A cyclist moves effortlessly at a constant speed of 12 m / s, but enters a muddy area where the coefficient of kinetic friction is 0.60. Will the rider leave the muddy area without having to pedal if the mud extends 11m? If so, how fast will it emerge?

Answers

Answer 1

Answer:

3.5 m/s

Explanation:

There are 3 forces on the cyclist:

Weight force mg pulling down,

Normal force N pushing up,

and friction force Nμ pushing left.

Sum of forces in the y direction:

∑F = ma

N − mg = 0

N = mg

∑F = ma

-Nμ = ma

-mgμ = ma

a = -gμ

a = -(10 m/s²)(0.60)

a = -6 m/s²

Velocity reached at end of 11 m:

v² = v₀² + 2aΔx

v² = (12 m/s)² + 2 (-6 m/s²) (11 m)

v = √12 m/s

v ≈ 3.5 m/s


Related Questions

A 590-turn solenoid is 12 cm long. The current in it is 36 A . A straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don't concern us).
What is the magnitude of the force on this wire assuming the solenoid's field points due east?

Answers

Complete Question

A 590-turn solenoid is 12 cm long. The  current in it is 36 A . A 2 cm straight wire cuts through the center of the solenoid, along a 4.5-cm diameter. This wire carries a 27-A current downward (and is connected by other wires that don't concern us).

What is the magnitude of the force on this wire assuming the solenoid's field points due east?

Answer:

The force is  [tex]F = 0.1602 \ N[/tex]

Explanation:

From the question we are told that

   The number of turns is  [tex]N = 590 \ turns[/tex]

   The  length of the solenoid is  [tex]L = 12 \ cm = 0.12 \ m[/tex]

   The current is  [tex]I = 36 \ A[/tex]

   The  diameter is  [tex]D = 4.5 \ cm = 0.045 \ m[/tex]

   The  current carried by the wire is  [tex]I = 27 \ A[/tex]

    The  length of the wire is  [tex]l = 2 cm = 0.02 \ m[/tex]

Generally the magnitude of the force  on this wire assuming the solenoid's field points due east is mathematically represented as

           [tex]F = B * I * l[/tex]

Here  B  is the magnetic field which is mathematically represented as

          [tex]B = \frac{\mu_o * N * I }{L}[/tex]

Here   [tex]\mu _o[/tex] is permeability of free space with value  [tex]\mu_ o = 4\pi *10^{-7} \ N/A^2[/tex]

substituting values

         [tex]B = \frac{4 \pi *10^{-7} * 590 * 36 }{ 0.12}[/tex]

           [tex]B = 0.2225 \ T[/tex]

So

      [tex]F = 0.2225 * 36 * 0.02[/tex]

      [tex]F = 0.1602 \ N[/tex]

Which does not account for the fact that fish can survive the winter in ponds in temperate climate zones? 1. the density of ice versus that of water 2. the unique properties of water 3. the intermolecular bonding of water 4. the tendency for water to freeze from the bottom up

Answers

Answer:

3. the intermolecular bonding of water

Explanation:

Anomalous behavior of water is an advantage in aquatic habitat during winter. Because of some unique properties of water, it behaves irregularly. Thus, a pond or river does not freeze completely during winter.

Water has its highest density when temperature is 4[tex]^{0}C[/tex] , and lowest volume at 4[tex]^{0}C[/tex]. Thus, the denser layers of water sink accordingly until the upper layer is the least dense during winter. This layer then freeze leaving the layers below it unfrozen.

Answer:

D. The tendency for water to freeze from the bottom up.

Explanation:

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

What is the current in milliamperes produced by the solar cells of a pocket calculator through which 4.2 C of charge passes in 2.7 h

Answers

Answer:

0.432mA

Explanation:

Current produced by the solar cells of the pocket calculator is expressed using the formula I = Q/t where;

Q is the charge (in Columbs)

t is the time (in seconds)

Given parameters

Q = 4.2C

t = 2.7 hrs

t = 2.7*60*60

t = 9720 seconds

Required

Current produced by the solar cell I

Substituting the given values into the formula;

I = 4.2/9720

I = 0.000432A

I = 0.432mA

Hence, the current in milliamperes produced by the solar cells of a pocket calculator is 0.432mA

1. What does the acronym LASER stand for? What characteristic of a laser makes it suitable for today's experiment?

Answers

Answer:Light Amplification by Stimulated Emission of Radiation. It is able to convert light or electrical energy into focused high energy beam to treat some sickness and diseases.

