Lab - Wave Properties in a Spring

11-05
The wave characteristics you will observe in this lab are common to all waves (water, light, sound,
etc.). Use your prior knowledge and the book to fill in the following blanks, then go in the hall and
perform the lab.
A wave is a disturbance that moves through (propagates) through empty space or through a
_____________. There are two types of waves. A _____________________ wave requires
matter to travel. List some examples of this type:
A _____________________ wave does not require a medium. Examples include:
In order to start and transmit a mechanical wave, a source of _____________ and an
_______________ medium are required. A single disturbance is referred to as a
_______________, and a series of disturbances is a wave __________.
The questions in bold are those you should observe directly. Others will be answered using the book.
A. TYPES OF MECHANICAL WAVES: In the hall, stretch the slinky on the floor until it is
stretched (but still loose). Practice sending single pulses down the slinky by popping your wrist
from the center to the side and back to the center. Then send a continuous wave train along as
your partner holds the other end still. A piece of ribbon should be tied to one coil. Watch the
motion of this ribbon (representing a particle) as the wave travels through the spring.
In this type of wave, the particles move (perpendicular, parallel)
to the direction the wave travels. This type of wave is called a __________________ wave.
Its pulses are called ________________ and ________________.
Now send a pulse by quickly pushing the spring forward and pulling
it back, as shown. This type of wave is called _______________. Watch the motion of the ribbon.
In this type, the particles move _____________ to the direction the wave travels. Its pulses
are called _____________ and _____________. Label each.
Note that all waves transfer _____________ without transferring _______________. In
mechanical waves, particles of the medium vibrate back and forth in simple harmonic motion while
the disturbance (or _____________) moves from one place to another.
B. WAVE SPEED
Send a large pulse, followed by a small one. Does one pulse catch up to the other? ______
(Hint: The person who sends these waves should watch how the waves look when they return. Make
sure that both pulses are large enough initially to make it back to the sender!) The size of the
pulse is called the __________________ of the wave. Did the size affect the speed? ______
Generate a single transverse pulse in the slinky, keeping the stretch constant. Using a stopwatch,
time the journey of the pulse from one end to the other and back again. Take the average of
several trials. _________
Without changing your positions on the floor (therefore keeping the _____________ the pulse
travels the same), pull the slinky tighter using only about 3/4 of the coils. This makes a completely
different medium through which the pulse will travel. Time the journey as before. ___________
Does the kind of medium affect the speed of the pulse? ___________
Lab – Wave Properties in a Spring ____________________
PHYSICSFundamentals
© 2004, GPB
11-06
C. WAVELENGTH AND FREQUENCY
Shake the slinky back and forth steadily to send a
transverse wave train while your partner holds the other end still. On the diagram, label wavelength
(- Greek letter lambda). The frequency of the wave depends on how fast you shake the slinky.
Shake it regularly but slowly, then regularly but rapidly.
Higher frequency waves are generated by shaking the spring (slowly, rapidly). High frequency
waves have (short, long) wavelengths, and low frequency waves have __________.
The speed of a wave in any medium is equal to the _______________ of the wave X
________________. This wave equation ___________________ shows that f and  are
______________ proportional. Write the units for each of the variables in this equation.

Answers

Answer 1

The exercise involves filling in the gaps with the possible wave

properties that can be obtained from a spring.

How is the Wave Properties in a Spring Lab exercise correctly completed?

The correctly completed exercise is presented as follows;

A wave is a disturbance that moves through a medium. There are two

types of waves. A mechanical wave requires matter to travel. List some

examples of this type: sound wave, water wave, spring waves.

A electromagnetic wave does not require a medium. Examples include: Light waves

In order to start and transmit a mechanical wave, a source of

disturbance and a physical medium are required. A single disturbance is

referred to as a pulse, and a series of disturbance is a wave train.

This type of wave is called transverse wave. Its pulses are called crest

and troughs.

