A nozzle with a radius of 0.22 cm is attached to a garden hose with a radius of 0.89 cm that is pointed straight up. The flow rate through hose and nozzle is 0.55 L/s.
Randomized Variables
rn = 0.22 cm
rh = 0.94 cm
Q = 0.55
1. Calculate the maximum height to which water could be squirted with the hose if it emerges from the nozzle in m.
2. Calculate the maximum height (in cm) to which water could be squirted with the hose if it emerges with the nozzle removed assuming the same flow rate.

Answers

Answer 1

Answer:

1. 0.2m

1. 66m

Explanation:

See attached file

A Nozzle With A Radius Of 0.22 Cm Is Attached To A Garden Hose With A Radius Of 0.89 Cm That Is Pointed
Answer 2

The expressions of fluid mechanics allows to find the result for the maximum height that the water leaves through the two points are;

1) The maximum height when the water leaves the hose is: Δy = 0.20 m

2) The maximum height of the water leaves the nozzle is: Δy = 68.6m

Given parameters

The flow rate  Q = 0.55 L/s = 0.55 10⁻³ m³ / s Nozzle radius r₁ = 0.22 cm = 0.22 10⁻² m Hose radius r₂ = 0.94 cm = 0.94 10⁻² m

To find

   1. Maximum height of water in hose

  2. Maximum height of water at the nozzle

Fluid mechanics studies the movement of fluids, liquids and gases in different systems, for this it uses two expressions:

The continuity equation. It is an expression of the conservation of mass in fluids.

           A₁v₁ = A₂.v₂

Bernoulli's equation. Establishes the relationship between work and the energy conservation in fluids.

          P₁ + ½ ρ g v₁² + ρ g y₁ = P₂ + ½ ρ g v₂² + ρ g y₂

Where the subscripts 1 and 2 represent two points of interest, P is the pressure, ρ the density, v the velocity, g the acceleration of gravity and y the height.

1, Let's find the exit velocity of the water in the hose.

Let's use subscript 1 for the nozzle and subscript 2 for the hose.

The continuity equation of the flow value that must be constant throughout the system.

      Q = A₁ v₁

      v₁ = [tex]\frac{Q}{A_1 }[/tex]  

The area of ​​a circle is:

     A = π r²

Let's calculate the velocity in the hose.

    A₁ = π (0.94 10⁻²) ²

    A₁ = 2.78 10⁻⁴ m²

    v₁ = [tex]\frac{0.55 \ 10^{-3}}{2.78 \ 10^{-4}}[/tex]

    v₁ = 1.98 m / s

Let's use Bernoulli's equation.

When the water leaves the hose the pressure is atmospheric and when it reaches the highest point it has not changed P1 = P2

      ½ ρ v₁² + ρ g y₁ = ½ ρ v₂² + ρ g v₂

      y₂-y₁ = ½  [tex]\frac{v_i^2 - v_2^2}{g}[/tex]  

At the highest point of the trajectory the velocity must be zero.

     y₂- y₁ = [tex]\frac{v_1^2}{2g}[/tex]

Let's calculate

     y₂-y₁ =  [tex]\frac{1.98^2}{2 \ 9.8}[/tex]  

     Δy = 0.2 m

 

2.  Let's find the exit velocity of the water at the nozzle

          A₁ = π r²

          A₁ = π (0.22 10⁻²) ²

          A₁ = 0.152 10⁻⁴ m / s

With the continuity and flow equation.

           Q = A v

            v₁ = [tex]\frac{Q}{A}[/tex]  

             v₁ = [tex]\frac{0.55 \ 10{-3} }{0.152 \ 10^{-4} }[/tex]  

             v₁ = 36.67 m / s

Using Bernoulli's equation, where the speed of the water at the highest point is zero.

           y₂- y₁ =  [tex]\frac{v^1^2}{g}[/tex]  

Let's calculate.

