Answer:
m = 2,759 kg
Explanation:
For this exercise we use the torque relationship
τ = I α
the moment is
τ= F r sin θ
since the force is tangential to the ring, the angle is 90º sin 90 = 1
τ = F r
the moment of inertia of a ring is given by
I = m r²
let's substitute
F r = m r²α
m = F / r α
let's calculate
m = 40 / (0.10 145)
m = 2,759 kg
A 0.5 kg basketball moving 5 m/s to the right collides with a 0.05 kg tennis
ball moving 30 m/s to the left. After the collision, the tennis ball is moving 34
m/s to the right. What is the velocity of the basketball after the collision?
Assume an elastic collision occurred.
A. 1.4 m/s to the left
B. 1.4 m/s to the right
C. 11.4 m/s to the right
D. 11.4 m/s to the left
SOMEBODY HELP ME
Answer:
A. 1.4 m/s to the left
Explanation:
To solve this problem we must use the principle of conservation of momentum. Let's define the velocity signs according to the direction, if the velocity is to the right, a positive sign will be introduced into the equation, if the velocity is to the left, a negative sign will be introduced into the equation. Two moments will be analyzed in this equation. The moment before the collision and the moment after the collision. The moment before the collision is taken to the left of the equation and the moment after the collision to the right, so we have:
[tex]M_{before} = M_{after}[/tex]
where:
M = momentum [kg*m/s]
M = m*v
where:
m = mass [kg]
v = velocity [m/s]
[tex](m_{1} *v_{1} )-(m_{2} *v_{2})=(m_{1} *v_{3} )+(m_{2} *v_{4})[/tex]
where:
m1 = mass of the basketball = 0.5 [kg]
v1 = velocity of the basketball before the collision = 5 [m/s]
m2 = mass of the tennis ball = 0.05 [kg]
v2 = velocity of the tennis ball before the collision = - 30 [m/s]
v3 = velocity of the basketball after the collision [m/s]
v4 = velocity of the tennis ball after the collision = 34 [m/s]
Now replacing and solving:
(0.5*5) - (0.05*30) = (0.5*v3) + (0.05*34)
1 - (0.05*34) = 0.5*v3
- 0.7 = 0.5*v
v = - 1.4 [m/s]
The negative sign means that the movement is towards left
Answer:
A. 1.4 m/s to the left
Explanation:
The volume of 2.0 kg of helium in a piston-cylinder device is initially 7 m3. Now the helium is compressed to 5 m3 while its pressure is maintained constant at 160 kPa. Determine: (a) The initial and final temperatures of helium. (b) The work required to compress it, in kJ.
Answer:
A) T1 = 269.63 K
T2 = 192.59 K
B) W = -320 KJ
Explanation:
We are given;
Initial volume: V1 = 7 m³
Final Volume; V2 = 5 m³
Constant Pressure; P = 160 KPa
Mass; m = 2 kg
To find the initial and final temperatures, we will use the ideal gas formula;
T = PV/mR
Where R is gas constant of helium = R = 2.0769 kPa.m/kg
Thus;
Initial temperature; T1 = (160 × 7)/(2 × 2.0769) = 269.63 K
Final temperature; T2 = (160 × 5)/(2 × 2.0769) = 192.59 K
B) world one is given by the formula;
W = P(V2 - V1)
W = 160(5 - 7)
W = -320 KJ
1. Consider the point exactly halfway between the two wires. Can you adjust the current so that, with current passing through each of the wires, the net field is zero here? 2. Note the distance between the field lines from one of the wires. Are they equally space? Explain. 3. With currents of different magnitudes passing through the two wires, how do the forces on the two wires compare? What is the reason for this?
