The electric field at the center of uniformly charge ring is zero if the charge were placed at a point on the axis of the ring other than the center.
if the charge were placed at a point on the axis of the ring other than the center the electric field then there is no motion.
electric field - It is a force produced by any charge near its surrounding.
If the charge were placed at a point on the axis of the ring other than the center then there is no motion because at the center of the ring E.F ( Electric field is zero.
electric filter the point on the axis of uniformly charged ring depend on the distance of the point from the center of the ring.
.
E.F = 0 because each half cancels each other . the negative charge will be in equilibrium as every part of the ring is uniformly attracted by it.
The electric field at the center of uniformly charge ring is zero if the charge were placed at a point on the axis of the ring other than the center.
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the more matter an object has, the more ________ it has
Answer:
Mass
No matter where an object is in the universe, the more matter an object has, the more mass it will have as well.
starting from rest, a boulder rolls down a hill with constant acceleration and travels 4.00 m during the first second.
Starting from rest, a boulder rolls down a hill with constant acceleration and travels 4.00 m during the first second, 8 m first it is moving.
What is acceleration?Speed increase is the name we provide for any cycle where the speed changes. Since speed is a speed and a heading, there are simply two different ways for you to speed up: change your speed or shift your course — or change both. In mechanics, speed increase is the pace of progress of the speed of an item as for time. Speed increases are vector amounts (in that they have size and heading). The direction of an article's speed increase is given by the direction of the net power following up on that item. The extent of an articles. An item's typical speed increase throughout some undefined time frame is its adjustment of speed, partitioned by the span of the period. Numerically, Momentary speed increase, in the meantime, is the constraint of the typical speed increase over a microscopic time period. In the terms of analytics, immediate speed increase is the subsidiary of the speed vector regarding time.
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3. According to Newton's First Law, unless acted upon by an unbalanced force, an object at rest will
Answer:
An object at rest remains at rest
Explanation:
An object at rest stays at rest and an object in motion stays in motion with the same speed
PLEASE HELP HURRY a penny dropped out a window reaches a speed of 35 m/s. how far did it fall? initial velocity is 0 m/s and gravity is 9.8 m/s
Answer:
use v²=u²+2as
v=35
u=0
a=9.8
35²=0²+2*9.8s
1225/2=9.8s
s = 62.5m
answer=62.5m
Answer:
The height from which the coin fell:
H = 63 m
Explanation:
Given:
V₀ = 0 m/s
V = 35 m/s
g = 9.8 m/c²
___________
H - ?
The height from which the coin fell is calculated by the formula:
H = (V² - V₀²) / (2·g)
H = (35² - 0²) / (2·9.8) ≈ 63 m
A generator produces 24.0V when turning at 900 rev/min. What emf does it produce when turning at 500 rev/min?
The EMF that will be produced is 13.3 V
What does EMF mean ?EMF is an acronym for Electromotive Force which can be defined as the potential difference between the terminals of a cell when it is not delivering any current in an external circuit.
Given that a generator produces 24.0V when turning at 900 rev/min. What EMF does it produce when turning at 500 rev/min?
The EMF produced by a generator is proportional to the angular speed at which the generator turns. That is, E [tex]\alpha[/tex] w
E1 = 24 VE2 = ?W1 = 900 rev/minW2 = 500 rev/minE1/E2 = W1/W2
24/E2 = 900/500
Reciprocate both side.
E2/24 = 500/900
E2 = 0.5556 x 24
E2 = 13.33 V
Therefore, the EMF that will be produced is 13.3 V
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M A uniformly charged ring of radius 10.0cm has a total charge of 75.0μC . Find the electric field on the axis of the ring at from the center of the ring.(d) 100 cmat from the center of the ring
The electric field on the axis of the ring at the center of the ring at 100.00cm is 6.64×〖10〗^5N/C
Where a positive charge (q) is given, always note that the charges are pointing away from the electric field. The expression used to calculate the electric field away from the center of the ring is given by the equation below,
E=(K_e×b)/(r^3×Q)
Given that;
E = Electric field
Ke = electric field constant
a = radius of the charge q from the center of the circle
Q = electric charge = 75.0μC
The radius of the charge is a = 10cm= 0.1m
Distance from the electric field is 100cm = 1m
Ke is a constant given as 8.99×〖10〗^(9) Nm^2/c^2
E=(8.99×〖10〗^(9 ) Nm^2/c^2×1)/(〖0.1〗^3×75×〖10〗^(-6) )
E=6.64×〖10〗^5N/C
Therefore, the electric field away from the center of the ring is 6.64×〖10〗^5N/C away from the center of the ring.
