"A parcel moving in a horizontal direction with speed v0 = 13 m/s breaks into two fragments of weights 1.4 N and 1.9 N, respectively. The speed of the larger piece remains horizontal immediately after the separation and increases to v1.9 = 29 m/s. Find the necessary speed and direction of the smaller piece immediately after the separation. (Assume the initial direction of the parcel is positive. Indicate the direction with the sign of your answer.)"

Answer:

The highest rms voltage will be 8.485 V

Explanation:

For alternating electric current, rms (root means square) is equal to the value of the direct current that would produce the same average power dissipation in a resistive load

If the peak or maximum voltage should not exceed 12 V, then from the relationship

[tex]V_{rms} = \frac{V_{p} }{\sqrt{2} }[/tex]

where [tex]V_{rms}[/tex] is the rms voltage

[tex]V_{p}[/tex] is the peak or maximum voltage

substituting values into the equation, we'll have

[tex]V_{rms} = \frac{12}{\sqrt{2} }[/tex] = 8.485 V

The two types of electromagnetic waves that have higher frequencies than the waves that make up ultraviolet light are gamma rays and X-rays.

Electromagnetic waves are components of the electromagnetic spectrum, which is made up of the following:

Each electromagnetic wave have a specific frequency and wavelength.

However, the two types of electromagnetic waves that have higher frequencies than the waves that make up ultraviolet light are gamma rays and X-rays.

Learn more about electromagnetic waves at: https://brainly.com/question/8553652

Answer:

gamma rays and X-rays

Explanation:

d on edge I got 100%

Answer:

The  acceleration is  [tex]a = 6.2 m/s^2[/tex]

Explanation:

From the question we are told that

    The  angle which the inclined plane make with horizontal is  [tex]\theta = 0.44 \ rad[/tex]

     The frictional coefficients are  [tex]\mu_{\mu s} = 0.61[/tex] and  [tex]\mu_{\mu k} = 0.23[/tex]

The force acting on the  crate  is mathematically represented as  

        [tex]f = F_w + F_N[/tex]

Here f is the net force at which the crate is sliding down the plane which is mathematically represented as

      [tex]f = ma[/tex]

        [tex]F_w[/tex] is the force due to weight which is mathematically represented as

        [tex]F_w = mg sin (\theta)[/tex]

       and  [tex]F_N[/tex] the force due to friction which is mathematically represented as

       [tex]F_N = \mu_{\mu k } * mg cos(\theta )[/tex]

So  

     [tex]ma = mgsin(\theta ) + \mu_{\mu k} mg cos(\theta )[/tex]

      [tex]a = gsin(\theta ) + \mu_{\mu k } * g cos(\theta)[/tex]

substituting values

      [tex]a = 9.8 sin(0.44 ) + 0.23 * 9.8* cos(0.44)[/tex]

      [tex]a = 6.2 m/s^2[/tex]

Answer:

velocity = distance/time

velocity = wavelength × frequency

Both of these are commonly known equations to calculate velocity with different variables.

Answer:

The magnetic field strength : 0.372 T

Explanation:

The equation of the induced emf is given by the following equation,

( Equation 1 ) emf = - N ( ΔФ / Δt ) - where N = number of turns of the coil, ΔФ = change in the magnetic flux, and Δt = change in time

The equation for the magnetic flux is given by,

( Equation 2 ) Ф = BA( cos( θ ) ) - where B = magnetic field, A = area, and θ = the angle between the normal and the magnetic field

The area of the circular coil is a constant, as well as the magnetic field. Therefore the change in the magnetic flux is due to the angle between the normal and the magnetic field. Therefore you can expect the equation for the change in magnetic flux to be the same as the magnetic flux, but only that there must be a change in θ.

