What is the role of the part in the diagram labeled Y?
modulate, amplify, and send out waves
O capture, amplify, and demodulate waves
change the amplitude and frequency of waves
O change the pulse and phase of waves

Answers

Answer 1

Question is incomplete and image is not attached ti the question. The required image is attached below, so the complete question is:

The diagram shows a device that uses radio waves.

What is the role of the part in the diagram labeled Y?

modulate, amplify, and send out waves capture, amplify, and demodulate waves change the amplitude and frequency of waves change the pulse and phase of waves

Answer:

2. capture, amplify, and demodulate waves

Explanation:

The part Y labeled in the diagram refers to radio receiver which capture, amplify and demodulate the radio waves.

The radio receiver seperates required radio frequency signals through antenna and consist of an amplifier that amplify or increase the power of receiving signal. At the end, demodulators present in receivers recover the information from the modulated wave.

Hence, the correct option is 2.

What Is The Role Of The Part In The Diagram Labeled Y?modulate, Amplify, And Send Out WavesO Capture,
Answer 2

Answer:

B

Explanation:

edge 2020


Related Questions

Refer the attached photo

Answers

Answer:

A

Explanation:

since the wooden bat is an opaque object placed after a translucent object, light will come through the plastic sheet but will be unable to go through the bat. hence the dark shadow of the bat on a lit sheet

If you're swimming underwater and knock two rocks together, you will hear a very loud noise. But if your friend above the water knocks two rocks together, you'll barely hear the sound.

Match the words.

The air-water interface is an example of boundary. The( )portion of the initial wave energy is way smaller than the( )portion. This makes the( ) wave hard to hear.

When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can( ) .

1. reflect more efficiently
2. transmitted
3. travel directly to your ears
4. boundary
5. reflected
6. discontinuity

Answers

Answer:

The air-water interface is an example of boundary. The transmitted portion of the initial wave energy is way smaller than the reflected portion. This makes the boundary  wave hard to hear.

When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can travel directly to your ear.

Explanation:

The air-to-water sound wave transmission is inhibited because more of reflection than transmission of the wave occurs at the boundary. In the end, only about 30% of the sound wave eventually reaches underwater. For sound generated underwater, all the wave energy is transmitted directly to the observer. Sound wave travel faster in water than in air because, the molecules of water are more densely packed together, and hence can easily transmit their vibration to their neighboring molecules, when compared to air.

Answer: The air-water interface is an example of boundary. The (transmitted) portion of the initial wave is way smaller than the (reflected) portion. This makes the (transmitted) wave hard to hear.

When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can (travel directly to your ears.)

Explanation:

The part of the sound wave that is transmitted across the boundary between air and water is much smaller than the part of the wave that is reflected. This is what makes it hard to hear your friend knocking two rocks together above the surface.

When you and the rocks are underwater, the sound that comes from knocking the rocks together can travel directly to your ears rather than having to be transmitted across mediums.

If an object is determined to have a negative charge of 1.6 micro Coulomb, you can conclude that the object has an excess of

Answers

Answer:

The object has an excess of [tex]10^{13}[/tex] electrons.

Explanation:

When an object has a negative charge he has an excess of electrons in its body. We can calculate the number of excessive electrons by dividing the charge of the body by the charge of one electron. This is done below:

[tex]n = \frac{\text{object charge}}{\text{electron charge}}\\n = \frac{-1.6*10^{-6}}{-1.6*10^{-19}} = 1*10^{-6 + 19} = 10^{13}[/tex]

The object has an excess of [tex]10^{13}[/tex] electrons.

