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denis23 [38]
3 years ago
7

Dos cargas Q1=2pc y Q2=4pc estan separadas por una distancia de 6cm ¿con que fuerza se atraen?

Physics
1 answer:
noname [10]3 years ago
3 0

Here we can use coulomb's law to find the force between two charges

As per coulombs law

]tex]F = \frac{kq_1q_2}{r^2}[/tex]

here we have

k = 9 * 10^9

q_1 = 2pC

q_2 = 4pC

r = 6cm = 0.06 m

now by using the above equation we have

F = \frac{9*10^9 * 2*10^{-12} * 4*10^{-12}}{0.06^2}

F = 2 * 10^{-11} N

so here the force between two charges is of above magnitude and this will be repulsive force between them as both charges are of same sign.

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Hello, I wanted an answer from a mathematician. The number 1.04 is closer to the number 1, 2, 1.25 or 1.5.
babymother [125]

Answer:

Explanation:

Of the 4 numbers given, the answer is 1 or A

If you take the absolute value of abs(1 - 1.04) you get 0.04.

(2 - 1.04) = 0.96

1.25 - 1.04 = .21

1.5 - 1.04 = 0.46

The last three are all larger than 0.04

Note: absolute value means the positive difference between 2 numbers (even  though it is negative). If it is negative, absolute value makes it positive.

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2 years ago
What two kinds of motion combine to produce projectile motion?
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Horizontal and Vertical are the two motions that combine to produce projectile motion.

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Suppose that a comet that was seen in 563 A.D. by Chinese astronomers was spotted again in year 1951. Assume the time between ob
Mars2501 [29]

Answer:

a=2.77*10^{13}m

R_a=5.49*10^{13}m

Explanation:

The period of the comet is the time it takes to do a complete orbit:

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writen in seconds:

2514years*\frac{3,154*10^7s}{1year}=7.93 *10^{10}s

Since the eccentricity is greater than 0 but lower than 1 you can know that the trajectory is an ellipse.

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R_a=a(1+e)\\R_a=2.77*10^{13}(1.986)\\R_a=5.49*10^{13}m

8 0
2 years ago
The 3.00 kg cube in fig. 15-47 has edge lengths d 6.00 cm and is mounted on an axle through its center. a spring (k 1200 n/m con
ira [324]

The Period of the resulting shm will be T=39.7

<u>Explanation:</u>

<u>Given data</u>

m=3kg

d=.06m

k=1200 N/m

Θ=3 °

T=?

we have the formulas,

I = (1/6)Md2

F = ma

F = -kx = -(mω2x)

k = mω2 τ = -d(FgsinΘ)

T=2 x 3.14/ √(m/k)

Solution for the given problem would be,

F=-Kx (where x= dsin Θ)

F=-k dsin Θ

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F=-10.16N

<u>By newton's second law.</u>

F = ma

a= F/m

a=(-10.16N)/3

a=3.38

<u>using the k=mω value</u>

k=mω

ω=k/m

ω=1200/3

ω=400

<u>Using F = -kx value</u>

x = F/-k

x=(-10.16)/1200

x=0.00847m

<u>Restoring the  torque value </u>

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α =(.06)(4)(9.81)sin(4°)

α=-1.781

<u>Rotational to linear form</u>

a = αr  

r = .1131 m

a=-1.781 x .1131 m

a=-0.2015233664

<u>Time Period</u>

T=2 x 3.14/ √(m/k)

T=6.28/√(3/1200)

T=6.28/0.158

T=39.7

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3 years ago
What does the law of conservation of energy say
HACTEHA [7]

Answer: states that energy can neither be created nor destroyed - only converted from one form of energy to another.

Explanation:

6 0
3 years ago
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