( Equation 3 ) ΔФ = BA( Δcos( θ ) )

Now as the coil rotates one-fourth of a revolution, θ changes from 0 degrees to 90 degrees. The " change in cos θ " should thus be the following,

Δcos( θ ) = cos( 90 ) - cos( 0 )

= 0 - 1 = - 1

Let's substitute that value in the third equation,

( Substitution of Δcos( θ ) previously, into Equation 3 )

ΔФ = BA( - 1 ) = - BA

Remember the first equation? Well if the change in the magnetic flux = - BA, then through further substitution, the emf should = - N( - BA ) / Δt. In other words,

emf =  - N( - BA ) / Δt,

emf = NBA / Δt,

B = ( emf )Δt / NA

Now that we have B, the magnetic field strength, isolated, let's solve for the area of the circular coil and substitute all known values into this equation.

Area ( A ) = πr²,

= π( 0.275 )² = 0.2376 m²,

B = ( 10,000 V )( 4.42 [tex]*[/tex] 10⁻³ s ) / ( 500 )( 0.2376 m² ) = ( About ) 0.372 T

The magnetic field strength : 0.372 T

Answer:

The tension in the string at that point is 90.75 N

Explanation:

Given;

mass of the object, m = 3 kg

length of string, r = 0.25 m

the angular velocity, ω = 11 rad/s

The tension on string can be equated to the centrifugal force on the object;

T = mω²r

Where;

T is the tension in the string

m is mass of the object

ω is the angular velocity

r is the radius of the circular path

T = 3 x (11)² x 0.25

T = 90.75 N

Therefore, the tension in the string at that point is 90.75 N

Explanation:

ω = 45 rev/min × (2π rad/rev) × (1 min / 60 s) = 4.71 rad/s

r = 10 cm = 0.10 m

1) The linear speed is:

v = ωr

v = (4.71 rad/s) (0.10 m)

v = 0.471 m/s

2) The linear acceleration in the tangential direction is 0.

The linear acceleration in the radial direction is:

a = v² / r

a = (0.471 m/s)² / (0.10 m)

a = 2.22 m/s²

Answer:

A) V_rms = 29 V

B) Vav = 0 V

Explanation:

A) We are told that;

V = V_o cos ωt

voltage amplitude; V = V_o = 41.0V

Now, the formula for the root-mean-square potential difference Vrms is given as;

V_rms = V/√2

Thus plugging in relevant values, we have;

V_rms = 41/√2

V_rms = 29 V

B) Due to the fact that the voltage is sinusoidal from the given V = V_o cos ωt, we can say that the average potential difference Vav between the two terminals of the power supply would be zero.

Thus; Vav = 0 V

A. The root-mean-square potential difference ([tex]V_{rms}[/tex]) is equal to 28.99 Volts.

B. For this voltage with a sinusoidal waveform (sine wave), the average potential difference ([tex]V_{ave}[/tex]) between the two terminals of the power supply is equal to zero (0).

Given the following data:

The voltage across the terminals of an alternating current (AC) power supply varies directly with time according to the equation:

[tex]V_0 = V_0cos(t)[/tex]

A. To find the root-mean-square potential difference ([tex]V_{rms}[/tex]):

Mathematically, root-mean-square for voltage in an alternating current (AC) power supply (circuit) is given by the formula:

[tex]V_{rms} = \frac{V}{\sqrt{2} }[/tex]

Substituting the given parameter into the formula, we have;

[tex]V_{rms} = \frac{41}{\sqrt{2} }\\\\V_{rms} = \frac{41}{1.4142 }\\\\V_{rms} = 28.99\; Volts[/tex]

B. To find the average potential difference ([tex]V_{ave}[/tex]) between the two terminals of the power supply:

For this voltage with a sinusoidal waveform (sine wave), the average potential difference ([tex]V_{ave}[/tex]) between the two terminals of the power supply is equal to zero (0).

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Answer:

a. The speed of the wave is 319.1m

b. The tension in the string is 41.74N

Explanation:

Please see the attachments below

The correct answer is C. The level of competition is not very high in most Masters  programs.

Explanation:

In sports, the word "master" is used to define athletes older than 30 and that usually are professional or have trained for many years, although novates are also allowed. This means in most cases in Master programs and teams a high level of competition can be expected due to the experience and extensive training of Master athletes. Indeed, many records in the field of sport belong to Master athletes rather than younger athletes. According to this, the incorrect statement is "The level of competition is not very high in most Masters  programs".

