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iogann1982 [59]
3 years ago
9

At a distance D from a very long (essentially infinite)uniform line of charge, the elecric field is 1000 N/C. Forthe field stren

gth to be 2000 N/C, the distance from the line wouldhave to be:a. D/√2b. √2Dc. D/2d. 2De. D/4
Physics
2 answers:
jolli1 [7]3 years ago
8 0

Answer:

A

Explanation:

The electric field = 1000N/C

Using the formula

F = q/4πeD^2 where F is the force experience by a charge body brought from infinity to D and e is the field constant in C/N.m^2 and D is the distance from the uniform field

1000 = q/4πeD^2

Make q subject of the formula

1000*(4πeD^2) = q

If the force changes to 2000 then equation changes to:

2000 = q /4πeD1^2 where D1 is the new distance

Substitute q in equation 2

2000 = 1000*(4πeD^2) / 4πeD1^2

Cancel 4πe

2000 = 1000 * D^2 / D1^2

Cross multiply and make D1^2 subject of the formula

D1^2 = 1000D^2/ 2000

D1^2 = D^2/2

Take square root of both side

D1 = √(D^2/2) = D / √2

Vsevolod [243]3 years ago
8 0

Answer:

The answer is A

Explanation:

Electrical field formula is:

F=2*k*q/D[tex]F=2*k*q/D^2=1000\\F=2*k*q/(D/\sqrt{2}) ^2=2000

To increase the electrical field from 1000 N/C to 2000 N/C when the q is constant the distance would have to be:

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Two balls have their centers 2.0 m apart. One ball has a mass of m1 = 7.9 kg. The other has a mass of m2 = 6.1 kg. What is the g
maks197457 [2]

Answer:

3.036×10⁻¹⁰ N

Explanation:

From newton's law of universal gravitation,

F = Gm1m2/r² .............................. Equation 1

Where F = Gravitational force between the balls, m1 = mass of the first ball, m2 = mass of the second ball, r = distance between their centers.

G = gravitational constant

Given: m1 = 7.9 kg, m2 = 6.1 kg, r = 2.0 m, G = 6.67×10⁻¹¹ Nm²/C²

Substituting into equation 1

F = 6.67×10⁻¹¹×7.9×6.1/2²

F = 321.427×10⁻¹¹/4

F = 30.36×10⁻¹¹

F = 3.036×10⁻¹⁰ N

Hence the force between the balls = 3.036×10⁻¹⁰ N

8 0
3 years ago
A professional cyclist rides a bicycle that is 92 percent efficient. For every 100 joules of energy he exerts as input work on t
emmainna [20.7K]
Efficiency =  Work Output / Work Input

92%  =  Work Output / 100

0.92 =   Work Output / 100

Work Output = 0.92 * 100

Work Output  = 92 joules.
8 0
2 years ago
A certain first-order reaction is 58% complete in 95 s. What are the values of the rate constant and the half-life for this proc
guajiro [1.7K]

Answer:

0.005734 s^{-1} and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..

Explanation:

Expression for rate law for first order kinetics is given by:

a_o=a\times e^{-kt}

where,

k = rate constant  

t = age of sample

a_o = let initial amount of the reactant  

a = amount left after decay process  

We have :

a_o=x

a=58\%\times x=0.58x

t = 95 s

0.58x=x\times e^{-k\times 95 s}

\k= 0.005734 s^{-1}

Half life is given by for first order kinetics::

t_{1/2}=\frac{0.693}{k}

=\frac{0.693}{0.005734 s^{-1}}=120.86 s

0.005734 s^{-1} and 20.86 seconds are the values of the rate constant and the half-life for this process respectively..

3 0
2 years ago
1. Calculate the average velocity of the following trip. You walk to Pershing Square 58
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Explanation:

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v = √((58 m)² + (135 m)²) / (12 min × 60 s/min)

v = 0.20 m/s

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