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Assume that the variable name has the value 33. What is the value of name after the assignment name = name * 2 executes?
When the variable name has been assigned with the value 33, then the value of the name after the assignment name will be 66.
In the programming language the sign ' = ', is referred to as the assignment operator which is used to assign values to a particular variable, where the variable referred to the name of the memory location. As it is given the assignment name = name * 2, it means that name on the right-hand side is the assignment name, so whatever value it gets is replaced by twice that previous value according to the operation.
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What happens to the period when we change the​ amplitude​ (drop point) of the pendulum? question 2 options: the period does not change when we change the amplitude of the pendulum the period increases when we change the amplitude of the pendulum the period decreases when we change the amplitude of the pendulum
When we change the amplitude of the pendulum, the period does not change when we change the amplitude of the pendulum.
What is a pendulum?A pendulum is device that could be used to show oscillatory motion. The motion of the pendulum is that of an object that is tossed from one side to another. Typically, the simple pendulum is used in most school labs to show a motion that is regular and repeating.
Now we know that the amplitude refers to the point of highest displacement of the pendulum. The point of the highest displacement would depend on the angle at which the pendulum was displaced. The period has to do with the time that it takes for a particular pendulum to complete a full cycle.
We know that the period of the pendulum does not have an effect on the amplitude of the pendulum. It then follows that; When we change the amplitude of the pendulum, the period does not change when we change the amplitude of the pendulum.
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what does it mean by the term self demagnetization.
Answer:
Self demagnetization means when magnets' poles are free (suspended) for too long, their magnetic attraction becomes weaker and thus, self demagnetizes.
__ which is a acidic mass of pratially decomped organic matter
Answer: partially decomposed
Explanation:
a test charge of 1µc is placed halfway between a charge of 4.7µc and another of 7.7 µc separated by 10 cm. what is the magnitude of the force (in newtons) on the test charge? your answer should be a number with two decimal places, do not include the unit.
The magnitude of the force on the test charge is 1.
What is magnitude?Greatness is the quantitative worth of seismic energy. It is a particular worth having no connection with distance and heading of the focal point. The idea of extents traces all the way back to the Ancient Greeks, when stars overhead was sorted into six sizes. Extent alludes to the evaluating of the splendor of stars, the first being the most splendid. It has been moved to different issues since basically the seventeenth 100 years. In Physics, greatness is characterized as the most extreme degree of size and the course of an item. Greatness is utilized as a typical consider vector and scalar amounts. By definition, we realize that scalar amounts are those amounts that have size as it were.
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4. A student has 500 identical, rectangular sheets of paper. The mass of 1.0m² of the paper is 0.080 kg. (2) Using a metre rule, she measures the length of one sheet of paper and its width. The length is 0.300 m and the width is 0.210m. (1) Calculate the mass of one sheet of paper.
mass of one sheet of paper is 0.00504 kg or 5.04g.
First we will find the surface area of a single sheet of paper
surface area = length ×width
surface area =0.3×0.21
surface area =0.063m²
now 0.063m² is the surface area of a single sheet of paper .
so now we will find number of papers required to form 1m² of area
which will be given by dividing 1m² by surface area of a single sheet
i.e.
number of sheets required = 1/0.063m²
number of sheets required =15.87
mass of one sheet = mass of 1 m^2 of paper / number of times narrower
mass of one sheet = 0.080 kg / 15.873
mass of one sheet = 0.00504 kg
mass of one sheet of paper is 0.00504 kg or 5.04g.
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A toy snake was horizontally thrown from a height of 115 m. It lands 595 m away. What is the toy snake’s initial speed? (TWO decimal places!!)
Initial speed for the toys snake will be 122.88 m/s
Given,
Height from where it has been dropped, h = 115 m
Height where the toy lands, s = 595 m
As we know that [tex]t = \sqrt{\frac{2h}{g} }[/tex]
here 'g' is acceleration due tp gravity = 9.81 m/s²
therefore [tex]t = \sqrt{\frac{2 * 115}{9.81\\ }[/tex]
[tex]t = \sqrt{23.445}[/tex]
t = 4.842 s
Now, According to the Second Equation of the Motion which relates distance with the acceleration and time, we can say that
[tex]S = ut + \frac{1}{2}at^{2}[/tex]
The toy snake is thrown horizontally, therefore acceleration acting on the toy will be zero (0).
s = 595 m
t = 4.842 s
therefore [tex]u = \frac{s}{t}[/tex]
[tex]u = \frac{595}{4.842}[/tex]
u = 122.88 m/s
The initial speed of the toy snake will be 122.88 m/s.
