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4 years ago

When electrical energy is being used by an electric light, what really happens to the energy?

Physics

2 answers:

vitfil [10]4 years ago

7 0

Energy is not created and not destroyed it will only change form

So heres your answer ; It is given off as other forms of energy/light and heat !!

=answer 2nd one

Ronch [10]4 years ago

5 0

A. it is given off as other forms of energy

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A big olive (* - 0.50 kg) lies at the origin of an xy coordinate system, and a big BrazlI nut (M - 1.5^kg) lie^s at the point (1

Afina-wow [57]

The <em>estimated</em> displacement of the center of mass of the olive is $\overrightarrow{\Delta r} = -0.046\,\hat{i} -0.267\,\hat{j}\,[m]$ .

<h3>Procedure - Estimation of the displacement of the center of mass of the olive</h3>

In this question we should apply the definition of center of mass and difference between the coordinates for <em>dynamic</em> ( $\vec r$ ) and <em>static</em> conditions ( $\vec r_{o}$ ) to estimate the displacement of the center of mass of the olive ( $\overrightarrow{\Delta r}$ ):

$\vec r - \vec r_{o} = \left[\frac{\Sigma\limits_{i=1}^{2}r_{i,x}\cdot(m_{i}\cdot g + F_{i, x})}{\Sigma \limits_{i =1}^{2}(F_{i,x}+m_{i}\cdot g)} ,\frac{\Sigma\limits_{i=1}^{2}r_{i,y}\cdot(m_{i}\cdot g + F_{i, y})}{\Sigma \limits_{i =1}^{2}(F_{i,y}+m_{i}\cdot g)} \right]-\left(\frac{\Sigma\limits_{i=1}^{2}r_{i,x}\cdot m_{i}\cdot g}{\Sigma \limits_{i= 1}^{2} m_{i}\cdot g}, \frac{\Sigma\limits_{i=1}^{2}r_{i,y}\cdot m_{i}\cdot g}{\Sigma \limits_{i= 1}^{2} m_{i}\cdot g}\right)$ (1)

Where:

$r_{i, x}$ - x-Coordinate of the i-th element of the system, in meters.
$r_{i,y}$ - y-Coordinate of the i-th element of the system, in meters.
$F_{i,x}$ - x-Component of the net force applied on the i-th element, in newtons.
$F_{i,y}$ - y-Component of the net force applied on the i-th element, in newtons.
$m_{i}$ - Mass of the i-th element, in kilograms.
$g$ - Gravitational acceleration, in meters per square second.

If we know that $\vec r_{1} = (0, 0)\,[m]$ , $\vec r_{2} = (1, 2)\,[m]$ , $\vec F_{1} = (0, 3)\,[N]$ , $\vec F_{2} = (-3, -2)\,[N]$ , $m_{1} = 0.50\,kg$ , $m_{2} = 1.50\,kg$ and $g = 9.807\,\frac{kg}{s^{2}}$ , then the displacement of the center of mass of the olive is:

<h3>Dynamic condition $\vec{r} = \left[\frac{(0)\cdot (0.50)\cdot (9.807)+(0)\cdot (0) + (1)\cdot (1.50)\cdot (9.807) + (1)\cdot (-3)}{(0.50)\cdot (9.807) + 0 + (1.50)\cdot (9.807)+(-3)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (0)\cdot (3) + (2)\cdot (1.50)\cdot (9.807) +(2) \cdot (-2)}{(0.50)\cdot (9.807) + (3)+(1.50)\cdot (9.807)+(-2)} \right]$ $\vec r = (0,704, 1.233)\,[m]$ </h3>

<h3>Static condition</h3><h3> $\vec{r}_{o} = \left[\frac{(0)\cdot (0.50)\cdot (9.807) + (1)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807) + (1.50)\cdot (9.807)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (2)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807)+(1.50)\cdot (9.807)} \right]$ </h3><h3> $\vec r_{o} = \left(0.75, 1.50)\,[m]$ </h3><h3 /><h3>Displacement of the center of mass of the olive</h3>

$\overrightarrow{\Delta r} = \vec r - \vec r_{o}$

$\overrightarrow{\Delta r} = (0.704-0.75, 1.233-1.50)\,[m]$

$\overrightarrow{\Delta r} = (-0.046, -0.267)\,[m]$

The <em>estimated</em> displacement of the center of mass of the olive is $\overrightarrow{\Delta r} = -0.046\,\hat{i} -0.267\,\hat{j}\,[m]$ . $\blacksquare$

To learn more on center of mass, we kindly invite to check this verified question: brainly.com/question/8662931

3 0

2 years ago

A force F at an angle θ above the horizontal is used to pull a heavy suitcase of weight mg a distance d along a level floor at c

Sati [7]

Answer:

option A

Explanation:

given,

Force = F

angle = θ

weight on suitcase = mg

distance = d

constant velocity so, acceleration a = 0

coefficient of friction = µ

Work done = ?

Work done is equal to force into displacement.

Friction act opposite to the force acting so, work done by frictional force will be negative.

frictional force will act into horizontal direction opposite to force.

here displacement is equal to d

now,

W = -F d cos θ

Hence,the correct answer is option A

3 0

3 years ago

A book is sitting on a table. jackie is pushing the book with a force of 9 N to the right. if the force of friction is 4 N to th

laila [671]

Answer:

Explanation:

4+5 =9

5n to the right

4 0

3 years ago

Read 2 more answers

How do you find the net force acting on an object?

yaroslaw [1]

It's D. By "net" they mean the overall force the object experiences, so sum all the force vectors, those in a negative direction (eg friction) should be subtracted.

6 0

3 years ago

2 kg<br> Use 10 m/s2 for g

Iteru [2.4K]

Answer:

See below

Explanation:

At point A the PE = mgh = 2 * 10 * 1 = 20 J

at point B, all of the PE , 20 J , is converted to Kinetic Energy

KE = 1/2 m v^2

20 = 1/2 (2)(v^2 )

20 = v^2 v = sqrt 20 = 4.47 m/s

for the friction part

vf = vo t + 1/2 a t^2 vf = final velocity = 0 (stopped)

vo = original velocity = 4.47 m/s

a = -1 m/s^2

0 = 4.47 t + 1/2 (-1) t^2

- .5t^2 + 4.47 t = 0

t ( -.5t+ 4.47) = 0 shows t = 4.47/.5 = 8.9 seconds

6 0

2 years ago