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3 years ago

Two resistances of 100ohms and 0 ohms were connected in parallel. What is the overall resistance

Physics

1 answer:

Inga [223]3 years ago

6 0

Answer:

20 ohms,100*2/10=20

Explanation:

since it as to be in parallel arrangement and connected in side by side due to same p.d in a cross each resistor

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Heat energy is produced when molecules move. True False

Dennis_Churaev [7]

Answer:

true

Explanation:

7 0

3 years ago

A potential energy function is given by u(x)=(3.00n)xâ(1.00n/m2)x3. at what position or positions is the force equal to zero?

qwelly [4]

I believe the correct form of the energy function is:

u (x) = (3.00 N) x + (1.00 N / m^2) x^3

or in simpler terms without the units:

u (x) = 3 x + x^3

Since the highest degree is power of 3, therefore there are two roots or solutions of the equation.

Since we are to find for the positions x in which the force equal to zero, u (x) = 0, therefore:

3 x + x^3 = u (x)

3 x + x^3 = 0

Taking out x:

x (3 + x^2) = 0

So one of the factors is x = 0.

Finding for the other two factors, we divide the two sides by x and giving us:

x^2 + 3 = 0

x^2 = - 3

x = sqrt (- 3)

x = - 1.732 i, 1.732 i

The other two roots are imaginary therefore the force is only equal to zero when the position is also zero.

Answer:

x = 0

5 0

3 years ago

Four point charges are individually brought from infinity and placed at the corners of a square. Each charge has the identical v

Brut [27]

To solve this problem we will apply the concept of voltage given by Coulomb's laws. From there we will define the charges and the distance, and we will obtain the total value of the potential difference in the system.

The length of diagonal is given as

$l = 2a$

The distance of the center of the square from each of the corners is

$r = \frac{2a}{2}= a$

The potential electric at the center due to each cornet charge is

$V_1 = \frac{kQ_1}{r_1}$

$V_2 = \frac{kQ_2}{r_2}$

$V_3 = \frac{kQ_3}{r_3}$

$V_4 = \frac{kQ_4}{r_4}$

The total electric potential at the center of the given square is

$V = V_1+V_2+V_3+V_4$

$V = \frac{kQ_1}{r_1}+ \frac{kQ_2}{r_2}+\frac{kQ_3}{r_3}+\frac{kQ_4}{r_4}$

Al the charges are equal, and the distance are equal to a, then

$V = \frac{kQ}{a}+ \frac{kQ}{a}+\frac{kQ}{a}+\frac{kQ}{a}$

$V = \frac{4kQ}{a}$

Therefore the correct option is E.

3 0

3 years ago

Dark matter has been detected by its

Zinaida [17]

Dark matter has been detected by its gravitational pull.

7 0

4 years ago

earth orbits the sun once every 365.25 days. Find the average angular. speed of Earth about the sun.Answer in units of rad/s.

anzhelika [568]

Answer:

$1.99\cdot 10^{-7} rad/s$

Explanation:

The average angular speed of the Earth about the Sun is given by:

$\omega = \frac{2 \pi}{T}$

where

$2 \pi$ rad is the total angle corresponding to one revolution of the Earth around the Sun

T is the orbital period of the Earth

The orbital period of the Earth is 365.25 d. We must convert it into seconds first:

$T=365.25 d \cdot (24 h/d) \cdot (60 min/h) \cdot (60 s/min)=3.16\cdot 10^7 s$

And by substituting into the equation above, we find the average angular speed:

$\omega=\frac{2 \pi rad}{3.16\cdot 10^7 m/s}=1.99\cdot 10^{-7} rad/s$

7 0

3 years ago

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