A standard 1 kilogram weight is a cylinder 48.5 mm in height and 49.0 mm in diameter. What is the density of the material? kg/m3

Answers

Answer:

Density = 10,933.93 kg/m^3

the density of the material is 10,933.93 kg/m^3

Explanation:

Density is the mass per unit volume

Density = mass/volume = m/V

Volume of a cylinder V = πr^2 h

Given;

Height h = 48.5mm = 0.0485 m

Radius r = diameter/2 = 49mm÷2 = 24.5mm = 0.0245m

Substituting the values;

Volume V = π×(0.0245^2)×0.0485

V = 0.000091458438030 m^3

V = 0.000091458 m^3

The mass is given as;

Mass = 1 kg

So, the density can be calculated as;

Density = 1/0.000091458

Density = 10933.92825785 kg/m^3

Density = 10,933.93 kg/m^3

the density of the material is 10,933.93 kg/m^3

A cylindrical shell of radius 7.00 cm and length 2.21 m has its charge uniformly distributed on its curved surface. The magnitude of the electric field at a point 15.2 cm radially outward from its axis (measured from the midpoint of the shell) is 36.0 kN/C. (a) Find the net charge on the shell.

Answers

Answer:

The net charge on the shell is 30x10^-9C

Explanation:

Pls see attached file

Which of the following is not considered a behavior?
A. eating
B. anxiety
C. sleeping
D. crying

Answers

━━━━━━━☆☆━━━━━━━

▹ Answer

B. Anxiety

▹ Step-by-Step Explanation

Anxiety isn't a behavior since it's a feeling. Behavior and feeling are different things therefore, anxiety is the correct answer.

Hope this helps!

- CloutAnswers ❁

Brainliest is greatly appreciated!

━━━━━━━☆☆━━━━━━━

Eating, sleeping, and crying all are considered as behaviors. However, anxiety cannot be considered as a behavior because it is a feeling. Thus, the correct option is B.

What is Anxiety?

Anxiety is an intense feeling of excessive, and persistent worry and the fear about everyday situations. This includes fast heart rate, rapid breathing, sweating, and feeling tired constantly may occur.

Behavior is the range of actions and mannerisms which are made by individuals, organisms, systems or the artificial entities in some environment. These systems can include other systems or organisms as well as the inanimate physical environment. Behaviors include eating, sleeping, and crying. Anxiety is not a behavior, it is a feeling.

Therefore, the correct option is B.

Learn more about Anxiety here:

https://brainly.com/question/28481974

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Suppose the ring rotates once every 4.30 s . If a rider's mass is 53.0 kg , with how much force does the ring push on her at the top of the ride?

Answers

The complete question is;

In an amusement park ride called The Roundup, passengers stand inside a 16-m-diameter rotating ring. After the ring has acquired sufficient speed, it tilts into a vertical plane.

Suppose the ring rotates once every 4.30 s . If a rider's mass is 53.0 kg , with how much force does the ring push on her at the top of the ride?

Answer:

F_top = 385.36 N

Explanation:

We are given;

mass;m = 52 kg

Time;t = 4.3 s

Diameter;d = 16m

So,Radius;r = 16/2 = 8m

The formula for the centrifugal force is given as;

F_c = mω²R

Where;

R = radius

Angular velocity;ω = 2πf

f = frequency = 1/t = 1/4.3 Hz

F_c = 53 × (2π × 1/4.3)² × 8 = 905.29 N.

The force at top would be;

F_top = F_c - mg

F_top = 905.29 - (9.81 × 53) N

F_top = 385.36 N

The force at the top of ride will be "385.36 N".

Force and mass:

According to the question,

Rider's mass, m = 52 kg

Time, t = 4.3 s

Diameter, d = 16 m

Radius, r = [tex]\frac{16}{2}[/tex] = 8 m

Frequency, f = [tex]\frac{1}{t}[/tex] = [tex]\frac{1}{4.3}[/tex] Hz

We know the formula,

Centrifugal force,  [tex]F_c[/tex] = mω²R

or,

Angular velocity, ω = 2πf

By substituting the values in the above formula,

[tex]F_c = 53(2\pi \times (\frac{1}{4.3})^2\times 8 )[/tex]

    [tex]= 905.29[/tex] N

hence,

The top force will be:

→ [tex]F_{top} = F_c[/tex] - mg

By substituting the values,

          [tex]= 905.29-(9.81\times 53)[/tex]

          [tex]= 385.36[/tex] N

Thus the above response is correct.  

