If an atom has 35 protons in the nucleus how many electrons will It have orbiting the nucleus ?

nignag [31]

That taxi will traveled 1500s by carrying the passenger.

3 0

3 years ago

How much heat energy must be added to 52kg Of water at 68°F to raise the temperature to 212°F? The specific heat capacity for wa

Anna [14]

Answer:

The amount of energy added to rise the temperature Q = 17413.76 KJ

Explanation:

Mass of water = 52 kg

Initial temperature $T_{1}$ = 68 °F = 20° c

Final temperature $T_{2}$ = 212 °F = 100° c

Specific heat of water $C = 4.186 \frac{KJ}{kg c}$

Now heat transfer Q = m × C × ( $T_{2}$ - $T_{1}$ )

⇒ Q = 52 × 4.186 × ( 100 - 20 )

⇒ Q = 17413.76 KJ

This is the amount of energy added to rise the temperature.

4 0

3 years ago

The age of the universe in seconds

Marianna [84]

Answer:

4.351968e+17 seconds

Explanation:

So I googled age of the universe and it says 13.8 billion years.

you'll have to do some conversions:

in one year, there are 365 days. in one day, there are 24 hours. in one hour, there are 3600 seconds.

$\frac{1 year}{365 days} x \frac{1 day}{24 hours} x \frac{1 hour}{3600 seconds}$

so you have to add in the current age of the universe to that equation:

$\frac{x seconds}{13.8 bil years} x \frac{1 year}{365 days} x \frac{1 day}{24 hours} x \frac{1 hour}{3600 seconds}$

arranged so that the proper "units" will read as "x seconds" at the end of the equation.

then you do the math to find x:

13.8 billion x 365 x 24 x 3600 = 4.351968e+17 seconds.

which would be 435,196,800,000,000,000 seconds

Hope this helps :)

7 0

4 years ago

Why is a rainbow considered to be an example of a continuous spectrum?

nika2105 [10]

It has many different colors without gaps.

3 0

4 years ago

Light takes 8 minutes to reach the Earth, and the speed of light is 3.0×10^8 m/s. a) What is the orbital speed of the Earth arou

spin [16.1K]

Answer:

(a) 28690 m/s (b) $2.46x10^{33}J$

Explanation:

The orbital speed is define as:

$v = \frac{2 \pi r}{T}$ (1)

Where r is the radius of the trajectory and T is the orbital period.

To determine the orbital speed of the Earth it is necessary to know the orbital period and the radius of the trajectory. That can be done by means of the Kepler's third law and average velocity equation.

The average velocity in a Uniform Rectilinear Motion is defined as:

$v = \frac{d}{t}$ (2)

Where v is the velocity, d is the covered distance and t is the time.

Equation 2 can be rewritten for d to get:

$d = vt$ (3)

In this case, v will be the speed of light and t, the 8 minutes that takes to reach the Earth.

The time will be converted to seconds so the units in equation 3 can match:

$8min . \frac{60s}{1min}$ ⇒ $480s$

$t = 480s$

Replacing all those values in equation 3 it is gotten:

$d = (3.0x10^{8}m/s)(480s)$

$d = 1.44x10^{11}m$

Kepler’s third law is defined as:

$T^{2} = r^{3}$

Where T is orbital period and r is the radius of the trajectory.

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

$T = \sqrt{(1.44x10^{11}m)^{3}}$

It is necessary to pass from meters to astronomical unit (AU), 1 AU is defined as the distance between the Earth and the Sun.

$T = \sqrt{1AU}$

$T = 1AU$

That can be expressed in units of years.

$T = 1AU . \frac{1year}{1AU}$

$T = 1year$

But there are 31536000 seconds in one year:

$T = 1year . \frac{31536000s}{1year}$

$T = 31536000s$

Finally, equation 1 can be used:

$v = \frac{2 \pi (1.44x10^{11}m)}{(31536000s)}$

$v = 28690 m/s$

So Earth orbital speed around the Sun is 28690 m/s.

b) What is its kinetic energy?

The kinetic energy is defined as:

$E = \frac{1}{2}mv^{2}$ (4)

Notice that it is necessary to found the mass of the Earth, that can be done combining the Universal law of gravity and Newton's second law:

$F = \frac{GMm}{r^{2}}$

$ma = \frac{GMm}{r^{2}}$ (5)

M will be isolated in equation 5:

$M = \frac{r^{2}a}{G}$

Where r is the radius of the Earth ( $6.38x10^{6}m$ )

$M = \frac{(6.38x10^{6}m)^{2}(9.8m/s^{2})}{6.67x10^{-11}kg.m/s^{2}.m^{2}/Kg^{2}})$

$M = 5.98x10^{24} Kg$

$E = \frac{1}{2}( 5.98x10^{24} Kg))(28.690m/s)^{2}$

$E = 2.46x10^{33}Kg.m^{2}/s^{2}$

$E = 2.46x10^{33}J$

Hence, the kinetic energy of Earth is $2.46x10^{33}J$ .

8 0

3 years ago