Answer:
0.546 [tex]\hat k[/tex]
Explanation:
From the given information:
The force on a given current-carrying conductor is:
[tex]F = I ( \L \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})[/tex]
where the length usually in negative (x) direction can be computed as
[tex]\L ^ {\to } = -x\hat i \\dL\limits ^ {\to }- dx\hat i[/tex]
Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:
[tex]\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})[/tex]
[tex]F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)[/tex]
[tex]F = I \int^3_1 - 9.0x^2 \ dx \hat k[/tex]
[tex]F = I (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k[/tex]
[tex]F = I (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ] \hat k[/tex]
where;
current I = 7.0 A
[tex]F = (7.0 \ A) (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ] \hat k[/tex]
[tex]F = (7.0 \ A) (9.0) \bigg [\dfrac{26}{3} \bigg ] \hat k[/tex]
F = 546 × 10⁻³ T/mT [tex]\hat k[/tex]
F = 0.546 [tex]\hat k[/tex]
Describe and give an example of mutualism.
Describe and give an example of commensalism.
Describe and give an example of parasitism.
Describe and give an example of competition.
Describe and give an example of predation.
Answer:
Mutualism, commensalism, parasitism, competition, and predation.
Explanation:
mutualism- relationship between two or more organisms where both are benefited. Example-oxpecker with rhino/zebra. They eat bugs off of them which means that they are getting food, while the rhino/zebra are getting cleaned up with pest control.
commensalism- relationship between two organisms where one benefits and the other isn't benefited or harmed. EX- tree frogs use plants as protectioin.he frog is benefited, and the plant is neither harmed nor benefited. Remora fish have a disk on their heads that they use to attach themselves to larger animals for protection. The animals they attach to are neither harmed nor benefited.
parasitism- in a relationship where an organism benefits at the expense of the other. (one is benefited while the other is harmed) ex- fleas and ticks that live on cats and dogs, tape worms that live in people and animals that eat the food which means that the people aren't getting the food or nutrition that they eat. lice, etc
competition- interaction within organisms/species in which both the organisms/species are harmed and is apart of natural selection. Examples may include two males fighting over a mate, animals competing over food, limited habitats that they are fighting over, territory, etc.
predation- the preying of one animal on another. It's where the predator hunts and eats another organism which is its prey. categorized within-(1) carnivory, (2) herbivory, (3) parasitism, and (4) mutualism. Each type of predation can by categorized based on whether or not it results in the death of the prey.ex- owls hunting mice, wolves hunting rabbits, lion hunting gazelle, etc.
We will now determine the indexes of refraction for two Mystery materials, A and B. These materials can be selected from the list of materials on the right. Be sure to set your laser pointer to a frequency of 589 nm. Questions:A. Devise an experiment for determining the indices of refraction for these. Explain your methodology. B. What are the indices of refraction for the two mystery materials, A and B?
Answer:
A) refraction experiment n = n₁ sin θ₁ / sin θ₂
B) n_A = 1.19 , n_B = 1.53
Explanation:
A) This exercise is a method to measure the refractive index of materials, a very useful and simple procedure is to create a plate of known thickness from each material, place the material on a paper with angle measurements (protractor), incline the laser beam and measure the angles of incidence and refraction (within the material), repeat for about three different angles of incidence and use the equation of refraction to determine the index
n₁ sin θ₁ = n₂ sinθ₂
n₂ = n₁ sin θ₁₁ /sin θ₂
If the medium surrounding the plate is air, its refractive index is n₁ = 1, the final expression is
n = n₁ sin θ₁ / sin θ₂
B) For this part, no data are given in the exercise, but we can take 50º as the angle of incidence and measure the angle of refraction. Suppose it is 40º for material A and 30º for material B, the refractive index would be
material A
n_A = sin 50 / sin 40
n_A = 1.19
material B
n_B = sin 50 / sin30
n_B = 1.53
Mr Jones launches an arrow horizontally at a rate of 40m/s off of a 78.4 m cliff towards the south, what direction and value is his acceleration air resistance is negligible.
A. 9.8 m/s/s west
b. 9.8m/s/s east
C. 9.8m/s's down
d 9.8m/s/s south
Answer:
9.8m/s^2 down (option C)
Explanation:
The only acceleration acting on this motion case in the acceleration due to gravity: 9.8 m/s^2 in the downwards direction.
A car traveling initially at a speed of 20 m/s accelerates to a speed of 31 m/s over a distance of 45 meters.