Now send a pulse by quickly pushing the spring forward and pulling it

back, as shown. This type of wave is called longitudinal wave. Watch the

motion of the ribbon. In this type, the particles move parallel to the

direction the wave travels. Its pulses are called compression and

rarefactions. Note that all waves transfer energy without transferring

matter. In mechanical waves, particle of the medium vibrate back and

forth in simple harmonic motion while the disturbance (or energy)

moves from one place to another.

B. Wave speed

Does the pulse catch up to the other? yes. The size of the pulse is called

the amplitude of the wave.

Did the size of the pulse affect the speed? No.

The average time wave it takes the wave to travel

Without changing your positions therefore keeping the distance the

pulse travels the same), pull the slinky tighter using only about 3/4 of

coils. This makes a completely different medium through which the

pulse will travel. Time the journey as before time record. Does the kind

of medium affect the speed of the pulse? Yes

C. Wavelength and Frequency

High frequency waves have short wavelengths and low frequency waves

have long wavelengths.

The speed of a wave in any medium is equal to the frequency of the wave × the wavelength. This wave equation [tex]\underline{f = \dfrac{v}{\lambda } }[/tex] shows that f and λ are

inversely proportional. The units of the variables are;

Units of the frequency, f is hertz unit HzUnits of the velocity, v, is m/sUnits of the wavelength, λ, is meters (m)

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Related Questions

Find the mass and weight of a body if a resultant force of 850 N causes its speed to increase from 6 m/s to 15 m/s in a time of 5s.

Answers

Hi! Let's see.

TOPIC: Mass and Weight.

Explanation:

Data:

Force(F) = 850 NInitial Velocity([tex]\bold{V_{0}}[/tex]) = 6 m/sFinal Velocity([tex]\bold{V_{f}}[/tex]) = 15 m/sTime(t) = 5 sGravity(g) = 9.8 m/s²Mass(m) = ?Weight(w) = ?

==================================================================

In order to calculate the mass, we must first find the acceleration.

to find the acceleration, we apply the acceleration formula in UARM:

[tex]\boxed{\boxed{\bold{a=\frac{V_{f}-V_{0}}{t}}}}[/tex]

Replace per data:

[tex]\boxed{a=\frac{15\frac{m}{s}-6\frac{m}{s}}{5s}}}}[/tex]

We subtract in the numerator:

[tex]\boxed{a=\frac{9\frac{m}{s}}{5s}}}[/tex]

It divides:

[tex]\boxed{\bold{a=1.8\frac{m}{s^{2}}}}[/tex]

==================================================================

Mass

Knowing it's acceleration, we can proceed to calculate the mass, applying it's formula:

[tex]\boxed{\boxed{\bold{m=\frac{F}{a}}}}[/tex]

Replace:

[tex]\boxed{m=\frac{850N}{1.8\frac{m}{s^{2}}}}}}[/tex]

Newton equivalents to kilograms:

[tex]\boxed{m=\frac{850kg*\frac{m}{s^{2}}}{1.8\frac{m}{s^{2}}}}}}[/tex]

Simplify the m/s²:

[tex]\boxed{m=\frac{850kg}{1.8}}}[/tex]

It divides:

[tex]\boxed{\bold{m=472.22\ kg}}[/tex]

Answer: The mass of the body is 472.22 kilograms.

==================================================================

Weight

To calculate the weight, we use it's formula:

[tex]\boxed{\boxed{\bold{W=m*g}}}}}[/tex]

Replace:

[tex]\boxed{W=472.22*9.8\frac{m}{s^{2}}}}}[/tex]

It multiplies:

[tex]\boxed{\bold{W=4627.75\ N}}}}[/tex]

Answer: The weight of the body is 4627.75 Newtons.

Final Answer:

Mass = 472.22 kg

Weight = 4627.75 N.

Cordially Alejanndraax. Greetings!