           Δy =  [tex]\frac{36.67^2 }{2 \ 9.8 }[/tex]  

           Δy = 68.6m

In conclusion using the expressions of fluid mechanics we can find the results the maximum height that the water leaves through the two cases are:

      1) The maximum height when the water leaves the hose is:

          Δy = 0.20 m

      2) The maximum height of the water when it leaves the nozzle is:

          Δy = 68.6 m

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Related Questions

Astronomers think planets formed from interstellar dust and gases that clumped together in a process called? A. stellar evolution B. nebular aggregation C. planetary accretion D. nuclear fusion

Answers

Answer:

C. planetary accretion

Explanation:

Astronomers think planets formed from interstellar dust gases that clumped together in a process called planetary accretion.

Answer:

[tex]\boxed{\sf C. \ planetary \ accretion }[/tex]

Explanation:

Astronomers think planets formed from interstellar dust and gases that clumped together in a process called planetary accretion.

Planetary accretion is a process in which huge masses of solid rock or metal clump together to produce planets.

A 4.00-Ω resistor, an 8.00-Ω resistor, and a 24.0-Ω resistor are connected together. (a) What is the maximum resistance that can be produced using all three resistors? (b) What is the minimum resistance that can be produced using all three resistors? (c) How would you connect these three resistors to obtain a resistance of 10.0 Ω? (d) How would you connect these three resistors to obtain a resistance of 8.00 Ω?

Answers

Answer:a) 4+8+24=36

B) 1/4+1/8+1/24=10

C) yu will connect them in parallel connection.

D) you will connect two in parallel then the remaining one in series to the ons connected in parallel.

Explanation:

(a)The maximum resistance that can be produced using all three resistors will be 36 ohms.

(b)The minimum resistance that can be produced using all three resistors will be 10 ohms.

(c)The three resistors to obtain a resistance of 10.0 Ω will be in the parallel connection.

(d) You connect these three resistors to obtain a resistance of 8.00 Ω will be in parallel. Two will be linked in parallel, and the last one will be connected in series to the two that are connected in parallel.

What is resistance?

Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Resistance is the enemy of the flow of current.

The maximum resistance that can be produced using all three resistors is obtained by adding all the given resistance;

[tex]\rm R_{max}=(4 +8+24 )\ ohms \\\\ R_{max}=36 \ ohms[/tex]

The minimum resistance that can be produced using all three resistors is obtained when connected in the parallel.

[tex]\rm R_{min}=\frac{1}{4} +\frac{1}{8} +\frac{1}{24} \\\\ R_{min}=10 \ ohm[/tex]

(c)The three resistors to obtain a resistance of 10.0 Ω will be in the parallel connection.

(d) You connect these three resistors to obtain a resistance of 8.00 Ω will be in parallel. Two will be linked in parallel, and the last one will be connected in series to the two that are connected in parallel.

Hence,the maximum resistance that can be produced using all three resistors will be 36 ohms.

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Consider a series RLC circuit where R=25.0 Ω, C=35.5 μF, and L=0.0940 H, that is driven at a frequency of 70.0 Hz. Determine the phase angle ϕ of the circuit in degrees.

Answers

Answer:

137.69°

Explanation:

The phase angle of an RLC circuit  ϕ is expressed as shoen below;

ϕ = [tex]tan^{-1} \dfrac{X_l-X_c}{R}[/tex]

Xc is the capacitive reactance = 1/2πfC

Xl is the inductive reactance = 2πfL

R is the resistance = 25.0Ω

Given C = 35.5 μF, L = 0.0940 H, and frequency f = 70.0Hz

Xl = 2π * 70*0.0940

Xl = 41.32Ω

For the capacitive reactance;

Xc = 1/2π * 70*35.5*10⁻⁶

Xc = 1/0.0156058

Xc = 64.08Ω

Phase angle ϕ = [tex]tan^{-1} \frac{41.32-64.08}{25} \\\\[/tex]

ϕ = [tex]tan^{-1} \frac{-22.76}{25} \\\\\\\\[/tex]

[tex]\phi = tan^{-1} -0.9104\\\\\phi = -42.31^0[/tex]

Since tan is negative in the 2nd quadrant;

