When the red barge collides with the blue barge, which has a mass of 1000000 kg, its final velocity with in collision elastic becomes 0.311 m/s.
Define the term collision elastic?A elastic collision is one in which the system does not experience a net loss of kinetic energy as a result of a collision. In elastic collisions, momentum as well as kinetic energy are both conserved.The given data:
m1= 150000 kg, v1= 0.25 m/s, m2= 1000000 kg, v2= 0.32 m/sUsing the law of conservation of momentum.
Following the formula, the red barge's final velocity (v3) is determined.
m1×v1+ m2×v2= (m1+m2)v3
Put the values;
150000 × 0.25+ 1000000×0.32= (150000+1000000)×v3
37500+ 320000= 1150000×v3
357500 = 1150000×v3
v3 = 0.311 m/s
Thus,
After one head-on elastic hit where the projectile is substantially more substantial than the target, the target particle's velocity will be nearly twice as fast as the projectile, but really the projectile's velocity will essentially be unaffected.
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a ping-pong ball covered with a conducting graphite coating has a mass of 5.0 x 10-3 kg and a charge of 4 uc. what electric field will balance exactly the weight of the ball? give magnitude and direction of the e.
The ball's weight will be precisely balanced by the electric field is E = 12.25 x 10³ N/C = 12.25 KN/C
The electrostatic force caused by the electric field must be equal to the weight of the body or charge in order to balance the object's weight. Therefore,
Weight = Electrostatic Force
E q = mg
where,
Electric field = (E) =?
5 x 103 kg is the mass of the charge, or m.
g = 9.8 m/s2 is the acceleration caused by gravity.
Q equals the charge's magnitude, which is 4 C (4 x 106 C).
Therefore,
E(4 x 10⁻⁶ C) = (5 x 10⁻³ kg)(9.8 m/s²)
E = 0.049 N/4 x 10⁻⁶ C
E = 12.25 x 10³ N/C = 12.25 KN/C
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A proton is at the origin. One electron is at the point (2m,4m) and the other is at the point (-2m,-4m). What is the net force on the proton. F=
The net force on the proton is 2.88 x 10⁻³⁰ N.
What is the net electric force between the particles?The net electric force between the particles is calculated by applying Coulomb's law as shown below.
F = ( kq₁q₂ ) / r²
where;
k is the Coulomb's constantq₁ is the charge of the electronq₂ is the charge of the protonr is the distance between the chargesThe distance between the charges is calculated as;
r = √ [ (-2 - 2)² + ( -4 - 4 )² ]
r = 8.94 m
F = ( 9 x 10⁹ x 1.6 x 10⁻¹⁹ x 1.6 x 10⁻¹⁹ ) / ( 8.94² )
F = 2.88 x 10⁻³⁰ N
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a hockey puck slides off the edge of a table with an initial velocity of 28.0 m/s. and experiences no air resistance. the height of the tabletop above the ground is 2.00 m. what is the angle below the horizontal of the velocity of the puck just before it hits the ground?
The angle which is calculated below the horizontal of the velocity of the puck just before it hits the ground is 12.60°.
Given the following data:
Initial velocity = 28.0 m/s
Acceleration due to gravity = 9.81
Displacement (height) = 2.00 meters.
Let us calculate and determine the angle below the horizontal of the velocity of the puck just before it hits the ground:
First of all, we need to determine the horizontal and vertical components of the hockey puck.
For horizontal component of this:
[tex]V^{2}= U^{2}+ 2 as\\V^{2}= 0^{2}+ 2 (9.8)(2)\\V^{2}=39.24\\ V=6.26 m/s[/tex]
For vertical component:
[tex]V_{x}=U_{x} \\V_{x}= 28.0 m/s[/tex]
Now, we can find the angle by using the following formula:
θ= [tex]tan^{-1} =\frac{Vx}{Vy}\\[/tex]
Substituting the values, we get.
Angle = 12.60 degrees.
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Two lab partners are doing a static electricity investigation. They begin with a
neutral pop can. After conducting the procedure, they observe that the pop can
has become charged positively. Which statement accurately explains how this
happened?
Electrons were added to the object.
Protons were added to the object.
Electrons were removed from the object.
Protons were removed from the object.
A) electrons were added to the object
B) protons were added to the object
C) electrons were removed from the object
D) protons were removed from the object
The electrons are transferred from one item to another during the conduction process.The electrons are particles with a free charge.There is no free movement of protons.