Explanation:

Answer:

Light amplification by stimulated emission of radiation

A concave mirror has a focal length of magnitude f. An object is palced in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear:______.
a) behind the mirror.
b) upright and reduced.
c) upright and enlarged.
d) inverted and reduced.
e) inverted and enlarged.

Answers

Answer:

D.

Inverted and reduced

If object is placed in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear upright and reduced.

What is a concave mirror?

When a hollow spherical is divided into pieces and the exterior surface of each cut portion is painted, it forms a mirror, with the inner surface reflecting the light.

A concave mirror is a name for this sort of mirror. An enlarged image is caused when the concave mirror is positioned too near to the object.

A concave mirror has a focal length of magnitude f. An object is placed in front of this mirror at a point 1/2 f from the face of the mirror. The image will appear upright and reduced.

Hence option B is correct.

To learn more about the concave mirror refer to the link;

https://brainly.com/question/25937699

Ratio of the speed of light in a vacuum to the speed of light in a medium Rule for how light is refracted at the boundary between two materials Process that occurs when the angle of incidence is greater than the critical angle

Answers

Answer:

TOTAL INTERNAL REFLECTION

Explanation:

Retraction is defined as the change in the direction of light rays as it moves from less dense medium to a denser medium.

For us to have a critical angle, the ray must be passing from the denser medium to the less dense medium. As the angle of refraction in the less dense medium is increasing, the angle of incidence in the less dense medium also increases. A point will reach when the refracted ray will be parallel to the interface i.e angle of refraction is 90°, the angle of incidence at this point is known as the critical angle. If the angle of refraction keeps increasing further, it will get to a point when the refracted ray becomes reflected into the denser medium. At this stage we say that the ray is internally reflected and this is the point when the angle of incidence is greater than the critical angle.

Hence it can be concluded that the process that occurs when the angle of incidence is greater than the critical angle is called TOTAL INTERNAL REFLECTION

A square loop, length l on each side, is shot with velocity v0 into a uniform magnetic field B. The field is perpendicular to the plane of the loop. The loop has mass m and resistance R, and it enters the field at t = 0s. Assume that the loop is moving to the right along the x-axis and that the field begins at x = 0m.

Required:
Find an expression for the loop's velocity as a function of time as it enters the magnetic field.

Answers

Answer:

v₀(1 + B²L²t/mR)

Explanation:

We know that the force on the loop is F = BIL where B = magnetic field strength, I = current and L = length of side of loop. Now the current in the loop I = ε/R where ε = induced e.m.f in the loop = BLv₀ where v₀ = velocity of loop and r = resistance of loop

F = BIL = B(BLv₀)L/R = B²L²v₀/R  

Since F = ma where a = acceleration of loop and m = mass of loop

a = F/m = B²L²v₀/mR

Using v = u + at where u = initial velocity of loop = v₀, t = time after t = 0 and v = velocity of loop after time t = 0

Substituting the value of a and u into v, we have

v = v₀ + B²L²v₀t/mR

= v₀(1 + B²L²t/mR)

So the velocity of the loop after time t is v = v₀(1 + B²L²t/mR)

The expression for the loop's velocity as a function of time as it enters the magnetic field is v = v₀(1 + B²L²t/mR).

Calculation of the loop velocity:

As we know that

Force on the loop

F = BIL

here

B = magnetic field strength,

I = current

and L = length of side of loop.

Now

the current in the loop I = ε/R

where

ε = induced e.m.f in the loop = BLv₀

where v₀ = velocity of loop

and r = resistance of loop

So,

F = BIL = B(BLv₀)L/R = B²L²v₀/R  

Also, F = ma where a = acceleration of loop and m = mass of loop

Now

a = F/m = B²L²v₀/mR

We have to use

v = u + at

where

u = initial velocity of loop = v₀,

t = time after t = 0

and v = velocity of loop after time t = 0

So, it be like

v = v₀ + B²L²v₀t/mR

= v₀(1 + B²L²t/mR)

Learn more about velocity here: https://brainly.com/question/332163

What is the observed wavelength of the 656.3 nm (first Balmer) line of hydrogen emitted by a galaxy at a distance of 2.40 x 108 ly

Answers

Answer:

λ = 667.85 nm

Explanation:

Let f be the frequency detected by the observer

Let v be the speed at which the observer is moving.