Find out more information about force here:

https://brainly.com/question/12970081

The bases of developing convective cumulus clouds will be relatively higher at a location with a relatively ______ difference between the surface temperature and surface dew point temperature.

Answers

Answer:

large

Explanation:

Cumulus clouds is a term in metrology that defines the type of clouds which are characterized by its low altitude, puffy appearance, and fair-weather nature. They are generally considered as low-level clouds, with less than than 2,000m in altitude except they are the more vertical cumulus congestus form.

Thus, it can be noted that, the difference between the surface dew point temperature and the surface temperature is related to relative humidity. Hence, in a situation when there is a LARGE difference between the surface temperature and the surface dewpoint temperature, then the relative humidity is very low (e.g., 10%).

Therefore, the bases of developing convective cumulus clouds will be relatively higher at a location with a relatively LARGE difference between the surface temperature and surface dew point temperature.

A small cylinder made of a diamagnetic material is brought near a bar magnet. One end of the cylinder is brought near the North pole of the magnet and is repelled. What happens when the other end of the cylinder is brought near the North pole of the magnet? Group of answer choices

Answers

Answer:

it attracts

Explanation:

since in a magnetic body there are two poles

(north and south poles)if the first pole was repeled when brought near the North Pole therefore the other end is going to attarct because the first end was also a North Pole while the second end will be a south pole

Transverse waves are sent along a 4.50 m long string with a speed of 85.00 m/s. The string is under a tension of 20.00 N. What is the mass of the string (in kg)?

Answers

Answer:

m = 0.0125 kg

Explanation:

Let us apply the formula for the speed of a wave on a string that is under tension:

[tex]v = \sqrt{\frac{F}{\mu} }[/tex]

where F = tension force

μ = mass per unit length

Mass per unit length is given as:

μ  = m / l

where m = mass of the string

l = length of the string

This implies that:

[tex]v = \sqrt{\frac{F}{m/l} }\\ \\v = \sqrt{\frac{F * l}{m} }[/tex]

Let us make mass, m, the subject of the formula:

[tex]v^2 = \frac{F * l}{m}\\\\m = \frac{F * l}{v^2}[/tex]

From the question:

F = 20 N

l = 4.50 m

v = 85 m/s

Therefore:

[tex]m = \frac{20 * 4.5}{85^2}\\\\m = \frac{90}{7225}\\ \\m = 0.0125 kg[/tex]

A projectile is launched from ground level with an initial speed of 47 m/s at an angle of 0.6 radians above the horizontal. It strikes a target 1.7 seconds later. What is the vertical distance from where the projectile was launched to where it hit the target.

Answers

Answer:

30.67m

Explanation:

Using one of the equations of motion as follows, we can describe the path of the projectile in its horizontal or vertical displacement;

s = ut ± [tex]\frac{1}{2} at^2[/tex]               ------------(i)

Where;

s = horizontal/vertical displacement

u = initial horizontal/vertical component of the velocity

a = acceleration of the projectile

t = time taken for the projectile to reach a certain horizontal or vertical position.

Since the question requires that we find the vertical distance from where the projectile was launched to where it hit the target, equation (i) can be made more specific as follows;

h = vt ± [tex]\frac{1}{2} at^2[/tex]               ------------(ii)

Where;

h = vertical displacement

v = initial vertical component of the velocity = usinθ

a = acceleration due to gravity (since vertical motion is considered)

t = time taken for the projectile to hit the target

From the question;

u = 47m/s, θ = 0.6rads

=> usinθ = 47 sin 0.6

=> usinθ = 47 x 0.5646 = 26.54m/s

t = 1.7s

Take a = -g = -10.0m/s   (since motion is upwards against gravity)

Substitute these values into equation (ii) as follows;

h = vt - [tex]\frac{1}{2} at^2[/tex]

h = 26.54(1.7) - [tex]\frac{1}{2} (10)(1.7)^2[/tex]

h = 45.118 - 14.45

h = 30.67m

Therefore, the vertical distance is 30.67m        

The voltage difference between the AA and AAA batteries should be quite small. What then might be the difference between them?