Which statement accurately explains how this happened?The electrons are transferred from one item to another during the conduction process.The electrons are particles with a free charge.There is no free movement of protons.The metal pie plate in the provided scenario is initially negatively charged.As a result of conduction, the plate becomes electrically neutral.It signifies the removal of extra electrons from the aluminum pie plate.At that point, the number of protons and electrons is equal.The plate becomes neutral electrically.Therefore, "Electrons were taken from the item" is the appropriate response. The transfer of electrons between two different materials occurs when they are rubbed together.As a result, one thing (the electron loss) becomes positively charged, and the other object (the gainer) becomes negatively charged (the electron gainer).To learn more about static electricity refer
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Four mass–spring systems oscillate in simple harmonic motion. Rank the periods of oscillation for the mass–spring systems from largest to smallest.m = 2 kg , k = 2 N/mm = 2 kg , k = 4 N/mm = 4 kg , k = 2 N/mm = 1 kg , k = 4 N/m
The order of the periods of oscillation from largest to smallest is:
T1 > T3 > T2 > T4
How to find the smallest?The period of oscillation for a mass-spring system is given by the formula:
T = 2π √(m/k)
Where T is the period, m is the mass of the oscillating object and k is the spring constant.
Using this formula, we can find the periods of oscillation for the four mass-spring systems:
m = 2 kg, k = 2 N/m: T = 2π √(2 kg / 2 N/m) = 2π √(1) = 2π secondsm = 2 kg, k = 4 N/m: T = 2π √(2 kg / 4 N/m) = 2π √(0.5) = π secondsm = 4 kg, k = 2 N/m: T = 2π √(4 kg / 2 N/m) = 2π √(2) = 2π √(2) = 2π √(2) secondsm = 1 kg, k = 4 N/m: T = 2π √(1 kg / 4 N/m) = 2π √(0.25) = π/ √(2) secondsSo the order of the periods of oscillation from largest to smallest is:
T1 > T3 > T2 > T4
Where T1 is the period of the mass-spring system with m = 2 kg and k = 2 N/m, T3 is the period of the mass-spring system with m = 4 kg and k = 2 N/m, T2 is the period of the mass-spring system with m = 2 kg and k = 4 N/m, and T4 is the period of the mass-spring system with m = 1 kg and k = 4 N/m.
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part 1 of 2
Objects with masses of 111 kg and 290 kg are separated by 0.396 m. A 23.9 kg mass is placed midway between them.
Find the magnitude of the net gravitational force exerted by the two larger masses on the 23.9 kg mass. The value of the universal gravitational constant is 6.672 × 10^-11 N • m^2 /kg^2.
Answer in units of N.
part 2 of 2
Leaving the distance between the 111 kg and the 290 kg masses fixed, at what distance from the 290 kg mass (other than infinitely remote ones) does the 23.9 kg mass experience et force of zero?
Answer in units of m.
(a) The gravitational force between the two large masses is 1.37 x 10⁻⁵ N.
(b) The distance from 290 kg mass in which 23.9 kg mass experiences a zero net force is 0.24 m.
What is the gravitational force between the two masses?The gravitational force between the two large masses is calculated as follows;
F = (Gm₁m₂ ) / R²
where;
G is universal gravitation constantm is the mass of the objectR is the distance between the massesF = ( 6.672 x 10⁻¹¹ x 111 x 290 ) / ( 0.396² )
F = 1.37 x 10⁻⁵ N
The distance from 290 kg mass in which 23.9 kg mass experiences a zero net force is calculated as;
Let the distance between 290 kg and 23.9 kg = d
F1 = ( 6.672 x 10⁻¹¹ x 23.9 x 290 ) / ( d² )
F1 = ( 4.62 x 10⁻⁷ ) / d²
F2 = ( 6.672 x 10⁻¹¹ x 23.9 x 111 ) / ( 0.396 - d )²
F2 = ( 1.77 x 10⁻⁷ ) / ( 0.396 - d )²
( 4.62 x 10⁻⁷ ) / d² = ( 1.77 x 10⁻⁷ ) / ( 0.396 - d )²
( 4.62 x 10⁻⁷ )(0.396 - d )² = ( 1.77 x 10⁻⁷ )d²
2.6(0.396 - d )² = d²
(0.396 - d )² = d²/2.6
0.396 - d = √ (d²/2.6)
0.396 - d = 0.62d
0.396 = 1.62d
d = 0.396 / 1.62
d = 0.24 m
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during a flight where the distance was 1200 km, an aircraft was slowed down due to bad weather. its average speed for the trip was reduced by 400 km/hr. and the time of flight increased by 60 minutes. the duration of the flight is:
The duration of the flight is 3 hours if its speed is reduced by 400 km/h, and the time of flight increased by 60 minutes.