Now, when the direction at which the observer is moving is away from the source, we have the frequency as;

f = f_o√((1 - β)/(1 + β))

From wave equations, we know that the wavelength is inversely proportional to the frequency. Thus, wavelength is now;

λ = λ_o√((1 + β)/(1 - β))

Where, β = Hr/c

H is hubbles constant which has a value of 0.0218 m/s • ly

c is speed of light = 3 × 10^(8) m/s

r is given as 2.40 x 10^(8) ly

Thus,

β = (0.0218 × 2.4 x 10^(8))/(3 × 10^(8))

β = 0.01744

Since we are given λ_o = 656.3 nm

Then;

λ = 656.3√((1 + 0.01744)/(1 - 0.01744))

λ = 667.85 nm

As the frequency of the ac voltage across a capacitor approaches zero, the capacitive reactance of that capacitor:_______.
a. approaches zero.
b. approaches infinity.
c. approaches unity.
d. none of the above.

Answers

Answer:

b. approaches infinity

Explanation:

Because Capacitive reactance is given as Xc = 1/ωC

So we can see that the value of capacitive reactance and therefore its overall impedance (in Ohms) decreases to zero as the frequency increases acting like a short circuit.

Same as the frequency approaches zero or DC, the capacitors reactance increases to infinity, acting like an open circuit which is why capacitors block DC

A microwave oven operates at 2.4 GHz with an intensity inside the oven of 2300 W/m2 . Part A What is the amplitude of the oscillating electric field

Answers

Answer:

The amplitude of the oscillating electric field is 1316.96 N/C

Explanation:

Given;

frequency of the wave, f = 2.4 Hz

intensity of the wave, I = 2300 W/m²

Amplitude of oscillating magnetic field is given by;

[tex]B_o = \sqrt{\frac{2\mu_o I}{c} }[/tex]

where;

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

I is intensity of wave

c is speed of light = 3 x 10⁸ m/s

[tex]B_o = \sqrt{\frac{2*4\pi *10^{-7}*2300}{3*10^8} } \\\\B_o = 4.3899 *10^{-6} \ T[/tex]

The amplitude of the oscillating electric field is given by;

E₀ = cB₀

E₀ = 3 x 10⁸ x 4.3899 x 10⁻⁶

E₀ = 1316.96 N/C

Therefore, the amplitude of the oscillating electric field is 1316.96 N/C

3. Which of the following accurately describes circuits?
O A. In a parallel circuit, the same amount of current flows through each part of the circuit
O B. In a series circuit, the amount of current passing through each part of the circuit may vary
O C. In a series circuit, the current can flow through only one path from start to finish
O D. In a parallel circuit, there's only one path for the current to travel.

Answers

Answer:

Option (c)

Explanation:

In a Series circuit, as the components are connected end-to-end ,the current can flow through only one path from start to finish.

(C.) is the only correct statement in the list of choices.

In a series circuit, the current can flow through only one path from start to finish.

A hydraulic lift raises a 2 000-kg automobile when a 500-N force is applied to the smaller piston. If the smaller piston has an area of 10 cm2, what is the cross-sectional area of the larger piston

Answers

Answer:

The cross-sectional area of the larger piston is 392 cm²

Explanation:

Given;

output mass of the piston, m₀ = 2000 kg

input force of the piston, F₁ = 500 N

input area of the piston, A₁ = 10 cm² = 0.001 m²

The output force is given by;

F₀ = m₀g

F₀ = 2000 x 9.8

F₀ = 19600 N

The cross-sectional area of the larger piston or output area of the piston will be calculated by applying the following equations;

[tex]\frac{F_i}{A_i} = \frac{F_o}{A_o} \\\\A_o= \frac{F_o A_i}{F_i} \\\\A_o = \frac{19600*0.001}{500} \\\\A_o = 0.0392 \ m^2\\\\A_o = 392 \ cm^2[/tex]

Therefore, the cross-sectional area of the larger piston is 392 cm²

When light travels from one medium to another with a different index of refraction, how is the light's frequency and wavelength affected

Answers

Answer:

The frequency does not change, but the wavelength does

Explanation:

Here are the options

A. When a light wave travels from a medium with a lower index of refraction to a medium with a higher index of refraction, the frequency changes and the wavelength does not.

B. The frequency does change, but the wavelength remains unchanged.

C. Both the frequency and wavelength change.

D. When a light wave travels from a medium with a lower index of refraction to a medium with a higher index of refraction, neither the wavelength nor the frequency changes.

E. The frequency does not change, but its wavelength does.

When light goes through one medium to the next, the frequency doesn't really change seeing as frequency is dependent on wavelength and light wave velocity. And when the wavelength shifts from one medium to the next.

