Answer:polarization pertains to the act or process of producing a positive electrical charge and a negative electrical charge such that between a nerve cell internal electrical charge..................
Explanation:
Which of the following statements is/are true for ultrasonic test?a. Equipment used for ultrasonic testing is portableb. Complicated shapes can be easily scannedc. Waves generated are health hazardousd. All the above statements are true
Answer:
Equipment used for ultrasonic testing is portable
Explanation:
An ultrasonic tester is used to detect flaws such as cracks and leaks. Hence, an ultrasonic tester is an integral tool used in material testing and engineering, manufacturing quality control, equipment condition monitoring, building and bridge inspection and much more.
Ultrasonic testing can be performed on steel and other metals and alloys, though it can also be used on concrete, wood and composites, although with less resolution.
A creature moves at a speed of 7.86 furlongs/fortnight (not a very common unit of speed). If 1.0 furlong = 220 yards 1 fortnight = 14 days 1 yard = 0.9144 meter and 1 day= 86400 sec find the speed of the creature (which is probably a snail).
1. 2.28745 × 109 m/s
2. 0.00130719 m/s
3. 1.16706 × 107 m/s
4. 47261.3 m/s
5. 0.25621 m/s
6. 39516.4 m/s
7. 5.29359 × 10−6 m/s
Answer:
2 0.00130719 m/s
Explanation:
Now we have
7.86 furlongs/fortnight
7.86 x 220 yards / fortnight
1729.2yards / fortnight
1647.8 yards / 14 days
123.5yards / day
So
123.5 x 0.9144 meter / day
112.meters / day
=112.9meters / 86400s
= 0.001307m/s
Alice drops a rock to a well. She hears the splash of the rock 4.1s later. The speed of sound is 340m/s. How deep is the well
Answer:
697 m
Explanation:
From the question above,
Note: Alice hears the splash of the rock in the well through the help of echo.
Applying,
v = 2x/t........................ Equation 1
Where v = speed of sound, x = depth of the well, t = time.
make x the subject of the formula in equation 1
x = vt/2................... Equation 2.
Given: v = 340 m/s, t = 4.1 s
Substitute these values into equation 2
x = 340(4.1)/2
x = 697 m.
Four students measured the same line with a ruler like the one shown below. The results were as follows: 5.52 cm, 6.63 cm, 5.5, and 5.93. Even though you cannot see the line they actually measured, which of the recorded measurements are possible valid measurements for this instrument, according to its precision? Select all that apply. 1. 5.52 2. 6.63 3. 5.5 4. 5.93
Answer:
correct answer is 1 and 3
Explanation:
In direct measurement with an instrument, the precision or absolute error of the instrument is given by its appreciation, in this case we see that the measurements have two decimal places, so the appreciation of the instrument must be 0.01 cm
Based on this appreciation, the valid measurements are 5.52 and 5.5.
the other two measurements have errors much higher than the assessment of the instrument, for which there must have been some errors in the measurement.
The correct answer is 1 and 3
Glycerin is poured into an open U-shaped tube until the height in both sides is 20 cm. Ethyl alcohol is then poured into one arm until the height of the alcohol column is 20 cm. The two liquids do not mix. What is the difference in height between the top surface of the glycerin and the top surface of the alcohol? Suppose that the density of glycerin is 1260 kg/m3and the density of alcohol is 790 kg/m3.
Answer:
Difference in height = 7.5 cm
Explanation:
We are given;.
Height of ethyl alcohol;h2 = 20 cm = 0.2 m
Density of glycerin: ρ1 = 1260 kg/m³
Density of ethyl alcohol; ρ2 = 790 kg/m³
To get the difference in height, the pressure at the top of the open end must be equal to the pressure at the point where the liquids do not mix since both points will be at different levels after the pouring.
