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3 years ago

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Which equation is correct according to Ohm’s law?

2 answers:

podryga [215]3 years ago

8 0

Where are the options ?

Korvikt [17]3 years ago

4 0

You didn't give us anything other than the question. No options are provided so I cannot answer. Nobody can.

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Define the law of universal gravitation in your own words.

Firdavs [7]

Newton's law of universal gravitation states that a particle attracts every other particle in the universe using a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers

7 0

3 years ago

What is the wavelength of violet light that has a frequency of 7.5x10^14

Crank

Λ= V/f
<span>but change it to represent the speed of light, c </span>
<span>λ= c/f </span>
<span>c = 3.00 x 10^8 m/s </span>
<span>Plug in your given info and solve for λ(wavelength) </span>
<span>λ= 3.00 x 10^8 m/s / 7.5 x 10^14 Hz
(3.00 x 10^8) / (7.5 x 10^14) = 300,000,000 / 750,000,000,000,000 = 0.0000004
Hope this helps :)
</span>

6 0

3 years ago

Read 2 more answers

A long, thin rod parallel to the y-axis is located at x = - 1 cm and carries a uniform positive charge density λ = 1 nC/m . A se

zheka24 [161]

Answer:

The electric field at origin is 3600 N/C

Solution:

As per the question:

Charge density of rod 1, $\lambda = 1\ nC = 1\times 10^{- 9}\ C$

Charge density of rod 2, $\lambda = - 1\ nC = - 1\times 10^{- 9}\ C$

Now,

To calculate the electric field at origin:

We know that the electric field due to a long rod is given by:

$\vec{E} = \frac{\lambda }{2\pi \epsilon_{o}{R}$

Also,

$\vec{E} = \frac{2K\lambda }{R}$ (1)

where

K = electrostatic constant = $\frac{1}{4\pi \epsilon_{o} R}$

R = Distance

$\lambda$ = linear charge density

Now,

In case, the charge is positive, the electric field is away from the rod and towards it if the charge is negative.

At x = - 1 cm = - 0.01 m:

Using eqn (1):

$\vec{E} = \frac{2\times 9\times 10^{9}\times 1\times 10^{- 9}}{0.01} = 1800\ N/C$

$\vec{E} = 1800\ N/C$ (towards)

Now, at x = 1 cm = 0.01 m :

Using eqn (1):

$\vec{E'} = \frac{2\times 9\times 10^{9}\times - 1\times 10^{- 9}}{0.01} = - 1800\ N/C$

$\vec{E'} = 1800\ N/C$ (towards)

Now, the total field at the origin is the sum of both the fields:

$\vec{E_{net}} = 1800 + 1800 = 3600\ N/C$

7 0

3 years ago

A very flexible helium-filled balloon is released from the ground into the air at 20. ∘C. The initial volume of the balloon is 5

koban [17]

Answer:

V = 38.0 L

Explanation:

As we know that number of moles will remains conserved inside the balloon

so we will have

$moles = \frac{PV}{RT}$

here we have

$\frac{P_1V_1}{RT_1} = \frac{P_2V_2}{RT_2}$

now we have

$P_1 = 760 mm Hg$

$P_2 = 76 mm Hg$

$V_1 = 5.00 L$

$T_1 = 20^o C = 293 K$

$T_2 = -50^o C = 223 K$

$\frac{(760mm Hg)(5L)}{R(293)} = \frac{(76mm Hg)(V)}{R(223)}$

$V = 38.0 L$

5 0

3 years ago

Read 2 more answers

Jay pushes the box with a force of 25 Newtons. There are 5 Newtons of friction present. In what direction will the box accelerat

yKpoI14uk [10]

Answer: To the right

Explanation:

8 0

1 year ago