Two motorcycles are traveling due east with different velocities. However, 3.63 seconds later, they have the same velocity. During this 3.63-second interval, motorcycle A has an average acceleration of 4.55 m/s2 due east, while motorcycle B has an average acceleration of 18.9 m/s2 due east. (a) By how much did the speeds differ at the beginning of the 3.63-second interval, and (b) which motorcycle was moving faster

Answers

Answer:

52.095 m/s

Motorcycle a was moving faster

Explanation:

We start by using one of the equations of motion

V = u + at

If the first motorcycle starts with an initial speed of u(a) and accelerates at a value of a(a) = 4.55 m/s², then the final speed after a time of 3.63 seconds is V(a). We then represent it as

V(a) = u(a) + a(a).t

If the second motorcycle starts with an initial speed of u(b) and accelerates at a value of a(b) = 18.9 m/s², then the final speed after a time of 3.63 seconds is V(b). We then represent it as

V(b) = u(b) + a(b).t

Assuming that the final speeds v(a) = v(b), and then subtract the equation of the second motorcycle from that of the first, we have

0 = u(a) - u(b) + a(a).t - a(b).t

-u(a) + u(b) = a(a).t - a(b).t, on rearranging, we have

u(b) - u(a) = [a(a) - a(b)]t

Since we have the values for acceleration and the time, we substitute so that

u(b) - u(a) = (4.55 - 18.9)3.63

u(b) - u(a) = -14.35 * 3.63

u(b) - u(a) = -52.095, or we rearrange to get

u(a) - u(b) = 52.095 m/s

A gas cylinder holds 0.10 mol of O2 at 150 C and a pressure of 3.0 atm. The gas expands adiabatically until the pressure is halved
Part A
What is the final volume?
Part B
What is the final temperature?

Answers

Answer:

V2 = 1.899*10^-3 m^3

T2 = 347.125 K

Explanation:

Using gas law, we know that

PV = nRT,

Where

V1 = 0.00115743 m^3.

gamma = 1.4

Now, when we solve for final volume, V2 we get

V2 = V1/((P2/P1)^(1/gamma))

V2 = 1.899*10^-3 m^3

Using the same law and method, when we try to solve for the temperature, we find that the final temperature, T2 is

T2 = T1*((V1/V2)^(gamma-1))

T2 = 347.125 K

The final volume is 1.899*10^-3 m^3

And, the final temperature is 347.125 K

Gas law:

here we used gas law,

we know that

PV = nRT,

Here

V1 = 0.00115743 m^3.

gamma = 1.4

Now final volume is

V2 = V1/((P2/P1)^(1/gamma))

V2 = 1.899*10^-3 m^3

Now the final temperature is

T2 = T1*((V1/V2)^(gamma-1))

T2 = 347.125 K

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An astronaut stands on the surface of an asteroid. The astronaut then jumps such that the astronaut is no longer in contact with the surface. The astronaut falls back down to the surface after a short time interval. Which of the following forces CANNOT be neglected when analyzing the motion of the astronaut?

Answers

Asteroids are known through the help of artificial gravity, to have small gravity. The forces that cannot be neglected when analyzing the motion of the astronaut is that the gravitational force between the astronaut and the asteroid.

The gravitational force between two objects is said to be inversely proportional to the distance between them when squared. Therefore, when an individual halve the distance then the force increases by four times.

Unbalanced forces are simply known to be brought about due to a change in motion, speed, and/or direction. If two forces act in the same direction on an object, the net force is said to be equal to the sum of the two forces.

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Given f(x)=2x+7, which of the following is the value of x when f(x)=13?

Answers

Answer:

The value of x is 3

Explanation:

Equation Solving

We are given the equation

f(x) = 2x + 7

Let's find the value of such that f(x) = 13.

2x + 7 = 13

Subtracting 7:

2x = 13 - 7

2x = 6

Dividing by 2:

x = 6/2

x = 3

The value of x is 3


Through what angle in degrees does a 33 rpm record turn in 0.32 s?
63°
35°
46°
74°

Answers

Answer:

1 rev = 2(pi) rad  pi(rad) = 180 degrees

so 33 rev/min * 1 min/60s * (2*pi)rad/1 rev *180 deg/ pi rad * .32 s = 63.36 degrees

Explanation:

63.36 estimated to 63 so 63

The angle in degrees where 33 rpm record turn in 0.32 s should be considered as the 63 degrees.