Thus;
P1 = P2
Formula for pressure is; P = ρgh
Thus;
ρ1 × g × h1 = ρ2 × g × h2
g will cancel out to give;
ρ1 × h1 = ρ2× h2
Making h1 the subject, we have;
h1 = (ρ2× h2)/ρ1
h1 = (790 × 0.2)/1260
h1 = 0.125 m
Difference in height will be;
Δh = h2 - h1
Δh = 0.2 - 0.125
Δh = 0.075 m = 7.5 cm
The difference in height between the top surface of the glycerin and the top surface of the alcohol as per Pascal's law is 7.5 cm.
What is Pascal's law?As per Pascal's law, "The pressure at any point of the vessel filled with incompressible liquid is same".
Given data -
The height of both section is, h2 = 20 cm = 0.20 m.
The density of glycerine is, [tex]\rho_{1} = 1260 \;\rm kg/m^{3}[/tex].
The density of alcohol is, [tex]\rho_{2} = 790 \;\rm kg/m^{3}[/tex].
The pressure at the top of the open end must be equal to the pressure at the point where the liquids do not mix since both points will be at different levels after the pouring. Therefore,
P1 = P2
The formula for pressure is;
P = ρgh
Thus;
ρ1 × g × h1 = ρ2 × g × h2
ρ1 × h1 = ρ2× h2
h1 = (ρ2× h2)/ρ1
h1 = (790 × 0.2)/1260
h1 = 0.125 m
Difference in height will be;
Δh = h2 - h1
Δh = 0.2 - 0.125
Δh = 0.075 m
Δh = 7.5 cm
Thus, we can conclude that the difference in height between the top surface of the glycerin and the top surface of the alcohol is 7.5 cm.
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When a piano tuner strikes both the A on the piano and a 440 Hz tuning fork, he hears a beat every 2 seconds. The frequency of the piano's A is
Answer:
The frequency is [tex]f = 439.5 \ Hz[/tex]
Explanation:
From the question we are told that
The frequency of the tuning fork is [tex]f_t = 440 \ Hz[/tex]
The beat period is [tex]T = 2 \ s[/tex]
Generally the beat frequency is mathematically represented as
[tex]f_b = \frac{1}{T}[/tex]
[tex]f_b = \frac{1}{2}[/tex]
[tex]f_b = 0.5 \ Hz[/tex]
The beat frequency is also represented mathematically as
[tex]f_b = f_t - f[/tex]
Where [tex]f[/tex] is the frequency of the piano
So
[tex]f = 440 - 0.5[/tex]
[tex]f = 439.5 \ Hz[/tex]
Calculate the location xcm of the center of mass of the Earth-Moon system. Use a coordinate system in which the center of the Earth is at x
Answer:
The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.
Explanation:
Let suppose that planet and satellite can be treated as particles. The masses of Earth and Moon ([tex]m_{E}[/tex], [tex]m_{M}[/tex]) are [tex]5.972\times 10^{24}\,kg[/tex] and [tex]7.349\times 10^{22}\,kg[/tex], respectively. The distance between centers is 384,403 kilometers. The location of the center of mass can be found by using weighted averages:
[tex]\bar x = \frac{x_{E}\cdot m_{E}+x_{M}\cdot m_{M}}{m_{E}+m_{M}}[/tex]
If [tex]x_{E} = 0\,km[/tex] and [tex]x_{M} = 384,403\,km[/tex], then:
[tex]\bar x = \frac{(0\,km)\cdot (5.972\times 10^{24}\,kg)+(384,403\,km)\cdot (7.349\times 10^{22}\,kg)}{5.972\times 10^{24}\,kg+7.349\times 10^{22}\,kg}[/tex]
[tex]\bar x = 4.673\,km[/tex]
The center of mass of the Earth-Moon system is 4.673 kilometers away from center of Earth.
The pendulum consists of two slender rods AB and OC which have a mass of 3 kg/m. The thin plate has a mass of 12 kg/m2 . a) Determine the location ӯ of the center of mass G of the pendulum, then calculate the mass moment of inertia of the pendulum about z axis passing through G. b) Calculate the mass moment of inertia about z axis passing the rotation center O.
