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alexgriva [62]
4 years ago
13

Determine the time needed for the load at b to attain a speed of 10 m/s, starting from rest.

Physics
1 answer:
quester [9]4 years ago
5 0
Unfortunately, I can't give a definite answer because I only know two information: the initial velocity which is 0 because it is at rest, and the final velocity which is 10 m/s. To solve this, I should know one other information to determine time. It could be the constant acceleration. The useful equation to be used is:


a = (vf - v₀)/t

So, assuming that the object was dropped from a certain height, the acceleration is due to gravity which is equal to 9.81 m/s². Then,  

9.81 = (10 - 0)/t
t = 1.02  seconds

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Answer:

\bigtriangledown.B=0 is proved.

Explanation:

The magnetic field in the long current carrying wire is,

B=\frac{\mu_{0}I }{2\pi r } \phi

Here, I is the current, B is the magnetic field.

Now, by using cylindrical coordinates for the divergence of B.

\bigtriangledown.B=\frac{1}{s} \frac{d}{d\phi} B

Put the value of B in above equation.

\bigtriangledown.B=\frac{1}{s} \frac{d}{d\phi}(\frac{\mu_{0}I }{2\pi r } \phi)\\\bigtriangledown.B=0

Hence, it is prove that for a long current I carrying wire magnetic field divergence that is \bigtriangledown.B=0.

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3 years ago
Need help in Psychology. Please help! Which of the following statements best describes abductive reasoning? A.I think,therefore
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3 years ago
A major limitation of using photovoltaic cells to generate electricity is that they
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Answer:

Explained below:

Explanation:

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What is true of both gravity and magnetism?
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Using the conversion factor from eV (electron volts) to joules, determine which energy line for an electron dropping from one en
Tom [10]

Explanation:

Wavelength in an emission spectrum,  \lambda=435\ nm=435\times 10^{-9}\ m  

The energy of an electron is given by :

E=\dfrac{hc}{\lambda}

Where

h is the Planck's constant

c is the speed of light

For 435 nm, the energy of the electron will be :

E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8\ m/s}{435\times 10^{-9}}

E=4.57\times 10^{-19}\ J

We know that 1\ eV=1.6\times 10^{-19}\ J

So, E=\dfrac{4.57\times 10^{-19}}{1.6\times 10^{-19}}

So, E = 2.86 eV

The energy of the electron dropping from one energy level is 2.86 eV. We know that,

\dfrac{hc}{\lambda}=E_{n_2}-E_{n_1}

From the given energy levels :

E_5-E_2=-0.544-(-3.403)

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So, the transition must be from E₅ to E₂. Hence, this is the required solution.                                             

5 0
4 years ago
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