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3 years ago

What things affect gravity

2 answers:

6 0

<span>The force of gravity = GMm / r^2, where G is gravitational constant, M is mass of one object, m is mass of another object, r is distance between them.

So, mass of objects affect gravity as well as the distance between the two objects.</span>

mamaluj [8]3 years ago

5 0

Distance and mass affect gravity. If an object has a greater mass, it's gravitational pull will be stronger. And if an object is further away from the one that is exhorting the gravitational pull, then the gravitational pull on the object will be weaker than on closer objects.

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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

3 years ago

Two forces act on an object. The first force has a magnitude of 17.0 N and is oriented 48.0° counterclockwise from the x ‑axis,

Ivanshal [37]

Answer:

Fn: magnitude of the net force.

Fn=30.11N , oriented 75.3 ° clockwise from the -x axis

Explanation:

Components on the x-y axes of the 17 N force(F₁)

F₁x=17*cos48°= 11.38N

F₁y=17*sin48° = 12.63 N

Components on the x-y axes of the the second force(F₂)

F₂x= −19.0 N

F₂y= 16.5 N

Components on the x-y axes of the net force (Fn)

Fnx= F₁x +F₂x= 11.38N−19.0 N= -7.62 N

Fny= F₁y +F₂y= 12.63 N +16.5 N = 29.13 N

Magnitude of the net force.

$F_{n} =\sqrt{(F_{nx})^{2} +(F_{ny}) ^{2} }$

$F_{n} =\sqrt{(-7.62)^{2} +(29.13) ^{2} }$

$F_{n} = 30.11N$

Direction of the net force (β)

$\beta =tan^{-1} (\frac{29.13}{7.62} )$

β=75.3°

Magnitude and direction of the net force

Fn= 30.11N , oriented 75.3 ° clockwise from the -x axis

In the attached graph we can observe the magnitude and direction of the net force

6 0

3 years ago

A bicyclist travels in a circle of radius 30.0 m at a constant speed of 8.00 m/s. the bicycle-rider mass is 82.0 kg calculate ne

IceJOKER [234]

F=m*(v^2/r)
F=82*(8^2/30)
F=174.9N

7 0

3 years ago

This involves convection currents.

Julli [10]

"Wind patterns" is the one among the following choices given in the question that <span>involves convection currents. The correct option among all the options that are given in the question is the second option or option "B". The other choices can be easily negated. i hope that this is the answer that has helped you.</span>

6 0

3 years ago

Read 2 more answers

What is the resultant force of 500g on abject accelerating at 5m/s2

max2010maxim [7]

Answer: 2.5 N

Explanation:

m = 500g = 0.5Kg

a = 5m/s2

F = ma = 0.5 x 5 = 2.5 N

6 0

3 years ago