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Why is the primary mirror in a telescope curved?
Answer:
The primary mirror in a telescope is curved so that it changes the path of the light hitting its surface.
Explanation:
Brainliest please
Answer:
the primary mirror in a telescope is curved because it changes the path of the light hitting its surface.
Why is the value of 'g' taken as negative when a body is thrown vertically upwards?
Answer:
See below
Explanation:
[tex]\displaystyle {\textsf {Acceleration is the rate of change of velocity with time:}}[/tex]
[tex]\boxed {g = \frac{dv}{dt}}[/tex]
[tex]\sf {where \;dv = \;change \;in \;velocity\; and \;dt \;the \;change \;in \;time}[/tex]
here g is the gravitational force
When a body is thrown vertically upward, it has an initial speed. As it keeps going up, it speed keeps decreasing until it becomes zero, then the ball starts dropping down
On the upward movement, the change in velocity is negative while the change in time is always positive
So dv/dt = a negative number divided by a positive number which is a negative value for g in the equation for g
in a parallel circuit with four 6-ohm resistors across a 24-volt battery, what is the total voltage across resistor-three (vr3) in the circuit?
The total voltage across resistor- three in the circuit is 24V
acc to ohm's law
the electrical current flowing through any circuit is directly proportional to the voltage applied across it.
i.e. V ∝ I
V= IR
I = current
v = voltage
R = resistance (proportionality constant)
R in parallel
(1/Rp = 1/R1 + 1/R2 + 1/R3 +1/R4)
on rearranging the above equation (V=IR)
I = V/R
I = 24/6 = 4A
I is the current in each resistor.
in the parallel circuit voltage across the component is same and total current is the sum of the current flowing through each component.
the total voltage across resistor-three (vr3) in the circuit is
V= IR₃ = 4×6= 24 volt
the total voltage across the resistor three (vr3) in the circuit is 24volt.
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when charge 2 is 3.0 m away from charge 1, the strength of the electric force on charge 2 by charge 1 is 0.80 n. if instead, charge 2 were 6.0 m away from charge 1, what would be the strength of the electric force on charge 2 by charge 1 in that case?
The strength of the electrostatic force is inversely proportional to the square of the distance
between the charges. If the distance is doubled, the force is 1/4.
The new force is 0.80/4 = 0.20 N.
What is electrostatic force?Electrostatic force is the attractive or repulsive force that exists between two charged particles. Also known as Coulomb interaction or Coulomb force. For example, the electrostatic forces between the protons and electrons of an atom are responsible for the stability of the atom.
The force acting along a line joining two point charges is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them.
[tex]F=K |\frac{q_{1} q_{2}}{r^{2} } |[/tex]
In the above formula, k is arbitrary and can be chosen to be any positive value. Since k is a constant, I chose to give the value of k as follows:
Therefore, with q₁ and q₂ values of 1 and r = 1 (two charges with 1 Coulomb charge each at a distance of 1 m), we get F = [tex]9 \times 10^9[/tex] N. In the above equation, ε₀ is given as the permittivity of free space and its value in SI units is [tex]8.854\times10^{-12} C^{2} N^{-1} m^{-2}[/tex].
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Which conclusion can be made from Gay-Lussac’s law?
For a constant volume, the pressure of a gas is directly proportional to the temperature in Kelvin is the conclusion which can be made from Gay-Lussac’s law and is denoted as option D.
What is Pressure?This is referred to as the perpendicular force applied on a body per unit area and the unit is Pascal.
Gay-Lussac was a scientist who discovered through numerous experiments and observations that at a constant volume, the pressure of a gas is directly proportional to the temperature in Kelvin.
This is denoted as the following equation below:
P ∝ T
P = kT where p is pressure, k is constant and t is temperature.
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The options are:
At a constant temperature, the pressure of a gas is directly proportional to the volume.At a constant temperature, the pressure of a gas is indirectly proportional to the volumeFor a constant volume, the pressure of a gas is indirectly proportional to the temperature in KelvinFor a constant volume, the pressure of a gas is directly proportional to the temperature in Kelvin6
Which of the following is NOT something that would lead astronomers to believe a black hole might be close-by in space?
A.
an object that has gas and dust appear to be funneling around an object
B.
an object that has spinning particles around an emptiness in space that emit x-rays
OC.
an object that has height and depth but no width
OD. an object in space with highly elevated temperatures
Reset
Next
The option that is NOT something that would lead astronomers to believe a black hole might be close-by in space is option C: an object that has height and depth but no width.
What is a black hole that can be close-by in space?There is said to be a wandering black hole that can be seen in an approximately 5,000 light-years away from earth.