[tex]\phi = 180-42.31^0\\\\\phi = 137.69^0[/tex]

Hence the phase angle ϕ of the circuit in degrees is 137.69°

The phase angle ϕ of the series RLC circuit that is driven at a frequency of 70.0 Hz is ϕ = 137.69°

Phase angle:

Given that:

capacitance C = 35.5 μF,

Inductance L = 0.0940 H,

The resistance R = 25.0Ω

and frequency f = 70.0Hz

The capacitive reactance is given by:

Xc = 1/2πfC

Xc = 1/2π × 70 × 35.5× 10⁻⁶

Xc = 1/0.0156058

Xc = 64.08Ω

The inductive reactance is given by:

Xl = 2πfL

Xl = 2π × 70 × 0.0940

Xl = 41.32Ω

The phase angle of an RLC circuit ϕ  is given by:

[tex]\phi=tan^{-1}\frac{X_l-X_c}{R}\\\\\phi=tan^{-1}\frac{41.32-64.08}{25}[/tex]

Ф = -42.31°

Since tan is negative in the 2nd quadrant, thus:

ϕ = 180° - 42.31°

ϕ = 137.69°

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A heat engine operates between 200 K and 100 K. In each cycle it takes 100 J from the hot reservoir, loses 25 J to the cold reservoir, and does 75 J of work. This heat engine violates the second law but not the first law of thermodynamics. Why is this true?

Answers

Answer:

It does not violate the first law because the total energy taken is what is used 100J = 25J + 75J

But violates 2nd lawbecause the engine has a higher energy after doing work than the initial for e.g A cold object in contact with a hot one never gets colder, transferring heat to the hot object and making it hotter confirming the second law

Which is one criterion that materials of a technological design should meet? They must be imported. They must be affordable. They must be naturally made. They must be locally produced.

Answers

Answer:

they must be affordable because they have to pay for it or they wont get the stuff they are bying.

Explanation:

need a brainliest please.

Answer: B, they must be affordable.

Explanation:

How could a country benefit from making it into space?

Answers

Answer:

space exploration pays off in goods, technology, and paychecks. The work is done by people who are paid to do it here on Earth. The money they receive helps them buy food, get homes, cars, and clothing. They pay taxes in their communities, which helps keep schools going, roads paved, and other services that benefit a town or city. The money may be spent to send things "up there", but it gets spent "down here." It spreads out into the economy.

What is the answer?​

Answers

Answer: i think it is d. none of them.

Explanation: The speed of light in a vacuum is 186,282 miles per second and so when you look and the answer choices and the question it doesnt make any since.

Terms to describe the opposition by a material.to being magnetised is

Answers

Answer:

Repulsion

Explanation:

Which of the following explains why a “control” is important in a case-control study of a disease? The researchers need to control the bias that those who contracted the disease may create when they talk to others. The researchers need to compare those who contracted the disease to those who did not. The researchers need to compare those who contracted the disease to those who contracted previous diseases. The researchers need to control the disease so that it is not spread further.

Answers

The researchers need to compare those who contracted the disease to those who did not.

The mass (M) of a piece of metal is directly proportional to its volume (V), where the proportionality constant is the density (D) of the metal. (1) Write an equation that represents this direct proportion, in which D is the proportionality constant. The density of lead metal is 11.3 g/cm3. (2) What is the mass of a piece of lead metal that has a volume of 17.3 cm3

Answers

Answer:

1) M = 11.3V2) 195.49 grams

Explanation:

1) If the mass (M) of a piece of metal is directly proportional to its volume (V), where the proportionality constant is the density (D) of the metal, this is expressed mathematically as shown;

M ∝ V

M = kV

For every proportionality sign, there will always be a proportionality constant 'k'

Since the proportionality constant is the density (D) of the metal, the equation will become;

M = DV

Given the density to be 11.3 g/cm3, the equation will become;

M = 11.3V

Hence, the equation that represents this direct proportion, in which D is the proportionality constant with metal density of 11.3g/cm³ is M = 11.3V

2) If the volume of the metal is 17.3cm³, on substituting this values into the equation in (1) to get the mass of the metal, we will have;

M = 11.3V

M = 11.3 * 17.3

M = 195.49 grams

Hence, the mass of a piece of lead metal that has a volume of 17.3 cm³ is 195.49 grams.