What is the magnitude of the car's acceleration?
Answer:abc defg hijk lmnop qrs tuv wx y and z
Explanation: now i know my abc's
Calculate the de Broglie wavelength of: a) A person running across the room (assume 180 kg at 1 m/s) b) A 5.0 MeV proton
Answer:
a
[tex]\lambda = 3.68 *10^{-36} \ m[/tex]
b
[tex]\lambda_p = 1.28*10^{-14} \ m[/tex]
Explanation:
From the question we are told that
The mass of the person is [tex]m = 180 \ kg[/tex]
The speed of the person is [tex]v = 1 \ m/s[/tex]
The energy of the proton is [tex]E_ p = 5 MeV = 5 *10^{6} eV = 5.0 *10^6 * 1.60 *10^{-19} = 8.0 *10^{-13} \ J[/tex]
Generally the de Broglie wavelength is mathematically represented as
[tex]\lambda = \frac{h}{m * v }[/tex]
Here h is the Planck constant with the value
[tex]h = 6.62607015 * 10^{-34} J \cdot s[/tex]
So
[tex]\lambda = \frac{6.62607015 * 10^{-34}}{ 180 * 1 }[/tex]
=> [tex]\lambda = 3.68 *10^{-36} \ m[/tex]
Generally the energy of the proton is mathematically represented as
[tex]E_p = \frac{1}{2} * m_p * v^2_p[/tex]
Here [tex]m_p[/tex] is the mass of proton with value [tex]m_p = 1.67 *10^{-27} \ kg[/tex]
=> [tex]8.0*10^{-13} = \frac{1}{2} * 1.67 *10^{-27} * v^2[/tex]
=> [tex]v _p= \sqrt{\frac{8.0 *10^{-13}}{ 0.5 * 1.67 *10^{-27}} }[/tex]
=> [tex]v = 3.09529 *10^{7} \ m/s[/tex]
So
[tex]\lambda_p = \frac{h}{m_p * v_p }[/tex]
so [tex]\lambda_p = \frac{6.62607015 * 10^{-34}}{1.67 *10^{-27} * 3.09529 *10^{7} }[/tex]
=> [tex]\lambda_p = 1.28*10^{-14} \ m[/tex]
An ideal gas occupies 0.4 m3 at an absolute pressure of 500 kPa. What is the absolute pressure if the volume changes to 0.9 m3 and the temperature remains constant?
Answer:
2.22 kPaExplanation:
The new volume can be found by using the formula for Boyle's law which is
[tex]P_1V_1 = P_2V_2[/tex]
Since we are finding the new volume
[tex]V_2 = \frac{P_1V_1}{P_2} \\[/tex]
From the question we have
[tex]V_2 = \frac{0.4 \times 500000}{0.9} = \frac{200000}{0.9} \\ = 222222.2222... \\ = 222222[/tex]
We have the final answer as
2.22 kPaHope this helps you
how does tom and jerry movie character influence your attitude
Answer:
it makes me wish I was a cartoon
Answer:
goofy and stupid and act like a kid
Explanation:
A penny is dropped from rest from a building 100m tall. what kind of motion is this
A. centripetal
B. Free fall
C. Linear
D. projectile
Answer:
this is a projectile
Answer:
D. projectile
Explanation:
Since it's dropped from a rest that means that it's velocity at the beginning is 0.
What is Newton's universal law of gravitation
Answer:
an object that is in motion wont go out of motion until there is another force pushing on it
Which is increased when the string of a stringed instrument is tightened?
timbre
pitch
wavelength
loudness
When the string of the instrument is tightened then the length of the string decreases hence the pitch and frequency will increase so, option B is correct.
What is pitch?The frequency at which the sound waves that create a sound vibrate determines its pitch. High-frequency sound waves produce high-pitched noises, and low-frequency sound waves make low-pitched noises. The ability to discern between harsh and flat sounds is known as pitch.
A string will vibrate at a varied frequency depending on its length. Higher frequency and higher pitch are produced by shorter strings.
The pitch rises as the anxiety does as well. A string's length is also crucial. A string vibrates and makes music when it is supported at two points and pulled. The pitch of this string will, however, rise if the length is shortened.
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Use the information below for the next five questions:
An open organ pipe emits B (494 Hz) when the temperature is 14°C. The speed of sound in air is v≈(331 + 0.60T)m/s, where T is the temperature in °C
Determine the length of the pipe.
What is the wavelength of the fundamental standing wave in the pipe?