Calculation of the angle:Since we know that

1 rev = 2(pi) rad

So here  pi(rad) = 180 degrees

Now for 33 rpm it should be like

=  33 rev/min * 1 min/60s * (2*pi)rad/1 rev *180 deg/ pi rad * .32 s

= 63.36 degrees

= 63 degrees

hence, The angle in degrees where 33 rpm record turn in 0.32 s should be considered as the 63 degrees.

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You are interested in a new sports car that can go from 0 m/s to 120 m/s in 12 s. What is the acceleration of the car?



Formula

image

A.
0.1
m
s
2
B.
10
m
s
2
C.
60
m
s
2
D.
1440
m
s
2

Answers

Answer:

A

Explanation:

Option B) 10m/s² is the correct answer.

Hence, the acceleration of the new sports car in the given period of time is 10m/s².

What is Motion?

Motion is simply the change in position of an object over time.

From the First Equation of Motion;

v = u + at

Where v is final velocity, u is initial velocity, a is acceleration and t is time elapsed.

Given the data in the question;

Initial velocity of the sport car u = 0m/sFinal velocity of the sport car v = 120m/sElapsed time t = 12sAcceleration of the sports car a = ?

To determine the acceleration of the new sports car, we substitute our given values into the expression above.

v = u + at

120m/s = 0m/s + (a × 12s)

a × 12s = 120m/s - 0m/s

a × 12s = 120m/s

a = 120m/s ÷ 12s

a = 10m/s²

Option B) 10m/s² is the correct answer.

Hence, the acceleration of the new sports car in the given period of time is 10m/s².

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two spheres A and B are projected off the edge of a 1.0 m high table with the same horizontal velocity . sphere A has a mass of 20.g and sphere B has a mass of 10.g.
     If both spheres leave the edge of the table at the same instant, sphere A will land
a.    at some time after sphere B.
b.    at the same time as sphere B.
c.    at some time before sphere B.
d.    There is not enough information to decide.​

Answers

A would land before since its heavier

A tsunami, an ocean wave generated by an earthquake, propagates along the open ocean at 700 km/hr and has a wavelength of 750 km. What is the frequency of the waves in such a tsunami?A. 6.8 HzB. 0.93 HzC. 0.00026 HzD. 1.1 HzE. 0.15 Hz

Answers

Answer:

C) 0.00026 Hz

Explanation:

In any wave, there exists a fixed relationship between the speed v, the frequency f and the wavelength λ, as follows:

       [tex]v = \lambda * f (1)[/tex]

Replacing by the givens in (1), and solving for f, we get:

       [tex]f = \frac{v}{\lambda} = \frac{700km/hr}{750 km} = 0.93 1/hr (2)[/tex]

Converting the units to Hz (1/sec), we get:

       [tex]f = 0.93 \frac{1}{hr} *\frac{1 hr}{3600sec} = 2.6e-4 = 0.00026 Hz (2)[/tex]

The answer C. is the right one.

Define conductor and insulator, including how the resistance is different in the two, and give at least one example of each.

Answers

Answer:

Those substances which can conduct electricity are called conductors, while those substances which don't conduct electricity are called insulators.

Resistance is the obstruction provided by the material through which the current passes,so since conductors conduct electricity and insulators don't,so the obstruction i.e resistance provided by the conductor must be less,while insulators being unable to conduct electricity,has very high resistance.

Example of conductor is copper

Example of insulator is plastic

A 10.0kg object is moving at 1 m/s when a force is applied in the direction of the objects motion, causing it to speed up to 4 m/s. If the force was applied for 5s what is the magnitude of the force

Answers

Answer:

F = 6[N].

Explanation:

To solve this problem we must use the principle of conservation of linear momentum, which tells us that momentum is conserved before and after applying a force to a body. We must remember that the impulse can be calculated by means of the following equation.