[tex]n= \frac{C}{V} \ and\ \frac{\lambda_o}{\lambda_m}[/tex]

where [tex]\lambda_o[/tex] indicates wavelength in vacuum

[tex]\lambda_m[/tex] indicates wavelength in medium

n indicates refractive index

v indicates velocity of light wave

c indicates velocity of light

And wavelength is medium-dependent. Frequency Here = v[tex]\lambda[/tex] and shift in wavelength and velocity, not shifts in overall frequency.

Therefore the correct option is E

A radar installation operates at 9000 MHz with an antenna (dish) that is 15 meters across. Determine the maximum distance (in kilometers) for which this system can distinguish two aircraft 100 meters apart.

Answers

Answer:

R = 36.885 km

Explanation:

In order to distinguish the two planes we must use the Rayleigh criterion that establishes two distinguishable objects if in their diffraction the central maximum of one coincides with the first minimum of the other

The diffraction equation for slits is

            a sin θ = m λ

the first minimum occurs for m = 1

             sin θ = λ a

as the diffraction experiments the angles are very small, we approximate

             sin θ = θ

 

             θ = λ / a

This expression is for a slit, in the case of circular objects, when solving the system in polar coordinates, a numerical constant appears, leaving the expression of the form

            θ = 1.22 λ / a

In this problem they give us the frequency, let's find the wavelength with the relation

           c = λ f

           λ = c / f

           θ = 1.22 c/ f a

since they ask us for the distance between the planes, we can use the definition of radians

          θ = s / R

if we assume that the distance is large, we can approximate the arc to the horizontal distance

          s = x

       

we substitute

             x / R = 1.22 c / fa

             R = x f a / 1.22c

Let's reduce the magnitudes to the SI system

            f = 9000 MHz = 9 109 Hz

            a = 15 m

           x = 100 m

let's calculate

            R = 100 10⁹ 15 / (1.22 3 108)

            R = 3.6885 10⁴ m

let's reduce to km

            R = 3.6885 10¹ km

            R = 36.885 km

What is the direction of the net gravitational force on the mass at the origin due to the other two masses?

Answers

Answer:

genus yds it's the

Explanation:

xmgxfjxfjxgdfjusufzjyhmfndVFHggssjtjhryfjftjsrhrythhrsrhrhsfhsgdagdah vhj

A person can see clearly up close but cannot focus on objects beyond 75.0 cm. She opts for contact lenses to correct her vision.
(a) Is she nearsighted or farsighted?
(b) What type of lens (converging or diverging) is needed to correct her vision?
(c) What focal length contact lens is needed, and what is its power in diopters?

Answers

Answer:

(a) nearsighted

(b) diverging

(c) the lens strength in diopters is 1.33 D, and considering the convention for divergent lenses normally prescribed as: -1 33 D

Explanation:

(a) The person is nearsighted because he/she cannot see objects at distances larger than 75 cm.

(b) the type of correcting lens has to be such that it counteracts the excessive converging power of the eye of the person, so the lens has to be diverging (which by the way carries by convention a negative focal length)

(c) the absolute value of the focal length (f) is given by the formula:

[tex]f=\frac{1}{d} =\frac{1}{0.75} = 1.33\,D[/tex]

So it would normally be written with a negative signs in front indicating a divergent lens.

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]

Suppose you are playing hockey on a new-age ice surface for which there is no friction between the ice and the hockey puck. You wind up and hit the puck as hard as you can. After the puck loses contact with your stick, the puck will

Answers

Answer:

Not slow down or speed up.

Explanation:

Hitting the puck accelerates the speed of the puck from zero to the speed with which it leaves at the instance they lose contact. Since there is no friction between the puck and the ice, there will be no force decelerating or accelerating the hockey puck, allowing the puck to move away and remain in motion without speeding up or slowing down indefinitely theoretically.

help... Please help!!!!!!!!!!!

Answers

Answer:

a) 6.8--5.10 thats equal 11.9

b) m=ris/run +10 equal 0.06/8 =7.5*10^-3

An electron moving in the direction of the +x-axis enters a magnetic field. If the electron experiences a magnetic deflection in the -y direction, the direction of the magnetic field in this region points in the direction of the:______

Answers

Answer:

-z axis

Explanation:

According to the left hand rule for an electron in a magnetic field, hold the thumb of the left hand at a right angle to the rest of the fingers, and the rest of the fingers parallel to one another. If the thumb represents the motion of the electron, and the other fingers represent the direction of the field, then the palm will push in the direction of the force on the electron. In this case, the left hand will be held out with the thumb pointing to the right (+x axis), and the palm facing your body (-y axis). The magnetic field indicated by the other fingers will point down in the the -z axis.