It is one that can be seen in the Carina-Sagittarius spiral aspect of our galaxy.
Note that, its discovery gives room for astronomers to state that the nearest form of isolated stellar-mass black hole to Earth can be as close as about 80 light-years away.
Therefore, The option that is NOT something that would lead astronomers to believe a black hole might be close-by in space is option C: an object that has height and depth but no width because it has no width, no depth and no height.
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1: The ball is on the 50 yard line. The ball travels west 5 yards before it's
handed off and ran forward (EAST) 15 yards where they're tackled.
What is the distance the ball traveled?
Your answer
This is a required question
* 1 point
The ball is on the 50-yard line. The ball travels west 5 yards. The distance ball traveled was 60 yards.
WHAT IS DIFFERENCE BETWEEN DISTANCE AND DISPLACEMENT ?Distance can be defined as the length of any path between any two locations. Displacement is the direct distance between any two points when calculated via the shortest path between them. When computing distance, the direction is disregarded. The displacement computation takes the direction into consideration.
CALCULATIONThe 5 yard gain would bring the ball to the 45 yard line (ball was tackled at the 50 yard line, so 50 - 5 = 45), but the 15 yards lost due to the tackle and push back would bring ball to the 60 yard line (45 + 15 = 60).
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When the speaker is lowered into water, does the frequency of the sound increase, decrease, or remain the same?.
Answer: it remains the same
Explanation:
the frequency of a sound wave cannot change as it crosses the water-air boundary
A flat surface of area 3.20m² is rotated in a uniform electric field of magnitude E=6.20 × 10⁵N . m²/C . Determine the electric flux through this area (a) when the electric field is perpendicular to the surface
The electric flux is 0 N.m²/C.
We need to know about electric flux to solve this problem. Electric flux is the measure of the electric field through a given surface, although an electric field in itself cannot flow. It can be determined as
Φ = E . S = E . Scosθ
where Φ is electric flux, E is electric field, θ is angle of surface and S is surface area.
From the question above, we know that
E = 6.20 × 10⁵ N/C
θ = 90⁰
S = 3.20 m²
By substituting the following parameter, we get
Φ = E . Scosθ
Φ = 6.20 × 10⁵ . 3.20cos(90⁰)
Φ = 0 N.m²/C
Hence, the electric flux is 0 N.m²/C
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2. A bookshelf is 200 cm and 40 N weight, with supports at its ends (X and Y).
a. A book weighing 12 N is placed in the middle of the shelf. What are the upward forces at X and Y?
b. The book is moved so that it is 40 cm from B. Use moments to calculate the forces at X and Y.
Answer:
force =m×a=40=200×a=a=200-40=160m/second sguaredExplanation:
always accerelation it measures in m/second sguared.
Two waves are traveling in the same direction along a stretched string. The waves are 90.0° out of phase. Each wave has an amplitude of 4.00cm . Find the amplitude of the resultant wave.
The resultant wave is calculated by adding the magnitude of both the waves traveling in same direction.The resulting wave is of the amplitude 5.66cm or 5.66 x 10⁻²m
Let us take A be the amplitude of both the waves that are travelling
We are given that waves are moving 90 degrees out of the phase
Let the equations of the waves be
y1= Asin(ωt-kx)
y2=Acos(ωt-kx)
Since the waves are ninety degrees travelling out of phase ,
The resultant wave is
y=y1+y2
= Asin(ωt-kx) +Acos(ωt-kx)
=√2 A sin(ωt-kx+π/4)
So, we can conclude resultant wave is √2A = √2 *4 =5.66cm
=5.66* 10 ⁻²m
Hence the resultant wave is 5.66cm
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The magnetic field 40.0 cm away from a long, straight wire carrying current 2.00 A is 1.00μT.(c) A t what distance is it one-tenth as large?
The distance should be 1.27 m from the wire in order to get magnetic field one-tenth as large .
The magnitude and direction of the magnetic field due to a straight wire carrying current can be calculated using the previously mentioned Biot-Savart law. Let "I" be the current flowing in a straight line and "r" be the distance. Then the magnetic field produced by the wire at that particular point is given by [tex]B=\frac{u_0I}{2\pi r}[/tex] ..(1)Since the wire is assumed to be very long, the magnitude of the magnetic field depends on the distance of the point from the wire rather than the position along the wire.Let [tex]B_1[/tex] be inside the plane and [tex]B_2[/tex] be outside the plane . It is required to calculate magnetic field at point A .
Direction of magnetic field at A is calculated using Right hand rule.