Answers

Answer:

The major difference is the capacity of both batteries. The AA battery has a higher capacity (a higher current) than the AAA battery.

Explanation:

The AA batteries and the AAA batteries are very similar in their voltage; both of them have 1.5 V.

The difference between these two batteries is their size and also the current that they have. The AAA battery is smaller than the AA battery, which means that the amount of electrochemical material is lower, so the AA battery has a higher capacity (a higher current) than the AAA battery. Generally, AA battery has 2400 mAh capacity and AAA battery has a capacity of 1000mAh; this means that AA battery has almost three times the capacity of an AAA battery.      

Furthermore, the size of the AA battery makes it more common than the AAA battery and therefore has higher commercial demand.                                  

I hope it helps you!

In a high school swim competition, a student takes 1.6 s to complete 1.5 somersaults. Determine the average angular speed of the diver, in rad/s, during this time interval.

Answers

Answer:

The  angular speed is [tex]w = 5.89 \ rad/s[/tex]

Explanation:

From the question we are told that

    The time taken is  [tex]t = 1.6 s[/tex]

    The number of somersaults  is n  =  1.5

The total angular displacement during the somersault is mathematically represented as

         [tex]\theta = n * 2 * \pi[/tex]

substituting values

        [tex]\theta = 1.5 * 2 * 3.142[/tex]

       [tex]\theta = 9.426 \ rad[/tex]

 The angular speed is mathematically represented as

         [tex]w = \frac{\theta }{t}[/tex]

substituting values

         [tex]w = \frac{9.426}{1.6}[/tex]

          [tex]w = 5.89 \ rad/s[/tex]

     

An 100 V/m electric field is directed along the x axis. If the potential at the origin is 300 V, what is potential at the point ( -2m, 0) point

Answers

Answer:

200volts

Explanation:

Pls see attached file

Answer:

100 V

Explanation:

Electric field E = 100 V/m

Potential at the origin = 300 V

Potential at point (-2m, 0) i.e 2 m behind the origin = ?

From the equation ΔV = EΔd,

ΔV = [tex]V_{0} - V_{x}[/tex]

where [tex]V_{0}[/tex] is the potential at origin,

and [tex]V_{x}[/tex] is the potential at point (-2, 0)

E = electric field

Δd = 0 - (-2) = 2 m

[tex]V_{0} - V_{x}[/tex] = 300 - [tex]x[/tex]

equating, we have

 300 - [tex]x[/tex] = 100 x 2

300 - [tex]x[/tex] = 200

[tex]x[/tex] = 100 V

A conducting sphere 45 cm in diameter carries an excess of charge, and no other charges are present. You measure the potential of the surface of this sphere and find it to be 14 kV relative to infinity. Find the excess charge on this sphere.

Answers

Answer:

The excess charge is [tex]Q = 3.5 *10^{-7} \ C[/tex]

Explanation:

From the question we are told that

      The diameter is [tex]d = 45 \ cm = 0.45 \ m[/tex]

       The potential of the surface is [tex]V = 14 \ kV = 14 *10^{3} \ V[/tex]

     

The radius of the sphere is  

           [tex]r = \frac{d}{2}[/tex]

substituting values

          [tex]r = \frac{0.45}{2}[/tex]

         [tex]r = 0.225 \ m[/tex]

The potential on the surface is mathematically represented as  

          [tex]V = \frac{k * Q }{r }[/tex]

Where k is coulomb's constant with value  [tex]k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]

  given from the question that there is no other charge the Q is the excess charge  

Thus  

        [tex]Q = \frac{V* r}{ k}[/tex]

substituting values

        [tex]Q = \frac{14 *10^{3} 0.225}{ 9*10^9}[/tex]

        [tex]Q = 3.5 *10^{-7} \ C[/tex]

         

In the photoelectric effect, if the intensity of light shone on a metal increases, what will happen?

