Answer:
The maximum torque on the loop is 395.80 N.m.
Explanation:
Given;
number of turns of the wire, N = 150 turns
length of the square loop, L = 18.0 cm = 0.18 m
current in the wire, I = 50.9 A
Magnetic field, B = 1.6 T
Maximum torque on the loop is given by;
τ = NIAB
τ = (150)(50.9)(0.18²)(1.6)
τ = 395.80 N.m
Therefore, the maximum torque on the loop is 395.80 N.m.
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...
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.
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
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:
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.HELP PLS!!
In a full/angled projectile, Vty is equal to the inverse of viy
O True
O False
A designer is creating an obstacle for an obstacle course where a person starts on a moveable platform of height H from the ground. The person grabs a rope of length L and swings downward. At the instant the rope is vertical, the person lets go of the rope and attempts to reach the far side of a water-filled moat. The left side of the moat is directly below the position where the person will let go of the rope. The designer runs several tests in which the rope has different lengths and moves the platform such that the rope is always initially horizontal. The designer notices that the person cannot land on the other side if the length L is very short. The designer also notices that the person also cannot land on the other side if the length L is very close to the height H.
Assume the size of the person is much smaller than the lengths L and H. Let D represent the horizontal distance from below the release point to where the person lands.
Required:
a. Why does the person land in the moat if the rope's length is very short?
b. Why does the person land in the moat if the length is nearly the same as the height of the platform?
Answer:
* when L → H chord too long
in this case we see that the speed to cross the well grows a lot (it goes towards infinity) therefore we do not have enough speed in the movement to cross
* when L → 0 very short string
the speed of the platform is very small, so we do not have the minimum required value
vox = √ (g / (2 (H)) D
Explanation:
For this exercise we are going to solve it using conservation of energy to find the velocity of the body and the launch of projectiles to find the velocity to cross the well.
Let's start with the projectile launch
as the body leaves the vertical its velocity must be horizontal
x = v₀ₓ t
y = y₀ + [tex]v_{oy}[/tex] t - ½ g t²
when reaching the ground its height of zero (y = 0) and the initial vertical velocity is zero
t = √ 2 y₀ / g
we substitute
x = vox √2y₀ / g
v₀ₓ = √(g / 2y₀) x
In the exercise, it tells us that the width of the well is D (x = D) and the initial height is the height of the platform minus the length of the rope (I = H - L)
v₀ₓ = √(g /(2 (H -L)) D
this is the minimum speed to cross the well.
Now let's use conservation of energy
starting point. On the platform
[tex]Em_{o}[/tex] = U = m g H
final point. At the bottom of the swing
Em_{f} = K + U = 1 / 2m v² + m g (H -L)
as there is no friction the mechanical energy is conserved
Em_{o} = Em_{f}
m g H = 1 / 2m v² + m g (H -L)
v = √ (2gL)
let's write our two equations
the minimum speed to cross the well
v₀ₓ = √ (g /(2 (H -L)) D
the speed at the bottom of the oscillatory motion
v = √ (2g L)
we analyze the extreme cases
* when L → H chord too long
in this case we see that the speed to cross the well grows a lot (it goes towards infinity) therefore we do not have enough speed in the movement to cross
* when L → 0 very short string
the speed of the platform is very small, so we do not have the minimum required value
vox = √ (g / (2 (H)) D
From this analysis we see that there is a range of lengths that allows us to have the necessary speeds to cross the well
V₀ₓ = v
g / (2 (H -L) D² = 2g L
4 L (H- L) = D²
4 H L - 4 L2 - D² = 0
L² - H L - D² / 4 = 0
let's solve the quadratic equation
L = [H ± √ (H2-D2)] / 2
we assume that H> D
L = ½ H [1 + - RA (1 - (D / H) 2)]
The two values of La give the range of values for which the two speeds are equal
A) The person lands in the moat if the rope's length is very short because :
The speed of the platform is less than the required minimum speedB) The person lands in the moat if the rope length is similar to the height of the platform because :
The speed required to cross the moat approaches infinityFollowing the assumptions;
size of the person is much smaller than L and H
D = horizontal distance
The conditions that will cause the person to land on the moatThe person will land in the moat when the rope's length is very short because as the rope reduces in length the speed reduces as well such that the speed of the platform goes below the required minimum speed which will enable the person cross over. while As the magnitude of the length tends towards the same magnitude of the height the speed required to cross the moat increases towards infinity and this speed cannot be attained by the person hence he will land in the moat.Hence we can conclude that The person lands in the moat if the rope's length is very short because The speed of the platform is less than the required minimum speed and The person lands in the moat if the rope length is similar to the height of the platform because,the speed required to cross the moat approaches infinity.
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color code of electrical resistors
Answer:
Tolerance: [tex]\pm 10\%[/tex]
Explanation:
Resistor Color Codes
Resistor Color Coding uses colored bands to quickly identify the resistive value or resistors and its percentage of tolerance.
Since the question does not provide a specific color table, we'll use the table attached below.
The colors of the resistor shown in the question are:
First band: orange
Second band: blue
Third band: brown
Fourth band: silver
The colors relate to the following numbers respectively:
3, 6, 10Ω, [tex]\pm 1\%[/tex]
The first two colors form the number 36
The third color is the multiplier: 36*10Ω = 360Ω
And the fourth color is the tolerance or the possible variation of the resistance [tex]\pm 1\%[/tex]
Resistance: 360Ω
Tolerance: [tex]\pm 10\%[/tex]
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.
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
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:
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
The bending of rocks due to the compression of tectonic plates is called
Ofaulting
O folding
subduction
plyometrics
Answer:
Folding
Explanation:
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
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
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.
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
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
How long would it take you to walk 3,962 km from New York to Los
Angeles?
Answer:
913 hours ur welcome :)
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]
how much min the basketball 1 player play
Answer:
A professional basketball game depends on the association presiding over the game. An NBA game lasts for 48 minutes whereas FIBA games take 40 minutes. The total time taken to play for any specialized game is over 2 hours 15 minutes. The time includes the time disruptions like fouls, timeouts, and breaks.
I hope it helps you...
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.
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
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
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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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!
What function do you think a flower that can stay warm at night might have for a plant?
Answer:
it keeps it growing because if it stays warm and yuh put it by a plant that needs sun yk
Explanation:
A flower of a plant that can stay warm at night as it keeps it growing.
What are the function of different parts of plant ?
The roots are the underground part of the plant which plays a major role in pulling the water and minerals, expands within the ground for better water absorption, anchors which helps in creating better stability.
A stem present above the ground bears leaves, fruits plus flowers. as it distributes the nutrients and minerals to the leaves, shields the plant and assists in asexual dissemination.
leaf is one of the most major parts of a plant as it contain chlorophyll pigment which assists the plants in preparation for food, the veins allow the flowing of nutrients plus water, it has the parts like petiole, leaf base and lamina help in photosynthesis.
Flower is the most bright and beautiful part of the plant show an important role in making food, used in fertilization process , the basic parts are petals, sepals, stamens, and pistil.
For more details plant, visit
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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]
What would be the speed of an object just before hitting the ground if dropped 100 meters?
We are given:
the initial height of the object (h) = 100 m
initial velocity (u) = 0 m/s
we will let the value of g = 10 m/s/s
Speed of the object just before hitting the ground:
From the third equation of motion:
v² - u² = 2ah (where v is the final velocity)
replacing the variables, we get:
v² - (0)² = 2(10)(100)
v² = 2000
v = 10√20 = 44.7 m/s
Therefore, the speed of the object just before hitting the ground is 44.7 m/s