·R]
Reasons:
The given combination in the question as obtained from a similar question online are;
1: m = 20 kg, r = 0.25·R
2: m = 10 kg, r = 1.0·R
3: m = 10 kg, r = 0.25·R
4: m = 15 kg, r = 0.75·R
5: m = 10 kg, r = 0.5·R
6: m = 40 kg, r = 0.25·R
According to the principle of conservation of angular momentum, we have;
[tex]I_i \cdot \omega _i = I_f \cdot \omega _f[/tex]
The moment of inertia of the merry-go-round, [tex]I_m[/tex] = 0.5·M·R²
Moment of inertia of the sandbag = m·r²
Therefore;
0.5·M·R²·[tex]\omega _i[/tex] = (0.5·M·R² + m·r²)·[tex]\omega _f[/tex]
Given that 0.5·M·R²·[tex]\omega _i[/tex] is constant, as the value of m·r² increases, the value of [tex]\omega _f[/tex] decreases.
The values of m·r² for each combination are;
Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²
Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²
Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²
Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²
Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²
Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²
Therefore, the rank from largest to smallest angular speed is as follows;
Combination 3 > Combination 1 > Combination 5 = Combination 6 >
Combination 2
Which gives;
[m = 10 kg, r = 0.25·R] > [m = 20 kg, r = 0.25·R] > [m = 10 kg, r = 0.5·R] > [m =
10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R] > [m = 10 kg, r = 1.0·R].
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The car on this ramp starts from rest. When released, it
accelerates at a constant rate. It has an initial position of 12 cm
from the top of the ramp, and has an average velocity of 1.20 m/s
for a total of 1.80 s. Which is the correct final position of the car?
Answer:
Explanation:
s 0.12 + 1.20(1.80) = 2.28 m from the top.
Please help. I'm mot sure what I need to do first...
Answer:
0.80 kN
Explanation:
Hope you understood it
A 400 g ball swings in a vertical circle at the end of
a 1.5-m-long string. When the ball is at the bottom
of the circle, the tension in the string is 10 N.
A) what is the speed of the ball at that point?
Answer:
a = 25 m/s2
Explanation:
A = f/m
A = Speed/Acceleration
F =‘Force
M = Mass
Several common barometers are built using a variety of fluids. For which fluid will the column of fluid in the barometer be the highest
Answer:
the one in which the fluid has the lowest density
A child's toy consists of a spherical object of mass 50 g attached to a spring. One end of the spring is fixed to the side of the baby's crib so that when the baby pulls on the toy and lets go, the object oscillates horizontally with a simple harmonic motion. The amplitude of the oscillation is 6 cm and the maximum velocity achieved by the toy is 3.2 m/s . What is the kinetic energy K of the toy when the spring is compressed 4.7 cm from its equilibrium position?
A)The following is a list of quantities that describe specific properties of the toy. Identify which of these quantities are known in this problem.
Select all that apply.
1. force constant k
2. total energy E
3. mass m
4. maximum velocity vmax
5. amplitude A
6. potential energy U at x
7. kinetic energy K at x
8. position x from equilibrium
B)What is the kinetic energy of the object on the spring when the spring is compressed 4.7 cm from its equilibrium position?
C)What is the potential energy U of the toy when the spring is compressed 4.7 cm from its equilibrium position?
Hi there!
Part A:
The only quantities explicitly given to us are:
3. mass (m)
4. Maximum velocity (vmax)
5. Amplitude (A)
8. Position x from equilibrium
Part B:
To solve, we must begin by calculating the force constant, 'k'.
We can use the following relationship:
[tex]v = \sqrt{\frac{k}{m}(A^2-x^2)[/tex]
We are given the max velocity which occurs at a displacement of 0 m, because the mass is the fastest at the equilibrium point. We can rearrange the equation for k/m:
[tex]\frac{v^2}{(A^2-x^2)} = \frac{k}{m}[/tex]
[tex]\frac{3.2^2}{(0.06^2-0)} = \frac{k}{m} = 2844.44[/tex]
Now, we can find the velocity at 4.7cm (0.047m) using the equation:
[tex]v = \sqrt{(2844.44)(0.06^2-0.047^2)} = 1.989 m/s[/tex]
Plug this value into the equation for kinetic energy:
[tex]KE = \frac{1}{2}mv^2\\\\KE = \frac{1}{2}(0.05)(1.989^2) = \boxed{0.0989 J}[/tex]
Part C:
The potential energy of a spring is given as:
[tex]U = \frac{1}{2}kx^2[/tex]
Find 'k' using the derived quantity above:
[tex]\frac{k}{m} = 2844.44\\\\k = 2844.44m = 142.22 N/m[/tex]
Now, calculate potential energy:
[tex]U = \frac{1}{2}(142.22)(0.047^2) = \boxed{0.157 J}[/tex]
Make one comparison between the moral condition of the world at the time of the Flood with our day. Only One Short explanation.