[tex]P=m*v\\or\\P=F*t[/tex]

where:

P = impulse or lineal momentum [kg*m/s]

m = mass = 10 [kg]

v = velocity [m/s]

F = force [N]

t = time = 5 [s]

Now we must be clear that the final linear momentum must be equal to the original linear momentum plus the applied momentum. In this way we can deduce the following equation.

[tex](m_{1}*v_{1})+F*t=(m_{1}*v_{2})[/tex]

where:

m₁ = mass of the object = 10 [kg]

v₁ = velocity of the object before the impulse = 1 [m/s]

v₂ = velocity of the object after the impulse = 4 [m/s]

[tex](10*1)+F*5=10*4\\10+5*F=40\\5*F=40-10\\5*F=30\\F=6[N][/tex]

Help me please,

A ball is thrown straight up in the air. What is the velocity and acceleration at the top of the path?

A) v 0m/s, = 0m/s/s

B) v = 0m/s, a 10m/s/s

C) v = 10m/s, a 10m/s/s

D) v = 10m/s, a = 0m/s/s

E) None of the above

Answers

Option B

Explanation:

no distance was given only the acceleration due to the fact that it went up (10m/s/s)

s0 it is

0 m/s and 10m/s/s (option B)

The US Patent and Trademark Office is responsible for protecting the intellectual property of inventors. A patent is granted if the idea is original and scientifically valid. A student creates a device that they claim will change the world by changing the way we understand energy. The student includes a table of energy measurements as part of his application.

Answers

Answer:

Deny the patent- the machine violates the law of conservation of mass

Explanation:

Took the test

Suppose a uniform solid sphere of mass M and radius R rolls without slipping down an inclined plane starting from rest. The linear velocity of the sphere at the bottom of the incline depends on?

Answers

Answer:

None of the mass or the radius of the sphere

Explanation:

When a uniform solid sphere of any given mass, say M and any given radius, say R, rolls without slipping downwards an inclined plane that starts from rest. The linear velocity of the sphere at about the bottom of the inclined happens not to depend on either of its mass or that of the radius of its sphere.



A popular ride at an amusment park lifts
customers up to a height of 50 m and then
drops them threw a displacement of 50 m
before slowing them to a stop. How fast
are the customers going at the 50 m
mark?

Answers

Answer:

[tex]31.32\ m/s[/tex]

Explanation:

[tex]We\ are\ given\ that:\\Height\ to\ which\ there're\ lifted=50m\\Displacement\ during\ the\ descent=50m\\Now,\\In\ order\ to\ find\ the\ velocity\ of\ the\ customers\ at\ 50\ m,\\We\ can\ use\ the\ Third\ Equation\ Of\ Motion,\ which is:\\2as=v^2-u^2\\As\ we\ know\ that,\\Acceleration\ due\ to\ gravity=9.81\ m/s^2\ or\ roughly\ 10\ m/s^2\\Displacement=50\ m\\Initial\ velocity=0\ m/s^2\\ [As\ they\ stop\ when\ they\ reach\ the\ maximum\ height\ of\ 50\ m\\ and\ begin\ their\ descent][/tex]

[tex]By\ reconstructing\ the\ Third\ Equation\ Of\ Motion,\ we\ have:\\2gs=v^2\\Hence,\\v^2=2*9.81*50 \\v^2=981\ m^2/s^2 \\v=\sqrt{981\ m^2/s^2} \\v=31.32\ m/s[/tex]

The compound formed from the elements calcium and chlorine is known as

Answers

Answer:

calcium chloride

Explanation:

an inorganic compound,a salt with the chemical formula CaCl2

One of two 25-year-old identical twins begins a trip on a spaceship traveling at 0.8 c while her twin remains on Earth. The twin on Earth tracks her twin's trip for 10 years. After ten years, what age is the twin on the spaceship according to the twin on Earth?
a. 41.7 years old.
b. 35 years old.
c. 31 years old.
d. 37.5 years old.

