All categories

3 years ago

How t calculate gravitational force

Physics

1 answer:

lions [1.4K]3 years ago

6 0

Answer:

F=G(m1m2)/Rsquare if radius is given

F=G(m1m2)/dsquare if distance is given

where,

f =gravitational force

G =gravitational constant

m1=mass of one object

m2=mass of another object

d=distance between two object from their center r=radius of earth/planet

You might be interested in

A rubber balloon has become negatively charged from being rubbed with a wool cloth, and the charge is measured as 1.00 × 10−14 C

inysia [295]

Answer:

1.60 x 10-10 C

Explanation:

5 0

3 years ago

Three small spheres, having masses m1 = 1 kg, m2 = 3 kg, and m3 = 4 kg, are held fixed on the x axis in deep space where the eff

Vlad [161]

Answer: $F_{net} = 0.7411 N$ towards the mass $m_{3}$

Explanation: We know that the gravitational force is a long range force which is always attractive in nature.

Given that:

mass $m_{1} = 1 kg$

mass $m_{2} = 3kg$

mass $m_{3} = 4kg$

The masses are positioned on X-axis at the following points:

Position of mass $m_{1}$ $x_{1} = 0$

Position of mass $m_{2}$ $x_{2} = 3$

Position of mass $m_{3}$ $x_{3} = 6$

Mathematically:

Gravitational force on mass $m_{2}$ due to mass $m_{1}$ is given by

$F_{21} = G \frac{m_{1}.m_{2}}{(r_{21})^2}$ ...................(1)

where: $(r_{21})^2$ = the radial distance between masses $m_{2}$ & $m_{1}$ =3

Similarly, gravitational force on mass $m_{2}$ due to mass $m_{3}$ is given by

$F_{23} = G \frac{m_{3}.m_{2}}{(r_{23})^2}$ ............................(2)

where: $(r_{23})^2$ = the radial distance between masses $m_{2}$ & $m_{3}$ =3

Now, put the respective values in the above equations.

$F_{21} = 6.67 \times 10^{-11 }\times \frac{1\times 3}{3^2}$

$F_{21} = 2.2233\times 10^{-11} N$

Again,

$F_{23} = 6.67 \times 10^{-11 }\times \frac{1\times 4}{3^2}$

$F_{23} = 2.9644\times 10^{-11} N$

∵Mass $m_{2}$ is in the middle of the masses $m_{3}$ & $m_{1}$ therefore the forces $F_{23}$ & $F_{21}$ will attract them in radially opposite direction.

∴ $F_{net} = F_{23} -F_{21} \\\\F_{net} = 2.9644-2.2233\\\\F_{net} = 0.7411 N$ towards the mass $m_{3}$

6 0

3 years ago

Read 2 more answers

A ladder rests against a vertical wall at a point 12 feet from the floor. The angle formed by the ladder and the floor is 63°. C

GenaCL600 [577]

Answer:

length of the ladder is 13.47 feet

base of wall to latter distance 6.10 feet

angle between ladder and the wall is 26.95°

Explanation:

given data

height h = 12 feet

angle 63°

to find out

length of the ladder ( L) and length of wall to ladder ( A) and angle between ladder and the wall

solution

we consider here angle between base of wall and floor is right angle

we apply here trigonometry rule that is

sin63 = h/L

put here value

L = 12 / sin63

L = 13.47

so length of the ladder is 13.47 feet

and

we can say

tan 63 = h / A

put here value

A = 12 / tan63

A = 6.10

so base of wall to latter distance 6.10 feet

and

we say here

tanθ = 6.10 / 12

θ = 26.95°

so angle between ladder and the wall is 26.95°

8 0

3 years ago

Three charges are enclosed inside a spherical closed surface. The net flux through the surface is −216 N · m2/C. If two of the c

sladkih [1.3K]

Answer:

q₃ = -4.81 nC

Explanation:

We can use the Gauss Law here:

∅ = q/∈₀

where,

∅ = Net Flux = - 216 N.m²/C

q = total charge enclosed inside sphere = ?

∈₀ = permittivity of free space = 8.85 x 10⁻¹² C/N.m²

Therefore,

- 216 N.m²/C = q / 8.85 x 10⁻¹² C²/N.m²

q = (-216 N.m²/C)(8.85 x 10⁻¹² C²/N.m²)

q = - 1.91 nC

So, the total charge will be sum of all three charges:

q = q₁ + q₂ + q₃

- 1.91 nC = 1.74 nC + 1.16 nC + q₃

q₃ = - 1.91 nC - 1.74 nC - 1.16 nC

q₃ = -4.81 nC

5 0

3 years ago

8. Noticing how much food is on your lunch tray is a quantitative observation because

zhenek [66]

Answer:

you are making an observation that uses numbers.

Explanation:

5 0

2 years ago