A"boat"is"moving"to"the"right"at"5"m/s"with"respect"to"the"water."A"wave"moving"to"the"left,"opposite"the"motion"of"the"boat."The"waves"have"2.0"m"between"the"top"of"the"crests"and"the"bottom"of"the"troughs."The"period"of"the"wave"is"8.3"s"and"their"wavelength"is"110"m."At"one"instant"the"boat"sits"on"a"crest"of"the"wave,"20"seconds"later,"what"is"the"vertical"displacement"of"the"boat

Answers

Answer 1

Answer:

Answer:

0.99m

Explanation:

Firs you calculate the relative velocity between the boat and the wave. The velocity of the boat is 5m/s and the velocity of the wave is given by:

$v=\lambda f=\lambda(1)/(T)=(110m)(1)/(8.3s)=13.25(m)/(s)$

the relative velocity is:

$v'=13.25m/s-5m/s=8.25(m)/(s)$

This velocity is used to know which is the distance traveled by the boat after 20 seconds:

$x'=v't=(8.25m/s)(20s)=165m$

Next, you use the general for of a wave:

$f(x,t)=Acos(kx-\omega t)=Acos((2\pi)/(\lambda)x-\omega t)$

you take the amplitude as 2.0/2 = 1.0m.

$\omega=(2\pi)/(T)=(2\pi)/(8.3s)=0.75(rad)/(s)$

by replacing the values of the parameters in f(x,t) you obtain the vertical displacement of the boat:

$f(165,20)=1.0m\ cos((2\pi)/(110m)(165)-(0.75(rad)/(s))(20s))\n\nf(165,20)=0.99m$

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If the solution described in the introduction is cooled to 0 ∘c, what mass of kno3 should crystallize? enter your answer numerically in grams.
A charge +Q is located at the origin and a second charge, +4Q, is at distance (d) on the x-axis.a. Where should a third charge, Q , be placed, so that all three charges will be in equilibrium? Express your answer in terms of d.b. What should be its sign, so that all three charges will be in equilibrium?c. What should be its magnitude, so that all three charges will be in equilibrium? Express your answer in terms of Q.

Suppose a boat moves at 16.4 m/s relative to the water. If the boat is in a river with the current directed east at 2.70 m/s, what is the boat's speed relative to the ground when it is heading east, with the current, and west, against the current? (Enter your answers in m/s.)

Answers

Answer:

with the current: 19.1 m/s eastern

against the current: 13.7 m/s western

Explanation:

The boat speed relative to ground is the sum of the boat speed relative to water + the water speed relative to ground.

Suppose eastern is positive, water speed due east is 2.7m/s.

When the boat is heading east, its speed relative to water is 16.4m/s, its speed relative to ground is 16.4 + 2.7 = 19.1 m/s eastern

When the boat is heading west, its speed relative to water is -16.4m/s, its speed relative to ground is -16.4 + 2.7 = -13.7 m/s western

A piccolo and a flute can be approximated as cylindrical tubes with both ends open. The lowest fundamental frequency produced by one kind of piccolo is 522.5 Hz, and that produced by one kind of flute is 256.9 Hz. What is the ratio of the piccolo's length to the flute's length?

Answers

Answer:

ratio of the piccolo's length to the flute's length is 0.4916

Explanation:

given data

frequency of piccolo = 522.5 Hz

frequency of flute = 256.9 Hz

to find out

ratio of the piccolo's length to the flute's length

solution

we get here length of tube that is express as

length of tube = velocity of sound ÷ fundamental frequency .......................1

so here ratio of Piccolo length to flute that is

$(L\ picco)/(L\ flute) = (f\ flute)/(f \ piccolo)$

$(l \ piccolo)/(L\ flute) = (256.9)/(522.5)$ = 0.4916

so ratio of the piccolo's length to the flute's length is 0.4916

The four tires of an automobile are inflated to a gauge pressure of 2.2 105 Pa. Each tire has an area of 0.023 m2 in contact with the ground. Determine the weight of the automobile.

Answers

Answer:

Force, F = 20240 N

Explanation:

It is given that,

Pressure exerted by the four tires of an automobile, $P=2.2* 10^5\ Pa$

Area of each tire, $A=0.023\ m^2$

Area of 4 tires, $A=0.092\ m^2$

We know that the pressure exerted by an object is equal to the force per unit area. Its formula is given by :

$P=(F)/(A)$

$F=P* A$

$F=2.2* 10^5* 0.092$

F = 20240 N

So, the weight of the automobile is 20240 N. Hence, this is the required solution.