6. What is the bulk modulus of oxygen if 32.0 g of oxygen occupies 22.4 L and the speed of sound in the oxygen is 317 m/s?

Answers

Answer:

[tex] \boxed{\sf Bulk \ modulus \ of \ oxygen \approx 143.5 \ kPa} [/tex]

Given:

Mass of oxygen (m) = 32.0 g = 0.032 kg

Volume occupied by oxygen (V) = 22.4 L = 0.0224 m³

Speed of sound in oxygen (v) = 317 m/s

To Find:

Bulk modulus of oxygen

Explanation:

[tex]\sf Density \ of \ oxygen \ (\rho) = \frac{m}{V}[/tex]

[tex]\sf \implies Bulk \ modulus \ of \ oxygen \ (B) = v^{2} \rho[/tex]

[tex]\sf \implies B = v^{2} \times\frac{m}{V}[/tex]

[tex]\sf \implies B = {(317)}^{2} \times \frac{0.032}{0.0224} [/tex]

[tex]\sf \implies B = {(317)}^{2} \times 1.428[/tex]

[tex]\sf \implies B = 100489 \times 1.428[/tex]

[tex]\sf \implies B = 143498.292 \: Pa[/tex]

[tex]\sf \implies B \approx 143.5 \: kPa[/tex]

A parallel-plate vacuum capacitor has 7.72 J of energy stored in it. The separation between the plates is 3.30 mm. If the separation is decreased to 1.45 mm, For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Stored energy. Part A what is the energy now stored if the capacitor was disconnected from the potential source before the separation of the plates was changed

Answers

Answer

3.340J

Explanation;

Using the relation. Energy stored in capacitor = U = 7.72 J

U =(1/2)CV^2

C =(eo)A/d

C*d=(eo)A=constant

C2d2=C1d1

C2=C1d1/d2

The separation between the plates is 3.30mm . The separation is decreased to 1.45 mm.

Initial separation between the plates =d1= 3.30mm .

Final separation = d2 = 1.45 mm

(A) if the capacitor was disconnected from the potential source before the separation of the plates was changed, charge 'q' remains same

Energy=U =(1/2)q^2/C

U2C2 = U1C1

U2 =U1C1 /C2

U2 =U1d2/d1

Final energy = Uf = initial energy *d2/d1

Final energy = Uf =7.72*1.45/3.30

(A) Final energy = Uf = 3.340J

Rod cells in the retina of the eye detect light using a photopigment called rhodopsin. 1.8 eV is the lowest photon energy that can trigger a response in rhodopsin. Part A What is the maximum wavelength of electromagnetic radiation that can cause a transition

Answers

Answer:

The maximum wavelength of the e-m wave is 6.9 x 10^-7 m

Explanation:

Energy required to trigger a response = 1.8 eV

we convert to energy in Joules.

1 eV = 1.602 x 10^-19 J

1.8 eV = [tex]x[/tex] J

[tex]x[/tex] = 1.8 x 1.602 x 10^-19 = 2.88 x 10^-19 J

The energy of an electromagnetic wave is gotten as

E = hf

where

h is the Planck's constant = 6.63 x 10^-34 J-s

and f is the frequency of the wave.

substituting values, we have

2.88 x 10^-19 = 6.63 x 10^-34 x f

f = (2.88 x 10^-19)/(6.63 x 10^-34)

f = 4.34 x 10^14 Hz

We know that the frequency of an e-m wave is given as

f = c/λ

where

c is the speed of light = 3 x 10^8 m/s

λ is the wavelength of the e-m wave

From this we can say that

λ = c/f

λ = (3 x 10^8)/(4.34 x 10^14)

λ = 6.9 x 10^-7 m

Find the rms current delivered by the power supply when the frequency is very large. Answer in units of A.

Answers

Answer:

The rms current is 0.3112 A.