Answer:
The answer is below
Explanation:
a) The location ӯ of the center of mass G of the pendulum is given as:
[tex]y=\frac{0+(\pi*(0.3\ m) ^2*12kg/m^2*1.8\ m-\pi*(0.1\ m) ^2*12kg/m^2*1.8\ m)+0.75\ m*1.5\ m *3\ kg/m}{(\pi*(0.3\ m) ^2*12kg/m^2-\pi*(0.1\ m) ^2*12kg/m^2)+3\ kg/m^2*0.8\ m+3\ kg/m^2*1.5\ m} \\\\y=0.88\ m[/tex]
b) the mass moment of inertia about z axis passing the rotation center O is:
[tex]I_G=\frac{1}{12}*3(0.8)(0.8)^2+ 3(0.8)(0.888)^2-\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8-\\0.888)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8-0.888)^2+\frac{1}{12}*3(1.5)(1.5)^2+\\3(1.5)(0.888-0.75)^2\\\\I_G=13.4\ kgm^2[/tex]
c) The mass moment of inertia about z axis passing the rotation center O is:
[tex]I_o=\frac{1}{12}*3(0.8)(0.8)^2+ \frac{1}{3}* 3(1.5)(1.5)^2+\frac{1}{2}*(12)(\pi)(0.3)^2(0.3)^2 +(12)(\pi)(0.3)^2(1.8)^2-\\\frac{1}{2}*(12)(\pi)(0.1)^2(0.1)^2 -(12)(\pi)(0.1)^2(1.8)^2\\\\I_o=13.4\ kgm^2[/tex]
A worker is pushing a crate of tools up a ramp into a truck. The crate has a mass of 120 kg and is accelerating at a rate of 1.05 m/s^2. Once he finishes pushing that crate up, he then takes another crate that is 3 times as massive and pushes it up the ramp with an acceleration of 0.71 m/s^2. What is the ratio of the force the worker used on the on the 120 kg crate and the more massive crate?
Answer:
B) 0.493
Explanation:
First Crate
Formula: F=ma
120kg*1.05m/s^2= 126N
Second Crate
Formula: F=ma
3*120kg=360kg
360kg*0.71m/s^2=255.6
Find the ratio by dividing
120kg/255.6kg=0.4929
Round 0.4929=0.493
Hope this Helps!!
Does anybody know to solve this?
Answer:
25600 cm
2.56 x 10⁴ cm
Explanation:
Step 1: Find conversion
There are 100 cm in 1 m.
Step 2: Set up dimensional analysis
[tex]256m(\frac{100cm}{1 m} )[/tex]
Step 3: Multiply and cancel out units
25600 cm
2.56 x 10⁴ cm
What will be the force if the particle's charge is tripled and the electric field strength is halved? Give your answer in terms of F
Answer:
F' = (3/2)F
Explanation:
the formula for the electric field strength is given as follows:
E = F/q
where,
E = Electric Field Strength
F = Force due to the electric field
q = magnitude of charge experiencing the force
Therefore,
F = E q ---------------- equation (1)
Now, if we half the electric field strength and make the magnitude of charge triple its initial value. Then the force will become:
F' = (E/2)(3 q)
F' = (3/2)(E q)
using equation (1)
F' = (3/2)F
When the charge of the particle is tripled, and magnetic field strength is halved, the force increases by factor of 1.5 (1.5 F)
The force experienced by a particle placed in magnetic field is given as;
F = qvB
where;
q is the charge of the particlev is the velocity of the particleB is the magnetic field strengthWhen the charge of the particle is tripled, q₂ = 3q and magnetic field strength is halved, B₂ = 0.5B.
The magnetic force of the particle is determined as;
[tex]F = qvB\\\\ v = \frac{F}{qB} \\\\ \frac{F_1}{q_1B}_1 = \frac{F_2}{q_2B_2}\\\\ F_2 = \frac{F_1 q_2B_2}{q_1 B_1} \\\\ F_2 = \frac{F \times 3q_1\times 0.5B_1}{q_1B_1} \\\\ F_2 = 1.5 F[/tex]
Thus, when the charge of the particle is tripled, and magnetic field strength is halved, the force increases by factor of 1.5 (1.5 F).