What is frequency of the fundamental standing wave in the pipe?
What is the frequency in the traveling sound wave produced in the outside air?
What is the wavelength in the traveling sound wave produced in the outside air?
How far from the mouthpiece of the flute should the hole be that must be uncovered to play D above middle C at 294 Hz? The speed of sound in air is 343 m/s.
Answer: Please see answer in explanation column.
Explanation:
Given that
v≈(331 + 0.60T)m/s
where Temperature, T = 14°C
v≈(331 + 0.60 x 14)m/s
v =331+ 8.4 = 339.4m/s
In our solvings, note that
f= frequency
λ=wavelength
L = length
v= speed of sound
a) Length of the pipe is calculated using the fundamental frequency formulae that
f=v/2L
Length = v/ 2f
= 339.4m/s/ 2 x 494Hz ( s^-1)= 0.3435m
b) wavelength of the fundamental standing wave in the pipe
L = nλ/2,
λ = 2L/ n
λ( wavelength )= 2 x 0.3435/ 1
= 0.687m
c) frequency of the fundamental standing wave in the pipe
F = v/ λ
= 339.4m/s/0.687m=
494.03s^-1 = 494 Hz
d) the frequency in the traveling sound wave produced in the outside air.
This is the same as the frequency in the open organ pipe = 494Hz
e)The wavelength of the travelling sound wave produced in the outside air is the same as the wavelength calculated in b above = 0.687m
f) To play D above middle c . the distance is given by
L =v/ 2 f
= 343/ 2 x 294
=0.583m
(1.5 pts) A woman pushes on a box to the left. If the box is accelerating, what forces are working on the
Question 2:
box? (Draw both y and x forces)
Answer:
Nope
Explanation:
what does the modal "must"indicate?
Answer:
The modal verb must is used to express obligation and necessity. The phrase have to doesn't look like a modal verb, but it performs the same function.
please help! What is the relationship between velocity and acceleration?
Answer:Acceleration implies any change in the velocity of the object with respect to time. Velocity is nothing but the rate of change of displacement. On the other hand, acceleration is the rate of change of velocity with respect to time.
Explanation:
Which element is used in the manufacture of mirrors and bronze?
Answer:
Silver
Explanation:
Silver is an important element in the manufacturing process of mirrors. Silver is used to make mirrors through the process we call "silvering". Silvering is a process in which a glass is coated with reflective substances so as to produce reflections, and then mirrors.
In Silvering process, Chlorine is also used. Stannous Chloride is the particular compound used to carry out the silvering, it has the chemical formula, SnCl₂
2. An ambulance traveling at 20 m/s emits a sound at 500 Hz. What frequency does a person standing on the corner of a street detect?
A circular coil of wire 8.6 cm in diameter has 15 turns and carries a current of 2.7 A. The coil is in a region where the magnetic field is 0.56 T. (a) What orientation of the coil gives the maximum torque on the coil, and what is this maximum torque? (b) For what orientation of the coil is the magnitude of the torque 71% of the maximum found in part (a)?
Answer:
(a) (i) The orientation of the coil which gives maximum torque on the coil is 90⁰
(a)(ii) The maximum torque is 0.132 Nm
(b) The orientation of the coil is 45⁰
Explanation:
Given;
diameter of the circular wire, d = 8.6 cm = 0.086 m
radius of the wire, r = d /2 = 0.043 m
number of turns, N = 15 turns
magnetic field, B = 0.56 T
The torque on the wire is given by;
τ = NIABsinθ
where;
θ is the orientation of the wire
(a) maximum torque occurs when the orientation of the wire is at 90⁰
The maximum torque is given by;
τ = NIABsin(90⁰)
τ = NIAB
τ = (15)(2.7)(π x 0.043²)(0.56)
τ = 0.132 Nm
(b)
71% of 0.132 = 0.71 x 0.132 = 0.0937 Nm
[tex]\tau = NIAB sin\theta\\\\sin\theta = \frac{\tau}{NIAB }\\\\ sin\theta = \frac{0.0937}{(15)(2.7)(\pi *0.043^2)(0.56)} \\\\sin\theta = 0.7111\\\\\theta = sin^{-1}(0.7111)\\\\\theta =45.32\\\\\theta = 45^0[/tex]
A sample of an ideal gas has a volume of 0.0100 m^3, a pressure of 100 x 10^3 Pa, and a temperature of 300K. What is the number of moles in the sample of gas?