Explanation:

Given that,

Suppose, The capacitance is 170 μF and the inductance is 2.94 mH. The resistance in the top branch is 278 Ohms, and in the bottom branch is 151 Ohms. The potential of the power supply is 47 V .

We know that,

When the frequency is very large then the capacitance can be treated as a short circuit and inductance as open circuit.

So,

We need to calculate the rms current

Using formula of current

[tex]I=\dfrac{V}{R}[/tex]

Where, V = voltage

R = resistance

Put the value into the formula

[tex]I=\dfrac{47}{151}[/tex]

[tex]I= 0.3112 \ A[/tex]

Hence, The rms current is 0.3112 A.

Light of wavelength 520 nm is incident a on a diffraction grating with a slit spacing of 2.20 μm , what is the angle from the axis for the third order maximum?

Answers

Answer:

θ = 45.15°

Explanation:

We need to use the grating equation in this question. The grating equation is given as follows:

mλ = d Sin θ

where,

m = order number = 3

λ = wavelength of light = 520 nm = 5.2 x 10⁻⁷ m

d = slit spacing = 2.2 μm = 2.2 x 10⁻⁶ m

θ = angle from the axis = ?

Therefore,

(3)(5.2 x 10⁻⁷ m) = (2.2 x 10⁻⁶ m) Sin θ

Sin θ = (3)(5.2 x 10⁻⁷ m)/(2.2 x 10⁻⁶ m)

Sin θ = 0.709

θ = Sin⁻¹(0.709)

θ = 45.15°

Intelligent beings in a distant galaxy send a signal to earth in the form of an electromagnetic wave. The frequency of the signal observed on earth is 2.2% greater than the frequency emitted by the source in the distant galaxy. What is the speed vrel of the galaxy relative to the earth

Answers

Answer:

Vrel= 0.75c

Explanation:

See attached file

2. The nuclear model of the atom held that
a. electrons were randomly spread through "a sphere of uniform positive
electrification."
b. matter was made of tiny electrically charged particles that were smaller than the
atom
C. matter was made of tiny, indivisible particles.
d. the atom had a dense, positively charged nucleus.​

Answers

Answer:

the atom had a dense, positively charged nucleus.​

Explanation:

Ernest Rutherford, based on the experiment carried out by two of his graduate students, established the authenticity of the nuclear model of the atom.

According to the nuclear model, an atom is made up of a dense positive core called the nucleus. Electrons are found to move round this nucleus in orbits. This is akin to the movement of the planets round the sun in the solar system.

It's nighttime, and you've dropped your goggles into a 3.2-m-deep swimming pool. If you hold a laser pointer 1.2 m above the edge of the pool, you can illuminate the goggles if the laser beam enters the water 2.0 m from the edge.
How far are the goggles from the edge of the pool?

Answers

Answer:

Explanation:

Laser angle with water surface is given by: Tan α = 1/2.0= 0.5/

α = 26.56°

Laser angle with Normal = 90 - 26.56 = 63.44 °

Assuming a red laser, refractive index in water is 1.331.

Angle of refraction in water is given by:

Ref Ind = Sin i / Sin r

1.331 = Sin 63.44 / Sin r

Sin r = 0.8945 / 1.331 = 0.6721

Angle r = 42.22°

For the path in water:

Tan 42.22 = x / 3.2

x = 2.9m where x is the lateral displacement of the laser ince it hits the water

So the goggles are 2.0 + 2.9 = 4.9 m from edge of pool

A flat, circular loop has 18 turns. The radius of the loop is 15.0 cm and the current through the wire is 0.51 A. Determine the magnitude of the magnetic field at the center of the loop (in T).

Answers

Answer:

The magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻ T.

Explanation:

Given;

number of turns of the flat circular loop, N = 18 turns

radius of the loop, R = 15.0 cm = 0.15 m

current through the wire, I = 0.51 A

The magnetic field through the center of the loop is given by;

[tex]B = \frac{N\mu_o I}{2R}[/tex]

Where;

μ₀ is permeability of free space = 4π x 10⁻⁷ m/A

[tex]B = \frac{N\mu_o I}{2R} \\\\B = \frac{18*4\pi*10^{-7} *0.51}{2*0.15} \\\\B = 3.846 *10^{-5} \ T[/tex]

Therefore, the magnitude of the magnetic field at the center of the loop is 3.846 x 10⁻ T.

