Answer:
Please find the answer in the explanation
Explanation:
Take the regular compass and hold it so the case is vertical. Now use it to investigate the direction of the coil’s magnetic field at locations other than the central axis.
What happens as you move away from the center axis toward the coil? The direction of the magnetic compass needle will move in an opposite direction since the direction of the induced voltage is reversed.
What happens above the coil?
the needle on the magnetic compass will be deflected. Since compasses work by pointing along magnetic field lines
Outside the coil? The magnetic compass needle will experience no deflection. Since there is no induced voltage or current.
Below the coil?
The needle will move in an opposite direction.
Which pair of objects would be most strongly attracted to each other?
A. A positively charged particle and a negatively charged particle
B. Two positively charged particles
C. Two negatively charged particles
D. A negatively charged particle and a neutral particle
Answer:
Its A
Explanation:
Just did the quiz
The pair of objects that should be strongly attracted is option A. A positively charged particle and a negatively charged particle.
Pair of objects that are most strongly attracted:When there is a positively charged particle & the negatively charged particle so due to this it should be strongly attracted. Also, when there are two positively charged particles, or a negative one or a include negative one or neutral one so it should not be strongly attracted. Therefore, the option A is correct.
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An object is dropped from a high building. It reaches the ground in two seconds.
a) Find the distance travelled by the object.
b) Find the speed of the object, when it hits the ground.
Answer:
a: 19.6 meters b: 9.8meters per second
Explanation:
speed and object falls is 9.8 meters per second. 9.8 meter times 2 is 19.8 meters.
1. Which statement is true of culture?
O Culture do not change or evolve
O Culture refers to the degree in which resources are used to address for problems.
O Culture has little affect on a persons life
Food can reflect culture
Why do you feel that you are being thrown upward out of your seat when going over an arced hump on a roller coaster
Answer: The options are not given.
Here are the options.
a) There is an additional force lifting up on you.
(b) At the top you continue going straight and the seat moves out from under you.
(c) You press on the seat less than when the coaster is at rest.Thus the seat presses less on you. (
d) Both b and c are correct.
(e) a, b, and c are correct.
The correct option Is D.
B.At the top you continue going straight and the seat moves out from under you. C.At the same time, you press on the seat less than when the coaster is at rest because the normal force expirienced will be less.
Explanation:
At the top you continue going straight and the seat moves out from under you.At the same time, you press on the seat less than when the coaster is at rest because the normal force expirienced will be less because it is as a result of a phenomenon called Weightlessness. This occur when there is no force or little force is acting on your body. At the top you continue going straight and the seat moves out from under you because there is no force acting on your body and when the body is in free fall i.e acceleration due to gravity , the person is not supported by any thing at.
That is the scenarion that occur...
A 30.0-kgkg box is being pulled across a carpeted floor by a horizontal force of 230 NN , against a friction force of 210 NN . What is the acceleration of the box?
Answer:
The acceleration of the box is 0.67 m/s²
Explanation:
Given that,
Mass of box = 30.0 kg
Horizontal force = 230 N
Friction force = 210 N
We need to calculate the acceleration of the box
Using balance equation
[tex]F-f_{k}=ma[/tex]
[tex]a=\dfrac{F-f_{k}}{m}[/tex]
Where, F = horizontal force
[tex]f_{k}[/tex] =frictional force
m= mass of box
a = acceleration
Put the value into the formula
[tex]a=\dfrac{230-210}{30}[/tex]
[tex]a=0.67\ m/s^2[/tex]
Hence, The acceleration of the box is 0.67 m/s²
in the figure shown if angle i increases slightly angle r will
Answer:
we need the image to do so.
Explanation:
sorry
Calculate the work WC done by the gas during the isothermal expansion. Express WC in terms of p0, V0, and Rv.