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Unit 1 Horizontal Kinematics Homework #3 Acceleration 1. A bowling ball starts from rest and moves 300 m down the gutter in 22.4 seconds. a) What is its speed at the end of the track? b) What is its acceleration?
Answer:
Explanation:
since the object start its motion from rest and ends its motion at 300 m so
ΔS=S2-S1=300m-0m=300 m
given time is 22.4 sec therefore
v=ΔS/t
v=300/22.4=14.7 m/s
so the speed of the ball at the end of the track is 14.7 m/s
now to find the acceleration we know that
a=v/t
a=14.7/2.4
a=0.65m/sec²
You throw a 20-N rock vertically into the air from ground level. You observe that when it is a height 14.8m above the ground, it is traveling at a speed of 25.0 m/s upward.
A) Use the work-energy theorem to find its speed just as it left the ground. What is it?
B) Use the work-energy theorem to find its maximum height. What is it?
Answer:
(A) The speed just as it left the ground is 30.25 m/s
(B) The maximum height of the rock is 46.69 m
Explanation:
Given;
weight of rock, w = mg = 20 N
speed of the rock at 14.8 m, u = 25 m/s
(a) Apply work energy theorem to find its speed just as it left the ground
work = Δ kinetic energy
F x d = ¹/₂mv² - ¹/₂mu²
mg x d = ¹/₂m(v² - u²)
g x d = ¹/₂(v² - u²)
gd = ¹/₂(v² - u²)
2gd = v² - u²
v² = 2gd + u²
v² = 2(9.8)(14.8) + (25)²
v² = 915.05
v = √915.05
v = 30.25 m/s
B) Use the work-energy theorem to find its maximum height
the initial velocity of the rock = 30.25 m/s
at maximum height, the final velocity = 0
- mg x H = ¹/₂mv² - ¹/₂mu²
- mg x H = ¹/₂m(0) - ¹/₂mu²
- mg x H = - ¹/₂mu²
2g x H = u²
H = u² / 2g
H = (30.25)² / 2(9.8)
H = 46.69 m
What is the largest wavelength that will give constructive interference at an observation point 141 m from one source and 355 m from the other source
Answer:
Explanation:
For constructive interference of two coherent light waves at a point , the condition is as follows
path difference = n x wavelength where n is any integer like 1 , 2 , 3 , .... etc .
Given in the question , path difference of light from the two sources
= 355 - 141 = 214 m
So applying the formula above ,
214 = n x wavelength
wavelength = 214 / n
For largest value of wavelength , n should be smallest . The smallest value of n is 1 .
so ,
largest wavelength possible for constructive interference
= 214 / 1
= 214 m .
The volume of a ball was measured at 500.0 cm3, and then mass was measured on this triple beam balance. What would be the most accurate reading for the mass of the ball?Using this mass, what would be the most accurate calculation of density for the ball? Round to the nearest ten thousandths.
Answer:
the answer is 0.8084
Explanation:
i just did it and got it correct
A 21.3 A current flows in a long, straight wire. Find the strength of the resulting magnetic field at a distance of 45.7 cm from the wire.
Answer:
The magnetic field strength due to current flowing in the wire is9.322 x 10⁻⁶ T.
Explanation:
Given;
electric current, I = 21.3 A
distance of the magnetic field from the wire, R = 45.7 cm = 0.457 m
The strength of the resulting magnetic field at the given distance is calculated as;
[tex]B = \frac{\mu_o I}{2\pi R}[/tex]
Where;
μ₀ is permeability of free space = 4π x 10⁻⁷ T.m/A
[tex]B = \frac{\mu_o I}{2\pi R}\\\\B = \frac{4\pi*10^{-7} *21.3}{2\pi(0.457)}\\\\B = 9.322 *10^{-6} \ T[/tex]
Therefore, the magnetic field strength due to current flowing in the wire is 9.322 x 10⁻⁶ T.
