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3 years ago

5

Dos cargas q_1=-8μC y q_2=13μC se encuentran a una distancia r=0.12 m. ¿Cuál es la fuerza resultante sobre una tercera carga q_3

=-4μC, situada en el centro de la distancia que separa las cargas q_1 y q_2?

1 answer:

KiRa [710]3 years ago

8 0

Answer:

209.53 N

Explanation:

To find the force on the third charge you use the Coulomb formula:

$F=k\frac{q_1q_2}{r^2}$

k: Coulomb's constant = 8.98*10^9 Nm^2/C^2

The force on the third charge is the contribution of the force between the third charge and the other ones:

$F=F_1+F_2$

By taking into account that the third charge is at the middle of the distance between charge 1 and charge 2 you have r = 0.12m/2 = 0.06m

Furthermore, you take into account that the first charge repels the third charge and the second charge attracts the third charge.

By replacing you have:

$F=k\frac{q_1q_3}{r^2}+k\frac{q_2q_3}{r^2}\\\\F=\frac{k}{r^2}q_3[q_1+q_2]\\\\F=\frac{(8.98*10^9Nm^2/C^2)(4*10^{-6}C)}{(0.06m)^2}[8*10^{-6}C+13*10^{-6}C]\\\\F=209.53\ N$

Hence, the force between on the third charge is 209.53 N

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Constructive interference of two coherent waves will occur if the path difference is:___.

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Constructive interference of two coherent waves will occur if the path difference is λ/2.

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2 years ago

Three masses (3 kg, 5 kg, and 7 kg) are located in the xy-plane at the origin, (2.3 m, 0), and (0, 1.5 m), respectively.

Artist 52 [7]

Answer:

a) C.M $=(\bar x, \bar y)=(0.767,0.7)m$

b) $(x_4,y_4)=(-1.917,-1.75)m$

Explanation:

The center of mass "represent the unique point in an object or system which can be used to describe the system's response to external forces and torques"

The center of mass on a two dimensional plane is defined with the following formulas:

$\bar x =\frac{\sum_{i=1}^N m_i x_i}{M}$

$\bar y =\frac{\sum_{i=1}^N m_i y_i}{M}$

Where M represent the sum of all the masses on the system.

And the center of mass C.M $=(\bar x, \bar y)$

Part a

$m_1= 3 kg, m_2=5kg,m_3=7kg$ represent the masses.

$(x_1,y_1)=(0,0),(x_2,y_2)=(2.3,0),(x_3,y_3)=(0,1.5)$ represent the coordinates for the masses with the units on meters.

So we have everything in order to find the center of mass, if we begin with the x coordinate we have:

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)}{3kg+5kg+7kg}=0.767m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)}{3kg+5kg+7kg}=0.7m$

C.M $=(\bar x, \bar y)=(0.767,0.7)m$

Part b

For this case we have an additional mass $m_4=6kg$ and we know that the resulting new center of mass it at the origin C.M $=(\bar x, \bar y)=(0,0)m$ and we want to find the location for this new particle. Let the coordinates for this new particle given by (a,b)

$\bar x =\frac{(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)}{3kg+5kg+7kg+6kg}=0m$

If we solve for a we got:

$(3kg*0m)+(5kg*2.3m)+(7kg*0m)+(6kg*a)=0$

$a=-\frac{(5kg*2.3m)}{6kg}=-1.917m$

$\bar y =\frac{(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)}{3kg+5kg+7kg+6kg}=0m$

$(3kg*0m)+(5kg*0m)+(7kg*1.5m)+(6kg*b)=0$

And solving for b we got:

$b=-\frac{(7kg*1.5m)}{6kg}=-1.75m$

So the coordinates for this new particle are:

$(x_4,y_4)=(-1.917,-1.75)m$

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3 years ago

A force vector F1 points due east and has a magnitude of 200 Newtons, A second force F2 is added to F1. The resultant of the two

Andrew [12]

A force vector F1 points due east and has a magnitude of 200 Newtons, A second force F2 is added to F1. The resultant of the two vectors has a magnitude of 400 newtons and points along the due east/west line. Find the magnitude and direction of F2. Note that there are two answers. <span>The given values are
F1 = 200 N</span> F2 =? Total = 400 N

Solution: F1 + F2 = T 200 N + F2 = 400N
F2 = 400 - 200
F2 = 200 N

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2 years ago

What is the initial velocity of a go-kart traveling at a uniform acceleration of 0.5 m/s^2 for 5s as it slows down to a stop?

Arlecino [84]

The go-kart's velocity $v$ after time $t$ is given by

$v=v_0+at$

where $v_0$ is its initial velocity and $a$ is its acceleration. After $t=5\,\mathrm s$ , the go-kart stops completely, so

$0\,\dfrac{\mathrm m}{\mathrm s}=v_0+\left(-0.5\,\dfrac{\mathrm m}{\mathrm s^2}\right)(5\,\mathrm s)\implies v_0=2.5\,\dfrac{\mathrm m}{\mathrm s}$

where $a$ because we know the go-kart is slowing down.

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3 years ago

A train moves at 300 km/h. How far will it go in 5 hours?

faust18 [17]

Answer:

1,500 km

Explanation:

300 km/h

300 kilometers PER <u><em>HOUR</em></u>

<u><em></em></u>

So you want to see how far it will go in 5 hours, all you need to do is multiply the kilometers by 5. Since we know that per hour it's 300 km.

300 * 5 = 1,500

The train moves at 1,500 km in 5 hours.

Hope this helped!

Have a supercalifragilisticexpialidocious day!

5 0

3 years ago