Complete Question
The complete question is shown on the first and second uploaded image
Answer:
The expression is [tex]W_c = P_o V_o ln (R_v)[/tex]
Explanation:
Generally smallest workdone done by a gas is mathematically represented as
[tex]dW = PdV[/tex]
Generally for an isothermal process
[tex]PV = nRT = constant [/tex]
=> [tex]P = \frac{nRT}{V}[/tex]
Generally the total workdone is mathematically represented as
[tex]W_c = \int\limits^{v_f}_{V_o} {\frac{nRT}{V} } \, dV[/tex]
=> [tex]W_c = nRT \int\limits^{V_f}_{V_o} {\frac{1}{V} } \, dV[/tex]
=> [tex]nRT [lnV] | \left \ {V_f}} \atop {V_o}} \right.[/tex]
=> [tex]W_c = nRT [ln(V_f) - ln(V_o)][/tex]
=> [tex]W_c = nRT ln \frac{V_f}{V_o}[/tex]
From the question [tex]\frac{V_f}{V_o } = R_v[/tex]
=> [tex]W_c = P Vln (R_v)[/tex]
at initial state
[tex]W_c = P_o V_o ln (R_v)[/tex]
Before digital filmmaking, what tool was used to control the speed of movement on the screen after filming?
Answer:
The appropriate response is "Optical printer ".
Explanation:
A photographic printer used mostly for optical aberrations, comprised simply of either a camera that captures the frame to expand, minimize, deform, respectively. through magnifying lenses. A projector that always, as distinct from some kind of touch printer, transferred the image to something like the printing supply.which two types of information are found in an elements box in the periodic table
Answer:
Each box represents an element and contains its atomic number, symbol, average atomic mass, and (sometimes) name.
Explanation:
Answer:
An element's period and group
Converting compound units
You would like to know whether silicon will float in mercury and you know that can determine this based on their densities. Unfortunately, you have the density of mercury in units of kilogram/meter3 and the density of silicon in other units: 2.33 gram/centimeter3. You decide to convert the density of silicon into units of kilogram/meter3 to perform the comparison. By which combination of conversion factors will you multiply 2.33 gram/centimeter3 to perform the unit conversion?
Answer:
Dividing the silicon density by 1000 and then multiply it by 1000000.
Explanation:
A kilogram equals 1000 grams and a cubic meter equals 1000000 cubic centimeters. Hence, we must divide the silicon density by 1000 and then multiply itby 1000000 to convert the value into kilograms per cubic centimeter. That is:
[tex]x = 2.33\,\frac{g}{cm^{3}}\times \frac{1\,kg}{1000\,g}\times \frac{1000000\,cm^{3}}{1\,m^{3}}[/tex]
[tex]x = 2330\,\frac{kg}{m^{3}}[/tex]
In a nutshell, we must multiply the density of silicon by 1000 to obtains its value in kilograms per cubic meter.
A certain superconducting magnet in the form of a solenoid of length 0.300 m can generate a magnetic field of 8.90 T in its core when its coils carry a current of 95 A. Find the number of turns in the solenoid.
Answer:
The number of turns in the solenoid is 22366.
Explanation:
The number of turns in the solenoid can be found using the following equation:
[tex] B = \mu_{0} I\frac{N}{L} [/tex]
Where:
B: is the magnetic field = 8.90 T
L: is the solenoid's length = 0.300 m
N: is the number of turns =?
I: is the current = 95 A
μ₀: is the magnetic constant = 4π×10⁻⁷ H/m
By solving equation (1) for N we have:
[tex] N = \frac{BL}{\mu_{0} I} = \frac{8.90 T*0.300 m}{4\pi \cdot 10^{-7} H/m*95 A} = 22366 turns [/tex]
Therefore, the number of turns in the solenoid is 22366.
I hope it helps you!
The horizontal surface on which the block slides is frictionless. The speed of the block before it touches the spring is 6.0 m/s. How fast is the block moving at the instant the spring has been compressed 15 cm
Answer:
The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.