Answers

Answer 41.7 years old

I answer this question already and it was correct!

What is the condition required of the phase difference (in radians) between two waves with the same wavelength if these waves interfere constructively?
a. (2m +1)π where m= 0, +1, +2, etc.
b. mπ where m = 0, +1, +2, etc.
c. 2mπ where m = 0, +1, +2, etc.
d. (m+1)π where m = 0, +1, +2, etc.

Answers

Answer:

c.

Explanation:

In order to two waves with the same wavelength can interfere constructively, their crests and valleys must coincide in space, so the phase difference must be equal to an integer number of wavelengths, i.e. m *(2 π rad), where m= 0, +1, +2, etc.This is equal to the stated by the answer c) , so c) it's the right answer.

A police officer standing at the side of the road uses a radar emitting frequency of 24.15 GHz. A car is going away from the officer at a speed of 50mph. What will be the difference in frequency of the beam reflected by the car received back by the radar?
a. 4.0 kHz (lower frequency)
b. 1.8 kHz (higher frequency)
c. 4.0 kHz (higher frequency)
d. 1.8 kHz (lower frequency)

Answers

Answer:

The correct answer is  C

Explanation:

  From the question we are told that  

       The frequency of the radar is [tex]f = 24.15 \ GHz = 24.15 *10^{9} \ Hz[/tex]

         The speed of the car is [tex]v = 50 mph = \frac{50}{2.237} = 22.35 \ m/s[/tex]

Generally the difference frequency reflected by the car and the frequency which the radar receives back is mathematically represented as

        [tex]\Delta f = \frac{f * 2 * v }{ c }[/tex]

Here  c is the speed of light with value  [tex]c = 3.0 *10^{8} \ m/s[/tex]

=>    [tex]\Delta f= \frac{2 * 24.15 *10^{9} * 22.35}{ 3.0*10^{8}}[/tex]      

=>    [tex]\Delta f = 4000 \ Hz[/tex]

=>     [tex]\Delta f = 4.0 \ kHz[/tex]  

Given that the value is positive then it a higher frequency

       

*
If a rock falls for 3 seconds off of a bridge, how far will the rock fall?
-30 m
-45m
-60m
-75m

Answers

It will fall for about 45m

what is mean by combination reaction ?​​

Answers

[tex] \underline{\purple{\large \sf Combination \: reaction :-}} [/tex]

Those reaction in which two or more substances combine to form a one new substance are called Combination reaction

In this reaction, We can add :

Two or more elements can combine to form a compound.Two or more compounds can combine to from a one new compound.An element and a compound can combine to form a new compound.

[tex] \underline{\green{\large \sf For\: example :}} [/tex]

[tex] \sf 2H_{2} + O_{2} \: \underrightarrow{Combination} \: 2H_{2}o[/tex]

In this, Hydrogen is an element and Oxygen is another element. Both are combined to form compound 'Hydrogen oxide'. Hydrogen oxide is commonly known as water.

A horizontal force of 350 NN is exerted on a 2.0-kgkg ball as it rotates (at arm's length) uniformly in a horizontal circle of radius 0.90 mm. Part A Calculate the speed of the ball. Express your answer to two significant figures and include the appropriate units.

Answers

Answer:

The value is  [tex]v = 12.6 \ m/s[/tex]

Explanation:

From the question we are told that

    The horizontal force is  [tex]F_h = 350\ N[/tex]

    The mass of the ball is  [tex]m = 2.0 \ kg[/tex]

     The radius is     [tex]r = 0.90 \ m[/tex]

     

Generally the speed of the ball is mathematically evaluated from  the formula for  centripetal force as

         [tex]F_ h = \frac{mv^2}{ r }[/tex]

=>       [tex]v = \sqrt{\frac{ F_h * r }{ m }}[/tex]

=>       [tex]v = \sqrt{\frac{ 350 * 0.9 }{ 2 }}[/tex]

=>      [tex]v = 12.6 \ m/s[/tex]

A carousel at the local carnival rotates once every 45 seconds.
(a) What is the linear speed of an outer horse on the carousel, which is 2.75 m from the axis of rotation?
(b) What is the linear speed of an inner horse that is 1.75 m from the axis of rotation?

