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4 years ago

How many liters are equivalent to 500 centiliters

Physics

2 answers:

andreev551 [17]4 years ago

4 0

5 Liters are equal to 500 centiliters

PilotLPTM [1.2K]4 years ago

4 0

1000 centiliters = 1 liter.

500 centiliters = x liter

1000 / 500 = 1/x Cross multiply

1000x = 500 Divide by 1000

x = 500/1000

x = 1/2

500 centiliters = 1/2 or 0.5 of a liter.

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Calculate the drop voltage for a battery of internal resistance 2 ohm and emf of 24 volt

Ivenika [448]

Answer:

<em>The drop voltage is 0.3 V</em>

Explanation:

Electromotive Force EMF

When connecting a battery of internal resistance Ri and EMF ε to an external resistance Re, the current through the circuit is:

$\displaystyle i=\frac{\varepsilon }{R_e+R_i}$

The battery has an internal resistance of Ro=2 Ω, ε=24 V and is connected to an external resistance of Re=158 Ω. Thus, the current is:

$\displaystyle i=\frac{24 }{158+2}$

$\displaystyle i=\frac{24 }{160}$

i = 0.15 A

The drop voltage is the voltage of the internal resistance:

$V_i = i.R_i$

$V_i = 0.15*2$

$\boxed{V_i = 0.3\ V}$

The drop voltage is 0.3 V

3 0

3 years ago

A string of length L, fixed at both ends, is capable of vibrating at 309 Hz in its first harmonic. However, when a finger is pla

Scorpion4ik [409]

Answer:

i = 0.3326 L

Explanation:

A fixed string at both ends presents a phenomenon of standing waves, two waves with the same frequency that are added together. The expression to describe these waves is

2 L = n λ n = 1, 2, 3…

The first harmonic or leather for n = 1

Wave speed is related to wavelength and frequency

v = λ f

λ = v / f

Let's replace in the first equation

2 L = 1 (v / f₁)

For the shortest length L = L-l

2 (L- l) = 1 (v / f₂)

These two equations form our equation system, let's eliminate v

v = 2L f₁

v = 2 (L-l) f₂

2L f₁ = 2 (L-l) f₂

L- l = L f₁ / f₂

l = L - L f₁ / f₂

l = L (1- f₁ / f₂)

Let's calculate

l / L = (1- 309/463)

i / L = 0.3326

4 0

3 years ago

A 9-μC positive point charge is located at the origin and a 6 μC positive point charge is located at x = 0.00 m, y = 1.0 m. Find

sukhopar [10]

Answer:

The coordinates of the point is (0,0.55).

Explanation:

Given that,

First charge $q_{1}=9\times10^{-6}\ C$ at origin

Second charge $q_{2}=6\times10^{-6}\ C$

Second charge at point P = (0,1)

We assume that,

The net electric field between the charges is zero at mid point.

Using formula of electric field

$E=\dfrac{kq}{r^2}$

$0=\dfrac{k\times9\times10^{-6}}{d^2}+\dfrac{k\times6\times10^{-6}}{(1-d)^2}$

$\dfrac{(1-d)}{d}=\sqrt{\dfrac{6}{9}}$

$\dfrac{1}{d}=\dfrac{\sqrt{6}}{3}+1$

$\dfrac{1}{d}=1.82$

$d=\dfrac{1}{1.82}$

$d=0.55\ m$

Hence, The coordinates of the point is (0,0.55).

3 0

3 years ago

Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2

Alinara [238K]

Answer:

a) $E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

b) $E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

$E=K\dfrac{q}{r^2}$

$E_1=K\dfrac{q_1}{r_1^2}$

$E_2=K\dfrac{q_2}{r_2^2}$

Now by putting the va;ues

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C$

$E_1=72.11\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C$

$E_2=21.58\times 10^{6}\ N/C$

The net electric field

$E=E_1-E_2$

$E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C$

$E_1=120.3\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C$

$E_2=147.92\times 10^{6}\ N/C$

The net electric field

$E=E_1+E_2$

$E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

7 0

3 years ago

Freckles are a dominant trait in humans. Both of the girls have the genotype FF for freckles. If either one marries a man with n

trapecia [35]

D. The man's traits (ff) would be recessive and with one of the girls (FF), all their offspring would have the genotype Ff, meaning all the offspring have freckles.

3 0

3 years ago

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