Answer:

3. Only accelerated motion can be sensed

Explanation:

Without windows on such a train, you'd have no frame of reference for your speed. By that I mean, without being able to see how fast you are moving past other things, it's almost as if you aren't moving at all... almost.

At rest you obviously aren't moving and in uniform motion, with a constant speed, it would feel as though you aren't moving. But during periods of acceleration you'll feel the force on your body (F=ma) and would be able to tell if you were moving in a particular direction.

You've probably felt this before. Maybe not on a windowless train but perhaps in a car or on a roller coaster. Speeding up makes you go back into your seat a bit and slowing down makes you lean forward a bit. Both speeding up and slowing down are examples of acceleration (just in different directions) and how fast you accelerate will affect how much force you experience.

So the answer would be option 3.

Side note: If the train wasn't smooth riding then there would be some amount of friction going on and you could probably tell if you were in motion by the products of that friction (like sound and vibrations) even at a constant speed.

Answer

aim directly at the image

Explanation

the light from the laser beam will also bend when it hits the air water interface , so aim directly at the fish

Answer:

Explanation:

For this exercise we will use that the potential is created by the charge inside the equinoctial surface and just like in Gauss's law we can consider all the charge concentrated in the center.

Therefore the potential on the ferric surface is

        V = k Q / r

where k is the Coulomb constant, Q the charge of the sphere and r the distance from the center to the point of interest

a) On the surface the potential

        V = 9 10⁹ Q / 0.5

        V = 18 10⁹ Q

Unfortunately you did not write the value of the load, suppose a value to complete the calculations Q = 1 10⁻⁷ C, with this value the potential on the surfaces V = 1800 V

b) The equipotential surfaces are concentric spheres, let's look for the radii for some potentials

for V = 1300V let's find the radius

             r = k Q / V

             r = 9 109 1 10-7 / 1300

             r = 0.69 m

other values ​​are shown in the following table

V (V)      r (m)

1800     0.5

1300     0.69

 800      1,125

 300     3.0

In other words, we draw concentric spheres with these radii and each one has a potential difference of 500V

C) The spacing of the spheres corresponds to lines of radii of the electric field that have the shape

         E = k Q / r²

the maximum magnitude is 5.5

Answer:

Option (e) = The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere.

Explanation:

So, we are given the following set of infomation in the question given above;

=> "spherical Gaussian surface of radius R centered at the origin."

=> " A charge Q is placed inside the sphere."

So, the question is that if we are to maximize the magnitude of the flux of the electric field through the Gaussian surface, the charge should be located where?

The CORRECT option (e) that is " The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere." Is correct because of the reason given below;

REASON: because the charge is "covered" and the position is unknown, the flux will continue to be constant.

Also, the Equation that defines Gauss' law does not specify the position that the charge needs to be located, therefore it can be anywhere.

Answer:

It will neither translate in the opposite direction nor .rotate so as to be at right angles, it will also neither rotate so as to be vertical direction

Answer:

Explanation:

The double slit interference phonemene is described for the case of constructive interference

          d sin θ= m λ                   (1)

let's use trigonometry to find the sinus

        tan θ = y / L

in general in interference phenomena the angles are small

       tan θ = sin θ / cos θ = sin θ

The double slit interference phonemene is described for the case of constructive interference

          d sin θ = m lam                    (1)

let's use trigonometry to find the sinus

        tan θ = y / L

in general in interference phenomena the angles are small

       tan θ = sin θ / cos θ = sin θ

we substitute

      sin θ = y / L

we substitute in equation 1

         d y / L = m λ

         λ = dy / L m

let's reduce the magnitudes to the SI system

  d = 0.44 mm = 0.44 10⁻³ m

  y = 5.5 mm = 5.5 10⁻³ m

  L = 4.2m

  m = 1

let's calculate

        λ = 0.44  10⁻³ 5.5 10⁻³ / (4.2 1)