Answers

Answer:

We know that the carousel does a complete rotation in 45 seconds.

Then the frequency of this carousel will be f =  1/45 seconds.

And the angular frequency will be 2*pi times the frequency, then we have:

angular frequency = w = 2*3.14*(1/45s) = 0.1396 s^-1

Now, the linear speed of an object that rotates with a radius R, and an angular frequency W is:

S = R*W

then:

a) in this case the radius is 2.75m, then the linear speed is:

S = 2.75m*0.1396 s^-1 = 0.3839 m/s

b) in this case the radius is 1.75m, then the linear speed here is:

S = 1.75m*0.1396 s^-1 = 0.2443 m/s

(a) The linear speed of an outer horse on the carousel is 0.384 m/s.

(b) The linear speed of an inner horse on the carousel is 0.244 m/s.

Given data:

The time interval for the rotation of carousel is, t = 45 s.

The distance of the outer horse from the axis of rotation is, r = 2.75 m.

The distance of an inner horse from the axis of rotation is, r' = 1.75 m.

(a)

The linear speed in this problem can be obtained from the concept of rotational mechanic, in which the ratio of the circumference and the time gives required linear speed. So,

v = 2 π r/t

Solving as,

v = 2 π (2.75) / 45

v = 0.384 m/s

Thus, we can conclude that the linear speed of an outer horse on the carousel is 0.384 m/s.

(b)

Now similarly the linear speed of an inner horse is calculated as,

v' = 2 π r' / t

Solving as,

v' = 2 π (1.75) / 45

v' = 0.244 m/s

Thus, we can conclude that the linear speed of an outer horse on the carousel is 0.244 m/s.

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20 pts.





Which of the following statements is true?
O Electromagnets use electrlcity and magnets.
O Magnetic fields are strongest around the poles of a magnet.
O The south pole of a magnet will repel the south pole of another magnet.
O all of the above

Answers

Answer:

all are true so d is right

Explanation:

Electromagnets use electrlcity and magnets is true.

Magnetic fields are strongest around the poles of a magnet is true.

The south pole of a magnet will repel the south pole of another magnet is true

and since all of them is true the answer is d all of the above

The surface tension of the alcohol propanol in air has a value of 23.70 units
and the surface tension of water in air is 72.80 units. Which two statements
are true?
A. Propanol will have a flatter meniscus than water because the
forces between the molecules are smaller.
B. Propanol will have a higher boiling point than water because the
forces between the molecules are greater.
C. Propanol will have a more curved meniscus than water because
the forces between the molecules are greater.
D. Propanol will have a lower boiling point than water because the
forces between the molecules are smaller.

Answers

Answer:

A & D your welcome

Propanol surface is more flattered and has less boiling point than that of water.

To determine the correct statements among all the statements, we need to know about the surface tension.

What is surface tension?Surface tension is defined as force per unit length.More the surface tension, more the force between the molecules to keep the molecules together.What does happen, if molecules have more force of attraction?If the molecules have large force of attraction, more heat will be required to break the bonds between them. So the boiling point increases. Also with more force of attraction, the surface of the liquid becomes less flattered.

Since, propanol has less surface tension than that of water, so its surface is more flattered and has less boiling point than that of water.

Learn more about the surface tension here:

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A 1.2-kg object moving with a speed of 8.0 m/s collides perpendicularly with a wall and emerges with a speed of 6.0 m/s in the opposite direction. If the object is in contact with the wall for 2.0 ms, what is the magnitude of the average force on the object by the wall?
a. 9.8 kN.
b. 8.4 kN.
c. 7.7 kN.
d. 9.1 kN.
e. 1.2 kN.

Answers

Given that,

Mass of the object, m = 1.2 kg

Initial speed of the object, u = 8 m/s

Final speed of the object, v = -6 m/s (in opposite direction)

Time, t = 2 ms

To find,

The average force on the object by the wall.