The length of the trip, d = 1200 km
Reduced average speed of the aircraft, v = 400 km/h
Increased time of flight, = 60 minutes = 1 hour
Let the initial time of the flight = t hours
Total time during reduced speed, = (t+1) hours
So, total distance, 1200 km = 400 × (t+1)
t+1 = 1200/400 = 3 hours
So total time of flight is 3 hours.
Initial time of flight, t = 3 - 1 = 2 hours
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how much oil (138,690 btu/gal) is needed to generate the electricity needed to keep a 100-watt (w) light bulb lit for one (1) year if the overall efficiency of the generating station is 30
7.2 gallons of oil is needed to generate the electricity needed to keep 100-watt light bulb lit for one year if the overall efficiency of generating station is 30%.
What is efficiency?Ratio of the work performed by machine or in process to the total energy expended / heat consumed is called as efficiency.
Energy consumed is 100 W * 8760 h = 876,000 Wh = 876 kWh.
Given, overall efficiency of the generating station is 30%, which means that only 30% of energy in the fuel is converted into usable electricity. So, for every 1 unit of energy in fuel, 0.3 units of energy are generated as electricity.
To find out how much oil is needed, we need to divide total energy consumed by the light bulb (876 kWh) by the overall efficiency of the generating station (0.3). So, 876 kWh / 0.3 = 2920 kWh
Now we need to convert the energy consumption in kWh to gallons of oil,
2920 kWh = 2920 * 3412 Btu/kWh = 1,000,944 Btu
1,000,944 Btu / 138,690 Btu/gal = 7.2 gallons
Therefore, 7.2 gallons of oil is needed to generate the electricity needed to keep 100-watt light bulb lit for one year.
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a 6.00-g bullet is fired with a velocity of 96 m/s toward a 10.00-kg stationary solid block resting on a frictionless surface. (a) what is the change in momentum of the bullet if it is embedded in the block?
When a 6.00-g bullet is fired at a target that weighs 10.00 kg and is lying on a frictionless surface at a speed of 96 m/s, the bullet's change in momentum if it penetrates the target is 0.576 m/s.
The momentum change is just equal because the final momentum is zero. Momentum is the phrase used to describe a substance's state of motion when its mass is not zero. Therefore, momentum is a property of all moving objects. The formula p = m v, where m denotes an object's mass and v denotes its rate of motion, describes momentum.
The bullet's momentum will alter and become equal to
momentum = 6/1000 * 96 = 0.576, if it is implanted in the block.
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the student explains how conduction is the charging by contact so the uncharged object takes
Electrons from the charged object as conduction are the charging by contact
What is Conduction?Conduction is indeed a method of charging by contact. In conduction, charges are transferred between two objects that are in contact with each other. When an uncharged object comes into contact with a charged object, electrons will flow from the charged object to the uncharged object, until both objects have the same charge.
In solids, conduction occurs when electrons flow through the material, moving from atom to atom. Metals, such as copper and aluminum, are good conductors of electricity because their electrons are not tightly bound to their atoms and can easily flow through the material. Insulators, such as rubber and plastic, are poor conductors of electricity because their electrons are tightly bound to their atoms and do not flow easily through the material.
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a satellite travels at constant speed in a circular orbit at a height just 100 miles above sea level. without doing any calculations, what is the acceleration of the satellite?
The acceleration of the satellite is just below the acceleration of gravity
i.e 9.8 m/s^2.
Given that,
the speed of the satellite in a circular orbit at a height just 100 miles above sea level.
The acceleration of the satellite is due to the force of gravity. The force of gravity is equal to the mass of the satellite times the acceleration of gravity, which is 9.8 m/s^2. The mass of the satellite is very small, so the force of gravity is also very small. However, the satellite is also moving very fast, so the centripetal force is also very small. As a result, the acceleration of the satellite is just a hair less than 9.8 m/s. The satellite is accelerating because it experiences a net force acting on it, and also because its velocity is changing.