Explanation:
The spring constant is 2000 newtons per meter. Let consider the spring-block system, from Principle of Energy Conservation we can represent it by the following model:
[tex]U_{k,1}+K_{1} = U_{k,2}+K_{2}[/tex]
[tex]K_{2} = K_{1}+(U_{k,1}-U_{k,2})[/tex] (Eq. 1)
Where:
[tex]K_{1}[/tex], [tex]K_{2}[/tex] - Initial and final kinetic energies of the block, measured in joules.
[tex]U_{k,1}[/tex], [tex]U_{k,2}[/tex] - Initial and final elastic potential energy, measured in joules.
And we expand the equation above by definitions of elastic potential energy and kinetic energy:
[tex]\frac{1}{2}\cdot m \cdot v_{2}^{2} = \frac{1}{2}\cdot m\cdot v_{1}^{2} + \frac{1}{2}\cdot k\cdot (x_{1}^{2}-x_{2}^{2})[/tex]
[tex]v_{2} = \sqrt{v_{1}^{2}+\frac{k}{m}\cdot (x_{1}^{2}-x_{2}^{2}) }[/tex] (Eq. 1b)
Where:
[tex]m[/tex] - Mass of the block, measured in kilograms.
[tex]k[/tex] - Spring constant, measured in newtons per meter.
[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Initial and final velocities of the block, measured in meters per second.
[tex]x_{1}[/tex], [tex]x_{2}[/tex] - Initial and final positions of spring, measured in meters.
If we know that [tex]v_{1} = 6\,\frac{m}{s}[/tex], [tex]k = 2000\,\frac{N}{m}[/tex], [tex]m = 2\,kg[/tex], [tex]x_{1} = 0\,m[/tex] and [tex]x_{2} = 0.15\,m[/tex], the final speed of the block moving at the instant the spring has been compressed is:
[tex]v_{2} = \sqrt{\left(6\,\frac{m}{s} \right)^{2}+\left(\frac{2000\,\frac{N}{m} }{2\,kg} \right)\cdot [(0\,m)^{2}-(0.15\,m)^{2}]}[/tex]
[tex]v_{2}\approx 3.674\,\frac{m}{s}[/tex]
The final speed of the block moving at the instant the spring has been compressed is approximately 3.674 meters per second.
What do you mean is a variable velocity and uniform velocity
Answer:
Uniformly accelerated rectilinear motion (MRUA), also known as uniformly varied rectilinear motion (MRUV), is one in which a mobile moves along a straight path being subjected to a constant acceleration.
Explanation:
A car traveling at 32.4 m/s skids to a stop in 4.55 s. Determine the skidding distance of the car (assume uniform acceleration).
Answer:
x = 73.71 [m]
Explanation:
In order to solve this problem we must use two formulas of kinematics. It is important to make it clear that these formulate are for uniformly accelerated motion, i.e. with constant acceleration.
[tex]v_{f }= v_{i}-(a*t)[/tex]
where:
Vf = fnal velocity = 0
Vi = initial velocity = 32.4 [m/s]
t = time = 4,55 [s]
a = acceleration or desacceleration [m/s^2]
0 = 32.4 - (a*4.55)
a = 7.12 [m/s^2]
Note: it is important to clarify that the negative sign in the above equation is because the car stops and decreases its speed to zero, thus its final velocity is equal to zero.
Now using the following equation:
[tex]x = x_{o} + (v_{i}*t)-(\frac{1}{2} )*a*t^{2}[/tex]
where:
xo = initial distance = 0
x = final distance [m]
Therefore we have:
x = 0 + (32.4*4.55) - (0.5*7.12*4.55^2)
x = 73.71 [m]
A car with a mass of 1500kg is traveling at a speed of 30m/s. What force must be applied to stop the car in 3 seconds?