Determine the next possible thickness of the film (in nm) that will provide the proper destructive interference. The index of refraction of the glass is 1.58 and the index of refraction of the film material is 1.48.

Answers

Answer:

I know the answer

Explanation:

We want to choose the film thickness such that destructive interference occurs between the light reflected from the air-film interface (call it wave 1) and from the film-lens interface (call it wave 2). For destructive interference to occur, the phase difference between the two waves must be an odd multiple of half-wavelengths.

You can think of the phases of the two waves as second hands on a clock; as the light travels, the hands tick-tock around the clock. Consider the clocks on the two waves in question. As both waves travel to the air-film interface, their clocks both tick-tock the same time-no phase difference. When wave 1 is reflected from the air-film boundary, its clock is set forward 30 seconds; i.e., if the hand was pointing toward 12, it's now pointing toward 6. It's set forward because the index of refraction of air is smaller than that of the film.

Now wave 1 pauses while wave two goes into and out of the film. The clock on wave 2 continues to tick as it travels in the film-tick, tock, tick, tock.... Clock 2 is set forward 30 seconds when it hits the film-lens interface because the index of refraction of the film is smaller than that of the lens. Then as it travels back through the film, its clock still continues ticking. When wave 2 gets back to the air-film interface, the two waves continue side by side, both their clocks ticking; there is no change in phase as they continue on their merry way.

So, to recap, since both clocks were shifted forward at the two different interfaces, there was no net phase shift due to reflection. There was also no phase shift as the waves travelled into and out from the air-film interface. The only phase shift occured as clock 2 ticked inside the film.

Call the thickness of the film t. Then the total distance travelled by wave 2 inside the film is 2t, if we assume the light entered pretty much normal to the interface. This total distance should equal to half the wavelength of the light in the film (for the minimum condition; it could also be 3/2, 5/2, etc., but that wouldn't be the minimum thickness) since the hand of the clock makes one revolution for each distance of one wavelength the wave travels (right?).

Select from the following for the next two questions:
A virtual, inverted and smaller than the object
B real, inverted and smaller than the object
C virtual, upright and smaller than the object
D real, upright and larger than the object
E virtual, upright and larger than the object
F real, inverted and larger than the object
G virtual, inverted and larger than the object
H real, upright and smaller than the object
An object is placed 46.9 cm away from a converging lens. The lens has a focal length of 10.0 cm. Select the statement from the list above which best describes the image an objesthse place 46.9 cm away from a spherical convex mirror. The radius of curvature of the mirror is 20.0 cm. Select the statement from the An object is placed 46.9 cm away from a spherical convex mirror. The radius of curvature of the mirror is 20.0 cm. Select the statement from the list above which best describes the image.

Answers

Answer:

Explanation:

1 )

An object is placed 46.9 cm away from a converging lens. The lens has a focal length of 10.0 cm.

Since the object is placed at a distance more than twice the focal length , its image will be inverted , real  and will be of the size less than the size of object . So option B is applied .

B)  real, inverted and smaller than the object.

2 )

An object is placed 46.9 cm away from a spherical convex mirror. The radius of curvature of the mirror is 20.0 cm.

The object is placed at a point beyond its radius of curvature, its image will be formed at a point between f and C   or between focal point and centre of curvature . Its size will be smaller than size of object and it will be real and inverted .

B)  real, inverted and smaller than the object.

Grocery store managers contend that there is less total energy consumption in the summer if the store is kept at a low temperature. Make arguments to support or refute this claim, taking into account that there are numerous refrigerators and freezers in the store.

Answers

Answer:

Argument in favor of less total energy consumption if the store is kept at a low temperature

Explanation:

Have in mind that if the store has numerous refrigerators and freezers, the energy consumption of those machines have to be included into the analysis.

