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IRISSAK [1]
3 years ago
11

The Earth's electric field creates a potential that increases 100 V for every meter of altitude. If an object of charge 4.5 mC a

nd mass 68 g falls a distance of 1.0 m from rest under the influence of the Earth's electric and gravitational fields, what is its final kinetic energy
Physics
1 answer:
castortr0y [4]3 years ago
8 0

Answer:

The final kinetic energy is  K =  1.1 \ J

Explanation:

From the question we are told that

    The electric field is  E =  100 \ V/m

    The charge on the object is  q =  4.5 mC  =  4.5 *10^{-3} \ C

    The mass of the object is  m_o  =  68 \ g  = 0.68 \ kg

     The distance moved by the object is d =  1.0 \ m

The workdone on the object by the fields  is  mathematically represented as

   W =  [qE + mg]d

Now this workdone is equivalent to the final kinetic energy so  

      K = W =  [qE + mg]d

substituting values

        K = W =  [4.5*10^{-3  } *100  + 0.68 * 9.8]* 1

        K =  1.1 \ J

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Answer:

8.60 g/cm³

Explanation:

In the lattice structure of iron, there are two atoms per unit cell. So:

\frac{2}{a^{3} }  = \frac{N_{A} }{V_{molar} } where V_{molar}  = \frac{A}{\rho } an and A is the atomic mass of iron.

Therefore:

\frac{2}{a^{3} } = \frac{N_{A} * p }{A}

This implies that:

A = (\frac{2A}{N_{A} * p)^{\frac{1}{3} }  }

  = \frac{4}{\sqrt{3} }r

Assuming that there is no phase change gives:

\rho = \frac{4A}{N_{A}(2\sqrt{2r})^{3}   }

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3 0
3 years ago
A simple harmonic oscillator completes 1550 cycles in 30 min. (a) Calculate the period. s (b) Calculate the frequency of the mot
nordsb [41]

Answer:

(a) 1.16 s

(b)0.861 Hz

Explanation:

(a) Period : The period of a simple harmonic motion is the time in seconds, required for a object undergoing oscillation to complete one cycle.

From the question,

If 1550 cycles is completed in (30×60) seconds,

1 cycle is completed in x seconds

x = 30×60/1550

x = 1.16 s

Hence the period is 1.16 seconds.

(b) Frequency : This can be defined as the number of cycles that is completed in one seconds, by an oscillating body. The S.I unit of frequency is Hertz (Hz).

Mathematically, Frequency is given as

F = 1/T ........................... Equation 1

Where F = frequency, T = period.

Given: T = 1.16 s.

Substitute into equation 1

F = 1/1.16

F = 0.862 Hz

Hence thee frequency = 0.862 Hz

6 0
3 years ago
The following is the longitudinal characteristic equation for an F-89 flying at 20,000 feet at Mach 0.638. The Short Period natu
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Answer:

hello your question is incomplete  attached below is the missing part  

answer : short period oscillations frequency  = 0.063 rad / sec

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Explanation:

first we have to state the general form of the equation

= ( S^2 + 2\alpha _{p} w_{np} S + w^{2} _{np} ) (S^{2} + 2\alpha _{s} w_{ns}S + w^{2} _{ns}  ) = 0

where :

w_{np}  = Natural frequency of plugiod oscillation

\alpha _{p} = damping ratio of plugiod  oscilations

comparing the general form with the given equation

w^{2} _{np}  = 18.2329

w^{2} _{ns} = 0.003969

hence the short period oscillation frequency ( w_{ns} ) =  0.063 rad/sec

phugoid oscillations natural frequency ( w_{np} ) = 4.27 rad/sec

8 0
3 years ago
Which astronomer supported the belief that earth was at the center fo the universe?
MissTica
B. Ptolemy believed that the earth was the center of the universe
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A horse runs a distance of 240 m in 20 s. Which of the following is a scalar quantity that can be determined from this
mariarad [96]

Explanation:

distance and time both are scaler quantity

6 0
3 years ago
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