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2 years ago

6

The strong nuclear force is an attractive force that binds _____.

1 answer:

Alja [10]2 years ago

3 0

Answer is: protons and neutrons in atomic nucleus.
Proton<span> is a </span>subatomic particle (p⁺ or p) <span>with a positive </span><span>electric charge.
</span>Neutron (n⁰ or n)<span> is a </span>subatomic particle <span>with no net </span>electric charge.
Protons and neutrons are nucleons.
The nuclear force is attractive force<span> between nucleons.</span>

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A force that needs to touch something to affect it

bija089 [108]

I would say friction because it requires to surfaces in order for the force to take place

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3 years ago

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3 years ago

What is the pressure exerted by a column of fresh water 10m high<br>(p=1000kg/m3 and g=10m/s)

alexira [117]

Height (h) = 10 m
Density (ρ) = 1000 Kg/m^3
Acceleration due to gravity (g) = 10 m/s^2
We know, pressure in a fluid = hρg
Therefore, the pressure exerted by a column of fresh water
= hρg
= (10 × 1000 × 10) Pa
= 100000 Pa

<u>Answer</u><u>:</u>

<u>1000</u><u>0</u><u>0</u><u> </u><u>Pa</u>

Hope you could understand.

If you have any query, feel free to ask.

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2 years ago

You throw a glob of putty straight up toward the ceiling, which is 3.60 m above the point where the putty leaves your hand. The

Sliva [168]

Answer:

Explanation:

Given

Height of ceiling is $h=3.6\ m$

Initial speed of Putty $u=9.5\ m/s$

Speed of Putty just before it strike the ceiling is given by

$v^2-u^2=2as$

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

$v^2-9.5^2=2\times (-9.8)\times 3.6$

$v^2=19.69$

$v=4.43\ m/s$

time taken by putty to reach the ceiling

$v=u+at$

$4.43=9.5-9.8\times t$

$t=\frac{5.07}{9.8}$

$t=0.517\ s$

8 0

3 years ago

What is the change in length of a 1400. m steel, (12x10^-6)/(C0) , pipe for a temperature change of 250.0 degrees Celsius? Remem

8090 [49]

Answer:

$\Delta L = 4.2 m$

Explanation:

As per the formula of thermal expansion we know that

$L = L_o(1 + \alpha\Delta T)$

so here we will have

$L_o = 1400 m$

$\alpha = 12 \times 10^{-6} per ^oC$

$\Delta T = 250 degree C$

so here change in the length of the rod is given as

$\Delta L = L - L_o$

$\Delta L = L_o \alpha \Delta T$

$\Delta L = 1400 (12 \times 10^{-6})(250)$

$\Delta L = 4.2 m$

8 0

3 years ago