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KatRina [158]
3 years ago
14

. A bead of mass m under the influence of gravity slides along a frictionless and massless wire whose height is given by a funct

ion y(x). Assume that at position (x, y) = (0, 0) the tangent of the wire is horizontal and the bead passes this point with a given speed v0. What should y(x) be so that the horizontal speed remains v0 at all times (that is what is the shape of the wire)? One simple solution is y = 0, find the other solution. (Hint: It should
Physics
1 answer:
eduard3 years ago
7 0

Answer:

y = constant

Explanation:

Bodies moving on the surface of Earth are subject to gravity, and they have a potential and kinetic energy. If there is no friction, the sum of the kinetic and potential energy remains constant.

Since potential energy depends on height, changes in altitude affect potential energy. Going higher increases this energy, this is accompanied by a reduction of kinetic energy and speed (since kinetic energy is related to speed). If the body goes down potential energy is reduced, but kinetic energy and speed increase.

For speed to remain constant the kinetic energy must remain constant. For the kinetic energy to remanin constant, the potential energy must remain constant, and for the potential energy to remain constant the height must remain constant.

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You wiggle a string,that is fixed to a wall at the other end, creating a sinusoidalwave with a frequency of 2.00 Hz and an ampli
FinnZ [79.3K]

Answer:

Explanation:

A general wave function is given by:

f(x,t)=Acos(kx-\omega t)

A: amplitude of the wave = 0.075m

k: wave number

w: angular frequency

a) You use the following expressions for the calculation of k, w, T and λ:

\omega = 2\pi f=2\pi (2.00Hz)=12.56\frac{rad}{s}

k=\frac{\omega}{v}=\frac{12.56\frac{rad}{s}}{12.0\frac{m}{s}}=1.047\ m^{-1}

T=\frac{1}{f}=\frac{1}{2.00Hz}=0.5s\\\\\lambda=\frac{2\pi}{k}=\frac{2\pi}{1.047m^{-1}}=6m

b) Hence, the wave function is:

f(x,t)=0.075m\ cos((1.047m^{-1})x-(12.56\frac{rad}{s})t)

c) for x=3m you have:

f(3,t)=0.075cos(1.047*3-12.56t)

d) the speed of the medium:

\frac{df}{dt}=\omega Acos(kx-\omega t)\\\\\frac{df}{dt}=(12.56)(1.047)cos(1.047x-12.56t)

you can see the velocity of the medium for example for x = 0:

v=\frac{df}{dt}=13.15cos(12.56t)

7 0
3 years ago
Which structures main function is to produce food (sugar) in a plant
leva [86]
Technically speaking, there are many "levels" of a plant for which this may hold true. I think the one you are referring to is the chloroplast. It takes in the light energy from the sun, water and carbon dioxide, and performs photosynthesis on them to produce sugar and oxygen. A leaf would also fit the description as this is a very general question.
8 0
3 years ago
Climatology is a subspecialty of what?
m_a_m_a [10]

 Climatology is a subspecialty of Meterology.
3 0
3 years ago
Read 2 more answers
If an amount of heat Q is needed to increase the temperature of a solid metal sphere of diameter D from 4°C to 7°C, the amount o
Hitman42 [59]

Answer:

Q = c M ΔT      where c is the heat capacity and M the mass present

Q2 / Q1 = M2 / M1    since the other factors are the same

M = ρ V     where ρ is the density

M = ρ Π (d / 2)^2           where d is the diameter of the sphere

M2 / M1 = (2 D/2)^2 / (D/2)^2 = 4

It will take 4Q heat to heat the second sphere

7 0
2 years ago
I would like help with this physics problem
Darina [25.2K]

(a) This is a freefall problem in disguise - when the ball returns to its original position, it will be going at the same speed but in the opposite direction. So the ball's final velocity is the negative of its initial velocity.

Recall that

v_f=v_i+at

We have v_f=-v_i, so that

-2v_i=at\implies-2\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)=\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\implies t=8\,\mathrm s

(b) The speed of the ball at the start and at the end of the roll are the same 8 m/s, so the average speed is also 8 m/s.

(c) The ball's average velocity is 0. Average velocity is given by \dfrac{v_i+v_f}2, and we know that v_f=-v_i.

(d) The position of the ball x_f at time t is given by

x_f=x_i+v_it+\dfrac12at^2

Take the starting position to be the origin, x_i=0. Then after 6 seconds,

x_f=\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2=42\,\mathrm m

so the ball is 42 m away from where it started.

We're not asked to say in which direction it's moving at this point, but just out of curiosity we can determine that too:

x_f-x_i=\dfrac{v_i+v_f}2t\implies42\,\mathrm m=\dfrac{8\,\frac{\mathrm m}{\mathrm s}+v_f}2(6\,\mathrm s)\implies v_f=6\,\dfrac{\mathrm m}{\mathrm s}

Since the velocity is positive, the ball is still moving up the incline.

8 0
3 years ago
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