The moral condition of the world today appears to be worse than it was in the antediluvian era.
The biblical account of the flood records that the world delved into apostasy in the days of Noah so much so that God regretted the fact that he created man. Some of the evils of the antediluvian world include; sodomy, drunkenness, lewdness and several forms of immorality.
We can see that these vices that led to the destruction of the world due to moral bankruptcy in the antediluvian era is still very much prevalent in our world today. The moral condition of the world today appears to be worse than it was in the antediluvian era.
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Jack sits in the chair of a Ferris wheel that is rotating at a constant 0.150 rev/srev/s . As Jack passes through the highest point of his circular part, the upward force that the chair exerts on him is equal to one-fourth of his weight.
What is the radius of the circle in which Jack travels? Treat him as a point mass.
Answer:
Explanation:
At the top of the arc, 3/4 of the acceleration of gravity is use to supply the necessary centripetal acceleration.
0.75g = ω²R
R = 0.75g/ω²
R = 0.75(9.81) / (0.15 rev/s)(2π rad/rev)²
R = 8.283006...
R = 8.28 m
(f) As the pions move away from each other, the ratio of the absolute value of electric potential energy to the final total kinetic energy of the two pions changes. At some point, the potential energy becomes negligible compared to the final total kinetic energy . We can consider that the value of the ratio is about 0.01 when , where is the final total kinetic energy of the two pions. What will the distance, , between the pions be when this criterion is satisfie
The distance r between the pions when the criteria are satisfied is 1.45 × 10⁻¹⁶ m
If we consider the potential energy (U) between the pions, then (U) can be expressed as:
[tex]\mathbf{U = \dfrac{kq^2}{r} ---- (1)}[/tex]
Given that at some instance, the potential energy becomes negligible compared to the final K.E.
As such the conservation of the total energy in the system can be given as:
E = U + KAgain, if we consider the ratio of the potential energy to the kinetic energy to be about 0.01, then:
[tex]\mathbf{\dfrac{U}{K}= 0.01} \\ \\ \mathbf{U = 0.01 K----(2)}[/tex]
∴
Equating both equations (1) and (2) together, we have:
[tex]\mathbf{\dfrac{kq^2}{r} = 0.01 K}[/tex]
[tex]\mathbf{\dfrac{kq^2}{r} = 0.01 \Bigg [ m_{o \pi}c^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{v_{\pi}^2}{c^2} }} \Big ] \Bigg] }[/tex]
[tex]\mathbf{r =\dfrac{kq^2}{ 0.01 \Bigg [ m_{o \pi}c^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{v_{\pi}^2}{c^2} }} \Big ] \Bigg] }}[/tex]
where:
r = distancek = Columb's constantq = charge on a protonm_o = rest mass of each pion in the previous questionc = velocity of light[tex]\mathbf{v_\pi}[/tex] = calculated velocity of proton in the previous questionReplacing their values in the above equation, the distance (r) between the pions is calculated as:
[tex]\mathbf{r =\dfrac{(9\times 10^9 \ N.m^2/C^2) (1.6022 \times 10^{-19} \ C)^2}{ 0.01 \Bigg [ (2.5 \times 10^{-28\ kg } )\times (3\times 10^8 \ m/s)^2 \Big [\dfrac{1}{\sqrt{1 - \dfrac{(2.97 \times 10^8 \ m/s)^2}{(3 \times 10^8 \ m/s)^2} }} \Big ] \Bigg] }}[/tex]
distance (r) = 1.45 × 10⁻¹⁶ m
Therefore, we can conclude that the distance r between the pions when the criteria are satisfied is 1.45 × 10⁻¹⁶ m
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Question: A NEO distance from the Sun is 1.18 AU. What is its relative speed compared to Earth (round your answer to 3 decimal places)
Its relative speed compared to Earth is 0.921
The speed of the object v = 2πr/T where r = radius of orbit and T = period of orbit.
Let v = speed of earth, r = radius of earth orbit = 1 AU and T = period of earth orbit.
So, v = 2πr/T
Also, v' = speed of NEO, r' = radius of NEO orbit = distance of NEO from sun = 1.18 AU and T' = period of NEO orbit.
So, v' = 2πr'/T'
v'/v = 2πr'/T' ÷ 2πr/T
v'/v = r'/r × T/T'
From Kepler's law, T² ∝ r³
So, T'²/T² = r'³/r³
(T'/T)² = (r'/r)³
T'/T = √[(r'/r)]³
T/T' = √[(r'/r)]⁻³
So, substituting this into the equation, we have
v'/v = r'/r × T/T'
v'/v = r'/r × √[(r'/r)]⁻³
v'/v = √[(r'/r)]⁻¹
Since r' = 1.18 AU and r = 1 AU, r'/r = 1.18
So, v'/v = √[(r'/r)]⁻¹
v'/v = √[(1.18)]⁻¹
v'/v = [1.0863]⁻¹
v'/v = 0.921
So, its relative speed compared to Earth is 0.921
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A stomp rocket takes 3.1 seconds to reach its maximum height.
