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3 years ago

When there is an equal amount of positive and negative charges on an object, the object is _____?

Physics

2 answers:

Gwar [14]3 years ago

8 0

Reached equilibrium, the positive and negative charges cancel or balance each other out.

pashok25 [27]3 years ago

4 0

Answer:

<h2>The object becomes neutral, has no charge.</h2>

Explanation:

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A jet leaves a runway whose bearing is N 3232degrees°E from the control tower. After flying 77 miles, the jet turns 90degrees°

OverLord2011 [107]

Answer:

$N(80.8^{\circ})E$

Explanation:

Bearing from control tower= $N(32^{\circ}+\theta})E$

But $tan \theta=\frac {opposite}{adjacent}$

$tan \theta=\frac {88}{77}$

$\theta=tan^{-1}(\frac {88}{77})=tan^{-1}(1.142857143)=48.81407483^{\circ}\approx 48.8^{\circ}$

Therefore, the bearing from control tower= $N(32^{\circ}+48.8^{\circ}})E$

Bearing= $N(80.8^{\circ})E$

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3 years ago

What is the magnitude of fs on an object lying on a flat surface without moving, on

Degger [83]

The magnitude of the force acting on the object lying on a flat surface without moving is 10 N.

The given parameters;

magnitude of force on the object, F = 10 N

angle between the object and the horizontal flat surface = 0⁰

Apply Newton's second law of motion to determine the magnitude of the force on the object.

Due to the position of the object, the magnitude of the force acting on it is calculated as;

$\Sigma F_{net} = F\sin(\theta ) + F cos(\theta)\\\\\Sigma F_{net} = 10 sin(0) + 10cos(0)\\\\\Sigma F_{net} = 10 \ N$

Therefore, the magnitude of the force acting on the object is 10 N.

Learn more here: brainly.com/question/19887955

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2 years ago

An astronaut is rotated in a horizontal centrifuge at a radius of 5.0 m. (a) What is the astronaut’s speed if the centripetal ac

sergeinik [125]

Explanation:

Given that,

Radius of circular path, r = 5 m

Centripetal acceleration, $a=7g\ m/s^2$

(a) Let v is the astronaut’s speed. The formula for the centripetal acceleration is given by :

$a=\dfrac{v^2}{r}$

$v=\sqrt{ar}$

$v=\sqrt{7\times 9.8\times 5}$

v = 18.5 m/s

(b) Let T denotes the time period. It is given by :

$T=\dfrac{2\pi r}{v}$

$T=\dfrac{2\pi \times 5}{18.5}$

T = 1.69 s

Let N is the number of revolutions. So,

$N=\dfrac{60}{1.69}=35.5\ rev/min$

So, the number of revolutions per minute is 35.5

Hence, this is the required solution.

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3 years ago

An object is measured 6.0cm in length, 3.0cm in width and 5.0cm in height. What is it’s volume?

sergey [27]

Answer:

Explanation:

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