NET FORCE ( PLEASE ANSWER ASAP )

This chart compares the different features of two planets in our solar system (Mercury and Mars). In this chart, the only numerical value or feature that is the same for both planets is gravity because for both planets gravity is 1.7 m/s2. This implies the gravitational force or the force that attracts objects towards the center of the planet is the same or that objects are pulled with the same force in both planets. Moreover, this factor depends on others such as mass, density, among others.

8 0

3 years ago

Develop an equation (with a proportionality constant) that describes the relationship between the gravitational force (fgrav), t

ivolga24 [154]

According to newton's law of gravitation, the gravitational force(F) is directly proportional to the product mass of the moon(Mm) and the mass of the planet (Mp) and it is inversely proportional to the square of the separation between them.

Fg ∝ (Mp)(Mm) →(1)

Fg ∝ 1/d²→(2)

Combining equation (1) and (2),

Fg ∝ (Mp)(Mm)/d²

Fg = G(Mp)(Mm)/d²

This is an equation that describes the relation between mass of moon (Mm) and mass of planet (Mp) and separation(d) between them.

To support the claim in favuor of this equation we use this equation to obtain the value of acceleration due to gravity on earth.

Let m be the mass of an object on earth then Fg between earth (Mp) and mass of an object is obtained by:

Fg = G(Mp)(m)/R², where R= Radius of earth

This force is equal to the weight of an object i.e.,

g= G(Mp)/R²

Putting the values of G, Mp and R , we get, g=9.81 m/s²

which is the value we obtained on earth for acceleration due to gravity.

To know more about gravitational constant, visit:

brainly.com/question/13959861

#SPJ4

5 0

1 year ago

Three packing crates of masses, M1 = 6 kg, M2 = 2 kg and M3 = 8 kg are connected by a light string of negligible mass that passe

never [62]

Answer:

39.81 N

Explanation:

I attached an image of the free body diagrams I drew of crate #1 and #2.

Using these diagram, we can set up a system of equations for the sum of forces in the x and y direction.

∑Fₓ = maₓ

∑Fᵧ = maᵧ

Let's start with the free body diagram for crate #2. Let's set the positive direction on top and the negative direction on the bottom. We can see that the forces acting on crate #2 are in the y-direction, so let's use Newton's 2nd Law to write this equation:

∑Fᵧ = maᵧ
T₁ - m₂g = m₂aᵧ

Note that the tension and acceleration are constant throughout the system since the string has a negligible mass. Therefore, we don't really need to write the subscripts under T and a, but I am doing so just so there is no confusion.

Let's solve for T in the equation...

T₁ = m₂aᵧ + m₂g
T₁ = m₂(a + g)

We'll come back to this equation later. Now let's go to the free body diagram for crate #1.

We want to solve for the forces in the x-direction now. Let's set the leftwards direction to be positive and the rightwards direction to be negative.

∑Fₓ = maₓ

F_f - F_g sinΘ = maₓ

The normal force is equal to the x-component of the force of gravity.

(F_n · μ_k) - m₁g sinΘ = m₁aₓ

(F_g cosΘ · μ_k) - m₁g sinΘ = m₁aₓ
[m₁g cos(30) · 0.28] - [m₁g sin(30)] = m₁aₓ
[(6)(9.8)cos(30) · 0.28] - [(6)(9.8)sin(30)] = (6)aₓ
[2.539595871] - [-58.0962595] = 6aₓ
60.63585537 = 6aₓ
aₓ = 10.1059759 m/s²

Now let's go back to this equation:

T₁ = m₂(a + g)

We have 3 known variables and we can solve for the tension force.

T = 2(10.1059759 + 9.8)
T = 2(19.9059759)
T = 39.8119518 N

The tension force is the same throughout the string, therefore, the tension in the string connecting M2 and M3 is 39.81 N.

3 0

3 years ago