Seismic attenuation and how spherical spreading affect amplitude, can anyone explain this please!

Answers

Answer:

Hey there!

This can be a confusing topic, so it's totally fine if you get confused...

First, Seismic Attenuation is how seismic waves lose energy as they expand and spread.

Secondly, when distance increases, amplitude decreases. This is because the distance (spherical spreading would mean radius) is inversely proportional to amplitude.

Let me know if this helps :)

Explain how surface waves can have characteristics of both longitudinal waves and transverse waves. Please use 3 content related sentences

Answers

Answer: Search Results

Featured snippet from the web

Answer: Surface waves can have characteristics of both longitudinal and transverse waves in the following way; The motion of the surface waves is up and down which is perpendicular to the direction of the wave. This is similar to the motion of transverse waves whereas the the motion of longitudinal.

Explanation:

Surface waves can exhibit characteristics of both longitudinal waves and transverse waves.

Surface waves are a type of mechanical wave that propagate along the interface between two different mediums, such as the ground and air or the surface of water. These waves combine properties of both longitudinal and transverse waves

Similar to longitudinal waves, surface waves involve particles oscillating in the same direction as the wave propagation. This creates compressions and rarefactions, leading to variations in density or pressure. These compressions and rarefactions are characteristic of longitudinal waves.

However, surface waves also exhibit transverse motion. As the wave propagates along the surface, particles move in a perpendicular direction to the wave's motion. This transverse motion causes particles to displace vertically or horizontally, similar to transverse waves.

By combining both longitudinal and transverse characteristics, surface waves possess a complex motion that allows them to travel along the surface while simultaneously causing particles to oscillate both parallel and perpendicular to the direction of wave propagation.

To know more about Surface waves, click here.

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What is the magnitude of the applied electric field inside an aluminum wire of radius 1.4 mm that carries a 4.5-A current

Answers

Answer:

Explanation:

From the question we are told that

    The radius is  [tex]r = 1.4 \ mm = 1.4 *10^{-3} \ m[/tex]

     The  current is  [tex]I = 4.5 \ A[/tex]

Generally the electric field is mathematically represented as

         [tex]E = \frac{J}{\sigma }[/tex]

Where [tex]\sigma[/tex] is the conductivity of  aluminum with value [tex]\sigma = 3.5 *10^{7} \ s/m[/tex]

J is the current density which mathematically represented as  

      [tex]J = \frac{I}{A}[/tex]

Here A is the cross-sectional area which is mathematically represented as  

       [tex]A = \pi r^2[/tex]

       [tex]A = 3.142 * (1.4*10^{-3})^2[/tex]

       [tex]A = 6.158*10^{-6} \ m^2[/tex]

So

    [tex]J = \frac{ 4.5 }{6.158*10^{-6}}[/tex]

    [tex]J = 730757 A/m^2[/tex]

So

       [tex]E = \frac{ 730757}{3.5*10^{7} }[/tex]

       [tex]E = 0.021 \ N/C[/tex]

Two imaginary spherical surfaces of radius R and 2R respectively surround a positive point charge Q located at the center of the concentric spheres. When compared to the number of field lines N1 going through the sphere of radius R, the number of electric field lines N2 going through the sphere of radius 2R is

Answers

Answer:

N2 = ¼N1

Explanation:

First of all, let's define the terms;

N1 = number of electric field lines going through the sphere of radius R

N2 = number of electric field lines going through the sphere of radius 2R

Q = the charge enclosed at the centre of concentric spheres

ε_o = a constant known as "permittivity of the free space"

E1 = Electric field in the sphere of radius R.

E2 = Electric field in the sphere of radius 2R.

A1 = Area of sphere of radius R.

A2 = Area of sphere of radius 2R

Now, from Gauss's law, the electric flux through the sphere of radius R is given by;

Φ = Q/ε_o

We also know that;

Φ = EA

Thus;

E1 × A1 = Q/ε_o

E1 = Q/(ε_o × A1)

Where A1 = 4πR²

E1 = Q/(ε_o × 4πR²)

Similarly, for the sphere of radius 2R,we have;

E2 = Q/(ε_o × 4π(2R)²)

Factorizing out to get;

E2 = ¼Q/(ε_o × 4πR²)

Comparing E2 with E1, we arrive at;

E2 = ¼E1

Now, due to the number of lines is proportional to the electric field in the each spheres, we can now write;

N2 = ¼N1

Find the momentum of a particl with a mass of one gram moving with half the speed of light.