Answer:
hello your question is incomplete attached below is missing part of the question
answer:
1 ) Magnetic field due to long current carrying wire : [tex]B = \frac{U_{0} I}{2\pi d}[/tex]
Therefore the net magnetic field due the both wires ; B = B[tex]_{1}[/tex] + B[tex]_{2}[/tex] . when we adjust the current I[tex]_{1}[/tex] = I[tex]_{2}[/tex] then the Netfield (B ) = zero
2) The distance between the field lines are not equally spaced and this is because the separation between field lines increases with the increase in the distance between the wires
3) Increase in current through the wire will lead to increase in force and this can be explained via this equation
[tex]F = \frac{U_{0}I_{1}I_{2} }{2\pi d }[/tex]
Explanation:
1 ) Magnetic field due to long current carrying wire : [tex]B = \frac{U_{0} I}{2\pi d}[/tex]
Therefore the net magnetic field due the both wires ; B = B[tex]_{1}[/tex] + B[tex]_{2}[/tex] . when we adjust the current I[tex]_{1}[/tex] = I[tex]_{2}[/tex] then the Netfield (B ) = zero
2) The distance between the field lines are not equally spaced and this is because the separation between field lines increases with the increase in the distance between the wires
3) Increase in current through the wire will lead to increase in force and this can be explained via this equation
[tex]F = \frac{U_{0}I_{1}I_{2} }{2\pi d }[/tex]
PLEASE HELP
A sharpshooter fires a 0.22 caliber rifle horizontally at 100 m/s at a target 75m away. How far does the
bullet drop by the time it reaches the target?
This question involves the concept of semi-projectile motion. It can be solved using the equations of motion in the horizontal and the vertical motion.
The bullet drops "2.76 m" by the time it reaches the target.
First, we will analyze the horizontal motion. We assume no air resistance, so the horizontal motion will be uniform. Hence, using the equation of uniform motion here to find the total time to reach the target:
[tex]s = vt\\\\t = \frac{s}{v}[/tex]
where,
s = distance = 75 m
v = velocity = 100 m/s
t = time = ?
Therefore,
[tex]t = \frac{75\ m}{100\ m/s}[/tex]
t = 0.75 s
Now, we will analyze the vertical motion of the bullet. We will use the second equation of motion in the vertical direction to find the height dropped by the bullet.
[tex]h = v_it+\frac{1}{2}gt^2[/tex]
where,
h = height dropped = ?
vi = initial vertical speed = 0 m/s
t = time interval = 0.75 s
g = acceleration due to gravity = 9.81 m/s²
therefore,
[tex]h = (0\ m/s)(0.75\ s)+\frac{1}{2}(9.81\ m/s^2)(0.75\ s)^2[/tex]
h = 2.76 m
Learn more about equations of motion here:
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The attached picture shows the equations of motion in the horizontal and vertical directions.
In a railroad switchyard, a rail car of mass 41,700 kg starts from rest and rolls down a 2.65-m-high incline and onto a level stretch of track. It then hits a spring bumper, whose spring compresses 79.4 cm. Find the spring constant.
Answer:
The spring constant is 3.44x10⁶ kg/s².
Explanation:
We cand find the spring constant by conservation of energy:
[tex] E_{i} = E_{f} [/tex]
[tex] mgh = \frac{1}{2}kx^{2} [/tex] (1)
Where:
m is the mass = 41700 kg
g is the gravity = 9.81 m/s²
h is the height = 2.65 m
x is the distance of spring compression = 79.4 cm
k is the spring constant =?
Solving equation (1) for k:
[tex]k = \frac{2mgh}{x^{2}} = \frac{2*41700 kg*9.81 m/s^{2}*2.65 m}{(0.794 m)^{2}} = 3.44 \cdot 10^{6} kg/s^{2}[/tex]
Therefore, the spring constant is 3.44x10⁶ kg/s².
I hope it helps you!
A stone is thrown horizontally with an initial speed of 10m/s from the edge of the cliff. A stop watch measures the stone’s trajectory with time from the top of the hill to the bottom to be 6.7s. What is the height of the cliff?
