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2 years ago

9

Quanto tempo deve ficar ligado um ferro eletrico de 1000 w para que tenha o mesmo consumo de energia que um chuveiro de 4400 w q

ue fica ligado 10 minutos

1 answer:

Cerrena [4.2K]2 years ago

8 0

Answer:

Thus, the time for the first lamp is 44 minutes.

Explanation:

Power of first lamp, P' = 1000 W

Power of second lamp, P'' = 4400 W

time for second lamp, t'' = 10 minutes

Let the time for first lamp is t'.

As the energy is same, so,

P' x t' = P'' x t''

1000 x t' = 4400 x 10

t' = 44 minutes

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A charge q1 = +5.00 nC is placed at the origin of an xy-coordinate system, and a charge q2 = -2.00 nC is placed on the positive

Ivahew [28]

Answer:

a

The x- and y-components of the total force exerted is

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

b

The magnitude of the force is

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction of the force is

$\theta =327.43 ^o$ Clockwise from x-axis

Explanation:

From the question we are told that

The magnitude of the first charge is $q_1 = +5.00nC = 5.00*10^{-9}C$

The magnitude of the second charge is $q_2 = -2.00nC = -2.00*10^{-9}C$

The position of the second charge from the first one is $d_{12} = 4.00i \ cm = \frac{4.00i}{100} = 4.00i *10^{-2} m$

The magnitude of the third charge is $q_3 = +6.00nC = 6.00*10^{-9}C$

The position of the third charge from the first one is $\= d_{31} = (4i + 3j) cm = \frac{ (4i + 3j)}{100} = (4i + 3j) *10^{-2}m$

$|d_{31}| =(\sqrt{4 ^2 + 3^2}) *10^{-2} m$

$|d_{31}| =5 *10^{-2} m$

The position of the third charge from the second one is

$\= d_{32} = 3j cm = 3j *10^{-2}m$

$|d_{32}| =(\sqrt{ 3^2}) *10^{-2} m$

$|d_{32}| =3 *10^{-2} m$

The force acting on the third charge due to the first and second charge is mathematically represented as

$F_{31 +32} = \frac{kq_3 q_1}{|d_{31}| ^3} *\= d_{31} + \frac{kq_3 q_2}{|d_{32}| ^3} *\= d_{32}$

Substituting values

$F_{31 +32} = \frac{9 *10^9 * 6 *10^{-9} * 5*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2} \\ \ + \ \ \ \ \ \ \ \ \ \frac{9 *10^9 * 6 *10^{-9} * -2*10^{-9} }{(5*10^{-2}) ^3} * (4i + 3j ) *10^{-2}$

$F_{31 +32} = 2.16 *10^{-5} (4i + 3j) - 12*10^{-5} j$

$F_{31 +32} = (8.64i - 5.52 j) *10^{-5}$

The magnitude of $F_{31 +32}$ is mathematically evaluated as

$|F_{31 +32}| = \sqrt{(8.64^2 + 5.52 ^2) } *10^{-5}$

$|F_{31 +32}| = 10.25 *10^{-5} N$

The direction is obtained as

$tan \theta = \frac{-5.52 *10^{-5}}{8.64 *10^{-5}}$

$\theta = tan ^{-1} [-0.63889]$

$\theta = - 32.57 ^o$

$\theta = 360 - 32.57$

$\theta =327.43 ^o$

5 0

3 years ago

Rank the following meteorites in terms of their age, from youngest to oldest:1) a meteorite containing 500 238U isotopes and 100

satela [25.4K]

Answer:

Explanation:

When 238U which is radioactive turns into 206Pb , it becomes stable and no further disintegration is done . Hence in the initial period ratio of 238U undecayed and 206Pb formed will be very high because no of atoms of 238U in the beginning will be very high. Gradually number of 238U undecayed will go down and number of 206Pb formed will go up . In this way the ratio of 238U and 206Pb in the mixture will gradually reduce to be equal to one or even less than one .

In the given option we shall calculate their raio

1 ) ratio of 238U and 206Pb = 5

2 ) ratio of 238U and 206Pb = 4

3 )ratio of 238U and 206Pb = 1

4 ) ratio of 238U and 206Pb = 20

5 )ratio of 238U and 206Pb = 3

lowest ratio is 1 , hence this sample will be oldest.

Ranking from youngest to oldest

4 , 1 , 2 , 5 , 3 .

4 0

3 years ago

A 1.0 kg ball falls from rest a distance of 15m. what was its change in potential energy

expeople1 [14]

Answer:

112.5 J

Explanation:

I calculated it by K/G BY M/S TO POTENTIAL ENERGY.

8 0

3 years ago

Use differentials to estimate the amount of metal in a closed cylindrical can that is 10 cm high and 4 cm in diameter if the met

ira [324]

The amount of metal in a closed cylindrical can that is 10 cm high and 4 cm in diameter if the metal on the top and the bottom is 0.1 cm thick and the metal on the sides is 0.05 cm thick is 8.8 cm.

The formula for calculating the volume of a cylinder is given below.

V = πr^2 h

Get the differential of the volume as shown:

dV = V/ h dh + V / r dr

V/ h = πr^2

V/ h = 2 πr h

Now, the differential becomes

dV = πr^2dh + 2πrh dr

Given the following parameters i.e. diameter and height

dh = 0.1 + 0.1 = 0.2 cm

dr = 0.05 cm

h = 10 cm

d = 4 cm

r = 2cm

Substituting the values in the above equation, we get

dV = 3.14(2)^2(0.2) + 2(3.14)(2)(10)(0.05)

dV = 2.512 + 6.28

dV = 8.792 cm

dV = 8.8 cm

If you need to learn more about diameter click here:

brainly.com/question/16813738

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4 0

1 year ago

When is kinetic energy transferred from one object to another? Question 24 options:

postnew [5]

When two objects collide is the answer...Hope this helps :)

7 0

3 years ago

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