We appreciate your visit to Objects with masses of 138 kg and 688 kg are separated by 0 387 m A 38 4 kg mass is placed midway between them. 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 :
Sure! To find the magnitude of the net gravitational force exerted by the two larger masses on the 38.4 kg mass, we begin by using Newton's law of universal gravitation, which states:
[tex]\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \][/tex]
where:
- [tex]\( G = 6.672 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \)[/tex] is the universal gravitational constant,
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses interacting, and
- [tex]\( r \)[/tex] is the distance between the centers of the two masses.
In this problem:
- Mass 1 ([tex]\( m_1 \)[/tex]) is 138 kg,
- Mass 2 ([tex]\( m_2 \)[/tex]) is 688 kg,
- The third mass ([tex]\( m_3 \)[/tex]) is 38.4 kg, placed midway between the other two masses,
- The total distance between Mass 1 and Mass 2 is 0.387 m, so the distance from Mass 3 to each of the other two masses is half of this, i.e., 0.387 m / 2 = 0.1935 m.
### Step 1: Calculate the gravitational force exerted by Mass 1 on Mass 3.
Using the formula:
[tex]\[ F_1 = \frac{G \cdot m_1 \cdot m_3}{r_1^2} \][/tex]
Substitute the known values:
[tex]\[ F_1 = \frac{6.672 \times 10^{-11} \cdot 138 \cdot 38.4}{(0.1935)^2} \][/tex]
This gives:
[tex]\[ F_1 = 9.44 \times 10^{-6} \, \text{N} \][/tex]
### Step 2: Calculate the gravitational force exerted by Mass 2 on Mass 3.
Using the formula:
[tex]\[ F_2 = \frac{G \cdot m_2 \cdot m_3}{r_2^2} \][/tex]
Substitute the known values:
[tex]\[ F_2 = \frac{6.672 \times 10^{-11} \cdot 688 \cdot 38.4}{(0.1935)^2} \][/tex]
This gives:
[tex]\[ F_2 = 4.71 \times 10^{-5} \, \text{N} \][/tex]
### Step 3: Determine the net gravitational force on Mass 3.
Since the forces exerted by the two larger masses on the 38.4 kg mass are opposite in direction, the net force is the difference between the two forces:
[tex]\[ F_{\text{net}} = F_2 - F_1 \][/tex]
So:
[tex]\[ F_{\text{net}} = 4.71 \times 10^{-5} \, \text{N} - 9.44 \times 10^{-6} \, \text{N} \][/tex]
This results in:
[tex]\[ F_{\text{net}} = 3.76 \times 10^{-5} \, \text{N} \][/tex]
Thus, the magnitude of the net gravitational force exerted on the 38.4 kg mass is approximately [tex]\( 3.76 \times 10^{-5} \, \text{N} \)[/tex].
[tex]\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \][/tex]
where:
- [tex]\( G = 6.672 \times 10^{-11} \, \text{N} \cdot \text{m}^2/\text{kg}^2 \)[/tex] is the universal gravitational constant,
- [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex] are the masses interacting, and
- [tex]\( r \)[/tex] is the distance between the centers of the two masses.
In this problem:
- Mass 1 ([tex]\( m_1 \)[/tex]) is 138 kg,
- Mass 2 ([tex]\( m_2 \)[/tex]) is 688 kg,
- The third mass ([tex]\( m_3 \)[/tex]) is 38.4 kg, placed midway between the other two masses,
- The total distance between Mass 1 and Mass 2 is 0.387 m, so the distance from Mass 3 to each of the other two masses is half of this, i.e., 0.387 m / 2 = 0.1935 m.
### Step 1: Calculate the gravitational force exerted by Mass 1 on Mass 3.
Using the formula:
[tex]\[ F_1 = \frac{G \cdot m_1 \cdot m_3}{r_1^2} \][/tex]
Substitute the known values:
[tex]\[ F_1 = \frac{6.672 \times 10^{-11} \cdot 138 \cdot 38.4}{(0.1935)^2} \][/tex]
This gives:
[tex]\[ F_1 = 9.44 \times 10^{-6} \, \text{N} \][/tex]
### Step 2: Calculate the gravitational force exerted by Mass 2 on Mass 3.
Using the formula:
[tex]\[ F_2 = \frac{G \cdot m_2 \cdot m_3}{r_2^2} \][/tex]
Substitute the known values:
[tex]\[ F_2 = \frac{6.672 \times 10^{-11} \cdot 688 \cdot 38.4}{(0.1935)^2} \][/tex]
This gives:
[tex]\[ F_2 = 4.71 \times 10^{-5} \, \text{N} \][/tex]
### Step 3: Determine the net gravitational force on Mass 3.
Since the forces exerted by the two larger masses on the 38.4 kg mass are opposite in direction, the net force is the difference between the two forces:
[tex]\[ F_{\text{net}} = F_2 - F_1 \][/tex]
So:
[tex]\[ F_{\text{net}} = 4.71 \times 10^{-5} \, \text{N} - 9.44 \times 10^{-6} \, \text{N} \][/tex]
This results in:
[tex]\[ F_{\text{net}} = 3.76 \times 10^{-5} \, \text{N} \][/tex]
Thus, the magnitude of the net gravitational force exerted on the 38.4 kg mass is approximately [tex]\( 3.76 \times 10^{-5} \, \text{N} \)[/tex].
Thanks for taking the time to read Objects with masses of 138 kg and 688 kg are separated by 0 387 m A 38 4 kg mass is placed midway between them. 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