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If the force shown in the diagram below does 1,110J of work while acting throughout the
10.0m distance, what is the magnitude of the force?
a. F = 139N
b. F = 111N
c. F = 8,860N
d. F = 904N
The force that is acting on the object is 139N. Option A
What is the magnitude of the force?Let us know that we can only say that there is work done if we have a force that has been applied in such a way that it can be able to move an object in the direction of the force.
As such, we are going to have that;
W = Fcosθ
W = work done
F = force applied
θ = angle turned
1110 = F cos 37 * 10
F = 1110/cos 37 * 10
F = 139N
There is the application of a force of about 139N on the object as have been shown here.
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A monkey climbs vertically on a vine. Its motion is shown on the following graph of vertical position yyy vs. time ttt. Graph of y (in meters) vs. t (in seconds) that starts at 6 m at 0sec, decreases linearly to 2m at 4 sec, stays constant at 2m from 4 sec to 7 sec, increases linearly to 5 m from 7 sec to 9 sec, then stays constant until 10 sec. Graph of y (in meters) vs. t (in seconds) that starts at 6 m at 0sec, decreases linearly to 2m at 4 sec, stays constant at 2m from 4 sec to 7 sec, increases linearly to 5 m from 7 sec to 9 sec, then stays constant until 10 sec. What is the instantaneous speed of the monkey at time t=5\text{ s}t=5 st, equals, 5, start text, space, s, end text?
The linear curve of the graphs shows an increase in distance of 5 m within 2 seconds. Then , the distance within 5 seconds will be 12.5m and the instantaneous speed is 2.5 m/s.
What is instantaneous speed ?Instantaneous speed of an object is the speed at a particular instant. It describes how far an object travelled at a particular moment in time.
It is given that, the graph shows linearity in curve , where for each 2 second time interval the monkey climbs 5m. Then, for a time interval of 5 seconds, the monkey will climb 12.5 m in the tree.
Instantaneous speed = distance/time
t= 5 s
d = 12.5 m
then v = 12.5/5 = 2.5 m/s
Therefore, the instantaneous speed of the monkey within the given time will be 2.5 m/s.
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Please help me! I will give brainliest!
1. Imagine 50 people walking in straight lines in a room. What would happen to the number of
collisions they would have with the walls if they moved to a room half the size of the original
room but walked at the same speed?
2. Imagine 50 people walking in straight lines in a room. If those 50 people start running, what needs to happen in order for the number of collisions they would have with the walls of the room to remain the same?
1) They would collide with the walls more frequently.
2) They would collide with the walked more frequently.
What is the collision?A collision is defined in physics as any event in which two or more bodies exert forces on each other in a relatively short period of time.In physics, a collision is the sudden, forceful coming together of two bodies, such as two billiard balls, a golf club and a ball, a hammer and a nail head, two railroad cars when coupled together, or a falling object and a floor.However, there are three types of collisions: elastic, inelastic, and completely inelastic. To summarize, momentum is conserved in all three types of collisions. What happens to the kinetic energy is what distinguishes the collisions.To learn more about collision refer to:
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PLSSSSSS HELP NOWWWWW
DO NOT GIVE ME THE WRONG ANSWER PLSSS
Question 1(Multiple Choice Worth 2 points)
(03.03 MC)
In which of the following situations would sonar be most practical?
Measuring the distance between objects in the ocean
Observing internal organs
Determining the frequency of sounds
Dampening the intensity of sounds
Answer:
A: Measuring the distance between objects in the ocean
Explanation:
A 50 kg painter stands on top of a 3 m platform on the roof of a 20000 kg shed. What is the painter's gravitational potential energy.
The painter's gravitational potential energy is 1470J
What is gravitational potential energy?
The energy that a body has or acquires when its location changes as a result of being in a gravitational field is known as gravitational potential energy. Gravitational potential energy can be defined as an energy that has a relation to gravitational force or gravity. Since the painter is on the top of a 3m platform his gravitational potential energy is given by W=mgh where m is the mass of the painter, g is the acceleration due to gravity and h is the height of the painter from the ground.
Now according to the question,
Mass of the painter=50Kg
Height of the platform=3m
Acceleration due to gravity=9.8
W=50×9.8×3
W=1470J
Therefore, the gravitational potential energy of the painter is 1470J
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a point charge is located on the -axis at and a second point charge is on the -axis at what is the total electric flux due to these two point charges through a spherical surface centered at the origin and with radius (a) 0.500 m, (b) 1.50 m, (c) 2.50 m?
