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sergij07 [2.7K]

4 years ago

12

The force acting between two charged particles a and b is 5.2 x 10^-5 newtons. Charges a and b are 2.4 x 10^-2 apart. If the cha

rge on particle a is 7.2 x 10^-8 coulombs, what is the charge of particle B?

1 answer:

Aneli [31]4 years ago

5 0

4.62x10^-11
hope this helped!

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A farmer with 600 ft of available fencing wants to enclose a rectangular are located adjacent to a farmhouse (with no fencing re

Goshia [24]

Answer:

The corresponding dimensions will be "x = 300 & y = 60".

Explanation:

Available fencing = 600 ft

Fencing,

⇒ $5y+x=600$

⇒ $x=600-5y$

As we know,

⇒ $Area =xy$

On substituting the given values, we get

$=(600-5y)y$

$=600y-5y^2$ ...(equation 1)

On differentiating with respect to y, we get

⇒ $\frac{dA}{dy} =600-10y$

$0 = 600-10y$

$600=10y$

$y=\frac{600}{10}$

$y=60$

On putting the values of y in equation 1, we get

⇒ $600(60)-5(60)^2$

⇒ $600(60)-5(3600)$

⇒ $36000 - 18000$

⇒ $18000$

Dimensions of the rectangular area:

x = 300

y = 60

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3 years ago

A 10kg wagon is pulled along the floor with a force of 5N to the North.

77julia77 [94]

Answer:

50N

Explanation:

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3 years ago

A single-turn plane loop of wire with a cross-sectional area 200 cm2 is perpendicular to a magnetic field that increases uniform

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Answer:

Induced current, $I=2.87\times 10^{-3}\ A$

Explanation:

Given that,

Area of cross section of the wire, $A=200\ cm^2=0.02\ m^2$

Time, t = 2.2 s

Initial magnetic field, $B_i=0.2\ T$

Final magnetic field, $B_f=2.8\ T$

Resistance of the coil, R = 8 ohms

The expression for the induced emf is given by :

$\epsilon=-\dfrac{d\phi}{dt}$

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$\epsilon=A\dfrac{d(B)}{dt}$

$\epsilon=A\dfrac{B_f-B_i}{t}$

$\epsilon=0.02\times \dfrac{2.8-0.2}{2.2}$

$\epsilon=-0.023\ volts$

So, the induced emf in the loop is 0.023 volts. The induced current can be calculated using Ohm's law as :

$\epsilon= IR$

$I=\dfrac{\epsilon}{R}$

$I=\dfrac{0.023}{8}$

$I=2.87\times 10^{-3}\ A$

So, the magnitude of the induced current in the loop of wire is $2.87\times 10^{-3}\ A$ . Hence, this is the required solution.

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