Answer:
Answer and steps in the pic
True/False
1. The Law of Superposition states that older rocks are on the top, younger rocks are on the botom.
Answer:
yeah
Explanation:
Geology. a basic law of geochronology, stating that in any undisturbed sequence of rocks deposited in layers, the youngest layer is on top and the oldest on bottom, each layer being younger than the one beneath it and older than the one above it.
Answer:
Found this answer off of google, "Geology. a basic law of geochronology, stating that in any undisturbed sequence of rocks deposited in layers, the youngest layer is on top and the oldest on bottom, each layer being younger than the one beneath it and older than the one above it."
Hope this helps, have a great day and stay safe!
Also Happy Halloween! :) :D :3
When driving at slower speeds you need to use what type of steering
wheel movements compared to when driving at faster speeds? *
Answer:
slower speeds = larger and faster steering wheel movements
faster speeds = small and slow steering wheel movements
Explanation:
When driving at slower speeds you need to use larger and faster steering wheel movements. This is because at slow speeds the car does not have enough momentum to make certain maneuvers with small steering wheel movements in a given amount of time, therefore making large and faster steering wheel movements gives the car enough time with the momentum it has to make the desired maneuver. At faster speeds only small and slow steering wheel movements are needed and while cause the car to quickly change to the desired direction due to the increased momentum of the car.
plz help me
what do you meant by significant figure?give different rules to clarify significant figure.
Scott travels north 2 miles and then goes west 5 miles before coming straight back south 7 miles what is his displacement
Answer:
Displacement is 5√2 miles
Explanation:
From south, Scott travelled 2 miles north and then 5 miles west then he came back south by 7 miles.
I've attached a diagram to indicate Scott's path travelled
Now, displacement is a vector quantity that tells us the overall change in position by Scott i.e distance from his initial position to his current final position while distance is a scalar quantity and tells us how much ground Scott covered in his motion over the entire trip.
Now, from the attached image, the displacement is denoted by "d".
d can be found from pythagoras theorem.
Thus;
d² = 5² + 5²
d = √50
d = 5√2 miles
A ball of mass m is found to have a weight Wx on Planet X. Which of the following is a correct expression for the gravitational field strength of Planet X?
A. The gravitational field strength of Planet X is mg.
B. The gravitational field strength of Planet X is Wx/m.
C. The gravitational field strength of Planet X is 9.8 N/kg.
D. The gravitational field strength of Planet X is mWx.
Answer: B. The gravitational field strength of Planet X is Wx/m.
Explanation:
Weight is a force, and as we know by the second Newton's law:
F = m*a
Force equals mass times acceleration.
Then if the weight is:
Wx, and the mass is m, we have the equation:
Wx = m*a
Where in this case, a is the gravitational field strength.
Then, isolating a in that equation we get:
Wx/m = a
Then the correct option is:
B. The gravitational field strength of Planet X is Wx/m.
true or false- When an atom's charge is balanced, the amount of negative charge equals the amount of neutral charge.
A 7.0 µF capacitor and a 5.0 µF capacitor are connected in series across a 5.0 kV potential difference. The charged capacitors are then disconnected from the source and connected to each other with terminals of like sign together. Find the charge on each capacitor (in mC) and the voltage across each capacitor (in V). (Due to the nature of this problem, do not use rounded intermediate values in your calculations—including answers submitted in WebAssign.)
Explanation:
Given two capacitors 7.0 µF and 5.0 µF connected in series, let s calculate the effective capacitance.
