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wariber [46]
3 years ago
11

If an atomic nucleus were the size of a dime how far away might one of its electrons be?

Physics
1 answer:
mihalych1998 [28]3 years ago
6 0
The radius of a nucleus of hydrogen is approximately r_{n1}=1\cdot 10^{-15}m, while we can use the Borh radius as the distance of an electron from the nucleus in a hydrogen atom: r_{e1}=5.3 \cdot 10^{-11}m

The radius of a dime is approximately r_{n2} = 9\cdot 10^{-3}m: if we assume that the radius of the nucleus is exactly this value, then we  can find how far is the electron by using the proportion
r_{n1}:r_{e1}=r_{n2}:r_{e2}
from which we find
r_{e2}= \frac{r_{e1} r_{n2}}{r_{n1}}= \frac{(5.3 \cdot 10^{-11}m)(9\cdot 10^{-3}m)}{1 \cdot 10^{-15}m}=477 m

So, if the nucleus had the size of a dime, we would find the electron approximately 500 meters away.
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A voltage V is applied to the primary coil of a step-up transformer with a 3:1 ratio of turns between its primary and secondary
Snezhnost [94]

Explanation:

Let N_p\ and\ N_s are the number of turns in primary and secondary coil of the transformer such that,

\dfrac{N_p}{N_s}=\dfrac{1}{3}

A resistor R connected to the secondary dissipates a power P_s=100\ W

For a transformer, \dfrac{N_s}{N_p}=\dfrac{V_s}{V_p}

V_s=(\dfrac{N_s}{N_p})V_p

V_s=3V_p...............(1)

The power dissipated through the secondary coil is :

P_s=\dfrac{V_s^2}{R}

100\ W=\dfrac{V_s^2}{R}

V_p^2=\dfrac{100R}{9}.............(2)

Let N_p'\ and\ N_s' are the new number of turns in primary and secondary coil of the transformer such that,

\dfrac{N_p'}{N_s'}=\dfrac{1}{24}

New voltage is :

V_s'=(\dfrac{N_s'}{N_p'})V_p'

V_s'=24V_p...............(3)

So, new power dissipated is P_s'

P_s'=\dfrac{V_s'^2}{R}

P_s'=\dfrac{(24V_p)^2}{R}

P_s'=24^2\times \dfrac{(V_p)^2}{R}

P_s'=24^2\times \dfrac{(\dfrac{100R}{9})}{R}

P_s'=6400\ Watts

So, the new power dissipated by the same resistor is 6400 watts. Hence, this is the required solution.

3 0
3 years ago
What is the gravitational potential energy of a 65.7 kg person standing on the roof of a 135 meter building?
horrorfan [7]
For the purpose we will use the following equation for potential energy:
U = m * g * h
In the above equation, m represents the mass of the object, h represents the height of the object and g represents the gravitational field strength (9.8 N/kg on Earth).
When we plug values into the equation, we get following:
U= 65.7kg * 9.8 N/kg *135m = 86921.1 J = 86.92 kJ

8 0
3 years ago
The pull of what is weaker when 2 objects are farther apart
9966 [12]
The gravitational pull is weaker.
7 0
3 years ago
If an IPS student weighs 115 lbs on Earth, what is their weight in Newtons?
Montano1993 [528]

Answer:

511.545 Newtons

Explanation:

So 1 pound=4.44822 Newton’s so 115 times 4.44822 is 511.5453, then round it to get 511.545 Newtons.

6 0
3 years ago
Read 2 more answers
In a house 5 bulbs of 25 watts each are used for 6 hours a day.Calculate the unit of electricity consumed in a month of 30 days.
Svetradugi [14.3K]

(5 bulbs) x (25 watt/bulb) x (6 hour/day) x (30 day/month) =

             (5 x 25 x 6 x 30) watt-hour/month =

                 22,500 watt-hour/month .

The most common unit of electrical energy used for billing purposes
is the 'kilowatt-hour' = 1,000 watt-hours .

               22,500 watt-hour/month =  <em>22.5 kWh/month</em>.

             (22.5 kWh/month) x (1.50 Rs/kWh) =  <em>33.75 Rs / month


</em>
4 0
3 years ago
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