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3 years ago

Help Me please!!!!!!!!!!!!!!!!!!!!! This My question on this test! I need This one! PLEASE!!!!!!!!

Physics

2 answers:

mojhsa [17]3 years ago

7 0

Answer:

I'll take your 5 points

faltersainse [42]3 years ago

3 0

Answer:

That is gold

Explanation:

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Plz answer #2 plz this is 8th grade science / physics

MrRissso [65]

All you need is your science book

5 0

3 years ago

Describe a scenario where a car's speed could stay the same, but the acceleration changes.

k0ka [10]

Answer:

An object's acceleration is the rate its velocity (speed and direction) changes. Therefore, an object can accelerate even if its speed is constant - if its direction changes.

Explanation:

8 0

3 years ago

A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this

wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

$r=r_o\times A^{\frac{1}{3}}$

$r_o=1.25 \times 10^{-15} m$ = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

$r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}$

$r=4.6656\times 10^{-15} m=4.6656 fm$

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) $F=k\times \frac{q_1q_2}{a^2}$

k= $9\times 10^9 N m^2/C^2$ = Coulombs constant

$q_1,q_2$ = charges kept at distance 'a' from each other

F = electrostatic force between charges

$q_1=+1.602\times 10^{-19} C$

$q_2=+1.602\times 10^{-19} C$

Force of repulsion between two protons on opposite sides of the diameter

$a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m$

$F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}$

$F=2.6527 N$

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

6 0

2 years ago

How was the formation of the outer planets affected by their distance from the sun?

djverab [1.8K]

The formation of the outer planets are affected by their distance from the sun with having them to maintain the lighter elements that they are composed of such as the hydrogen and helium, having them far away will also make their planet more cooler as the sun is distant from them.

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3 years ago

What are 3 adjectives that describe a super bowl crowd

Aleks04 [339]

Loud
Crazed
Herd-like
Irrational
Deafening
Primitive
Possessed
Tribal

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3 years ago