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Feliz [49]
2 years ago
5

You are exploring a planet and drop a small rock from the edge of a cliff. In coordinates where the +y direction is downward and

neglecting air resistance, the vertical displacement of an object released from rest is given by y − y0 = 1 2 gplanett2, where gplanet is the acceleration due to gravity on the planet. You measure t in seconds for several values of y − y0 in meters and plot your data with t2 on the vertical axis and y − y0 on the horizontal axis. Your data is fit closely by a straight line that has slope 0.400 s2/m. Based on your data, what is the value of gplanet?
Physics
1 answer:
Lelu [443]2 years ago
6 0

Answer:

value of the acceleration of gravity on the planet is 5.00 m/s²

Explanation:

The problem is similar to a free fall exercise, with another gravity value, the expression they give us is the following:

       y-yo = ½ gₐ t²       (1)

They tell us that they make a squared time graph with the variation of the distance, it is appropriate to clarify this in a method to linearize a curve, which is plotted the nonlinear axis to the power that is raised, specifically, the linearization of a curve The square is plotted against the other variable.

  Let's continue our analysis, as we have a linear equation, write the equation of the line.

     

        y1 = m x1 + b       (2)

where  “y1” the dependent variable, “x1” the independent variable, “m” the slope and “b” the short point

In this case as the stone is released its initial velocity is zero which implies that b = 0,

We plot on the “y” axis the time squared “t²” and on the horizontal axis we place “y-yo”.  To better see the relationship we rewrite equation 1 with this form

        t² = 2 /gₐ  (y-yo)

 

With the two expressions written in the same way, let's relate the terms one by one

        y1 = t²

        x1 = (y-yo)

        m = 2/gap

        b= 0

We substitute and calculate

        m = 2/gp

        gₐ = 2/m

        gₐ = 2/ 0.400

        gₐ = 5.00 m / s²

This is the value of the acceleration of gravity on the planet, note that the decimals are to keep the figures significant

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The strength of the magnetic field is 4.8\cdot 10^{-5} T

Explanation:

According to Faraday's Law, the magnitude of the induced emf in the coil is equal to the rate of changeof the flux linkage through the coil:

\epsilon = \frac{N\Delta \Phi}{\Delta t} (1)

where

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\Delta \Phi is the change in magnetic flux through the coil

\Delta t = 2.77 ms = 2.77\cdot 10^{-3} s is the time interval

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The coil is rotated from a position perpendicular to the Earth's magnetic field to a position parallel to it, so the final flux is zero, and the magnitude of the flux change is simply equal to the initial flux:

\Delta \Phi = B A cos \theta

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B is the strength of the magnetic field

A is the area of the coil

\theta=0^{\circ} is the angle between the normal to the coil and the field

The area of the coil can be written as

A=\pi r^2

where

r=\frac{15.5 cm}{2}=7.75 cm = 7.75\cdot 10^{-2} m is its radius

Substituting everything into eq.(1) and solving for B, we find:

\epsilon= \frac{NB\pi r^2 cos \theta}{\Delta t}\\B=\frac{\epsilon \Delta t}{\pi r^2 cos \theta}=\frac{(0.166)(2.77\cdot 10^{-3})}{(505)\pi (7.75\cdot 10^{-2})^2(cos 0^{\circ})}=4.8\cdot 10^{-5} T

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What is the temperature when a solid begins to liquefy
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Answer:

Explanation:

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2 years ago
The Lamborghini Huracan has an initial acceleration of 0.80g. Its mass, with a driver, is 1510 kg. If an 80 kg passenger rode al
irina [24]

Answer:

a = 15.1 g

Explanation:

The relation between mass and acceleration is given by :

m\propto \dfrac{1}{a}

If a₁ = 0.80g, m₁ = 1510 kg, m₂ = 80 kg, we need to find a₂

So,

\dfrac{m_1}{m_2}=\dfrac{a_2}{a_1}\\\\a_2=\dfrac{a_1m_1}{m_2}\\\\a_2=\dfrac{0.8g\times 1510}{80}\\\\a_2=15.1g

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A Micro –Hydro turbine generator is accelerating uniformly from an angular velocity of 610 rpm to its operating angular velocity
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Answer:

Angular displacement of the turbine is 234.62 radian

Explanation:

initial angular speed of the turbine is

\omega_i = 2\pi f_1

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similarly final angular speed is given as

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\alpha = 5.9 rad/s^2

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\theta = (63.88)(3.2) + (\frac{1}{2})(5.9)(3.2^2)

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