[tex]\frac{1}{C_T} = \frac{1}{7.0} +\frac{1}{5.0}\\ \frac{1}{C_T} = \frac{5+7}{35} \\C_T = \frac{35}{12}\\ C_T = 2.9166\mu F[/tex]
Given
V = 5.0kV = 5000Volts
To get the total charge in the circuit, we will use the formula:
[tex]Q = C_TV\\Q = 2.9166\times 10^{-6} \times 5000\\Q = 0.014583 C\\Q = 1.4583 \times 10^{-2}C[/tex]
Since the same charge flows in a series connected capacitors, the charge on them in mC is 14.583mC
For the voltage across 7.0 µF:
[tex]V = \frac{Q}{C}\\V = \frac{0.014583}{0.000007} \\V = 2,083.28V[/tex]
Voltage across the 7.0 µF is 2,083.28V
The voltage across the 5.0 µF:
V = 5000 - 2,083.28
V = 2,916.72Volts
Voltage across the 5.0 µF is 2,916.72V
Answer:
The charge on the first capacitor is 14.59 mC
The voltage across the first capacitor is 2084 V
The charge on the second capacitor is 14.59 mC
The voltage across the second capacitor is 2918 V
Explanation:
Given;
first capacitor, C₁ = 7.0 µF
second capacitor, C₂ = 5.0 µF
potential difference, V = 5000 V
The equivalent capacitance is given by;
[tex]\frac{1}{C_{eq}}= \frac{1}{C_1} +\frac{1}{C_2}\\\\\frac{1}{C_{eq}}= \frac{C_1 + C_2}{C_1C_2}\\\\C_{eq} = \frac{C_1C_2}{C_1 +C_2}\\\\ C_{eq} = \frac{(7*10^{-6})(5*10^{-6})}{(7*10^{-6} \ +\ 5*10^{-6})}\\\\C_{eq} = 2.917*10^{-6} \ F[/tex]
The charge on the first capacitor is given by;
[tex]Q_1 = C_{eq} V\\\\Q_1 = 2.917*10^{-6} *5000\\\\Q_1 = 0.01459 \ C\\\\Q_1 = 14.59 \ mC[/tex]
The voltage across the first capacitor is given by;
[tex]V_1 = \frac{Q_1}{C_1} \\\\V_1 = \frac{0.01459}{7*10^{-6}}\\\\ V_1 = 2084.286 \ V[/tex]
V₁ = 2084 V
The charge on the second capacitor is given by;
[tex]Q_2 = C_{eq} V\\\\Q_2 = 2.917*10^{-6} *5000\\\\Q_2 = 0.01459 \ C\\\\Q_2 = 14.59 \ mC[/tex]
The voltage across the second capacitor is given by;
[tex]V_2 = \frac{Q_2}{C_2}\\\\ V_2 = \frac{0.01459}{5*10^{-6}}\\\\V_2 = 2918 \ V[/tex]
Sometimes we will want to write vectors in terms of a coordinate grid. To show a vector points
horizontally (along the x-axis), place an x after the magnitude of the vector. To show a vector point
vertically (along the y-axis), place a y after the magnitude.
4) Using the notation above,
i. How would you write d1?
ii. How would you write d2?
iii. How would you write dtotal?
d1=(0,5)
d2=(5,5)
Answer:
III) [tex]d_{1}+ d_{2}=d_{t}[/tex]
Explanation:
I) coordinate (0,5) is the head for [tex]d_{1}[/tex] I will put the tail coordinate as (0,0) but it could be any other number in the x just not in the 5 with the the y being any other value.
II) coordinate (5,5) is the head for [tex]d_{2}[/tex] the tail needs to be in the head of [tex]d_{1}[/tex] being (0,5)
III) coordinates for [tex]d_{t}[/tex] is connecting the tail from [tex]d_{1}[/tex] and the head of [tex]d_{2}[/tex] making it (0,0)[tex](tail)[/tex] and (0,5)[tex](head)[/tex] and is written as [tex]d_{1}+ d_{2}=d_{t}[/tex]
(i) using coordinate grid notation to represent d₁, d₁ = 5y
(ii) using coordinate grid notation to represent d₂, d₂ = 5x + 5y
(ii) The sum of d₁ and d₂ is written as 5x + 10y
In order to show the horizontal direction of a vector, we will place x after the magnitude of the vector.
Also, to show the vertical direction of a vector, we will place a y after the magnitude of the vector.