- What is its initial velocity? (Do not use units. If the answer is negative, please put a
negative sign in front of the answer.)
- What is its maximum height? (Do not use units. If the answer is negative, please put a
negative sign in front of the answer.)
Answer:
Explanation:
Ignoring air resistance the time to rise will equal the time to fall and initial velocity will be the same magnitude as final velocity just before impact.
v = at
v = 9.8(3.1)
v = 30.38
v = 30 m/s
max height can be found knowing the velocity is zero at the top of its flight.
v² = u² + 2as
s = (v² - u²) / 2a
s = (0.00² - 30.38²) / (2(-9.8))
s = 47.089
s = 47 m
A stomp rocket takes 3.1 seconds to reach its maximum height then the initial velocity is given as v = 30 m/s and maximum height is 47.089 m.
What is Velocity?Velocity is defined as rate of change of position with respect to time.
SI unit of velocity is m/sec. Velocity is a vector quantity.
Given that in the question time taken by rocket to reach maximum height is 3.1 sec. Ignoring air resistance the time to rise will equal the time to fall and initial velocity will be the same magnitude as final velocity just before impact.
v = at
v = 9.8(3.1)
v = 30.38
v = 30 m/s
Max height can be found knowing the velocity is zero at the top of its flight.
v² = u² + 2as
s = (v² - u²) / 2a
s = (0.00² - 30.38²) / (2(-9.8))
s = 47.089
s = 47 m
So, the initial velocity is given as v = 30 m/s and maximum height is 47.089 m.
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I need help with this equation. 4 tutors so far on the math side are unable to help me solve the problem.
A boat is using echo-sounding equipment to measure the depth of the water underneath it, as illustrated in the first diagram.
The equipment in the boat sends a short pulse of sound downwards and detects the echo after a time interval of 0.80s. i Describe how an echo is caused. ii The speed of sound in water is 1500 m/s. Calculate the distance travelled (in metres) by the sound in 0.80 s.
Answer:
Explanation:
Echo is caused by sound energy reflecting off of "hard" surfaces. It could be as simple as a change in density of the material the sound is traveling through.
In 0.8 s, the sound has traveled 0.8(1500) = 1200 m.
That means the object that reflected the sound is 600 m below the boat. The sound took 0.4 s to reach the object and another 0.4 s to return the echo.
A 5 kg box is sitting on a rough wooden surface. The coefficient of static friction between the box and surface is 0.6. If the normal force on the box is 50 N, calculate the force of friction which must be overcome to move the box. Round your answer to the nearest whole number.
The force of friction needed to overcome to move the box is 29.4N
According to Newton's second law;
[tex]\sum F_x = ma_x\\[/tex]
Taking the sum of force along the plane;
[tex]F_m -F_f = ma\\F_m -F_f = 0\\F_m=F_f = \mu R[/tex]
This shows that the moving force is equal to the frictional force
Given that
[tex]\mu = 0.6\\R = mg = 49N[/tex]
Get the frictional force;
Since
[tex]F_f = \mu R\\F_f = 0.6 \times 49\\F_f = 29.4N[/tex]
Hence the force of friction needed to overcome to move the box is 29.4N
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how does the structure of compounds determines the properties of the compounds?
Answer:
The chemical structure of the molecule is responsible for each of these characteristics. The chemical structure is comprised of the bonding angle, the kind of bonds, the size of the molecule, and the interactions that occur among the molecules. Even little changes in the chemical structure of a molecule may have a significant impact on the characteristics of the substance.
Explanation:
Hope it helps:)
a coconut falls from the top of a tree and takes 3.5 seconds to reach the ground. How tall is the tree?
Hello!
To solve, we can begin by using the kinematic equation:
[tex]d = v_it + \frac{1}{2}at^2[/tex]
Where:
vi = initial velocity (m/s)
t = time (s)
a = acceleration (in this case, due to gravity. g = 9.8 m/s²)
Since the object falls from rest, the initial velocity is 0 m/s.
[tex]d = \frac{1}{2}at^2[/tex]
Plug in the given values:
[tex]d = \frac{1}{2}(9.8)(3.5^2) = \boxed{60.025 m}[/tex]
A mass vibrates back and forth from the free end of an ideal spring of spring constant 20 N/m with an amplitude of 0.30 m. What is the kinetic energy of this vibrating mass when it is 0.30 m from its equilibrium position?
Hi there!