Answer:
Explanation:
pV = nrT
n = PV/RT
n = (100*10^3)(.01)/(300*0.082057)
n = 40.62 moles
Which image illustrates the interaction of a light wave with a mirror?
t J
A
с
.
A. A
B. B
C. C
D. D
0
Answer:
I'm pretty sure its A
Explanation:
because its a reflection- Hope you get a good grade!
7) A moving object is rolling on a surface that is 5 m off the ground. The object is moving at a constant speed of 4 m/s. If the object is 3.2 kg, what is the final energy of the ball after rolling for 10 m, assuming friction is negligible?
156.8 J
131.2 J
182.4 J
25.6 J
8) A spreadsheet application is used to create a computational model of the energy experienced by a pendulum. How do the energy values of the pendulum relate?
The sum of the potential energy and the kinetic energy is always constant.
The sum of the potential energy and the kinetic energy is always 0.
The potential energy is always greater than the kinetic energy.
The kinetic energy is always equal to the potential energy.
131.2 J and The last one on number 8
I gave the same answer and it passed.
7) The final energy of the ball after rolling for 10 m is 182.4 J so, option C is correct.
8) When friction is negligible the total energy is the sum of kinetic and potential energy is constant so, option A is correct.
What is energy?Energy is the ability to perform work in physics. It could exist in several different forms, such as potential, kinetic, thermal, electrical, chemical, radioactive, etc. Additionally, there is heat and work, which is energy being transferred from one body to another.
Given:
A moving object is rolling on a surface that is 5 m off the ground,
The speed of the object, v = 4 m/s,
The mass of the object, m = 3.2 kg,
Calculate the kinetic energy after 10 meters as shown below,
KE = 1/2 × 4² × 3.2
KE = 25.6 J
Calculate the potential energy as shown below,
PE = 3.2 × 9.8 × 5
PE = 156.8 J
Thus, total energy = KE + PE
The total energy = 25.6 + 156.8
The total energy = 182.4 J
8) when there is no resistance. Combined mechanical energy I.e. the total amount of kinetic and potential energy is constant.
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that delivers oxygen to your body and In the video your blood is compared to a picks up CO2 to be released out when you breath. PLEASE I NEED A ANSWER
A T-shirt cannon mounted at the top of an arena needs to fire a t-shirt into the first row, a horizontal distance of 39 meters away. If the cannon launches t-shirts at 12 m/s, how high is the cannon mounted?
Question 1 options:
3.3 m
16.2 m
53.4 m
8.9 m
Answer:
h = 51.75 m
nearest answer is:
53.4 m
Explanation:
First we analyze the horizontal motion. Since, the air friction is assumed to be negligible. Hence, the horizontal motion shall be uniform. Therefore,
s = V₀ₓ t
where,
s = horizontal distance = 39 m
V₀ₓ = Horizontal Initial Velocity = 12 m/s
t = time = ?
Therefore,
39 m = (12 m/s)t
t = 39 m/12 m/s
t = 3.25 s
Now, we analyze the vertical motion. Applying newton's second equation of motion to vertical motion:
h = V₀y t + (1/2)gt²
where,
h = height of cannon = ?
V₀y = initial vertical velocity = 0 m/s
g = 9.8 m/s²
Therefore,
h = (0 m/s)(3.25 s) + (1/2)(9.8 m/s²)(3.25 s)²
h = 51.75 m
As a rough model of the impact of walking/running, consider that half the mass of the body falls from a height of 4.77-cm onto a single foot. (During a typical stride, an adult's center-of-mass moves approximately this distance vertically). Use the kinematic equations to calculate the speed of an object falling from this height at the moment of impact with the ground under the influence of gravity.A. As a rough model of the impact of walking, consider that half of the mass of the entire body strikes the ground with a downward velocity of 1.0 m/s and comes to a full vertical stop over an impact duration of 20 ms. Calculate the force associated with this single step for a person with a mass of 74.2 kg. B. Calculate the stress (solid pressure) of a force of 1880 N applied across the 0.4 cm^2 cross-sectional area of the typical Achilles tendon. For reference, the maximum rupture stress of tendons has been reported in the range of 100-150 MPa.