Recall that the efficiency (or Coefficient Of Performance - COP) of a frezzer or refrigerator is inversely proportional to the temperature difference between the inside of th machine and the environment where it is operation, therefore the smaller the difference, the highest their efficiency. Therefore, the cooler the environment (the temperature at which the store is kept) the better performance of the running refrigerators and freezers.

A steel bridge is 1000 m long at -20°C in winter. What is the change in length when the temperature rises to 40°C in summer? The average coefficient of linear expansion of this steel is 11 × 10-6 C-1.

Answers

Answer:

ΔL = 0.66 m

Explanation:

The change in length on an object due to rise in temperature is given by the following equation of linear thermal expansion:

ΔL = αLΔT

where,

ΔL = Change in Length of the bridge = ?

α = Coefficient of linear thermal expansion = 11 x 10⁻⁶ °C⁻¹

L = Original Length of the Bridge = 1000 m

ΔT = Change in Temperature =  Final Temperature - Initial Temperature

ΔT = 40°C - (-20°C) = 60°C

Therefore,

ΔL = (11 x 10⁻⁶ °C⁻¹)(1000 m)(60°C)

ΔL = 0.66 m

A 50kg block slides down a slope that forms an angle of 54 degrees if it is known that when descending it has a force of 40N and a coefficient of friction of 0.33. What is the acceleration in the block?

Answers

Answer:

The acceleration in the block is 2.1 m/s²

Explanation:

Given that,

Mass = 50 kg

Angle = 54°

Force = 40 N

Coefficient of friction = 0.33

We need to calculate the acceleration in the block

Using balance equation

[tex]F_{net}=F_{f}-F\cos\theta[/tex]

[tex]ma=\mu mg\sin\theta-F\cos\theta[/tex]

[tex]a=\dfrac{\mu mg\sin\theta-F\cos\theta}{m}[/tex]

Put the value into the formula

[tex]a=\dfrac{0.33\times50\times9.8\sin54-40\cos54}{50}[/tex]

[tex]a=2.1\ m/s^2[/tex]

Hence, The acceleration in the block is 2.1 m/s²

Two motorcycles are traveling in opposite directions at the same speed when one of the cyclists blasts her horn, which has frequency of 544 Hz. The other cyclist hears the frequency as 563 Hz. If the speed of sound in air is 344 m/s, what is the speed of the motorcycles

Answers

Answer:

6ms^-1

Explanation:

Given that the frequency difference is

( 563- 544) = 19

So alsoThe wavelength of each wave is = v/f = 344 /544

and there are 19 of this waves

So it is assumed that each motorcycle has moved 0.5 of this distance

in one second thus the speed of the motorcycles will be

=> 19/2 x 344/544 = 6.0 m/s

An emf is induced by rotating a 1060 turn, 20.0 cm diameter coil in the Earth's 5.25 ✕ 10−5 T magnetic field. What average emf (in V) is induced, given the plane of the coil is originally perpendicular to the Earth's field and is rotated to be parallel to the field in 10.0 ms? V †

Answers

Answer:

The average emf induced in the coil is 175 mV

Explanation:

Given;

number of turns of the coil, N = 1060 turns

diameter of the coil, d = 20.0 cm = 0.2 m

magnitude of the magnetic field,  B = 5.25 x 10⁻⁵ T

duration of change in field, t = 10 ms = 10 x 10⁻³ s

The average emf induced in the coil is given by;

[tex]E = N\frac{\delta \phi}{dt} \\\\E = N\frac{\delta B}{\delta t}A[/tex]

where;

A is the area of the coil

A = πr²

r is the radius of the coil = 0.2 /2 = 0.1 m

A = π(0.1)² = 0.03142 m²

[tex]E = \frac{NBA}{t} \\\\E = \frac{1060*5.25*10^{-5}*0.03142}{10*10^{-3}} \\\\E = 0.175 \ V\\\\E = 175 \ mV[/tex]

Therefore, the average emf induced in the coil is 175 mV

If you wish to observe features that are around the size of atoms, say 5.5 × 10^-10 m, with electromagnetic radiation, the radiation must have a wavelength of about the size of the atom itself.


