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3 years ago

Which statement is true for electric field lines?

Physics

2 answers:

KiRa [710]3 years ago

8 0

The best answer is D. field lines should always be crossing each other.

yawa3891 [41]3 years ago

3 0

The correct statement is

Field lines should always be perpendicular to the equipotential lines.

we know that, along equipotential lines, work done in moving a charge is zero. Since work done is zero, the force on the charge must be perpendicular to the displacement. the electric force on a charge is parallel to the direction of electric field. hence the electric field must be perpendicular to the displacement or equipotential lines.

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What is radioactive dating? How is it used to determine age of something?

Liono4ka [1.6K]

Answer:

Technique of comparing abundance ratio between radioactive isotopes to a reference isotope to determine the age of a material called radioactive dating. It determines the age by having a more abundance of isotopes in the cellular being.

6 0

3 years ago

The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is

stealth61 [152]

To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.

For mercury, density, gravity and height are defined as

$\rho_m = 846lb/ft^3$

$g = 32.17405ft/s^2$

$h_1 = 1in = \frac{1}{12} ft$

For the air the defined properties would be

$\rho_a = 0.0075lb/ft^3$

$g = 32.17405ft/s^2$

$h_2 = ?$

We have for equilibrium that

$\text{Pressure change in Air}=\text{Pressure change in Mercury}$

$\rho_m g h_1 = \rho_a g h_2$

Replacing,

$(846)(32.17405)(\frac{1}{12}) = (0.0075)(32.17405)(h_2)$

Rearranging to find $h_2$

$h_2 = \frac{(846)(32.17405)(\frac{1}{12}) }{(0.0075)(32.17405)}$

$h = 9400ft$

Therefore the elevation of the mountain top is 9400ft

7 0

3 years ago

2. An alternating current is represented by the equation I=20sin 100mt.

Sladkaya [172]

Explanation:

The general equation of an AC current is given by :

$I=I_o sin\omega t$

Where

I₀ is the peak value of current

$\omega$ is angular frequency

$As\ \omega=2\pi f$

So,

$f=\dfrac{\omega}{2\pi}\\\\f=\dfrac{100\pi}{2\pi}\\\\f=50\ Hz$

We know that,

$I_{rms}=\dfrac{I_0}{\sqrt{2}}\\\\=\dfrac{20}{\sqrt{2}}\\\\I_{rms}=14.14\ A$

So, the frequency is 50 Hz and the maximum rms value of current is 14.14 A.

3 0

3 years ago

Before the big bang, the universe was much ____ and _____ than it is now.

Dahasolnce [82]

The answer to the question is that before the big bang, the universe was much hotter and more dense than it is now. Letter B.
It is because after the big bag occurred, the universe became cooler and less dense.
a. - does not correspond in the answer because the universe became less dense after the big bang.
c - the universe became cool and less dense after the big bang so being cool and less dense does not correspond to the question.
d - cooler does not answer the question because it only became cooler after the big bang.

5 0

3 years ago

Read 2 more answers

An athlete competing in long jump leaves the ground with a speed of 9.14 m/s at an angle of 35.00 above the horizontal. What is

Lemur [1.5K]

Answer:

R = 8.01 m

Explanation:

We can solve this problem using the projectile launch equations. The jump length is the throw range

R = v₀² sin 2θ / g

in the exercise they give us the initial speed of 9.14 m / s and in the launch angle 35º

let's calculate

R = 9.14² sin (2 35) / 9.8

R = 8.01 m

this is the jump length

5 0

3 years ago