Light rays that are reflected twice inside a rain water droplet can produce a second-order rainbow. True False mim
Answer:
Yes
Explanation:
Choose the box as the system of interest. What objects in the surroundings exert significant forces on this system
Answer:
Hello your question is incomplete attached below is the complete question
Your are lifting a heavy box we will consider this process for different choices of system and surroundings
Choose the box as the system of interest. What objects in the surroundings exert significant forces on this system
The floor the box you nonethe earthanswer : The earth and YOU
Explanation:
The objects that will exert significant force on this system would be the force exerted by the earth and force exerted by you
This is because the force exerted by the earth is an internal force and it can be calculated as : m*g ( mass of the object * acceleration due to gravity ) also the force exerted by you are force due to normal reaction when lifting a load
An agent has just listed a power plant. This is an example of what type of real estate?
Answer: industrial properties
Explanation:
The power plant listed by the agent falls under industrial property. An industria property is a property which is either used in manufacturing, rentage, warehousing, processing e.t.c.
in this case the plant Would be used in the generation of power which would be sold. so yes the property falls under the category
An ion has a mass of 6.79 × 10-27 kg and a charge of +4e. What is the magnitude of the electric field that will balance the gravitational force on the particle?
Answer:
Explanation:
200.34
Consider position [x] = L, time [t] = T, velocity [v] = L/T and acceleration [a] = L/T 2 . Find the exponent A in the equation v = a^2 t^ A /x
Answer:
The exponent A in the equation is 3.
Explanation:
v = a^2 t^ A /x
[tex]v = \frac{a^2t^A}{x} \\\\vx = a^2t^A\\\\(\frac{L}{T})(L) = (\frac{L}{T^2})^2(T)^A\\\\ \frac{L^2}{T}= (\frac{L^2}{T^4})(T)^A\\\\ \frac{L^2}{T} *\frac{T^4}{L^2} = (T)^A\\\\T^3 = T^A\\\\\frac{T^3}{T^3} = \frac{T^A}{T^3}\\\\T^{3-3} = T^{A-3}\\\\3-3 = A-3\\\\0 = A-3\\\\A = 3[/tex]
Therefore, the exponent A in the equation is 3.
"For the lowest harmonic of pipe with two open ends, how much of a wavelength fits into the pipe’s length?"
Answer:
0.5 lambda(wavelength)
Explanation:
We know that
The first harmonic for both side open ended pipe is
L= 1/2lambda
So L = 0.5*wavelength
A security guard walks at a steady pace, traveling 190 m in one trip around the perimeter of a building. It takes him 260 s to make this trip.what is his speed. express answer in 2 significant figures
Answer:
0.7 m/s
Explanation:
The following data were obtained from the question:
Distance travalled (d) = 190 m
Time (t) = 260 secs
Speed (S) =..?
Speed is simply defined as the distance travelled per unit time. Mathematically, it is expressed as:
Speed = Distance (d) /time(t)
S = d/t
With the above formula, we can calculate the speed of the security guard as follow:
Distance travalled (d) = 190 m
Time (t) = 260 secs
Speed (S) =..?
S = d/t
S = 190/260
S = 0.7 m/s
Therefore, the speed of the security guard is 0.7 m/s
A security guard walks at a steady pace, traveling 190 m in one trip around the perimeter of a building. The speed of the security guard is approximately 0.73 m/s.
Given:
Distance (d) = 190 m
Time (t) = 260 s
To calculate the speed of the security guard, we can use the formula:
Speed = Distance / Time
Speed = 190 / 260 = 0.73 m/s
Therefore, A security guard walks at a steady pace, traveling 190 m in one trip around the perimeter of a building. The speed of the security guard is approximately 0.73 m/s (rounded to two significant figures).
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Mass and Weight: What is the mass of an object that experiences a gravitational force of 685 N near Earth's surface where g
Answer:
m = 69.9 kg
Explanation:
The mass and the weight of an object are two different quantities. Mass is basically the amount of matter that is present in a body. It remains same everywhere in the universe and measured in kilograms.
Weight is basically a force. It is the force by which earth attracts everything towards itself. The weight of an object changes from planet to planet, with the change in value of the gravitational acceleration (g).
Therefore, the relation between mass and weight of an object is given by the following formula:
W = mg
m = W/g
where,
m = mass = ?