Answer:
0.967 m/s
1855 N
[tex]46.375\ \text{MPa}[/tex]
Explanation:
v = Final velocity
u = Initial velocity = 0
s = Displacement = 4.77 cm
g = a = Acceleration due to gravity = [tex]9.81\ \text{m/s}^2[/tex]
From the kinematic equations
[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 0.0477+0}\\\Rightarrow v=0.967\ \text{m/s}[/tex]
The velocity of the object at the moment of impact is 0.967 m/s
Now
[tex]\Delta v[/tex] = Change in velocity = 1 m/s
t = Time taken = 20 ms
m = Half mass of the person = [tex]\dfrac{74.2}{2}=37.1\ \text{kg}[/tex]
[tex]F=\dfrac{m}{t}\\\Rightarrow F=\dfrac{37.1\times 1}{20\times 10^{-3}}\\\Rightarrow F=1855\ \text{N}[/tex]
The force associated with a single step of the person is 1855 N
A = Area = [tex]0.4\ \text{cm}^2[/tex]
Stress is given by
[tex]\sigma=\dfrac{F}{A}\\\Rightarrow \sigma=\dfrac{1855}{0.4\times 10^{-4}}\\\Rightarrow \sigma=46375000\ \text{Pa}=46.375\ \text{MPa}[/tex]
The stress on the tendon is [tex]46.375\ \text{MPa}[/tex]
The speed of object during falling is 0.967 m/s.
(A) The magnitude of force associated with this single step for a person is 1855 N.
(B) The required value of stress at tendons is [tex]4.70 \times 10^{7} \;\rm Pa[/tex].
Given data:
The height of fall is, h = 4.77 cm = 0.0477 m.
The magnitude of downward velocity is, v' = 1.0 m/s.
The duration of impact is, [tex]t = 20 \;\rm ms =20 \times 10^{-3} \;\rm s[/tex].
The mass of person is, m = 74.2 kg.
The magnitude of force is, F' = 1880 N.
The cross-sectional area is, [tex]A =0.4 \;\rm cm^{2} = 0.4 \times 10^{-4} \;\rm m^{2][/tex].
The problem has several parts using different concepts. First obtain the final speed of object to fall by using the second kinematic equations of motion as,
[tex]v^{2}=u^{2}+2gh[/tex]
Solving as,
[tex]v^{2}=0^{2}+(2 \times 9.8 \times 0.0477)\\\\v = \sqrt{(2 \times 9.8 \times 0.0477)} \\v = 0.967 \;\rm m/s[/tex]
Thus, the speed of object during falling is 0.967 m/s.
(A)
Now coming to next part, the half of mass means, m' = m/2 = 74.2/2 = 37.1 kg.
Apply the expression of average force as,
[tex]F =\dfrac{m'v'}{t}[/tex]
Solving as,
[tex]F =\dfrac{37.1 \times 1}{20 \times 10^{-3}}\\\\F = 1855 \;\rm N[/tex]
Thus, the magnitude of force associated with this single step for a person is 1855 N.
(B)
The expression for the stress is given as,
[tex]\sigma = \dfrac{F'}{A}[/tex]
Solving as,
[tex]\sigma = \dfrac{1880}{0.4 \times 10^{-4}}\\\\\sigma =4.70 \times 10^{7} \;\rm Pa[/tex]
Thus, the required value of stress at tendons is [tex]4.70 \times 10^{7} \;\rm Pa[/tex].
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A television of mass 15 kg sits on a table. The coefficient of static friction
between the table and the television is 0.35. What is the minimum applied
force that will cause the television to slide?
A) 38 N
B) 147 N
C) 51 N
D) 79 N
Answer:
more than 51.45 N
__________________________________________________________
We are given:
Mass of the television = 15 kg
Coefficient of Static friction = 0.35
Minimum force required to move the television:
Normal Force:
We know that the normal force is equal and opposite to the Weight of the television
Weight of the television = Mg
[where m is the mass and g is the acceleration due to gravity]
Weight = 15 * 9.8
Weight = 147 N
Force of Friction:
We are given the coefficient of Friction = 0.35
We know that coefficient of Friction = Force of friction / Normal Force
replacing the variables
0.35 = Force of Friction / 147
Force of Friction = 147 * 0.35 [multiplying both sides by 147]
Force of Friction = 51.45 N
Since a force of 51.45 N is will be applied opposite to the direction of application of Force, the television will only move when we apply more force than 51.45 N
Answer:
it is C
Explanation:
an object of mass 4kg moving with initial velocity if 20m/s accelerates for 10s and attaind a final velocity of 60m/s calculate the acceleration
Answer:
given us,
mass= 4×9.8gm m(9.8) formula
= 39.2
final velocity (v)= 60m/s
initial velocity (u)= 20m/s
time(t)= 10s
acleration(a)=?
now,
accleration(a)= v-u/t=60- 20/10
=40/10
=4m/s
:. the acceleration is 4 m/s
Explanation:
first we have to calculate mass and we can use acceleration formula
A car has a mass of 850 kg. By pushing on the car, Evan increases its speed
from 3.5 m/s to 5 m/s. What impulse did Evan apply to the car?
