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Lunna [17]
4 years ago
12

The nucleus of a hydrogen atom is a single proton, which has a radius of about 1.2 × 10-15 m. The single electron in a hydrogen

atom orbits the nucleus at a distance of 5.3 × 10-11 m. What is the ratio of the density of the hydrogen nucleus to the density of the complete hydrogen atom?
Physics
1 answer:
Nutka1998 [239]4 years ago
6 0

Answer: 0.86 × 10^14

Explanation:

Given the following :

Radius of proton = 1.2 × 10-15 m

Radius of hydrogen atom = 5.3 × 10-11 m

Density of proton could be calculated thus:

Mass of proton = 1.67 × 10^-27 kg

Using the formula :

(4/3) × pi × r^3

(4/3) × 3.142 × (1.2 × 10^-15)^3 = 7.24 × 10^-45

Density = mass / volume

Density = (1.67 × 10^-27) / ( 7.24 × 10^-45)

= 0.2306 × 10^18

Density of hydrogen atom:

Mass of hydrogen atom= 1.67 × 10^-27 kg

Using the formula :

(4/3) × pi × r^3

(4/3) × 3.142 × (5.3 × 10^-11)^3 = 6.24 × 10^-31

Density = mass / volume

Density = (1.67 × 10^-27) / ( 6.24 × 10^-31)

= 0.2676 × 10^4

Ratio is thus:

Density of proton / density of hydrogen atom

0.2306 × 10^18 / 0.2676 × 10^4 = 0.8617 × 10^14

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If Star A is magnitude 1.0 and Star B is magnitude 9.6 , which is brighter and by what factor?
ludmilkaskok [199]

Answer:

Star A is brighter than Star B by a factor of 2754.22

Explanation:

Lets assume,

the magnitude of star A = m₁ = 1

the magnitude of star B = m₂ = 9.6

the apparent brightness of star A and star B are b₁ and b₂ respectively

Then, relation between the difference of magnitudes and apparent brightness of two stars are related as give below: (m_{2} - m_{1}) = 2.5\log_{10}(b_{1}/b_{2})

The current magnitude scale followed was formalized by Sir Norman Pogson in 1856. On this scale a magnitude 1 star is 2.512 times brighter than magnitude 2 star. A magnitude 2 star is 2.512 time brighter than a magnitude 3 star. That means a magnitude 1 star is (2.512x2.512) brighter than magnitude 3 bright star.

We need to find the factor by which star A is brighter than star B. Using the equation given above,

(9.6 - 1) = 2.5\log_{10}(b_{1}/b_{2})

\frac{8.6}{2.5}  = \log_{10}(b_{1}/b_{2})

\log_{10}(b_{1}/b_{2}) = 3.44

Thus,

(b_{1}/b_{2}) = 2754.22

It means star A is 2754.22 time brighter than Star B.

3 0
3 years ago
Identify global climate zones and characteristics of each
IgorC [24]

research the different time zone around the world, and characteristics of each

3 0
3 years ago
5. What is the velocity of a Nolan Catholic football player if he runs 200.0 m in 24.0 s?
Degger [83]

Answer:

\boxed{ \bold{ \huge{ \boxed {\sf{8.33 \: m \:/s \: }}}}}

Explanation:

Distance travelled = 200 metre

Time taken = 24 second

Velocity = ?

<u>Finding </u><u>the</u><u> </u><u>velocity</u><u> </u>

\boxed{ \sf{velocity =  \frac{distance \: travelled \: }{time \: taken}}}

\dashrightarrow{ \sf{velocity =  \frac{200 \: m}{24 \: s} }}

\dashrightarrow{ \sf{velocity = 8.33 \: m/s}}

Hope I helped!

Best regards!

6 0
3 years ago
A 800 kg safe is 2.1 m above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compresses
stellarik [79]

Answer:

k = 17043.5 N/m = 17.04 KN/m

Explanation:

First we need to find the force applied by safe pn the spring:

F = Weight of Safe

F = mg

where,

F = Force Applied by the safe on the spring = ?

m = mass of safe = 800 kg

g = 9.8 m/s²

Therefore,

F = (800 kg)(9.8 m/s²)

F = 7840 N

Now, using Hooke's Law:

F = kΔx

where,

K = Spring Constant = ?

Δx = compression = 46 cm = 0.46 m

Therefore,

7840 N = k (0.46 m)

k = 7840 N/0.46 m

<u>k = 17043.5 N/m = 17.04 KN/m</u>

5 0
4 years ago
When nuclear fission occurs some mass is lost where does that mass go
horrorfan [7]
The loss of matter is called the mass defect. The missing matter is converted into energy. You can actually calculate the amount of energy produced during a nuclear reaction with  fairly simple equation developed by Albert Einstein; E = mc^2. In this equation, E is the amount of energy produced, m is the missing mass, or the mass defect, and c is the speed of light, which is a rather large number. The speed of light is squared, making that part of the equation a very large number that, even when multiplied by a small amount of mass, yields a large amount of energy.
4 0
3 years ago
Read 2 more answers
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