Answer:
The answer is 15,000 NExplanation:
To find the force given the mass , velocity and time can be found by using the formula
[tex]f = \frac{m \times v}{t} \\ [/tex]
where
m is the mass
v is the velocity
t is the time
From the question
m = 1500 kg
v = 30 m/s
t = 3 s
We have
[tex]f = \frac{1500 \times 30}{3} = \frac{45000}{3} \\ [/tex]
We have the final answer as
15,000 NHope this helps you
HELP PLEASE!!!
If we have a sample of silicon (Si) atoms that has 14 protons, 14 electrons, and 18 neutrons
What is the name of this specific silicon isotope?
si-14
si-32
si-46
si-153
Answer:
It is si-32
Explanation:
Answer:
silicon-32
Explanation:
just took the quiz and got it right
Calculate the ratio of the mechanical energy at B and mechanical energy at a (eb,ea) and (ec,ea). What do these ratios tell you about the conservation of energy?
A) is the mechanical conserved between a and b? explain
B) is the mechanical energy conserved between b and c ?explain
Answer:
Yes at A the mechanical energy is conserved.
Yes at B the part of mechanical energy is conserved potential energy and kinetic energy and some is lost as frictional force.
Explanation:
Ratio = Eb/ Ea= 1058.3 J/2940 J= 0.3599
Ratio = Ec/ Eb= 0J/ 1058.3 J= 0
At point A the skater is at rest or it is the starting point and the whole energy is due to the position of the skater i.e= mgh = 50 *9.8*6= 2940 J
Since there's no movement there is no Kinetic energy = 0 J
Yes at A the mechanical energy is conserved.
At point B the skater has traveled for some of the distance . It has potential energy and kinetic energy.
Yes at B the part of mechanical energy is conserved as potential energy and kinetic energy.
The total Mechanical energy = 1058.3 J
At point B Total Mechanical energy = PE+ KE
1058.3J = 980 J + 78.3 J
1058.3 J = mgh + 1/2mv²
= 50*2*9.8 + 1/2 *50*(8.85)²
= 980 J + 78.3 J
As the total energy of the system must remain the same some of the mechanical energy is lost as frictional force at point B .
2940 J-1058.3 J= 1881.7
At Point C the skater has arrived at the end point and the height , speed, PE, KE and ME all are zero.
(a) The ratio of the mechanical energy at B and mechanical energy at A is 0.36.
(b) The ratio of the mechanical energy at C and mechanical energy at A is 0.
(c) mechanical energy is conserved between a and b.
(d) mechanical energy is not conserved between b and c.
The given parameters;
mechanical energy at A, [tex]E_a = 2,940 \ J[/tex]mechanical energy at B, [tex]E_b =1,058.3 \ J[/tex]mechanical energy at C, [tex]E_c = 0[/tex]The ratio of the mechanical energy at B and mechanical energy at A;
[tex]ratio = \frac{E_b}{E_a} = \frac{1058.3}{2940} = 0.36[/tex]
The ratio of the mechanical energy at C and mechanical energy at A;
[tex]ratio = \frac{E_c}{E_a} = \frac{0}{2940} = 0[/tex]
The change mechanical energy between A and B from the given position;
[tex]\Delta E = mg(h_b - h_a) - \frac{1}{2}m(v_b^2 - v_a^2)\\\\ \Delta E = 50\times 9.8(2-6) \ - \ \frac{1}{2} \times 50(8.85^2 - 0)\\\\\Delta E =- 1960 + 1960\\\\\Delta E = 0 \ J[/tex]
Thus, we can conclude that mechanical energy is conserved between a and b.
The change mechanical energy between A and B from the given position;
[tex]\Delta E = mg(h_c - h_b) - \frac{1}{2}m(v_c^2 - v_b^2)\\\\ \Delta E = 50\times 9.8(0-2) \ - \ \frac{1}{2} \times 50(0^2 - 8.85^2)\\\\\Delta E = -980 + 1960 \\\\\Delta E = 980 \ J[/tex]
Thus, we can conclude that mechanical energy is not conserved between b and c.
