The statement that is false is "there are places on earth where all forces are absent." Forces, such as gravity, are always present on Earth.
All the given statements are true except the statement "there are places on earth where all forces are absent." is false.
Forces change the momentum of a body,forces always occur in equal and opposite pairs (Newton's third law), where there is no force, objects continue to move the way they were moving, and forces cause an acceleration to take place are all true statements about forces.
Newton's third law simply states that there is an equal and opposite reaction to every action. So, if object A exerts a force on object B, object B will exert an equal but opposite force on object A.
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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 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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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 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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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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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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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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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 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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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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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.
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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cody's car acce;rates from 0m/s to 45 m/s northward in 15 seconds. what is the acceleration of the car
The acceleration of the Cody's car when moving from 0 m/s to 45 m/s is calculated to be 3 m/s².
The acceleration is described to be the rate of change of velocity. It is a vector quantity. Its S I unit is m/s². If an object in a straight line accelerates up or slows down, it is said to be accelerated. Mathematically, a = Δv/t
where, a = acceleration
Δv = change of velocity = 45 - 0 = 45 m/s
t = time = 15 sec
After plugging the values into the equation above, we have,
a = Δv/t = 45/15 = 3 m/s²
Thus, the required acceleration of the car is calculated to be 3 m/s².
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A car drives 40 miles at an angle of 35 degrees north of east then drives for 50 miles due north and finally 10 miles at an angle of 20 degrees north of west what is the cars resultant magnitude and direction
Answer:
Below
Explanation:
FIND TOTAL NORTH DISPLACEMENT
40 sin 35 + 50 + 10 sin 20 = 76.363 mi
FIND TOTAL E-W DISPLACEMENT
40 cos 35 - 10 cos 20 = 23.369 mi
Total resultant displacement ( using pythag theorem)
sqrt ( 76.363^2 + 23.369^2 ) = 79.86 mi
angle = arctan ( 79.23 / 23.37) = 73 ° North of East
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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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]
An airplane travels down a
runway for 4.0 seconds with an
acceleration of 9.0 m/s2. What
its change in velocity during
this
time?
Someone helpppp!!
Answer:
We can use the equation of motion of an object under constant acceleration to calculate the change in velocity during this time. The equation is:
Δv = at
Where:
Δv = change in velocity (m/s)
a = acceleration (m/s^2)
t = time (s)
Given that the acceleration of the airplane is 9.0 m/s^2 and the time it travels down the runway is 4.0 seconds we can substitute these values into the equation:
Δv = 9.0 m/s^2 * 4.0 s = 36 m/s
Therefore, the change in velocity of the airplane during this time is 36 m/s.
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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How does genetics affect our chance of developing certain diseases and disorders?
Answer:
A genetic predisposition results from specific genetic variations that are often inherited from a parent. These genetic changes contribute to the development of a disease but do not directly cause it. Some people with a predisposing genetic variation will never get the disease while others will, even within the same family.
Explanation:
Hope this helps!
A car that weighs 10,500N drives at a constant speed of v = 49.0m/s for a distance of d =
5,110m. What is the net work done on the car during this drive?
a. Wnet = 0
b. Wnet = 1.28 · 106J
c. Wnet = 2.05J
d. Wnet = 5.37 · 107J
The work that has been done in driving the car is 5.37 * 107J.
What is the net work done?We know that the force that moves the car in the forward direction is the force that is provided by the engine of the car and as such we have to look at the work that is done as the product of the force and the distance that is covered here.
We know that;
Work done = force * distance
Work done = 10,500N * 5,110m
Work done = 5.37 * 107J
A total work that has a magnitude of 5.37 * 107J has been done by the car.
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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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PLEASE ANSWER MY OTHER QUESTION
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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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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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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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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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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a constant force fa is applied to an object of mass m, initially at rest. the object moves in the horizontal x-direction, and the force is applied in the same direction. after the force has been applied, the object has a speed of vf. which mathematical routines can be used to determine the time in which the force is applied to the object of mass m? select two answers.
t=Δpx/FA
Δt=Mvf/FA
F = ma, Newton's second rule of motion, may be used to calculate when a force is applied to an object of mass m.
The time may also be calculated using the equation of motion, vf = vi + at, where an is the acceleration and t is the time.
There are two mathematical procedures that may be used to figure out when a force is applied to an object with mass m: Newton's Second Law of Motion states that F = ma, where ma is the object's mass and an is its acceleration, and F is the applied force.
The acceleration may be found by rearranging the equation, which can then be used to calculate how long it will take the item to reach its final speed vf.
Kinematic Equation for Constant Acceleration: where vi is the starting velocity, t is the time, and an is the acceleration, vf = vi + at The time t may be calculated from the final velocity vf and the acceleration a by replacing the initial velocity with 0 (because the object is initially at rest).
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The complete question is:
A constant force FA is applied to an object of mass M, initially at rest. The object moves in the horizontal x-direction, and the force is applied in the same direction. After the force has been applied, the object has a speed of vf. Which mathematical routines can be used to determine the time in which the force is applied to the object of mass M? Select two answers.
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