[tex]B_{net}=B_2-B_1\\\\B_{net}=\frac{u_0I}{2\pi (0.4-\frac{3\times10^{-3}}{2}) } -\frac{u_0I}{(0.4+\frac{3\times10^{-3}}{2})} \\\B_{net}=7.5\times 10^{-9}T[/tex]....(1)
Let a distance "R" from the wire so that the magnetic field is [tex]\frac{B_{net}}{10}[/tex]
Using the magnetic field equation , we get
[tex]\frac{B_{net}}{10} =B_2-B_1\\\\\\frac{7.5\times 10^{-9}}{10} =\frac{u_0I}{2\pi (R-\frac{3\times10^{-3}}{2}) } -\frac{u_0I}{(R+\frac{3\times10^{-3}}{2})} \\\\\R^2-2.25\TIMES10^{-6}=1.6\\\\R=1.27m[/tex]
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Two boxes are at rest…
The pair of forces include;
Normal force of mass 1 on mass 2(F₁₂) and the reaction of mass 2 on mass 1 (F₂₁)
Normal force of mass 2 on the table A (F₂A) and the reaction of the table on the mass 2 (FA2).
What is normal force?
The normal force is an everyday force that is felt when a surface pushes against an object that is placed on that surface.
Normal force acts in opposite direction to the weight of the object.
Based on Newton's third law of motion, for every action, there is an equal and opposite reaction.
The downward force exerted on the table, is equal to the reaction of the table on the masses.
The pair of forces acting on the masses stacked together include the following;
Normal force of mass 1 on mass 2(F₁₂) and the reaction of mass 2 on mass 1 (F₂₁)Normal force of mass 2 on the table A (F₂A) and the reaction of the table on the mass 2 (FA2).Learn more about normal forces here: https://brainly.com/question/14486416
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Doug, who runs track for his high school, was challenged to a race by his younger brother, matt. Matt started running first, and doug didn’t start running until matt had finished a quarter-mile lap on the school track. Doug passed matt as they both finished their sixth lap. If both boys ran at a constant speed, with doug running 2 miles an hour faster than matt, what was matt’s speed?.
The speed at which Matt ran is 10 miles per hour. Therefore, option (B) is the correct answer.
Matt runs at a speed of x miles per hour, whereas Doug does it at a speed of 2 miles per hour faster. Doug's speed will thereafter be equal to x + 2 miles per hour. Since there are four quarter-mile laps, Doug completes 1.5 miles in the time it requires Matt to complete 1.25 miles.
Matt:
Speed = x
Distance = 1.25
Time = [tex]\frac{1.25}{x}[/tex]
Doug:
Speed = x+2
Distance = 1.5
Time = [tex]\frac{1.5}{x+2}[/tex]
Making an equation by putting the two timings on the chart equal to one another and then solving for x will reveal that the two boys took the same amount of time from the moment Doug started.
[tex]\frac{1.5}{x+2}[/tex] = [tex]\frac{1.25}{x}[/tex]
1.5x = 1.25 (x+2)
0.25x = 2.5
x = 10
Therefore, Matt’s speed is 10 miles per hour.
The complete question is:
Doug, who runs track for his high school, was challenged to a race by his younger brother, Matt. Matt started running first, and Doug didn’t start running until Matt had finished a quarter-mile lap on the school track. Doug passed matt as they both finished their sixth lap. If both boys ran at a constant speed, with Doug running 2 miles an hour faster than Matt, what was Matt’s speed?
A. 10.5 miles per hour
B. 10 miles per hour
C. 9 miles per hour
D. 8 miles per hour
E. 7.5 miles per hour
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I need help with this ASAP... Please and thank you
Answer: 1) D & 2) H
Explanation:
D. Tropical rainy &
H. Long-term average temperature and precipitation for the area
Perform the calculation and report your answer using sig figs. 3.42 + 4 + 5.2 + 12
Answer:
3.42 + 4 + 5.2 + 12= 24.62
So since we have 24.62 it will be 4 sig figs
Explanation:
In the electron-transport chain, as electrons move along a series of carriers, they release energy that is used to do what?.
In the electron-transport chain, as electrons move along a series of carriers, they release energy that is used to pump protons across a membrane.
In plants and animals, the electron transport chain refers to the consistent streaming of electrons between cells where electron donors donate electrons and electron acceptors accept electrons through a redox reaction.
An electrochemical gradient is created as a result of the redox reaction and ATPs are produced.
In this chain, there are 4 major protein complexes that are transferred across the plasma membrane.
This cellular process occurs in mitochondria and is responsible for photosynthesis and cellular respiration. Its primary steps are Krebs' cycle, glycolysis, oxidative phosphorylation, and pyruvate oxidation.
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