        λ = 5.76190 10-7 m

let's reduce to num

  lam = 5.56190 10-7 m (109 nm / 1m)

  lam = 556,190 nmtea

we substitute

      without tea = y / L

we substitute in equation 1

         d y / L = m lam

         lam = dy / L m

let's reduce the magnitudes to the SI system

  d = 0.44 me = 0.44 10-3 m

  y = 5.5 mm = 5.5 10-3

  L = 4.2m

  m = 1

let's calculate

        lam = 0.44 10⁻³  5.5 10⁻³ / (4.2 1)

        lam = 5.76190 10⁻⁷ m

let's reduce to num

  lam = 5.56190 10⁻⁷ m (109 nm / 1m)

  lam = 556,190 nm

Answer:

3m

Explanation:

89.1 MHz means

89.1×10^6 cycles/second.

Electromagnetic radiation (including radio waves) travel at

3.0×10^8meters/second

Wavelength = Speed/Frequency

The wavelength of a

89.1MHz radio signal is

3.0×10^8/89.1x10^6

= 0.03x10^2

= 3meters

Answer:

3.18 m/s

Explanation:

Given that

Initial speed of the ball, u = 20 m/s

Angle of inclination, θ = 45°

Distance from the ball, h = 50 m

Using equations of projectile to solve this, we have

We start by finding the time of flight, T

T = 2Usinθ/g

T = (2 * 20 * sin45)/9.8

T = (40 * 0.7071) / 9.8

T = 28.284/9.8

T = 2.89 s

Next we find the Range, R

R = u²sin2θ/g

R = (20² * sin 90) / 9.8

R = (400 * 1) / 9.8

R = 400/9.8 = 40.82 m

Distance the gk must cover

40.82 - 50 m

-9.18 m or 9.18 m in the opposite direction.

Speed of the GK = d/t

9.18 / 2.89 = 3.18 m/s

Answer: Weight truck+load = 29.4×[tex]10^{3}[/tex] N

Explanation: When an object floats in a fluid, there is an upward force, caused by the liquid, acting on the object that opposes the weight of the immersed object. This force is called Buyoyant Force and is determined by:

B = d*V*g

where

d is density of the fluid;

V is volume of liquid displaced due to the immersed object;

g is acceleration due to gravity;

For the truck, the system is in equilibrium, which means buyoyant force is equal weight. Then:

Volume displaced is

V = 10*8*0.0375

V = 3 [tex]m^{3}[/tex]

Density of water: 1000kg/[tex]m^{3}[/tex]

[tex]F_{P} = F_{B}[/tex]

[tex]F_{P}[/tex] = 1000*3*9.8

[tex]F_{P}[/tex] = 29.4×[tex]10^{3}[/tex] N

The weight of the truck and the load is 29.4×[tex]10^{3}[/tex] Newtons

Answer:

Rotate a piece of polaroid film about an axis perpendicular to the ray while looking through it in that sky direction.

Explanation:

Polarization involves constraining a transverse wave e.g light waves to vibrate in one phase only. Since unpolarized light vibrates in all direction during propagation. Polarization can be achieved by a polaroid.

A polaroid is a material the make transverse waves to vibrate in one direction after passing through it. It has various applications in sun glasses, wind shield of a car etc.

If the slit of the polaroid is perpendicular to the polarized light from a particular direction in the sky, there would be no propagation of the light. But when it is parallel to the polarized light from the direction, the light would propagate through the polaroid.

Answer:

Explanation:

Using the formula for calculating the resistivity of an object to find the diameter.

Resistivity P = RA/L

R is the resistance of the material

A is the cross sectional area

L is the length of the material

Since A = πd²/4

P = R( πd²/4)/L

P = Rπd²/4L ... 1

If the second wire of the same material and length is found to have resistance R/9, the resistivity of the second material will be;

P₂ = (R/9)A₂/L₂

P₂ = (R/9)(πd₂²/4)/L₂

P₂ = (Rπd₂²/36)/L₂

P₂ = (Rπd₂²)/36L₂

Since the length and resistivity are the same;

P = P₂  and L =L₂

Equating 1 and 2;

Rπd²/4L =  (Rπd₂²)/36L₂

Rπd²/4L =  (Rπd₂²)/36L

d² = d₂²/9

d₂² = 9d²

Taking the square root of both sides;

√d₂² = √9d²

d₂ = 3d

Therefore the diameter of the second wire is 3 times that of the initial wire

Answer:

Infrared telescope and camera

Explanation:

An infrared telescope uses infrared light to detect celestial bodies. The infrared radiation is one of the known forms of electromagnetic radiation. Infrared radiation is given off by a body possessing some form of heat. All bodies above the absolute zero temperature in the universe radiates some form of heat, which can then be detected by an infrared telescope, and infrared radiation can be used to study or look into a system that is void of detectable visible light.