Answers

Answer:

C) There will be more electrons ejected

Explanation:

The number of electrons ejected whenever a photoelectric effect is identified it is proportional to the intensity of the incident light

Nevertheless, the photoelectrons' maximal kinetic energy is independent of their light intensity

Therefore, the maximum speed of the electron ejected doesn't really depend on the light intensity

So,  if the intensity rises, only the number of electrons ejected will rised

Therefore the option c is correct

Answer:

C) There will be more electrons ejected

Explanation:

In the photoelectric effect, photons with an energy of E are shone upon a piece of metal, and if the energy of the photons overcome the work function ϕ of the metal, then electrons with will be ejected from the metal with a kinetic energy KE.

E_photon = Φ + KE

Each photon is capable of ejecting one electron from the metal. Therefore, increasing the intensity of the light (the number of photons shone on the metal) will increase the number of electrons ejected from the metal.

I attach a 4.1 kg block to a spring that obeys Hooke's law and supply 3.8 J of energy to stretch the spring. I release the block and it oscillates with a period of 0.13 s. What is the amplitude of oscillation

Answers

Answer:

The amplitude of the oscillation is 2.82 cm

Explanation:

Given;

mass of attached block, m = 4.1 kg

energy of the stretched spring, E = 3.8 J

period of oscillation, T = 0.13 s

First, determine the spring constant, k;

[tex]T = 2\pi \sqrt{\frac{m}{k} }[/tex]

where;

T is the period oscillation

m is mass of the spring

k is the spring constant

[tex]T = 2\pi \sqrt{\frac{m}{k} } \\\\k = \frac{m*4\pi ^2}{T^2} \\\\k = \frac{4.1*4*(3.142^2)}{(0.13^2)} \\\\k = 9580.088 \ N/m\\\\[/tex]

Now, determine the amplitude of oscillation, A;

[tex]E = \frac{1}{2} kA^2[/tex]

where;

E is the energy of the spring

k is the spring constant

A is the amplitude of the oscillation

[tex]E = \frac{1}{2} kA^2\\\\2E = kA^2\\\\A^2 = \frac{2E}{k} \\\\A = \sqrt{\frac{2E}{k} } \\\\A = \sqrt{\frac{2*3.8}{9580.088} }\\\\A = 0.0282 \ m\\\\A = 2.82 \ cm[/tex]

Therefore, the amplitude of the oscillation is 2.82 cm

A rubber ball is attached to a string and whirled around in a circle. If the string is 1.0 m long (measured from the center of the baseball to the far end of the string) and the ball’s speed is 10 m/s, what is the ball’s centripetal acceleration?

Answers

Centripetal acceleration = (speed squared) / (radius)

Centripetal acceleration = (10 m/s)² / (1.0 m)

Centripetal acceleration = (100 m²/s²) / (1.0 m)

Centripetal acceleration = 100 m/s²

Four identical charges particles of charge 1Uc, 2Uc,
3Uc and 4Uc
are placed at x = lm, x=2m,
x=3m and
x=5m. The electric field intensity
at origin is?

Answers

Answer:

17.94 kN/C is the electric field intensity at the origin due to the charges.