Final answer:

The vehicle's weight can be calculated by rearranging the definition of pressure (Pressure = Force/Area) to solve for Force (Force = Pressure * Area), then multiplying by four to account for all four tires. Remember that the result will be in newtons, so to convert it to kilograms, divide by gravitational acceleration.

Explanation:

Your question revolves around the concept of pressure. Tire pressure is a type of air pressure which is a part of physics. To determine the weight of the automobile in which each of the four tires has an area of 0.023 m2 and is inflated to a gauge pressure of 2.2 x 105 Pa, we need to utilize the fundamental equation of pressure:

P = F/A

Where P is the pressure, F is the force (which in this case will be the weight of the car), and A is the area of each of the tires where they are in contact with the ground. Solving for the weight (F) results in:

F = P * A

In your case, because there are four tires we multiply the result by four, therefore:

F = 4 * (2.2 x 105 Pa) * (0.023 m2)

We have to multiply this by 4 to account for all four tires. Finally, your weight will be in newtons, to convert it to kg you will divide by gravitational acceleration (approx 9.8 m/s2).

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An electric generator contains a coil of 140 turns of wire, each forming a rectangular loop 71.2 cm by 22.6 cm. The coil is placed entirely in a uniform magnetic field with magnitude B = 4.32 T and initially perpendicular to the coil's plane. What is in volts the maximum value of the emf produced when the loop is spun at 1120 rev/min about an axis perpendicular to the magnetic field?

Answers

Answer:

11405Volt

Explanation:

To solve this problem it is necessary to use the concept related to induced voltage or electromotive force measured in volts. Through this force it is possible to maintain a potential difference between two points in an open circuit or to produce an electric current in a closed circuit.

The equation that allows the calculation of this voltage is given by,

$\epsilon = BAN \omega$

Where

B = Magnetic field

A= Area

N = Number of loops

$\omega$ = Angular velocity

Our values previously given are:

$N = 140$

$A = 71.2*10^(-2)m*22.6*10^(-2)m=0.1609m^2$

$B = 4.32 T$

$\omega = 1120 rev / min$

We need convert the angular velocity to international system, then

$\omega = 1120 rev/min$

$\omega = 1120rev/min*(2\pi)/(1rev)*(1min)/(60sec)$

$\omega = 117.2rad/s$

Applying the equation for emf, we replace the values and we will obtain the value.

$\epsilon = BAN \omega$

$\epsilon = (4.32)(0.1609)(140)*117.2$

$\epsilon = 11405Volt$

Electrons with energy of 25 eV have a wavelength of ~0.25 nm. If we send these electrons through the same two slits (d = 0.16 mm) we use to produce a visible light interference pattern what is the spacing (in micrometer) between maxima on a screen 3.3 m away?

Answers

Answer:

The spacing is 5.15 μm.

Explanation:

Given that,

Electron with energy = 25 eV

Wave length = 0.25 nm

Separation d= 0.16 mm

Distance D=3.3 m

We need to calculate the spacing

Using formula of width

$\beta=(\lambda D)/(d)$

Put the value into the formula

$\beta=(0.25*10^(-9)*3.3)/(0.16*10^(-3))$

$\beta=5.15*10^(-6)\ m$

$\beta=5.15\ \mu m$

Hence, The spacing is 5.15 μm.

Final answer:

To calculate the spacing between maxima in a double slit interference pattern, we use the formula x = L * λ / d. Converting the given units to meters and plugging the values into the formula, we find that the spacing between maxima on the screen is approximately 5.14 micro meters.

Explanation:

To calculate the spacing between maxima, we can utilize the formula for double slit interference, θ = λ/d where λ represents the wavelength of the electron, d is the distance between the two slits, and θ is the angle of diffraction. Considering the small angle approximation for tan θ ≈ θ, we get x = L * λ / d, where x is the distance between maxima on the screen, and L is the distance from the slits to the screen.

Firstly, the electron's wavelength needs to be converted from nm to m, resulting in λ = 0.25 * 10^-9 m. Similarly, the slit separation d should be converted from mm to m, giving d = 0.16 * 10^-3 m. Inserting these values into the formula along with L = 3.3 m, we can solve for x.

x = (3.3 m * 0.25 * 10^-9 m) / 0.16 * 10^-3 m =~ 5.14 μm

So, the spacing between maxima on the screen is approximately 5.14 micrometers.

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Here ΔI/Δt characterizes the rate at which the current I through the inductor is changing with time t.
Based on the equation given in the introduction, what are the units of inductance L in terms of the units of E, t, and I (respectively volts V, seconds s, and amperes A)?
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