Answers

Answer:

129900

Explanation:

Given that

Mass of the particle, m = 1 g = 1*10^-3 kg

Speed of the particle, u = ½c

Speed of light, c = 3*10^8

To solve this, we will use the formula

p = ymu, where

y = √[1 - (u²/c²)]

Let's solve for y, first. We have

y = √[1 - (1.5*10^8²/3*10^8²)]

y = √(1 - ½²)

y = √(1 - ¼)

y = √0.75

y = 0.8660, using our newly gotten y, we use it to solve the final equation

p = ymu

p = 0.866 * 1*10^-3 * 1.5*10^8

p = 129900 kgm/s

thus, we have found that the momentum of the particle is 129900 kgm/s

A simple pendulum takes 2.20 s to make one compete swing. If we now triple the length, how long will it take for one complete swing?

Answers

Answer:

Time taken for 1 swing = 3.81 second

Explanation:

Given:

Time taken for 1 swing = 2.20 Sec

Find:

Time taken for 1 swing , when triple the length(T2)

Computation:

Time taken for 1 swing = 2π[√l/g]

2.20 = 2π[√l/g].......Eq1

Time taken for 1 swing , when triple the length (3L)

Time taken for 1 swing = 2π[√3l/g].......Eq2

Squaring and dividing the eq(1) by (2)

4.84 / T2² = 1 / 3

T2 = 3.81 second

Time taken for 1 swing = 3.81 second

"Can we consider light wave as a single frequency wave? Either Yes or No, explain the reason of your answer. "

Answers

Answer:

Well, yes.

We can have an isolated light wave that is defined by only one frequency (and one wavelenght). But this is not a really common situation, most of the light that we can see in nature, is actually a composition of different waves with different frequencies.

Even if we have, for example, a red laser, the actual frequency of the light that comes from the laser may be in a range of frequencies, so the actual wave is a composition of different waves with really close frequencies.

An example of a light wave defined by only one frequency can be, for example, the photon that comes out of a change in energy of an electron.

Here we have a single photon, with a single frequency, that is modeled as a single frequency wave.

A long, horizontal hose of diameter 5.4 cm is connected to a faucet. At the other end, there is a nozzle of diameter 1.2 cm. Water squirts from the nozzle at velocity 20 m/sec. Assume that the water has no viscosity or other form of energy dissipation.
a) What is the velocity of the water in the hose?
b) What is the pressure differential between the water in the hose and water in the nozzle?
c) How long will it take to fill a tub of volume 120 liters with the hose?

Answers

Answer:

a) 0.988 m/s

b) 199512 Pa

c) 57.52 s

Explanation:

given that

A

A1 v1 = A2 v2

d1² v1 = d2² v2

v2 = [d1/d2]² v1

v2 = (1.2/5.4)² * 20

v2 = 0.049 * 20

v2 = 0.988 m/s

B

P + 1/2 ρ v² = K.

[p2 - p1] = 1/2 ρ [v1² - v2²]

[p2 - p1] = 1/2 * 1000 [20² - 0.988²]

[p2 - p1] = 500 * (400 - 0.976)

[p2 - p1] = 500 * 399

[p2 - p1] = 199512 Pa

C

Flow rate = AV = π [d²/ 4 ] * v

= π [0.012² / 4 ] * 20 = 0.00226 m³ /s

= π [0.054² / 4 ] * 0.988 = 0.00226 m³ /s

130 liters = 0.13 m³

t = 0.13/ 0.00226 = 57.52 s

a transformer changes 95 v acorss the primary to 875 V acorss the secondary. If the primmary coil has 450 turns how many turns does the seconday have g

Answers

Answer:

The number of turns in the secondary coil is 4145 turns

Explanation:

Given;

the induced emf on the primary coil, [tex]E_p[/tex] = 95 V

the induced emf on the secondary coil, [tex]E_s[/tex] = 875 V

the number of turns in the primary coil, [tex]N_p[/tex] = 450 turns

the number of turns in the secondary coil, [tex]N_s[/tex] = ?

The number of turns in the secondary coil is calculated as;

[tex]\frac{N_p}{N_s} = \frac{E_p}{E_s}[/tex]

[tex]N_s = \frac{N_pE_s}{E_p} \\\\N_s = \frac{450*875}{95} \\\\N_s = 4145 \ turns[/tex]

Therefore, the number of turns in the secondary coil is 4145 turns.

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