We appreciate your visit to Objects with masses of 135 kg and 435 kg are separated by 0 350 m a Find the net gravitational force exerted by these objects. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!
Answer :
Final answer:
(a) The net gravitational force exerted on the 67.0 kg object is [tex]\( 2.73 \times 10^{-8} \, \text{N}\)[/tex] towards the 435 kg mass.
(b) The 67.0 kg object can be placed 1.45 m from the 435 kg mass to experience a net force of zero.
Explanation:
(a) To find the net gravitational force on the 67.0 kg object, we use Newton's law of universal gravitation: [tex]\( F = \frac{{G \cdot m_1 \cdot m_2}}{{r^2}} \)[/tex]. Substituting the given values and summing the forces from both masses yields the net force.
(b) To find the position where the net force is zero, we set the forces exerted by each mass equal and solve for the distance from the larger mass.
Thanks for taking the time to read Objects with masses of 135 kg and 435 kg are separated by 0 350 m a Find the net gravitational force exerted by these objects. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!
- Why do Businesses Exist Why does Starbucks Exist What Service does Starbucks Provide Really what is their product.
- The pattern of numbers below is an arithmetic sequence tex 14 24 34 44 54 ldots tex Which statement describes the recursive function used to..
- Morgan felt the need to streamline Edison Electric What changes did Morgan make.
Rewritten by : Barada
Answer: a) 3.94. 10⁻⁵ N b) 0.23 m. from 435 Kg. mass.
Explanation:
a) In order to find the net gravitational force on the 67.0 Kg object, we need to apply the superposition principle to the forces due to gravity, exerted by m1 and m2 on the mass located between them.
These forces are explained by the Universal Gravitation Law, discovered by Isaac Newton, and can be written as follows:
F = G m₁. m₂ / (r₁₂)², which is a vector equation.
If we assume that the three masses are aligned over a horizontal line, we can write the vector equation as an algebraic one, with components only on the x axis, taking the direction towards the greater mass as the positive one, so we can write the following equation, derived from the expression for the Universal Gravitation Law.
Fnet = F1 + F2 = G. mₓ / (.35/2) m² . (405 kg - 135 Kg)
Replacing mx= 67.0 Kg, and G = 6.67 . 10⁻¹¹ N. kg⁻².m⁻², we find the gravitational net force on the 67.0 Kg, solving for Fnet:
Fnet = 6,67. 10⁻¹¹ N. Kg⁻² . m⁻². 67.0 Kg /(. 175)² m². 300 Kg = 3.94.10⁻⁵ N, directed towards the bigger mass, 405 Kg.
b) In order to find a point between both masses where the net force on the third mass be zero, we need to make equal each other the magnitude of the force exerted by m₁ and m₂, as follows:
F₁ = F₂
Now, if we call x to the distance between the bigger mass and the third mass, as we know that the distance between m₁ and m₂ is 0.35 m, the distance between m1 and the 67.0 Kg mass, will be (0.35 - x).
So, we can write the following expression:
F₁ = F₂ ⇒ G m₁ mₓ / (0.35-x)² = G m₂ mₓ / x²
Simplifying common terms, and solving the 2nd degree equation for x, we get only one point between both masses that satisfy the equation, as follows:
X = 0.23 m from the 435 Kg. mass.