W = Weight = 685 N
g = 9.8 m/s²
Therefore,
m = (685 N)/(9.8 m/s²)
m = 69.9 kg
The mass of an object that experience a gravitational force of 685N near the earth is approximately 69.90 kg
The mass and weight are 2 different quantities.
The mass of a substance is the amount of matter present in it. Mass is measured in kilograms
The Weight is the measurement of gravitational pull of planet towards the object. Therefore, weight is a force and measured in Newton.
Mathematically, the weight can be express as follows
W = mg
where
m = mass
g = acceleration due to gravity = 9.8 m/s²
Therefore,
685 = 9.8 × m
m = 685 / 9.8
m = 69.8979591837
m = 69.90 kg
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The aqueous humor in a person's eye is exerting a force of 0.242 N on the 1.21 cm2 area of the cornea. What pressure is this in mm Hg?
Answer:
15.00124mmHgExplanation:
Pressure is defined as the ratio of force applied to an object to its area.
Pressure = Force/Area
Given parameters
Force = 0.242N
Area = 1.21cm²
Required parameters
Pressure = 0.242/1.21
Pressure = 0.2N/cm²
Using the conversion to convert the pressure to mmHg
1N/cm² = 75.0062mmHg
0.2N/cm² = y
y = 0.2 * 75.0062
y = 15.00124mmHg
Hence the pressure in mmHg is 15.00124mmHg
How many valence electrons does each Cholrine atom have?
Answer:
7 valence electrons is right answer
Explanation:
2 elections in first shell and 5 electrons in second shell which total makes seven electrons.
hope it helped you:)
A ball is thrown straight up in the air. For which situations are both the instantaneous velocity and the acceleration zero? (Select all that apply.)
Answer:
Therefore, the situation in which both the instantaneous velocity and acceleration become zero, is the situation when the ball reaches the highest point of its motion.
Explanation:
When a ball is thrown upward under the free fall action of gravity, it starts to loose its Kinetic Energy as it moves upward. As the ball moves in upward direction, its kinetic energy gradually converts into its potential energy. As a result the speed of the ball starts to decrease as it moves up. Therefore, at the highest point during its motion, the velocity of ball becomes zero and it stops at the highest point for a moment, and then it starts to fall back down, under the influence of gravitational force.
Therefore, the situation in which both the instantaneous velocity and acceleration become zero, is the situation when the ball reaches the highest point of its motion.
In the park, there are 25 pigeons, 15 squirrels, 5 rabbits, and 5 stray cats. Why
can't you draw a line graph using only this information?
Answer: Line graph should be used to show how one variable changes over time not to show multiple categories or variables are at one specific point in time.
Explanation:
In maths, statistics, and related fields, graphs are used to visually display variables and their values. In the case of line graphs, these are mainly used to display evolution or change of a variable over time. For example, a line graph can show how the number of divorces changed from 1920 to 2010.
In this context, the number of different animals in the park cannot be represented through a line graph because this situation does not imply a variable changing over time. Moreover, this situation includes multiple variables or categories of animals and the data shows only one specific point in time, which can be better represented through a bar graph.
Any body moving with simple harmonic motion is being acted on by a force that is:__________.
a) proportional to a sine or cosine function of the displacement
b) directly proportional to the displacement
c) proportional to the inverse square of the displacement
d) constant
Answer:
B
Explanation:
Because this oscillations occur when the restoring force is directly proportional to displacement, given as
F=-kx
Where k= force constant
X= displacement
A point source emits electromagnetic energy at a rate of 100 W. The intensity 10 m from the source is
Answer:
[tex]I=0.0795\ W/m^2[/tex]
Explanation:
Given that,
Power of electromagnetic energy, P = 100 W
We need to find the intensity at a distance of 10 m from the source. Intensity is equal to the power per unit area. So,
[tex]I=\dfrac{P}{4\pi r^2}\\\\I=\dfrac{100}{4\pi (10)^2}\\\\I=0.0795\ W/m^2[/tex]
So, the intensity at a distance of 10 m is [tex]0.0795\ W/m^2[/tex]