We can begin by using the work-energy theorem in regards to an oscillating spring system.
Total Mechanical Energy = Kinetic Energy + Potential Energy
For a spring:
[tex]\text{Total ME} = \frac{1}{2}kA^2\\\\\text{KE} = \frac{1}{2}mv^2\\\\PE = \frac{1}{2}kx^2[/tex]
A = amplitude (m)
k = Spring constant (N/m)
x = displacement from equilibrium (m)
m = mass (kg)
We aren't given the mass, so we can solve for kinetic energy by rearranging the equation:
ME = KE + PE
ME - PE = KE
Thus:
[tex]KE = \frac{1}{2}kA^2 - \frac{1}{2}kx^2\\\\[/tex]
Plug in the given values:
[tex]KE = \frac{1}{2}(20)(0.3^2) - \frac{1}{2}(20)(0.3^2) = \boxed{0 \text{ J}}[/tex]
We can also justify this because when the mass is at the amplitude, the acceleration is at its maximum, but its instantaneous velocity is 0 m/s.
Thus, the object would have no kinetic energy since KE = 1/2mv².
A crane is lifting a 500 lb car. If the power of the crane 1.82 hp, find the velocity of the car.
Answer:
Explanation:
550 ft•lb/s / hp•(1.82 hp) / 500 lb = 2.00 ft/s
How does friction help us in walking.
The_____ scale is called an absolute temperature scale, and its zero point is called absolute zero.
Kelvin
Fahrenheit
Celsius
The Celcius Scale is called an absolute Temperature scale,and it's zero point called absolute zero
_______________
A 1.1 kg ball drops vertically onto a floor, hitting with a speed of 23 m/s. It rebounds with an initial speed of 5.0 m/s. (a) What impulse acts on the ball during the contact
Hi there!
We know that:
I = Δp = m(vf - vi)
Plug in the given values. Remember to take into account direction ⇒ let the rebound velocity be positive and initial be negative.
I = 1.1(5 - (-23)) = 30.8 Ns
EXAM ENDS IN 30 MINS
PLSSS HELPPP ILL MAKE U BRAINLIEST
Explanation:
F = Icurrent×length×Bfieldstrength×sin(angle field to wire)
in our case
Icurrent = 10 A
length = 0.02km = 20 meters
B = 10^-6 T
angle = 30 degrees.
F = (20 A)(20m)(10^-6 T)×sin(30) = 400× 10^‐6 ×0.5 N =
= 200 × 10^-6 = 2 × 10^‐4 N
What is a list of all the states of matter?
Answer:
3
Explanation:
state of matter are solid
liquid and
gases
Answer:
3
Explanation:
state of a matter are solid liquid and gas
The amount of work done in example B is:
Answer:
Explanation:
20 n is an unknown amount
If that is supposed to be 20 N(ewtons)
then W = Fd = 20(15) = 300 J
Answer: it will be 300 newton meters
Explanation:
How fast would a(n) 75 kg man need to run in order to have the same kinetic energy as an 8.0 g bullet fired at 390 m/s
Answer:
Explanation:
½(75)v² = ½(0.008)390²
v² = (0.008)390²/75
v² = 16.224
v = 4.027...
v = 4.0 m/s
One of the great challenges of cosmology today is to ---
A
determine the amount of matter in the Universe
B
find intelligent signals emanating from outer space
С
look backward in time to before the Big Bang
D
locate wormholes to help define the structure of the Universe
Answer:
Steady-state theory, in cosmology, a view that the universe is always expanding but maintaining a constant average density, with matter being continuously created to form new stars and galaxies at the same rate that old ones become unobservable as a consequence of their increasing distance and velocity of recession.
Cosmic inflation is a faster-than-light expansion of the universe that spawned many others. ... Cosmic inflation solves these problems at a stroke. In its earliest instants, the universe expanded faster than light (light's speed limit only applies to things within the universe).
A 2.0 kg particle moving along the z-axis experiences the
force shown in (Figure 1). The particle's velocity is
3.0 m/s at x = 0 m.
A) At what point on the x axis does the particle have a turning point?
At point x = 0, the particle accelerates. Since there will be change of velocity at that point. The the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.
Given that a 2.0 kg particle moving along the z-axis experiences the force shown in a given figure.
Force is the product of mass and acceleration. While acceleration is the rate of change of velocity. Both the force and acceleration are vector quantities. They have both magnitude and direction.
If the particle's velocity is 3.0 m/s at x = 0 m, that mean that the particle experience change of velocity at point x = 0. Since the the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.
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Standing at a crosswalk, you hear a frequency of 550 Hz from the siren of an approaching ambulance. After the ambulance passes, the observed frequency of the siren is 475 Hz. Determine the ambulance's speed from these observations. (Take the speed of sound to be 343 m/s.)