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3 years ago

8

a circuit contains 2 ohm and 4 ohm resistors connected in parallel, find the total resistance & the amount of current in eac

h resistor if it was connected to a 12 v battery

1 answer:

Lynna [10]3 years ago

7 0

Answer:

Explanation:

Resistance

R = R1 * R2 / (R1 + R2)

R1 = 2

R2 = 4

R = 2*4 / (2 + 4)

R = 8 / 6

R = 1.3333

Current R1

R1 = 2 ohms

V = 12 volts

I = ?

V = I * R

12 = I * 2

12/2 = I

I = 6 amperes.

Current R2

V = 12 volts

R = 4 ohms

I = ?

V = I * R

12 = I * 4

12/4 = I

I = 3 amps

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Find the 24th term of the arithmetic sequence 11, 14, 17, … . Express the 24 terms of the series of this sequence using sigma no

Anna71 [15]

Answer:

The 24th term is 80 and the sum of 24 terms is 1092.

Explanation:

Given that,

The arithmetic series is

11,14,17,........24

First term a = 11

Difference d = 14-11=3

We need to calculate the 24th term of the arithmetic sequence

Using formula of number of terms

$t_{n}=a+(n-1)d$

Put the value into the formula

$t_{24}=11+(24-1)\times3$

$t_{24}=80$

$t_{24}=u_{24}=80$

We need to calculate the sum of the first 24 terms of the series

Using formula of sum,

$S_{n}=\dfrac{n}{2}(a+u_{24})$

Put the value into the formula

$S_{n}=\dfrac{24}{2}\times(11+80)$

$S_{n}=1092$

Hence, The 24th term is 80 and the sum of 24 terms is 1092.

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4 years ago

A star has a surface temperature of 30,000 K. At what wavelength does the peak radiation occur? 100 nm

tangare [24]

Use the displacement law, peak wavelength = 0.0029/T =0.0029/30000 = 97nm

8 0

3 years ago

Read 2 more answers

The mean distance of an asteroid from the Sun is 2.98 times that of Earth from the Sun. From Kepler's law of periods, calculate

lutik1710 [3]

Answer:

The asteroid requires 5.14 years to make one revolution around the Sun.

Explanation:

Kepler's third law establishes that the square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit:

$T^{2} = a^{3}$ (1)

Where T is the period of revolution and a is the semi-major axis.

In the other hand, the distance between the Earth and the Sun has a value of $1.50x10^{8} Km$ . That value can be known as well as an astronomical unit (1AU).

But 1 year is equivalent to 1 AU according with Kepler's third law, since 1 year is the orbital period of the Earth.

For the special case of the asteroid the distance will be:

$a = 2.98(1.50x10^{8}Km)$

$a = 4.47x10^{8}Km$

That distance will be expressed in terms of astronomical units:

$4.47x10^{8}Km.\frac{1AU}{1.50x10^{8}Km}$ ⇒ $2.98AU$

Finally, from equation 1 the period T can be isolated:

$T = \sqrt{a^{3}}$

$T = \sqrt{(2.98)^{3}}$

$T = \sqrt{26.463592}$

$T = 5.14AU$

Then, the period can be expressed in years:

$5.14AU.\frac{1yr}{1AU} ⇒ 5.14 yr$

$T = 5.14 yr$

Hence, the asteroid requires 5.14 years to make one revolution around the Sun.

8 0

3 years ago

A student is about to sled down a hill. He stands on a hill which is 30 meters tall and he weighs 40kg what is his GPE

PolarNik [594]

Answer:

11760J

Explanation:

Given parameters:

Height of hill = 30m

Weight = 40kg

Unknown:

Gravitational potential energy = ?

Solution:

To find the gravitational potential energy, it is the energy due to the position of a body;

G.PE = mgh

m is the mass

g is the acceleration due to gravity

h is the height

Now insert the parameters and solve;

G.PE = 40 x 9.8 x 30 = 11760J

3 0

3 years ago

A solid sphere of uniform density has a mass of 3.0 × 104 kg and a radius of 1.0 m. What is the magnitude of the gravitational f

elena55 [62]

Answer:

a) Fg = 9.495x10⁻⁶N

b) Fg = 3.908x10⁻⁶N

c)

$F_{g} =\frac{Gm_{1}m_{2}R }{r^{3} }$

Explanation:

Given:

m₁ = mass = 3x10⁴kg

r = radius = 1 m

m₂ = 9.3 kg

Questions:

a) What is the magnitude of the gravitational force due to the sphere located at R = 1.4 m, Fg = ?

b) What is the magnitude of the gravitational force due to the sphere located at R= 0.21 m, Fg = ?

c) Write a general expression for the magnitude of the gravitational force on the particle at a distance r ≤ 1.0 m from the center of the sphere.

a) Since R > r, the equation for the gravitational force is:

$F_{g} =\frac{Gm_{1}m_{2} }{R^{2} }$

Here,

G = gravitational constant = 6.67x10⁻¹¹m³/s² kg

Substituting values:

$F_{g} =\frac{6.67x10^{-11}*3x10^{4}*9.3 }{1.4^{2} } =9.495x10^{-6} N$

b) Since R < r, the equation for the gravitational force is:

$F_{g} =\frac{Gm_{1}m_{2}R }{r^{3} } =\frac{6.67x10^{-11}*3x10^{4}*9.3*0.21 }{1^{3} } =3.908x10^{-6} N$

c) The general expression for the magnitude of the gravitational force on the particle at a distance r ≤ 1.0 is the same to b)

$F_{g} =\frac{Gm_{1}m_{2} R}{r^{3} }$

8 0

3 years ago