Stars are celestial bodies that are constantly radiating heat. In order to see a clearer picture of the these bodies, Infrared images is better used, since they are able to penetrate the surrounding clouds of dust, and have located many more stellar components than any other types of telescope, especially in dusty regions of star clusters like the Trapezium cluster.

Answer:

The peak electric field is  [tex]E_o = 3593.6 V/m[/tex]

Explanation:

From the question we are told that

     The power is  [tex]P = 1.2 \ kW = 1.2 *10^{3} \ W[/tex]

     The cross-sectional area is  [tex]A = 700 \ cm^2 = 700 *10^{-4} \ m^2[/tex]

Generally the average intensity of the  microwave is mathematically represented as

      [tex]I = \frac{c * \epsilon _o * E_o^2 }{2}[/tex]

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

     and  [tex]\epsilon_o[/tex] is the permitivity of free space with value [tex]\epsilon_o = 8.85*10^{-12} \ m^{-3} \cdot kg^{-1}\cdot s^4 \cdot A^2[/tex]

   also [tex]E_o[/tex] is the peak electric field.

Now making [tex]E_o[/tex] the subject [tex]E_o = \sqrt{\frac{2 * I }{ c * \epsilon _o } }[/tex]

But this intensity of the  microwave can also be represented mathematically as

       [tex]I = \frac{ P }{A }[/tex]

substituting values

      [tex]I = \frac{ 1.2 *10^{3} }{700 *10^{-4} }[/tex]]  

      [tex]I = 17142.85 \ W/m^2[/tex]

So

      [tex]E_o = \sqrt{\frac{2 * 17142.85 }{ 3.0*10^{8}] * 8.85*10^{-12} } }[/tex]

      [tex]E_o = 3593.6 V/m[/tex]

The peak electric field of the microwave is 3,593.1 V/m.

The given parameters;

The intensity of the wave is calculated as follows;

[tex]I = \frac{P}{A} \\\\I = \frac{1,200}{700 \times 10^{-4}} \\\\I = 17,142.86 \ W/m^2[/tex]

The peak electric field is calculated as follows;

[tex]E_o = \sqrt{\frac{2I}{c \varepsilon _o} } \\\\E_o = \sqrt{\frac{2\times 17,142.86}{3\times 10^8 \times 8.85 \times 10^{-12}} } \\\\E_o = 3,593.1 \ V/m[/tex]

Thus, the peak electric field of the microwave is 3,593.1 V/m.

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Answer:

Total Force that the air inside the house exerts upward on the ceiling is 5.25 × 10⁶ lb

Explanation:

Force =  Atmospheric Pressure × Floor Area

Where; Standard Atmospheric Pressure = 2100 lb/ft²

            Floor Area = 2500 ft²

Substitute the data

∴ Total Force = 2100 lb/ft² × 2500 ft²

  Total Force = 5.25 × 10⁶ lb

Answer:

The  amount by which the second filter reduces the intensity of light emerging from the first filter is

     z =  0.60

Explanation:

From the question we are told that

    The angle between the axes is  [tex]\theta = 41^o[/tex]

The intensity of polarized light that emerges from the second filter is  mathematically represented as

         [tex]I= I_o cos^2 \theta[/tex]

 Where [tex]I_o[/tex] is the intensity of light emerging  from the first filter

        [tex]I = I_o [cos(41.0)]^2[/tex]

      [tex]I =0.60 I_o[/tex]

This means that the second filter reduced the intensity by z =  0.60

Answer:

1) on the surface you can see the slits with equal spacing, on the one hand and on the other hand it is smooth.

