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3 years ago

5

Ayudaaa no se como resolverlo con procedimiento :(

1 answer:

Lubov Fominskaja [6]3 years ago

6 0

This is not Spanish bro

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When a 20-V emf is placed across two resistors in series, a current of 2.0 A is present in each of the resistors. When the same

ehidna [41]

Answer:

7.24 ohm

Explanation:

Let R1 and R2 are resistance of two resistors.

Emf=E=20 V

Current,I=2 A

Current,I'=10 A

We have to find the magnitude of the greater of the two resistances.

In series

$R=R_1+R_2$

$V=IR$

By using the formula

$20=2(R_1+R_2)$

$R_1+R_2=\frac{20}{2}=10$ ...(1)

In parallel

$\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}$

$\frac{1}{R}=\frac{R_2+R_1}{R_1R_2}$

$R=\frac{R_1R_2}{R_1+R_2}$

$20=10(\frac{R_1R_2}{R_1+R_2}$

$2=\frac{R_1R_2}{10}$

$R_1R_2=20$

$R_2=\frac{20}{R_1}$

Substitute the value

$\frac{20}{R_1}+R_1=10$

$R^2_1+20=10R_1$

$R^2_1-10R_1+20=0$

$R_1=\frac{10\pm\sqrt{(-10)^2-4(20)}}{2}$

By using quadratic formula

$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$R_1=\frac{10\pm 2\sqrt 5}{2}$

$R_1=\frac{10+2\sqrt 5}{2}=5+\sqrt 5=7.24 ohm$

$R_1=\frac{10-2\sqrt 5}{2}=2.76 ohm$

Substitute the value

$R_2=\frac{20}{7.24}=2.76 ohm$

$R_2=\frac{20}{2.76}=7.24 ohm$

Hence, the magnitude of the greater of the two resistance=7.24 ohm

8 0

3 years ago

Read 2 more answers

Two tiny particles having charges 20.0 μC and 8.00 μC are separated by a distance of 20.0 cm What are the magnitude and directio

Alecsey [184]

Answer:

The magnitude and direction of electric field midway between these two charges is $10.8\times10^{5}\ N/C$ along AB.

Explanation:

Given that,

First charge $q_{1}= 20\mu C$

second charge $q_{2}= 8\mu C$

Distance = 20 cm

We need to calculate the electric field

For first charge,

Using formula of electric field

$E_{1}= \dfrac{kq_{1}}{r^2}$

Put the valueinto the formula

$E_{1}=\dfrac{9\times10^{9}\times20\times10^{-6}}{10\times10^{-2}}$

$E_{1}=18\times10^{5}\ N/C$

Direction of electric field along AB

We need to calculate the electric field

For second charge,

Using formula of electric field

$E_{2}= \dfrac{kq_{2}}{r^2}$

Put the valueinto the formula

$E_{2}=\dfrac{9\times10^{9}\times8\times10^{-6}}{10\times10^{-2}}$

$E_{2}=7.2\times10^{5}\ N/C$

Direction of electric field along AO

We need to calculate the net electric field at midpoint

$E_{net}=E_{1}-E_{2}$

$E_{net}=(18-7.2)\times10^{5}\ N/C$

$E_{net}=10.8\times10^{5}\ N/C$

Direction of net electric field along AB

Hence, The magnitude and direction of electric field midway between these two charges is $10.8\times10^{5}\ N/C$ along AB.

8 0

3 years ago

Alexandra drops an egg from 30 m above the ground to hit Vanessa on the head. If Vanessa stands 3.5 meters tall. How fast is the

I am Lyosha [343]

This is problem of free falling objects, which can be solved using the formula:

V = sqrt(2gy)

Where v is the velocity upon impact

G is the acceleration due to gravity ( 9.81 m/s2)

Y is the height

Since Venessa is 3.5 m

Y = 30 -3.5 = 26.5 m

V = sqrt(2 (9.81 m/s2) ( 26.5 m))

<span>V = 22.8 m/s</span>

7 0

3 years ago

A planet is 10 light years away from Earth. What speed would you need to go for a trip to the planet and back to take only 5 yea

viva [34]

Answer:

a. speed, v = 0.97 c

b. time, t' = 20.56 years

Given:

t' = 5 years

distance of the planet from the earth, d = 10 light years = 10 c

Solution:

(a) Distance travelled in a round trip, d' = 2d = 20 c = L'

Now, using Length contraction formula of relativity theory:

$L'' = L'\sqrt{1 - \frac{v^{2}}{c^{2}}}$ (1)

time taken = 5 years

We know that :

time = $\frac{distance}{speed}$

5 = $\frac{L''}{v}$ (2)

Dividing eqn (1) by v on both the sides and substituting eqn (2) in eqn (1):

$\frac{L'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5$

$\frac{20'\sqrt{1 - \frac{v^{2}}{c^{2}}}}{v} = 5$

Squaring both the sides and Solving above eqution, we get:

v = 0.97 c

(b) Time observed from Earth:

Using time dilation:

$t'' = \frac{t'}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}$

$t'' = \frac{5}{\sqrt{1 - \frac{(0.97c)^{2}}{c^{2}}}}$

Solving the above eqn:

t'' = 20.56 years

4 0

3 years ago

A scale contains a spring with a spring constant of 291 N/m. Placing a mass on the scale causes the spring to be compressed by 2

WINSTONCH [101]

E = 1/2*k*x^2 = 0.5*291*0.0289^2 = 0.12 J

6 0

3 years ago