Required:

a. What is its frequency?

b. What type of electromagnetic radiation might this be?

Answers

Answer:

a) 5.5×10^17 Hz

b) visible light

Explanation:

Since the wavelength of the electromagnetic radiation must be about the size of the about itself, this implies that;

λ= 5.5 × 10^-10 m

Since;

c= λ f and c= 3×10^8 ms-1

f= c/λ

f= 3×10^8/5.5 × 10^-10

f= 5.5×10^17 Hz

The electromagnetic wave is visible light

A converging lens 7.50 cm in diameter has a focal length of 330 mm . For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of resolving power of the human eye. Part A If the resolution is diffraction limited, how far away can an object be if points on it transversely 4.10 mm apart are to be resolved (according to Rayleigh's criterion) by means of light of wavelength 600 nm

Answers

Answer:

D Is 430m

Explanation:

See attached file

A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several times. A 5.0 μg dust particle is suspended in midair just above the center of the carpet.

Required:
What is the charge on the dust particle?

Answers

Answer:

The  charge on the dust particle is  [tex]q_d = 6.94 *10^{-13} \ C[/tex]

Explanation:

From the question we are told that

    The length is  [tex]l = 2.0 \ m[/tex]

    The width is  [tex]w = 4.0 \ m[/tex]

   The charge is  [tex]q = -10\mu C= -10*10^{-6} \ C[/tex]

    The mass suspended in mid-air is [tex]m_a = 5.0 \mu g = 5.0 *10^{-6} \ g = 5.0 *10^{-9} \ kg[/tex]

   

Generally the electric field on the carpet is mathematically represented as

           [tex]E = \frac{q}{ 2 * A * \epsilon _o}[/tex]

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

substituting values

           [tex]E = \frac{-10*10^{-6}}{ 2 * (2 * 4 ) * 8.85*10^{-12}}[/tex]

           [tex]E = -70621.5 \ N/C[/tex]

Generally the electric force keeping the dust particle on the air  equal to the force of gravity acting on the particles

        [tex]F__{E}} = F__{G}}[/tex]

=>     [tex]q_d * E = m * g[/tex]

=>      [tex]q_d = \frac{m * g}{E}[/tex]

=>      [tex]q_d = \frac{5.0 *10^{-9} * 9.8}{70621.5}[/tex]

=>     [tex]q_d = 6.94 *10^{-13} \ C[/tex]

6. If you wanted to develop a telescope, what kind of lenses would you use for the objective lens (the lens that collects the light) and the eyepiece? Explain your reasoning. Draw a picture with ray tracing of your setup.

Answers

Answer:

objetive: a converging lens for large diameter lenses

eyepiece you must select a lens with a small focal length and the diameter is not important

The selected lenses should decrease chromatic aberration.

Explanation:

A telescope is an instrument that collects light from very distant objects, therefore very weak.

Therefore you should select a converging lens for large diameter lenses, to collect magnanimous light and with a large focal length.

For the eyepiece you must select a lens with a small focal length and the diameter is not important

the telescope magnification is

                 m = f_objective / F_ocular

The selected lenses should decrease chromatic aberration.

In general, these lenses are heavy, so refractory telescopes were imposed, so it uses a concave mirror instead of an objective lens.

Answer: this the real answer try it objetive: a converging lens for large diameter lenseseyepiece you must select a lens with a small focal length and the diameter is not importantThe selected lenses should decrease chromatic aberration.Explanation:A telescope is an instrument that collects light from very distant objects, therefore very weak.Therefore you should select a converging lens for large diameter lenses, to collect magnanimous light and with a large focal length.For the eyepiece you must select a lens with a small focal length and the diameter is not importantthe telescope magnification is                 m = f_objective / F_ocularThe selected lenses should decrease chromatic aberration.In general, these lenses are heavy, so refractory telescopes were imposed, so it uses a concave mirror instead of an objective lens.

Explanation:

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