Explanation:

From the question, we are told that

The distance of 1 μC from origin = 1 m

The distance of 2 μC from origin = 2 m

The distance of 3 μC from origin = 3 m

The distance of 4 μC from origin = 5 m

Therefore, for us to find the electric field intensity, we'll solve below:

The formula for Electric field intensity = ( k * q ) / ( r * r )

where , r is distance ,

k = 9 * 10^9 ,

and , q is charge .

now ,

electric field intensity at the origin = [ k * 10^(-6) / 1 * 1 ] +[ k * 2 * 10^(-6) / 2 * 2 ] + [ k * 3 * 10^(-6) / 3 * 3 ] + [ k * 4 * 10^(-6) / 5 * 5 ]

=> electric field intensity at the origin = k * 10^(-6) [ 1 + 1/2 + 1/3 + 4/25 ] N/C

=> electric field intensity at the origin = 9 * 10^9 * 10^(-6) * 1.99 N/C

=> electric field intensity at the origin = 17.94 kN/C

How many excess electrons must be distributed uniformly within the volume of an isolated plastic sphere 20.0 cm in diameter to produce an electric field of 1450 N/C just outside the surface of the sphere

Answers

Answer:

1.007 × 10^(10) electron

Explanation:

We are given;

Electric Field;E = 1450 N/C

Diameter;d = 20 cm = 0.2 m

So, Radius: r = 0.2/2 = 0.1 m

Formula for Electric field just outside the sphere is given by the formula;

E = kq/r²

Where;

E is the magnitude of the electric field. q is the magnitude of the point charge r is distance from the point charge

k is a constant with a value of 9 x 10^(9) N.m²/C²

Making q the subject, we have;

q = Er²/k

Thus,

q = 1450 × 0.1²/(9 × 10^(9))

q = 1.61 × 10^(-9) C

Now, total charge q is also given by the formula;

q = Ne

Where;

e is charge on electron which is 1.6 × 10^(-19)

N is number of excess electrons

Making N the formula, we have;

N = q/e

N = (1.61 × 10^(-9))/(1.6 × 10^(-19))

N = 1.007 × 10^(10) electron

symbol of science hhshsjsiwtwwisjzhJava​

Answers

Answer:

is this a company name.? java is a computer software right..

Determine the maximum height and range of a projectile fired at a height of 6 feet above the ground with an initial velocity of 100 feet per second at an angle of 40 degrees above the horizontal.Maximum heightRange Question 20 options:a) 70.56 feet183.38 feet b) 92.75 feet310.59 feet c) 92.75 feet183.38 feet d) 70.56 feet314.74 feet e)

Answers

Answer:

C is the correct answer

Explanation:

g The force of kinetic friction for a particular pair of interacting objects is always _____ the force of static friction. less than greater than equal to None of the above

Answers

Answer:

less than

Explanation:

The force of kinetic friction for a particular pair of interacting objects is always less than the force of static friction.

The force of static friction between two surfaces is always higher than the force of kinetic friction.

Two long, parallel wires carry currents in the same direction. If I1 = 10 A, and I2 = 20 A, and they are d = 1.0 m apart, what is the magnetic field at a point P midway between them?

Answers

Answer:

The magnetic field at mid point between two parallel wires is 1.2 x 10⁻⁵ T

Explanation:

Given;

current in the first wire, I₁ = 10 A

current in the second wire, I₂ = 20 A

distance between the two wires, d = 1.0 m

Magnetic field at mid point between two parallel wires is calculated as;

[tex]B = \frac{\mu_o I_1}{2\pi r} + \frac{\mu_o I_2}{2\pi r} \\\\B = \frac{\mu_o }{2\pi r}(I_1 +I_2)[/tex]

where;

r is the midpoint between the wires, = 0.5 m

μ₀ is the permeability of free space, = 4π x 10⁻⁷

[tex]B = \frac{\mu_o }{2\pi r}(I_1 +I_2)\\\\B = \frac{4\pi*10^{-7} }{2\pi *0.5}(10 +20)\\\\B = \frac{4\pi*10^{-7} *30}{2\pi *0.5}\\\\B = 1.2 *10^{-5} \ T[/tex]

Therefore, the magnetic field at mid point between two parallel wires is 1.2 x 10⁻⁵ T

Two cylindrical resistors are made from the same material. The shorter one has length L, diameter D, and resistance R1. The longer one has length 8L, diameter 4D, and resistance R2. How do the resistances of these two resistors compare

Answers

Answer:

the resistance of the longer one is twice as big as the resistance of the shorter one.