(i) Using coordinate grid to represent d₁ = (0, 5)
[tex]d_1 = 0(x) + 5(y)\\\\d_1 = 5y[/tex]
(ii) Using coordinate grid to represent d₂ = (5, 5)
[tex]d_2 = 5x + 5y[/tex]
(iii) The total vector is written as;
[tex]d_1 + d_2 = 5y + (5x + 5y)\\\\d_1 + d_2 = 5y + 5x + 5y\\\\d_1 + d_2 = 5x + 10y[/tex]
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The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the center of the earth. (a) What magnitude and sign of charge would a 60-kg human have to acquire to overcome his or her weight by the force exerted by the earth’s electric field? (b) What would be the force of repulsion between two people each with the charge calculated in part (a) and separated by a distance of 100 m? Is use of the earth’s electric field a feasible means of flight? Why or why not?
Answer:
a) The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.
b) The force of repulsion between two people is [tex]13.851\times 10^{6}[/tex] newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.
Explanation:
a) From Second Newton's Law, we form this equation of equilibrium:
[tex]\Sigma F = F_{E}-W = 0[/tex] (Eq. 1)
Where:
[tex]F_{E}[/tex] - Electrostatic force exerted on human, measured in Newton.
[tex]W[/tex] - Weight of the human, measured in Newton.
If we consider that human can be represented as a particle and make use of definitions of electric field and weight, the previous equation is expanded and electric charge is cleared afterwards:
[tex]q\cdot E-m\cdot g = 0[/tex]
[tex]q = \frac{m\cdot g}{E}[/tex] (Eq. 2)
[tex]E[/tex] - Electric field, measured in Newtons per Coloumb.
[tex]m[/tex] - Mass, measured in kilograms.
[tex]g[/tex] - Gravity acceleration, measured in meters per square second.
[tex]q[/tex] - Electric charge, measured in Coulomb.
As electric field of the Earth is directed in toward the center of the planet, that is, in the same direction of gravity, electric field must be a negative value. If we know that [tex]m = 60\,kg[/tex], [tex]g = 9.807\,\frac{m}{s^{2}}[/tex] and [tex]E = -150\,\frac{N}{C}[/tex], the charge that a 60-kg human must have to overcome weight is:
[tex]q = \frac{(60\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}{-150\,\frac{N}{C} }[/tex]
[tex]q = -3.923\,C[/tex]
The magnitude of the electric charge that a 60-kg human must have to overcome weight is 3.923 coulombs and its sign is negative.
b) The electric force of repulsion between two people with the same charge calculated in part (a) is determined by Coulomb's Law, whose definition we proceed to use:
[tex]F = \kappa \cdot \frac{q^{2}}{r^{2}}[/tex] (Eq. 3)
Where:
[tex]\kappa[/tex] - Electrostatic constant, measured in Newton-square meter per square Coulomb.
[tex]q[/tex] - Electric charge, measured in Coulomb.
[tex]r[/tex] - Distance between two people, measured in meters.
If we know that [tex]\kappa = 9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}[/tex], [tex]q = -3.923\,C[/tex] and [tex]r = 100\,m[/tex], then the force of repulsion between two people is:
[tex]F = \left(9\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot \left[\frac{(-3.923\,C)^{2}}{(100\,m)^{2}} \right][/tex]
[tex]F = 13.851\times 10^{6}\,N[/tex]
The force of repulsion between two people is [tex]13.851\times 10^{6}[/tex] newtons. The use of the earth's electric field a feasible means of flight is not feasible since electric force of repulsion would destroy human body before taking advantage of any possible flight skill.
What is a personified statement?
Answer:
A statement about you
Explanation:
Answer:
Personification is when you give an animal or object qualities or abilities that only a human can have. This creative literary tool adds interest and fun to poems or stories. Personification is what writers use to bring non-human things to life. It helps us better understand the writer's message.
