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Ad libitum [116K]
4 years ago
7

A positive charge, q1, of 5 µC is 3 × 10–2 m west of a positive charge, q2, of 2 µC. What is the magnitude and direction of the

electrical force, Fe, applied by q1 on q2?
Physics
1 answer:
EastWind [94]4 years ago
4 0
The law you need is :
lets call the constant c = 4pi*epsilon naught*r2.
where r is radius ( or seperation of the charges). epsilon naught is a constant of about 8.85*10^-12 or something like that. and 4 pi is just 4 times pi.
the actual equation is 
FORCE= (q1*q2)/c

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A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp
babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

4 0
3 years ago
Which is true of electricity generated both from coal and from nuclear reactions?
aleksklad [387]
They both release greenhouse gases, I think
8 0
4 years ago
Read 2 more answers
an object is tossed upwards with the same inital velocity on the earth then on the moon . will the max
vodomira [7]

The one tossed upward on the Moon will rise to a greater maximum height before starting to fall.  

It'll also spend more total time in flight before returning to the hand that tossed it. (I almost said that it'll spend "more time in the air".  That would be silly on the Moon.)

6 0
3 years ago
What is the length of the x-component of the vector shown below?
ollegr [7]

Answer:

Option B. 8.1

Explanation:

From the question given above, the following data were obtained:

Angle θ = 71°

Hypothenus = 25

Adjacent = x

Thus, we can obtain the x component of the vector by using the cosine ratio as illustrated below:

Cos θ = Adjacent /Hypothenus

Cos 71 = x/25

Cross multiply

x = 25 × Cos 71

x = 25 × 0.3256

x = 8.1

Therefore, the x component of the vector is 8.1

4 0
3 years ago
A resistor R, inductor L, and capacitor C are connected in series to an AC source of rms voltage ΔV and variable frequency. If t
Alexandra [31]

Answer:E=\frac{\pi R\left ( \Delta V\right )^2\sqrt{LC}}{\left ( R^2+\frac{9}{4}\left (\frac{L}{C} \right )\right )}

Explanation:

We know resonant frequency is given by

\omega_0=\frac{1}{\sqrt{LC}}

and the operating frequency is given by

\omega =2\omega_0=\frac{2}{\sqrt{LC}}

The capacitance reactance is given by

X_c=\frac{1}{\omega C}=\frac{\sqrt{LC}}{2C}=\frac{1}{2}\sqrt{\frac{L}{C}}

inductive reactance is given by

X_L=\omega L=\left ( \frac{2}{\sqrt{LC}}\right )L=2\sqrt{\frac{L}{C}}

Thus impedance is

Z=\left ( R^2+\left (X_L-X_C \right )^2 \right )^\frac{1}{2}

Z=\left ( R^2+\left (2\sqrt{\frac{L}{C}}-\frac{1}{2}\sqrt{\frac{L}{C}} \right )^2 \right )^\frac{1}{2}

Z=\left ( R^2+\frac{9}{4}\left ( \frac{L}{C} \right ) \right )^\frac{1}{2}

The average power delivered is

P_{avg.}=\frac{\Delta V^2}{Z}cos\phi =\frac{\left ( \Delta V\right )^2}{Z}\left (\frac{R}{Z} \right )

P_{avg.}=\frac{\left (\Delta V \right )^2R}{\left ( R^2+\frac{9}{4}\left (\frac{L}{C} \right )\right )}

Energy Delivered in one cycle is given by

E=P_{avg}T

E=\frac{\left (\Delta V \right )^2R}{\left ( R^2+\frac{9}{4}\left (\frac{L}{C} \right )\right )}\left ( \frac{2\pi }{\frac{2}{\sqrt{LC}}}\right )

E=\frac{\pi R\left ( \Delta V\right )^2\sqrt{LC}}{\left ( R^2+\frac{9}{4}\left (\frac{L}{C} \right )\right )}              

3 0
3 years ago
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