The total electric flux due to these two point charges through a spherical surface centred at the origin and with radius 0.500 m, 1.50 m and 2.50 m are respectively Ф₁ = 0, Ф₂ = -519.8 N m²/ C, Ф₃ = -67.8 N m²/ C.
Charge q₁ = 4 nC = 4 × 10⁻⁹ C
Charge q₂ = -4.6 nC = -4.6 × 10⁻⁹ C
The radius of various spheres is R₁ = 0.5 m, R₂ = 1.5 m, R₃ = 2.5 m
∈₀ is the electric permittivity of free space
It is observed that the sphere of radius R₁ does not contain any charges. Therefore, the flux through that sphere is zero.
Ф₁ = 0
The sphere of radius R₂ contains only charge q₂. So, the flux through that sphere is given as, Ф₂ = q₂/∈₀ = (-4.6 × 10⁻⁹)/(8.85 × 10⁻¹²) = -519.8 N m²/ C
The sphere of radius R₃ contains both charges q₁, q₂. So, the flux through that sphere is
Ф₃ = (q₁ + q₂)/∈₀ = (-4.6 × 10⁻⁹ + 4 × 10⁻⁹)/(8.85 × 10⁻¹²) = (-0.6 × 10⁻⁹)/(8.85 × 10⁻¹²) = -67.8 N m²/ C
The given question is incomplete. The complete question is 'A point charge q₁ = 4 nC is located on the x-axis at x = 2 m and a second point charge q₂ = -4.6 nC is on the y-axis at y = 1. What is the total electric flux due to these two point charges through a spherical surface centred at the origin and with radius (a) 0.500 m, (b) 1.50 m, (c) 2.50 m?'
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A 1650 kg car is going a street at 42 m/s east. What is the momentum of the car
Formula for momentum:
[tex]p=mv[/tex]
momentum(measured in kg*m/s) = mass(measured in kg) * velocity(measured in m/s)
__________________________________________________________
Given:
[tex]m=1650kg[/tex]
[tex]v=42m/s[/tex]
[tex]p=?[/tex]
__________________________________________________________
Finding momentum:
[tex]p=mv[/tex]
[tex]p=1650\times42[/tex]
__________________________________________________________
Answer:
[tex]\fbox{p = 69,300 kg*m/s}[/tex]
A hollow, conducting sphere with an outer radius of 0. 240 m and an inner radius of 0. 200 m has a uniform surface charge density of +6. 47 × 10−6 C/m2. A charge of -0. 900 μC is now introduced into the cavity inside the sphere.
a. What is the new charge density σ on the outside of the sphere? Express your answer with the appropriate units.
b. Calculate the strength of the electric field E just outside the sphere. Express your answer with the appropriate units.
c. What is the electric flux through a spherical surface just inside the inner surface of the sphere? Express your answer with the appropriate units
Consider the following explanation for defining surface charge density:
= Q/surface sphere's area.Answers: (a) 5.73 * 10-6 C/m2, (b) 648 * 10 N/C, and (c) 56.5 * 10 3 Nm2/C.
What is the exterior of the sphere's new charge density?Consider the following explanation for defining surface charge density:
= Q/surface sphere's area
Q thus equals * * 4* * R2.
The new Q is as follows: Q initial -0.5 microC =5 micro C at the outer radius.
The new outside charge on the electric is equal to:
E=k*Q/R2outer=9 109*5 microC/(0.25m)2=648*103 N/C.
Last but not least, the flux equivalent to the internal charge is
The total flux is equal to net change divided by 0 according to Gauss's law.
0.5 microC/0 = 56.5 * 10 3 Nm2/C, therefore.
Answers: (a) 5.73 * 10-6 C/m2, (b) 648 * 10 N/C, and (c) 56.5 * 10 3 Nm2/C.
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The period of the sound waves produced by the motor is 8. 3 milliseconds. Calculate the frequency of the sound waves
please help hahhah
The sound waves produced by the engine have a frequency of 120 Hz.
How to calculate frequency?To calculate the frequency of the sound waves produced by the engine, we can use the following formula:
f = 1/T
where f is the frequency and T is the period of the sound wave.