A. 4250 kg•m/s
B. 1275 kg•m/s
C. 850 kg•m/s
D. 2975 kg•m/s
Answer:
B. 1275 kg*m/s
Explanation:
I = F(deltaT) = (deltaP) = mv2- mv1
Therefore,
I = mv2-mv1
m = 850 kg
v2 = 5 m/s
v1 = 3.5 m/s
I = (850)(5)-(850)(3.5)
I = 1275 kg* m/s
The particle accelerator at CERN can accelerate an electron through a potential
difference of 80 kilovolts. Calculate
(a) The kinetic energy (in keV) of the electron
Answer:
K.E = 1.28 × 10^-17 KeV
Explanation:
Given that a particle accelerator at CERN can accelerate an electron through a potentialdifference of 80 kilovolts.
To Calculate the kinetic energy (in keV) of the electron, let us first find the electron charge which is 1.60 × 10^-19C
The kinetic energy = work done
K.E = e × kV
Substitute e and the voltage into the formula
K.E = 1.60 × 10^-19 × 80
K.E = 1.28 × 10^-17 KeV
Therefore, the kinetic energy is approximately equal to 1.28 × 10^-17 KeV
Two billiard balls (each with mass equal to 170 g) collide head-on along the same line. Billiard ball A originally traveled eastward at 8 m/s while billiard ball B originally traveled westward at 2 m/s. Calculate the speed and direction of each ball after the collision.
Answer:
lucky mauld mauldgomary was an british poet...
A train is traveling at 55m/s begins to slow down as it approaches a bend in the tracks. If it travels around the bend at a speed of9 m/s and it takes 49 s to properly slow down, what distance does the train travel while slowing down?
Answer:
x = 1127 [m]
Explanation:
In order to solve this problem, we must use the equations of kinematics. With the first equation, we must find the acceleration and with the second equation we must find the distance.
[tex]v_{f} =v_{o} -a*t[/tex]
where:
Vf = final velocity = 9 [m/s]
Vo = initial velocity = 55 [m/s]
a = acceleration o desacceleration [m/s²]
t = time = 49 [s]
Now replacing:
9 = 55 - a*49
a*49 = 55 + 9
a = 1.306 [m/s²]
Note: The negative sign in the above equation means that the speed decreases.
Now using the second equation.
[tex]v_{f}^{2} =v_{o}^{2} -2*a*x[/tex]
(9)² = (55)² - 2*(1.306)*x
2944 = 2.612*x
x = 1127 [m]
A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns. They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) (4.00 m/s2 ) .At that instant and in unit-vector notation, what is the acceleration of the wallet
Complete Question:
A purse at radius 2.00 m and a wallet at radius 3.00 m travel in uniform circular motion on the floor of a merry-go-round as the ride turns.
They are on the same radial line. At one instant, the acceleration of the purse is (2.00 m/s2 ) i + (4.00 m/s2 ) j .At that instant and in unit-vector notation, what is the acceleration of the wallet
Answer:
aw = 3 i + 6 j m/s2
Explanation:
Since both objects travel in uniform circular motion, the only acceleration that they suffer is the centripetal one, that keeps them rotating.It can be showed that the centripetal acceleration is directly proportional to the square of the angular velocity, as follows:[tex]a_{c} = \omega^{2} * r (1)[/tex]
Since both objects are located on the same radial line, and they travel in uniform circular motion, by definition of angular velocity, both have the same angular velocity ω.∴ ωp = ωw (2)
⇒ [tex]a_{p} = \omega_{p} ^{2} * r_{p} (3)[/tex]
[tex]a_{w} = \omega_{w}^{2} * r_{w} (4)[/tex]
Dividing (4) by (3), from (2), we have:[tex]\frac{a_{w} }{a_{p}} = \frac{r_{w} }{r_{p}}[/tex]
Solving for aw, we get:[tex]a_{w} = a_{p} *\frac{r_{w} }{r_{p} } = (2.0 i + 4.0 j) m/s2 * 1.5 = 3 i +6j m/s2[/tex]