Explanation:

Given that :

For the shorter cylindrical resistor

Length = L

Diameter = D

Resistance = R1

For the longer cylindrical resistor

Length = 8L

Diameter = 4D

Resistance = R2

So;

We all know that the resistance of a given material can be determined by using the formula :

[tex]R = \dfrac{\rho L }{A}[/tex]

where;

A = πr²

[tex]R = \dfrac{\rho L }{\pi r ^2}[/tex]

For the shorter cylindrical resistor ; we have:

[tex]R = \dfrac{\rho L }{\pi r ^2}[/tex]

since 2 r = D

[tex]R = \dfrac{\rho L }{\pi (\frac{2}{2 \ r}) ^2}[/tex]

[tex]R = \dfrac{ 4 \rho L }{\pi \ D ^2}[/tex]

For the longer cylindrical resistor ; we have:

[tex]R = \dfrac{\rho L }{\pi r ^2}[/tex]

since 2 r = D

[tex]R = \dfrac{ \rho (8 ) L }{\pi (\frac{2}{2 \ r}) ^2}[/tex]

[tex]R = \dfrac{32\rho L }{\pi \ (4 D) ^2}[/tex]

[tex]R = \dfrac{2\rho L }{\pi \ (D) ^2}[/tex]

Sp;we can equate the shorter cylindrical resistor to the longer cylindrical resistor as shown below :

[tex]\dfrac{R_s}{R_L} = \dfrac{ \dfrac{ 4 \rho L }{\pi \ D ^2}}{ \dfrac{2\rho L }{\pi \ (D) ^2}}[/tex]

[tex]\dfrac{R_s}{R_L} ={ \dfrac{ 4 \rho L }{\pi \ D ^2}}* { \dfrac {\pi \ (D) ^2} {2\rho L}}[/tex]

[tex]\dfrac{R_s}{R_L} =2[/tex]

[tex]{R_s}=2{R_L}[/tex]

Thus; the resistance of the longer one is twice as big as the resistance of the shorter one.

At what temperature will silver have a resistivity that is two times the resistivity of iron at room temperature? (Assume room temperature is 20° C.)

Answers

Answer:

The temperature of silver at this given resistivity is 2971.1 ⁰C

Explanation:

The resistivity of silver is calculated as follows;

[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\[/tex]

where;

Rt is the resistivity of silver at the given temperature

Ro is the resistivity of silver at room temperature

α is the temperature coefficient of resistance

To is the room temperature

T is the temperature at which the resistivity of silver will be two times the resistivity of iron at room temperature

[tex]R_t = R_o[1 + \alpha(T-T_o)]\\\\\R_t = 1.59*10^{-8}[1 + 0.0038(T-20)][/tex]

Resistivity of iron at room temperature = 9.71 x 10⁻⁸ ohm.m

When silver's resistivity becomes 2 times the resistivity of iron, we will have the following equations;

[tex]R_t,_{silver} = 2R_o,_{iron}\\\\1.59*10^{-8}[1 + 0.0038(T-20)] =(2 *9.71*10^{-8})\\\\\ \ (divide \ through \ by \ 1.59*10^{-8})\\\\1 + 0.0038(T-20) = 12.214\\\\1 + 0.0038T - 0.076 = 12.214\\\\0.0038T +0.924 = 12.214\\\\0.0038T = 12.214 - 0.924\\\\0.0038T = 11.29\\\\T = \frac{11.29}{0.0038} \\\\T = 2971.1 \ ^0C[/tex]

Therefore, the temperature of silver at this given resistivity is 2971.1 ⁰C

A woman is standing at the rim of a nonuniform cylindrical horizontal platform initially at rest. The platform is free to rotate about frictionless orthogonal axle that goes through its center and has 4 m in diameter and moment of inertia of 500 kgm2. The woman then starts walking along the rim in clockwise direction at a constant speed of 1.50 m/s relative to the Earth. If a woman has 60 kg, how much work does she do to set herself and the platform into motion?

