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mariarad [96]
4 years ago
10

Atomic nuclei are made of protons and neutrons. This fact by itself shows thatthere must be some other type of force in addition

to electrical and gravitationalforces. Explain.
Physics
1 answer:
Stells [14]4 years ago
5 0

Protons and neutrons are packed together in a very small region called nucleus. Protons are positively charged and we know that like charges repel. Then how is it that protons are not repelling each other and flying away from nucleus?

You may think that gravitational force is holding all the protons together but it is not so. Gravitational force is many times weaker than repulsive force.

It is actually strong force which holds proton together. At this short distance, strong force comes into play and is several times stronger than the repulsive force.

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A handheld glass rod can be charged by rubbing it with silk or a plastic bag while holding it in your hands. Would you conclude
WITCHER [35]

Answer:

a conductor is an object or type of material that allows the flow of charge (electrical current) in one or more directions.

Explanation:

. Materials made of metal are common electrical conductors.

3 0
3 years ago
A golden-colored cube is handed to you. The person wants you to buy it for $100, saying that is a gold nugget. You pull out your
kenny6666 [7]

Answer:

Explanation:

Volume of cube=side³

=2*2*2

=8 cm³

Mass of cube=40g

Its density=mass/volume

=5g/cm³

which differs the value given in the text hence it is not gold and u should not bye it...

Instead u should call the police for arresting the person lol...

6 0
3 years ago
Read 2 more answers
Air at 80 kPa, 27°C, and 220 m/s enters a diffuser at a rate of 2.5 kg/s and leaves at 42°C. The exit area of the diffuser is
9966 [12]

Answer:

exit velocity = 62 m/s

exit pressure is 98.52332 kPa

Explanation:

given data

pressure p1 = 80 kPa

temperature t1 =  27°C = 300 K

velocity v1 = 220 m/s

temperature t2 =  42°C = 315K

area exit = 370 cm²

lose heat at a rate = 18 kJ/s

gas constant of air = 0.287 kPa·m^3/kg·K.

h1 = 300.19 kJ/kg

h2 = 315.27 kJ/kg

to find out

exit velocity and exit pressure

solution

first we apply here energy balance equation to find out outlet velocity

Einlet = Eoutlet

m(h1 + v1²/2)  = m (h2+v2²/2) +Q

m(v1²-v2²) /2  =  m (h2 -h1) +Q

v2² = v1² - 2Cp Δt - 2Q/m

here  Δt = h2 - h1 and put all value

v2² = 220² - 2(1.005)10³ (15) - 2(18)10³ /2.5

exit velocity = 62 m/s

so

now find outlet pressure that is

p2 = mRT / A2 v2

put value

p2 = 2.5 (287) 315 / (370 10^{-4} 62)

p2 = 98.52332 kPa

exit pressure is 98.52332 kPa

3 0
3 years ago
A boy finds an abandoned mine shaft in the woods, and wants to know how deep the hole is. He drops in a stone, and counts 4 seco
oee [108]
S=(0x4)+(0.5x4.81x4x4)
S=0.78.48

The depth is approximately 78 meters.
(My brain hurts now) :P Good Luck!
4 0
3 years ago
A small, 2.00-mm-diameter circular loop with R = 1.00×10^−2 Ω is at the center of a large 100-mm-diameter circular loop. Both lo
Assoli18 [71]

Answer:

i=2.1\times 10^{-8}}\ A

Explanation:

Given that

Diameter of small loop d= 2 mm  or r=1 mm

Diameter of large loop D= 100 mm  or R=50 mm

We know that induce emf given as

\varepsilon =-\dfrac{d\phi }{dt}

\varepsilon =-\dfrac{BA }{dt}

B=\dfrac{\mu _oI}{2\pi R}

\varepsilon =-\pi \times r^2\times \dfrac{4\pi \times 10^{-7}(I_2-I_1)}{2\pi Rdt}

\varepsilon =-\pi \times 0.001^2\times \dfrac{4\pi \times 10^{-7}(-1-1)}{2\pi \times 0.05\times 0.1}

\varepsilon =2.1\times 10^{-10}\ V

So induce current

i=emf/R

i=\dfrac{2.1\times 10^{-10}}{10^{-2}}\ A

i=2.1\times 10^{-8}}\ A

5 0
3 years ago
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