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3 years ago

75 kilometers per hour is an example of: A. speed. B. velocity C. acceleration

Physics

2 answers:

kherson [118]3 years ago

5 0

The answer is A. speed
hope this helps! :D

ozzi3 years ago

4 0

It is A. Speed as speed is how fast you are traveling and 75 kilometers per hour is how fast you are traveling.

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For your final exam in electronics, you’re asked to build an LC circuit that oscillates at 10 kHz. In addition, the maximum curr

Marina CMI [18]

Answer:

L= 2 mH

$C=1.26\times 10^{-7}\ F$

Explanation:

Given that

Frequency , f= 10 kHz

Maximum current ,I = 0.1 A

Maximum energy stored ,E= 1 x 10⁻⁵ J

The maximum energy stored in the inductor is given as follows

$E=\dfrac{1}{2}LI^2$

Where ,L= Inductance

I=Current

E=Energy

Now by putting the values in the above equation

$10^{-5}=\dfrac{1}{2}\times L\times 0.1^2$

$L=\dfrac{2\times 10^{-5}}{0.1^2}\ H$

L=0.002 H

L= 2 mH

We know that frequency f is given as

$2\pi f=\dfrac{1}{\sqrt{LC}}$

C=Capacitance , f=frequency ,L=Inductance

Now by putting the values

$2\pi \times 10\times 10^3=\dfrac{1}{\sqrt{0.002\times C} }$

$62831.85=\dfrac{1}{\sqrt{0.002\times C}}$

$\sqrt{0.002\times C=\dfrac{1}{62831.85}$

$0.002\times C=0.0000159^2$

$C=\dfrac{0.0000159^2}{0.002}\ F$

$C=1.26\times 10^{-7}\ F$

Therefore the inductance and capacitance will be 2 mH and 1.26 x 10⁻⁷ F respectively.

6 0

3 years ago

The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This

erica [24]

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

$g = \frac{GM}{(d-R_{CM})^2}$

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

$R_{CM} =$ Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

$g = \frac{GM}{(d-R_{CM})^2}$

$g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}$

$g = 3.4*10^{-5}m/s^2$

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

$\omega = \frac{2\pi}{T}$

At the same time we have that centripetal acceleration is given as

$a_c = \omega^2 r$

Replacing

$a_c = (\frac{2\pi}{T})^2 r$

$a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)$

$a_c = 3.34*10^{-5}m/s^2$

3 0

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