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One student runs with a velocity of +10 m/s while a second student runs with a velocity of –10 m/s. Which student has the faster velocity? Why?
Answer:
The one with the faster velocity is the one with a velocity of -10m/s
A car traveling at 27 m/s slams on its brakes to come to a stop. It decelerates at a rate of 8 m/s2 . What is the stopping distance of the car?
v² - u² = 2 a ∆x
where u = initial velocity (27 m/s), v = final velocity (0), a = acceleration (-8 m/s², taken to be negative because we take direction of movement to be positive), and ∆x = stopping distance.
So
0² - (27 m/s)² = 2 (-8 m/s²) ∆x
∆x = (27 m/s)² / (16 m/s²)
∆x ≈ 45.6 m
The stopping distance of car achieved during the braking is of 45.56 m.
Given data:
The initial speed of car is, u = 27 m/s.
The final speed of car is, v = 0 m/s. (Because car comes to stop finally)
The magnitude of deacceleration is, [tex]a = 8\;\rm m/s^{2}[/tex].
In order to find the stopping distance of the car, we need to use the third kinematic equation of motion. Third kinematic equation of motion is the relation between the initial speed, final speed, acceleration and distance covered.
Therefore,
[tex]v^{2}=u^{2}+2(-a)s[/tex]
Here, s is the stopping distance.
Solving as,
[tex]0^{2}=27^{2}+2(-8)s\\\\s = 45.56 \;\rm m[/tex]
Thus, we can conclude that the stopping distance of car achieved during the braking is of 45.56 m.
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A person has a mass of 1000g and an acceleration of 20 m/s/s. What is the force on the person
Answer:
20000
Explanation:
Newtons Second law states that the force acting on an object is equal to its mass times its acceleration, f=ma. To solve for force, plug in your values for m and a, and then solve. f = (1000)*(20) = 20000
Peter is running laps around a circular track with a diameter of 100 meters. If it takes Peter 12 minutes to run 4 laps, how quickly is he running (in meters per second)?
Answer:
v = 1.74 m/s
Explanation:
Given that,
Diameter of a circular track, d = 100 m
Distance covered for the 4 laps,
[tex]D=4\pi d\\\\D=4\pi \times 100\\\\D=1256.63\ m[/tex]
Time, t = 12 minutes = 720 s
We need to find the velocity of the peter. It can be calculated as follows :
[tex]v=\dfrac{D}{t}\\\\v=\dfrac{1256.63\ m}{720\ s}\\\\v=1.74\ m/s[/tex]
So, the speed is running with a velocity of 1.74 m/s.
Peter is running at 1.7453 m/sec.
Given to us,
Diameter of the circular track, D = 100 meters,
Number of laps Peter run, L = 4 laps,
Time taken by Peter, t = 12 minutes,
1 lap = circumference of the circle,
4 laps = 4 x circumference of the circle,
As we know, the circumference of a circle is given by πD.
So, 4 laps = 4 x circumference of the circle,
[tex]\begin{aligned}4 laps &= 4\times \pi \times D\\&= 4 \times \pi \times 100\\& = 1,256.6370\ meters\\\end{aligned}[/tex]
Also, we know that 1 minute has 60 sec.
so, 4 minutes = (4 x 60) seconds
Further, speed is given [tex]\bold{(\dfrac{Distance}{Time} )}[/tex]
Thus,
[tex]\begin{aligned}speed &= \dfrac{Distance\ coverd\ by\ Peter}{Time\ taken\ by\ Peter}\\&=\dfrac{1,256.6370}{12\times 60}\\&=1.7453\ m/sec \end{aligned}[/tex]
Hence, Peter is running at 1.7453 m/sec.