Answers

Answer:

e

Explanation:

i took it myself and got it right

Besides the gravitational force, a 2.80-kg object is subjected to one other constant force. The objectstarts from rest and in 1.20 s experiences a displacement of (4.20 i - 3.30 j) m, where the direction of jis the upward vertical direction. Determine the other force.

Answers

Answer:

the other force= (16.3i + 14.6j)N

EXPLANATION:

Given:

Mass=2.80-kg

t= 1.2s

Since the object started from rest, the origin is (0,0) which symbolize the the object's initial position.

We will need to calculate the magnitude of the displacement using the below formula;

d = (1/2)at2 + v0t + d0

But note that

d0 = 0,( initial position)

v0 = 0( initial position)

a is the net acceleration

d = √[4.202 + (-3.30)2] m = 5.34 m

Hence, the magnitude of the displacement is 5.34 m, then we can make 'a' the subject of formula in the above expression in order to calculate the value for acceleration, note that d0 = 0,( initial position) and v0 = 0( initial position)

d = (1/2)at2

a = 2d/t2 = 2(5.34)/(1.20)2 m/s2 = 7.42 m/s2

the net acceleration is 7.42 m/s2

Acceleration in terms of the vector can be calculated as

a=2(ri - r0)/t^2

Where t =1.2s which is the time

a= 2(4.2i - 3.30j)/ 1.2^2

a=( 5.83i - 4.58j)m/s

now the net force can now be calculated since we have known the value of acceleration, using the formula below;

F(x) = ma - mg

Where a = 5.83i - 4.58j)m/s and g= 9.8m/s

2.8(5.83i - 4.58j)m/s - (2.80 × 9.8)m/s^2

Therefore, the other force= (16.3i + 14.6j)N

A rod has length 0.900 mm and mass 0.500 kgkg and is pivoted at one end. The rod is not uniform; the center of mass of the rod is not at its center but is 0.500 mm from the pivot. The period of the rod's motion as a pendulum is 1.49 ss. What is the moment of inertia of the rod around the pivot

Answers

Answer:

The moment of inertia is  [tex]I =0.14 \ kg \cdot m^2[/tex]

Explanation:

From the question we are told that

    The length of the rod is  [tex]l = 0.900 \ m[/tex]

     The mass of the rod is  [tex]m = 0.500 \ kg[/tex]

      The distance of the center of mass from the pivot is  [tex]d = 0.500 \ m[/tex]

      The period of the rod's motion is  [tex]T = 1.49 \ s[/tex]

Generally the period of the motion is mathematically represented as

       [tex]T = 2 \pi * \sqrt{\frac{I}{m* g * d} }[/tex]

Where [tex]I[/tex] is the moment of inertia about the pivot so making [tex]I[/tex] the subject of formula

      [tex]I = [\frac{T}{2\pi } ]^2 * m * g * d[/tex]

substituting values

        [tex]I = [\frac{1.49}{2* 3.142 } ]^2 * 0.5 * 9.8 * 0.5[/tex]

       [tex]I =0.14 \ kg \cdot m^2[/tex]

(5 pt) You tie a cord to a pail of water, and your swing the pail in a vertical circular 0.700 m. What is the minimum speed must you give the pail at the highest point of the circle if no water is to spill from it

Answers

Answer:

The minimum speed required  is 2.62m/s

Explanation:

The value of  gravitational acceleration = g = 9.81 m/s^2

Radius of the vertical circle = R = 0.7 m

Given the mass of the pail of water = m

The speed at the highest point of the circle = V

The centripetal force will be needed must be more than the weight of the pail of water in order to not spill water.

Below is the calculation:

[tex]\frac{mV^{2}}{R} = mg[/tex]

[tex]V = \sqrt{gR}[/tex]

[tex]V = \sqrt{9.81 \times 0.7}[/tex]

[tex]V = 2.62 m/s[/tex]

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