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Triss [41]
3 years ago
12

Two point charges of magnitudes +5.00 μC, and +7.00 μC are placed along the x-axis at x = 0 cm and x = 100 cm, respectively. Whe

re must a third charge be placed along the x-axis so that it does not experience any net force because of the other two charges? Two point charges of magnitudes +5.00 μC, and +7.00 μC are placed along the x-axis at x = 0 cm and x = 100 cm, respectively. Where must a third charge be placed along the x-axis so that it does not experience any net force because of the other two charges? 50 cm 45.8 cm 9.12 cm 91.2 cm 4.58 cm
Physics
1 answer:
joja [24]3 years ago
8 0

Answer:

45.8 cm

Explanation:

To solve this, we will use the formula

5 / x² = 7/(1 - x)²

5 / x² = 7 / (1 - 2x + x²)

5 / 7 = x² / (1 - 2x + x²)

x = 0.5 * (√(35) - 5) meters

x = 0.5 * (5.916 - 5)

x = 0.5 * (0.916)

x = 0.458 or x = 45.8

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azamat

Answer:

Answer D : about 1067 meters

Explanation:

There are two steps to this problem:

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a=\frac{Vf-Vi}{t}

Where Vf is the final velocity of the plane (in our case: zero )

Vi is the initial velocity of the plane (in our case: 80 m/s)

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a=\frac{Vf-Vi}{t}\\-3=\frac{0-80}{t}\\t=\frac{-80}{-3} = \frac{80}{3}

with units corresponding to seconds given the quantities involved in the calculation.

2) Second knowing the time it took the plane to stop, now use that time in the equation for the distance traveled under accelerated motion:

Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi= 80 (\frac{80}{3}) +\frac{1}{2} (-3) (\frac{80}{3}) ^{2}=1066.666666...

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7 0
3 years ago
An object is placed 18 cm in front of spherical mirror.if the image is formed at 4cm to the right of the mirror, calculate it's
ivolga24 [154]
1) Focal length

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d_o is the distance of the object from the mirror
d_i is the distance of the image from the mirror

In this case, d_o = 18 cm, while d_i=-4 cm (the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
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3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.

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r=2f=2 \cdot 5.1 cm=10.2 cm
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4 years ago
An x-ray photon is scattered by an originally stationary electron. how does the frequency of the scattered photon compare relati
Viefleur [7K]

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U_{0}=K_{f}

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K_{f}=\frac{1}{2}mv_{f}^{2}

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This new equation can be simplified if we multiplied both sides of the equation by a 2, so we get:

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so now we can solve this for the final velocity, so we get:

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Answer:8.3m/sec 30 sec,

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5 0
3 years ago
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