2)If the angle is zero we see a bright light called undispersed light

For different angles we see the colors of the spectrum

3) must be able to see the well-collimated light emission source

Explanation:

1) A diffraction grating (diffraction grating) is a surface on which a series of indentations are drawn evenly spaced.

These crevices or lines are formed by copying a standard metal net when the plastic is melted and after hardening is carefully removed, or if the nets used are a copy of the master net.

The network can be of two types of transmission or reflection, in teaching work the most common is the transmission network, on the surface you can see the slits with equal spacing, on the one hand and on the other hand it is smooth.

The number of lines per linear mm determines which range of the spectrum a common value can be observed to observe the range of viable light is 600 and 1200 lines per mm.

2) when looking through the diffraction grating what we can observe depends on the relative angle between the eye and the normal to the network.

If the angle is zero we see a bright light called undispersed light

For different angles we see the colors of the spectrum, if it is an incandescent lamp we see a continuum with all the colors in the visible range and if it is a gas lamp we see the characteristic emission lines of the gas.

3) Before mounting the grid on the spectrometer, we must be able to see the well-collimated light emission source, this means that it is clearly observed.

The spectrometers have several screws to be able to see the lamp clearly, this is of fundamental importance in optical experiments.

Answer: 117.8 nm

Explanation:

Given,

Nonreflective coating refractive index : n = 1.21

Index of refraction: [tex]n_0[/tex] = 1.52

Wave length of light = λ = 570 nm = [tex]570\times10^{-9}\ m[/tex]

[tex]\text{ Thickness}=\dfrac{\lambda}{4n}[/tex]

[tex]=\dfrac{570\times10^{-9}\ m}{4\times1.21}\\\\\approx\dfrac{117.8\times 10^{-9}\ m}{1}\\\\=117.8\text{ nm}[/tex]

Hence, the minimum thickness of the coating that will accomplish= 117.8 nm

Answer:

 m = 3,265 10⁻²⁰  E

Explanation:

For this exercise we can use Newton's second law applied to our system, which consists of a capacitor that creates the uniform electric field and the drop of oil with two extra electrons.

             ∑ F = 0

             [tex]F_{e}[/tex] - W = 0

the electric force is

             F_{e} = q E

as they indicate that the charge is two electrons

             F_{e} = 2e E

The weight is given by the relationship

             W = mg

we substitute in the first equation

               2e E = m g

               m = 2e E / g

let's put the value of the constants

              m = (2 1.6 10⁻¹⁹ / 9.80) E

               m = 3,265 10⁻²⁰  E

 The value of the electric field if it is a theoretical problem must be given and if it is an experiment it can be calculated with measures of the spacing between plates and the applied voltage, so that the system is in equilibrium

Answer:

A) [tex]V = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}[/tex]

B) [tex] d = 4181.49 kg/m^{3} = 4.18 g/cm^{3} [/tex]

Explanation:

A) Using the Archimedes' force we can find the weight of water displaced:

[tex] W_{d} = W_{a} - W_{w} [/tex]

Where:

[tex]W_{a}[/tex]: is the weight of the block in the air = 20.1 N

[tex]W_{w}[/tex]: is the weight of the block in the water = 15.3 N

[tex] W_{d} = W_{a} - W_{w} = 20.1 N - 15.3 N = 4.8 N [/tex]

Now, the mass of the water displaced is:

[tex] m = \frac{W_{d}}{g} = \frac{4.8 N}{9.81 m/s^{2}} = 0.49 kg [/tex]

The volume of the block can be found using the mass of water displaced and the density of the water:

[tex]V = \frac{m}{d} = \frac{0.49 kg}{997 kg/m^{3}} = 4.92 \cdot 10^{-4} m^{3} = 492 cm^{3}[/tex]

B) The density of the block can be found as follows:

[tex] d = \frac{W_{a}}{g*V} = \frac{20.1 N}{9.81 m/s^{2}*4.92 \cdot 10^{-4} m^{3}} = 4181.49 kg/m^{3} = 4.18 g/cm^{3} [/tex]

I hope it helps you!