Sound waves have a period of 8.3 milliseconds, so we can plug this value into the equation to find the frequency.
f = 1 / (8.3 x 10^-3s)
f = 120 hertz
Therefore, the sound waves produced by the engine have a frequency of 120 Hz.
It is important to note that frequency is the number of wave oscillations or cycles that occur in a unit of time and is measured in hertz (Hz). The period is the time of one oscillation and is the reciprocal of the frequency. The lower the frequency, the longer the period and vice versa.
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if the contacts on a manual starter cannot be closed immediately after a motor overload has tripped them open, what is the probable reason?
If the contacts on a manual starter cannot be closed immediately after a motor overload has tripped them open, the probable reason is that the thermal overload device (a type of overload relay) has not had enough time to cool down.
The thermal overload device is designed to protect the motor from damage by sensing the temperature of the motor windings and tripping the starter if the temperature becomes too high.
The contacts of the starter will remain open until the thermal overload device has cooled down enough to reset, which typically takes several minutes.
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A 1. 0 kg toy car is released at the top of a frictionless track on the left and rolls off of the track from its right
side ramp. The car starts at a height of 0. 80 m, goes through a 0. 50 m diameter loop, and exits the ramp at a
height of 0. 25 m.
The change in gravitational potential energy is 5.39 J.
Describe gravitational potential energy?Gravitational potential energy is a type of potential energy that an object possesses due to its position in a gravitational field. It is the energy that an object has due to its height and the force of gravity acting upon it. The formula for gravitational potential energy is given by mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object above a reference point. When an object falls from a height, it converts its gravitational potential energy into kinetic energy.
To find the change in gravitational potential energy, we can use the formula: ΔU = mgh, where m is the mass of the object (1.0 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the change in height.
The change in height is 0.80 m - 0.25 m = 0.55 m. Plugging in the values, we get:
ΔU = 1.0 kg * 9.8 m/s^2 * 0.55 m = 5.39 J
So the change in gravitational potential energy is 5.39 J.
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The complete question is:
A 1.0kg toy car is released at the top of a frictionless track on the left and rolls off of the track from its right side ramp. The car starts at height of 0.80 m, goes through a 0.50 m diameter loop, and exits the ramp at a height of 0.25 m 0.80 m 0.50 mi 0.25 m What is the change in the car's gravitational potential energy from A to B? Round answer to two significant digits.
if we assume that all of the u 235 and u 238 undergo fission and roughly 200 mev of energy is emitted per fission, for how many days can this fuel power a 3 gw power plant, assuming the plant is able to convert fuel energy to power with roughly 30% efficiency?
The plant can convert fuel energy into power with an efficiency of about 30% for 241 days .
As per the data given in the above question are as bellow,
We need to know how much fuel is in the plant in order
To calculate how many days a 3 GW power plant can run on fuel containing U-235 and U-238.
Assume the facility contains 1 metric tonne of U-235 and U-238-containing fuel.
The total energy released from the fission if all of the U-235 and U-238 split would be
200 MeV/fission * 2.51021 fissions/metric tonne = 51023 MeV.
The total energy transferred to electricity would be
51023 MeV * 0.3 = 1.51023 MeV = 1.51016 J
if we assume a 30% efficiency.
It would require 1.51016 J / 3109 J/s or 5106 s or 5,800 hours or 241 days to generate 3 GW at the power plant.
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HELPP ME ASPP PLEASE
Point A represents the b.crest.
Transverse wavePoint A in a transverse wave represents the wavelength. The wavelength is the distance from a point on a wave to the next identical point on the wave. It is the measure of the length of the wave, and is measured in meters (m). A wave's wavelength is the distance between two identical points on the wave, such as peak-to-peak, or crest-to-crest. The crest of the wave is the highest point of the wave. It is the point where the wave reaches its maximum displacement from its undisturbed position. The amplitude of the wave is the maximum distance the wave deviates from its undisturbed position. It is the measure of the maximum displacement of the wave and is measured in meters (m). The trough of the wave is the lowest point of the wave. It is the point where the wave reaches its minimum displacement from its undisturbed position. The particles move at right angles to the direction of the wave. This means that they move up and down as the wave moves forward.To learn more about Transverse wave refer to:
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two cars in an inelastic collision . A small car with a mass of 1600kg is moving east at 55 km/hr a van with a mass of 3300 kg is moving west 65 km/hr . calculate the speed and direction of the combined vehicles just after impact
The speed and direction of the combined vehicles just after impact is 7.2 m/s west.