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A light wave in space is described by the general function: u(r, t) = 1 (2π) 3/2 Z dk A(k) e i(k·r−ωt) . (1) (a) Find the particular solution knowing that at t = 0 the wave is given by
Answer:
hello your question is incomplete attached below is the complete question
Answer : attached below
Explanation:
A) finding the particular solution at t = 0
attached below is the detailed solution of the particular solution knowing that t = 0
PLS HELP 3. Which graph best represents the relationship between acceleration due to gravity and mass for
objects near the surface of Earth? [Neglect air resistance.)
Acceleration
This question involves the concepts of mass, acceleration due to gravity, weight, Newton's Gravitational Law, gravitational force, and graphs.
Graph "A" re[resents the best relationship between acceleration due to gravity, and mass for objects.
The relationship between the mass of earth and the acceleration due to gravity can be found by equating the weight of the object and the gravitational force, from Newton's Gravitational Law on it:
[tex]Weigth = Gravitational\ Force\\\\mg = \frac{GmM}{r^2}\\\\g = \frac{GM}{r^2} ---------- eqn(1)[/tex]
where,
g = acceleration due to gravity
G = universal gravitational constant
M = mass of Earth
r = radius of Earth
Hence, it is clear from equation (1), that mass of the Earth and the acceleration due to gravity have a direct relationship with each other. Therefore the graph between them will be a straight line, which is Graph A.
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The attached picture illustrates Newton's Law of Gravitation.
There are 5.5 L of a gas present at -38.0 C. What is the temperature if the volume of the gas has changed to 1.30 L?
Answer:
We are given:
V1 = 5.5L T1 = -38 C or 235 k
V2 = 1.3L T2 = T
From the gas equation:
PV = nRT
Since the pressure (P) , number of moles (n) and the universal gas constant (R) are constants, we can write the same equation as:
V / T = k (where k is a constant)
so a bit more insight, since the values noted above are constant, when multiplied by each other, they will provide us with a constant number irrespective of the value of the variables
Changing the variables for the first case:
V1 / T1 = k (where k is the same constant) ----------------(1)
Similarly,
V2 / T2 = k (again, k has the same value)----------------(2)
From (1) and (2):
k is the common value
V1 / T1 = V2 / T2
Replacing the variables
5.5 / 235 = 1.3 / T
T = 1.3 * 235 / 5.5
T = 55.54 k
Therefore, at 55.54 K the gas will have a volume of 1.3L
How long would it take you to walk 3,962 km from New York to Los
Angeles?
Answer:
913 hours ur welcome :)
Two particles are separated by 0.38 m and have charges of -6.25x 10 C and 2.91 x 10 C. Use Coulomb's law to predict the force between the particles if the distance is doubled. The equation for Coulomb's law is Fe = g, and the constant, k, equals 9.00 x 10° Nm/C A. -1.13 x 10-6 N OB. 1.13x 106N O C. 2.83 x 10-7 N OD.-2.83x 10N sUBMIT
Answer:
I do not understand what you are asking
If an object is moving with a constant velocity to the right, what direction is the net force.
Group of answer choices
A.To the right
B.To the left
C.Net force is 0
D.Not enough information
Answer:
At constant velocity, his weight equals the force of friction. In other words, there is no net force. If however, he loosens his grip and decreases the friction force, he will accelerate downward.
Explanation:
Compare and contrast the CONFLICT (choose one) in the short story you read with the elements appearing in The Watsons Go to Birmingham—1963. Explain how they are similar or different in a few sentences.
Answer:
they were in two places in flint and Birmingham and in Birmingham it is hot but flint of cold the Simi is they both have Sunday school for Joetta
Explanation:
use in your own words teachers know when your not trust me.
what is the volume of an object that has a density of 65g/cm3 and a mass of 130g.
Density ρ is mass m per unit volume v, or
ρ = m / v
Solving for v gives
v = m / ρ
So the given object has a volume of
v = (130 g) / (65 g/cm³) = 2 cm³
Question 4
Which of the following is unique for any given element?
O the mass of a neutron
o the number of neutrons
o the charge on the electons
O the number of protons