Explanation:
A force of 41 N acts on an object which has a mass of 2.4 kg. What acceleration (in m/s2) is produced by the force
Answer:
The acceleration is [tex] a = 17.083 \ m/s^2 [/tex]
Explanation:
From the question we are told that
The force is [tex]F = 41 \ N[/tex]
The mass of the object is [tex]m = 2.4 \ kg[/tex]
Generally the force is mathematically represented as
[tex]F = m* a[/tex]
=> [tex] 41 = 2.4* a[/tex]
=> [tex] a = 17.083 \ m/s^2 [/tex]
Which of the following is NOT a benefit of cool down activities?
Answer:
Blood is moved away from the brain
Explanation:
Took the test
Homeboy Joe is riding his skateboard and playing Among Us. Because he's distracted he doesn't notice he's about to skate right off the 69.0 m cliff he's on. If he lands 22.0 m from the cliff, how fast was he skating?
Complete this problem so that it can be solved using the concept listed Angela is playing basketball
A. Acceleration
B. Speed
While you were at the skate park, what did you notice about potential and kinetic energy
Answer: I only know that the skater riding down the ramp increeses it's kinetic energy.
Explanation:
A steel ball moves from a position of +125 meters to a position of -75 meters. This motion takes 90.0 seconds. What is the velocity of the steel ball?
Answer:
2.22m/s to the left
Explanation:
Given parameters:
Initial position = +125m
Final position = -75m
Motion time = 90s
Unknown:
Velocity of the steel ball = ?
Solution:
The velocity of the steel ball is given as the displacement divided by the time;
Velocity = [tex]\frac{displacement}{time}[/tex]
The net displacement of the ball = 125- (-75) = 200m to the left
Input the parameters and solve for the velocity;
Velocity = [tex]\frac{200}{90}[/tex] = 2.22m/s to the left
What is the effect of applying an unbalanced force on an object?
A. Object moves
B. Object remains stationary
C. Mass of object increases
D. Inertia of object increases
Object will start moving in the direction where comparatively high force is directing to move
What is an unbalance force ?When the resultant force acting on a body is not equal to zero , the force acting on the body are known as unbalance forces . The body acted upon by unbalance forces changes its state of motion.
According to Newton if a body is experiencing net force =0 (which means that all horizontal and vertical forces are cancelling each other ), then the body will acquire an equilibrium state and will not move because forces are balancing each other and making net force on the body zero .
But unbalance force can tend the body to move in that direction in which a comparatively high force is acting due to which cancellation will not occur ( equilibrium state will not occur )
hence correct option A)Object moves
learn more about unbalance forces
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The average speed of an object, S SS, is calculated using the formula S = D T S= T D S, equals, start fraction, D, divided by, T, end fraction, where T TT is the time it takes to travel a distance of D DD units. Rearrange the formula to solve for time ( T ) (T)
Answer:
T = D/S
Explanation:
Given the formula for calculating the average speed of an object as:
average speed = Distance/Time
Let average speed = S
Distance = D
Time = T
Substitute
S = D/T
Make T the subject of the formula:
From S = D/T
Cross multiply
ST = D
Divide both sides by S
ST/S = D/S
T = D/S
Hence the expression for he time T after rearrangement is T = D/S
A lamp of mass m hangs from a spring scale which is attached to the ceiling of an elevator. When the elevator is stopped at the fortieth floor, the scale reads mg. What does it read as the elevator slows down to stop at the ground floor?
a. more than mg
b. mg
c. less than mg
d. zero
e. can't tell
Answer:
The correct answer is (a)
Explanation:
A spring scale measures the weight of an object not the mass because according to hooke's law the extension of a spring is directly proportional to the load or force attached/applied to it. The force of gravity acting on the mass of any substance as it goes up actually reduces and increases as it comes down.
