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3 years ago

2) If the initial velocity of an object is equal to a final velocity, what is the acceleration of the object?

Physics

1 answer:

Arlecino [84]3 years ago

8 0

Answer:

The acceleration is 0.

Explanation:

The formula for acceleration is:

a = (v-u)/t

If v and u are equal, thus it would be 0 when subtracted, and anything divided by 0 is 0.

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Two organ pipes, both open at one end but closed at the other, are each 1.14 m long. One is now lengthened by 2 cm. Find the fre

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Answer: 1.3 Hz

Explanation:

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3 years ago

An electric circuit through which no current flows is known as?

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The answer is : Open circuit

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4 years ago

Read 2 more answers

the mass of one water drop is 0.0008kg and the gravitational field strength is 10N/kg what is its weight

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Weight = (mass) x (gravity)

Weight = (8 x 10⁻⁴ kg) x (10 N/kg) = 0.008 Newton

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3 years ago

A satellite of mass M = 270kg is in circular orbit around the Earth at an altitude equal to the earth's mean radius (6370 km). A

zubka84 [21]

To solve this problem we will apply the concepts related to Orbital Speed as a function of the universal gravitational constant, the mass of the planet and the orbital distance of the satellite. From finding the velocity it will be possible to calculate the period of the body and finally the gravitational force acting on the satellite.

PART A)

$V_{orbital} = \sqrt{\frac{GM_E}{R}}$

Here,

M = Mass of Earth

R = Distance from center to the satellite

Replacing with our values we have,

$V_{orbital} = \sqrt{\frac{(6.67*10^{-11})(5.972*10^{24})}{(6370*10^3)+(6370*10^3)}}$

$V_{orbital} = 5591.62m/s$

$V_{orbital} = 5.591*10^3m/s$

PART B) The period of satellite is given as,

$T = 2\pi \sqrt{\frac{r^3}{Gm_E}}$

$T = \frac{2\pi r}{V_{orbital}}$

$T = \frac{2\pi (2*6370*10^3)}{5.591*10^3}$

$T = 238.61min$

PART C) The gravitational force on the satellite is given by,

$F = ma$

$F = \frac{1}{4} mg$

$F = \frac{270*9.8}{4}$

$F = 661.5N$

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3 years ago

A beam of red light is made to pass through two slits that are 3.55 E-3 meters apart. On a screen 2.25 meters away from the slit

JulsSmile [24]

I am assuming you know the relation obtained between slit width, distance of screen from slits, distance of interference pattern obtained on the screen from the center and the wavelength of monochromatic light used in Young's Double Slit experiment.
λ = $\frac{y*d}{D} = \frac{3.55*10^{-3}*1.25*10^{-4} }{2.25} = 1.97*10^{-7} m$
λ ~ 1.97 ×10⁻⁷m

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3 years ago