What is final speed of the cars after the collision?
The final speed of the cars after the collision is obtained by applying the principle of conservation of linear momentum.
m₁u₁ + m₂u₂ = v ( m₁ + m₂ )
where;
m₁ is the mass of the first carm₂ is the mass of the second caru₁ is the initial velocity of the first car = 55 km/h = 15.28 m/su₂ is the initial velocity of the second car = 65 km/h = 18.1 m/sv is the final velocity of the two cars(1600 x 15.28) - ( 3300 x 18.1) = v ( 1600 + 3300)
-35,282 = 4,900v
v = -35,282 / 4,900
v = -7.2 m/s
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30. a signal with an input power of 15 mw is transmitted along an optic fiber which has an attenuation per unit length of 0.30 db km-1.the power at the receiver is 2.4 mw. calculate the length of the fiber.
The length of the fiber is 75.7 km.
What is length?Length is a measurement of distance or size. It is commonly used to describe the linear extent of an object, such as the size of a room, the length of a river, or the size of a person. It can also be used to describe the duration of an event, such as the length of a movie or the length of a football game. Length is typically measured in units such as meters, feet, or inches.
The power at the receiver can be calculated using the formula:
P_r = P_i * 10^(-α*L/10)
where P_r is the power at the receiver, P_i is the input power, α is the attenuation per unit length and L is the length of the fiber.
Rearranging the formula, we obtain:
L = 10*log(P_i/P_r)/α
Substituting the values, we get:
L = 10*log(15/2.4)/0.30 = 75.7 km
Therefore, the length of the fiber is 75.7 km.
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Two children pull a 320 kg sled on an ice rink - Child A pulls with a force of 450 N [S 15° E]
while child B pulls with 380 N [W].
If the sled starts from rest, how far will it go in 2.0s, and in what direction?
Please add a diagram!
Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2).
How do you calculate frictional force pulling?The resistive force of friction (Fr) divided by the normal or perpendicular force (N) pushing the objects together yields the coefficient of friction (fr), which is a numerical value. The formula fr = Fr/N serves as a representation of it.
The force of gravity acting on an object is known as its weight, and it can be determined using the formula w = mg, which equals the mass times the acceleration of gravity. The newton is the SI unit for weight since it is a force.
The formula F = m a, where F the is the force acting on the object, m is the object's mass, and an is the acceleration, can be used to express Newton's second equation of motion. This formula's subject, m, must be changed in order to get the object's mass.
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a block attached to a spring with unknown spring constant oscillates with a period of 2.7 s s . parts a to d are independent questions, each referring to the initial situation. part a what is the period if the mass is doubled?
The Doubling the mass does not affect the period of oscillation.
Oscillation is a repetitive motion that occurs when a force causes an object to move back and forth between two positions. This motion is often seen in physical systems such as pendulums, springs, and electrical circuits.
The period of a block attached to a spring with unknown spring constant is determined by the mass of the block and the spring constant. Doubling the mass of the block will not change the period, so the period will remain 2.7 s. This is because doubling the mass of the block will double both its inertia and its gravitational pull, but these two forces are balanced by the spring force, which remains unchanged. Therefore, doubling the mass does not affect the period of oscillation.
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If you were to go into outer space, your ______ would stay the same, but your ______ would change question 7 options: a. age, spaceship b. weight, mass c. mass, weight d. inertia, magnitude
B is the answer ur welcome
what is the line charge density on a long wire if the electric field 45 cm from the wire has magnitude 240 kn/c and points toward the wire?
The line charge density on a long wire if the electric field 45 cm from the wire has magnitude 240 kn/c and points toward the wire is 2.08 x 10-7 C/m.
Calculating the line charge density of a long wire is a simple process. To begin, we need to know the magnitude of the electric field at a given distance from the wire. In this case, the electric field 45 cm from the wire has a magnitude of 240 kn/c and points toward the wire.
Using this information, we can calculate the line charge density of the wire using the equation:
Line charge density (λ) = Electric field (E) * Permittivity of free space (ε0) / 2π * distance (r)
Plugging in the values, we get:
λ = 240 kn/c * 8.85 x 10-12 C2/Nm2 / 2π * 0.45m
Solving this equation, we get a line charge density of 2.08 x 10-7 C/m.
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