If F = ma, as a increases, F will also increase and vice versa
Where F = force
m = mass
a = acceleration (due to gravity in this case)
From the above explanation, it can be deduced that the scale will read more than mg as it gets to the ground because of an increase in the force of gravity (which also increases a) as it approaches the ground.
A projectile is fired straight up with an initial velocity of 40.0 m/s . Approximately how high will the projectile ?
Answer:
it depends on the wind and any other conditions but if you have a controlled environment it should take 1 second to get 40 meters but it could go higher in which it could take about 5 seconds to go 200 meters
Explanation:
hope it helped
:)
what chemical diverged from trees
a resin and turpentine
b sodium
c lead
d marcotting
Answer:
a. resin and turpentine
Explanation:
The chemicals that diverged from trees are resin and turpentine.
Resin are produced by special cells in trees, most times we see them when a tree is damaged or cut. They are usually derived from pines and firs.
Turpentine is obtained by distilling resin.
Turpentine has an antiseptic property that has different uses. They are used as cleansing agents and for producing sanitary materials.
Consider a double Atwood machine constructed as follows: A mass 4m is suspended from a string that passes over a massless pulley on frictionless bearings. The other end of this string supports a second similar pulley, over which passes a second string supporting a mass of 3m at one end and m at the other. Using two suitable generalized coordinates, set up the Lagrangian and use the Lagrange equations to find the acceleration of the mass 4m when the system is released. Explain why the top pulley rotates even though it carries equal weights on each side.
Answer:
Hello your question is incomplete attached below is the complete question
Answer : x ( acceleration of mass 4m ) = [tex]\frac{g}{7}[/tex]
The top pulley rotates because it has to keep the center of mass of the system at equilibrium
Explanation:
Given data:
mass suspended = 4 meters
mass suspended at other end = 3 meters
first we have to express the kinetic and potential energy equations
The general kinetic energy of the system can be written as
T = [tex]\frac{4m}{2} x^2 + \frac{3m}{2} (-x+y)^2 + \frac{m}{2} (-x-y)^2[/tex]
T = [tex]4mx^2 + 2my^2 -2mxy[/tex]
also the general potential energy can be expressed as
U = [tex]-4mgx-3mg(-x+y)-mg(-x-y)+constant=-2mgy +constant[/tex]
The Lagrangian of the problem can now be setup as
[tex]L =4mx^2 +2my^2 -2mxy +2mgy + constant[/tex]
next we will take the Euler-Lagrange equation for the generalized equations :
Euler-Lagrange equation = [tex]4x-y =0\\-2y+x +g = 0[/tex]
solving the equations simultaneously
x ( acceleration of mass 4m ) = [tex]\frac{g}{7}[/tex]
The top pulley rotates because it has to keep the center of mass of the system at equilibrium
Compare the kinetic energy of a 21,000 kg truck moving at 130 km/h with that of an 84.0 kg astronaut in orbit moving at 28,000 km/h.
Answer:
K.E of the Astronaut is greater
Explanation:
Mathematically;
K.E = 1/2 * m * v^2
For the truck, the kinetic energy will be
1/2 * 21,000 * 130^2 = 177,450,000 J
while for the astronaut, his K.E will be
1/2 * 84 * 28,000^2 = 3.2928 * 10^10
This shows that the K.E of the astronaut is greater
When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the sky diver and, thus, slows him down. Suppose the weight of the sky diver is 915 N and the drag force has a magnitude of 1061 N. The mass of the sky diver is 93.4 kg. Take upward to be the positive direction. What is his acceleration, including sign
Explanation:
According to newton's second law of motion.
[tex]\sum Fx = ma\\\\\sum Fx = 1061 - 915\\\\\sum Fx = 146N[/tex]
m is the mas of the sky diver = 93.4kg
a is the acceleration of the skydiver
From the formula above;
[tex]a = \frac{\sum Fx}{m}\\ \\a = \frac{146}{93.4}\\\\a = 1.563m/s^2[/tex]
Hence the acceleration of the sky diver is 1.563m/s²