Solution,

Let F be the force. Using Newton's second law of motion,

F = ma, a is acceleration

[tex]F=\dfrac{m(v-u)}{t}\\\\F=\dfrac{1.2\times ((-6)-8)}{2\times 10^{-3}}\\\\=8400\ N[/tex]

or

F = 8.4 N

So, the magnitude of average force in the object by the wall is 8.4 N.

Which is NOT a form of erosion?


a

Rivers moving dirt


b

Water running downhill, carving out the land and pushing dirt to the side

c

Water evaporating and condensing

d

Heavy winds blowing soil to other locations

HELP ASAP

Answers

Answer:

water evaporating and condensing, because that's the water cycle and it isnt eroding anything

A tennis ball moves 16 meters northward, then 22 meters southward, then 12 meters northward, and finally 32 meters southward.

Answers

U didn’t ask a question so I’ll just say that’s cool that it moves like that

For an n- channel JFFT, IDSS = 8 mA, and VP = - 6 V. If VGS = - 2 V. What is the value of the drain current ID?

Answers

Answer: the value of the drain current ID is 3.56 mA

Explanation:

Given that;

IDSS = 8 mA

VP(VGS_OFF) = - 6 V.

VGS = - 2 V

value of the drain current ID = ?

Drain current ID is expressed as;

ID = IDSS [ 1 - (VGS/VP(VGS_OFF)) ]²

we substitute our value

ID = 8 [ 1 - (-2 / -6) ]²

ID = 8 [ 1 - 0.3333 ]²

ID = 8 [ 0.6667 ]²

ID = 8 [ 0.4444 ]

ID = 3.5552 ≈ 3.56 mA

Therefore the value of the drain current ID is 3.56 mA

A ping pong ball of mass 0.0027 kg and radius 0.020 m is completely submerged 0.3 m in water. How fast is it moving when it emerges from the water?

Answers

Answer:

The value speed  is  [tex]v = 8.192 \ m/s[/tex]

Explanation:

From the question we are told that

   The mass of the ping pong ball is [tex]m = 0.0027 \ kg[/tex]

    The  radius is  [tex]r = 0.020 \ m[/tex]

     The depth of the ping pong inside the water is  [tex]s = 0.3 \ m[/tex]

Gnerally the force with which the ball will emerge will from the water is mathematically represented as

            [tex]F = B - m g[/tex]

Here B  is the Buoyant force on the ball which is mathematically represented as

         [tex]B = \rho_w * V_b * g[/tex]

Here  [tex]V_b[/tex] is the volume of the ball which is mathematically represented as

       [tex]V_b = \frac{4}{3} * \pi * r^3[/tex]

=>     [tex]V_b = \frac{4}{3} * 3.142 * 0.020 ^3[/tex]

=>     [tex]V_b = 3.35*10^{-5} \ m^3[/tex]

        [tex]\rho_w[/tex]  is the density of water with value  [tex]\rho_w = 1000 \ kg/m^3[/tex]

So  

       [tex]B = 1000 * 3.35*10^{-5} * 9.8[/tex]

=>    [tex]B = 0.3284 \ N[/tex]

So

     [tex]F = 0.3284 - 0.0027 * 9.8[/tex]

=>  [tex]F = 0.3284 - 0.02646[/tex]

=>  [tex]F = 0.3020 \ N[/tex]

Generally force is mathematically represented as

     [tex]F = m * a[/tex]

So

      [tex]0.3020 = 0.0027 * a[/tex]

=>  [tex]a = 111.84 \ m/s^2[/tex]

Generally  from kinematic equation

     [tex]v^2 = u^2 + 2as[/tex]

Here u  is the velocity of the ping pong at depth of  0.3 m and the value is zero given that at that point there was no motion

So

      [tex]v^2 = 0^2 + 2 * 111. 84 * 0.3[/tex]

=>    [tex]v = 8.192 \ m/s[/tex]

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