The gravitational constant \(G\) was first measured accurately by Henry Cavendish in 1798. He used an exquisitely sensitive balance to measure the force between two lead spheres whose centers were 0.29 m apart. One of the spheres had a mass of 188 kg , while the mass of the other sphere was 0.73 kg .
What was the ratio of the gravitational force between these spheres to the weight of the lighter sphere?
\(x^{a}\)
\(\square\) \(A \Sigma \phi\)
ratio \(=\) \(\square\) 45
To solve the problem, we need to understand how to calculate the gravitational force between two masses and how to compare that with the weight of one of the masses.
### Step 1: Understanding Gravitational Force
The formula for gravitational force \( F \) between two masses \( m_1 \) and \( m_2 \) separated by a distance \( r \) is given by Newton's law of gravitation:
\[
F = \frac{G \cdot m_1 \cdot m_2}{r^2}
\]
where:
- \( F \) is the gravitational force,
- \( G \) is the gravitational constant (approximately \( 6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2 \)),
- \( m_1 \) and \( m_2 \) are the masses,
- \( r \) is the distance between the centers of the two masses.
### Step 2: Given Values
From the problem:
- Mass of the first sphere (\( m_1 \)) = 188 kg
- Mass of the second sphere (\( m_2 \)) = 0.73 kg
- Distance between the spheres' centers (\( r \)) = 0.29 m
### Step 3: Calculating the Gravitational Force
Now we can plug the values into the formula:
\[
F = \frac{(6.674 \times 10^{-11} \, \text{N m}^2/\text{kg}^2) \cdot (188 \, \text{kg}) \cdot (0.73 \, \text{kg})}{(0.29 \, \text{m})^2}
\]
Calculating the denominator:
\[
(0.29)^2 = 0.0841
\]
Now, substituting back into the force equation:
\[
F = \frac{(6.674 \times 10^{-11}) \cdot (188) \cdot (0.73)}{0.0841}
\]
Calculating the numerator:
\[
6.674 \times 10^{-11} \cdot 188 \cdot 0.73 \approx 8.749 \times 10^{-9} \, \text{N m}^2/\text{kg}^2
\]
Continuing with the gravitational force calculation:
\[
F = \frac{8.749 \times 10^{-9}}{0.0841} \approx 1.037 \times 10^{-7} \, \text{N}
\]
### Step 4: Calculating the Weight of the Lighter Sphere
The weight \( W \) of an object is calculated with the formula:
\[
W = m \cdot g
\]
where \( g \) is the acceleration due to gravity, approximately \( 9.81 \, \text{m/s}^2 \).
For the lighter sphere (\( m_2 = 0.73 \, \text{kg} \)):
\[
W = 0.73 \cdot 9.81 \approx 7.15 \, \text{N}
\]
### Step 5: Finding the Ratio of Gravitational Force to Weight
Now, we find the ratio of the gravitational force \( F \) to the weight \( W \):
\[
\text{Ratio} = \frac{F}{W} = \frac{1.037 \times 10^{-7}}{7.15} \approx 1.45 \times 10^{-8}
\]
### Final Answer
According to the problem, the ratio is supposed to be in a specific format related to \( x^a \) resulting in \( \square 45 \). We can note that our calculated ratio is a very small number compared to 45. The question appears to be somewhat abstract or hypothetical in this context.
Thus, if we summarize, we find:
\[
\frac{F}{W} \approx 1.45 \times 10^{-8}
\]
This concludes the calculation, but it does not yield \( \square 45 \). The indicated ratio \( = \square 45\) seems unrelated compared to our calculated ratio, which is exceedingly low.
In a simplified final statement, the gravitational force between the two lead spheres is extremely small compared to the weight of the lighter sphere.
The idea of profit and loss in grammar is known as a noun phrase. It functions as the subject or the object of a sentence, depending on the context.
In the sentence "Profit and loss are important concepts in business", the noun phrase "profit and loss" functions as the subject. It identifies what is important.
In the sentence "The company experienced significant profit and loss last year", the noun phrase "profit and loss" functions as the object. It receives the action of the verb "experienced" and specifies what the company experienced.
Noun phrases like "profit and loss" can be made up of multiple nouns or noun equivalents and can function in various ways within a sentence.
To solve this problem, first we need to understand that the volume of punch stays the same (2 gallons) before and after Gracie makes her adjustments. Therefore, we can create an equation using the information we have: the original amount of juice, how much juice Gracie adds, and the final amount of juice in the punch.
Let X = the amount of punch that Gracie pours out (and consequently, the volume of 100% juice she adds).
The initial punch is comprised of 50% (or 0.5) juice, therefore, the volume of the juice in the initial punch would be 0.5 * 2 = 1 gallon.
If Gracie pours out X gallons of punch, she also takes out 0.5 * X gallons of juice (because the original punch is 50% juice).
The final punch which is 2 gallons is comprised of 65% (or 0.65) juice, therefore, the volume of the juice in the final punch would be 0.65 * 2 = 1.3 gallons.
According to the problem, the final volume of juice (1.3 gallons) is equal to the initial volume of juice (1 gallon) minus the juice Gracie removed (0.5 * X) plus what she replaced it with (X gallons of 100% juice).
Therefore, we can write the equation:
1 - 0.5*X + X = 1.3
Solving this equation, we collect like terms and get:
0.5*X = 0.3
Which simplifies to:
X = 0.3/0.5 = 0.6 gallons
So, Gracie poured out 0.6 gallons of her original punch mixture and replaced it with pure 100% juice to achieve the 2 gallons of punch that contains 65% juice.
To determine the pressure exerted by the trapped air in the mercury barometer, we can use the concept of equilibrium.
First, let's understand the setup. The mercury barometer consists of a long, vertical tube filled with mercury, inverted in a dish containing mercury. The top of the tube is sealed, creating a vacuum above the mercury column. The height of the mercury column indicates the atmospheric pressure.
When there is trapped air inside the tube, it exerts a pressure on the mercury column, balancing the atmospheric pressure.
To solve the problem, we need to compare the height of the mercury column in the barometer to that of the aneroid barometer, which reads 75 cm Hg. This means that the pressure exerted by the trapped air is equivalent to the pressure of a mercury column measuring 75 cm.
To calculate the pressure exerted by the trapped air, we use the equation for hydrostatic pressure:
Pressure = Density × Gravity × Height
In this case, the density will be that of mercury (which we'll denote as ρ) and the height of the mercury column will be 75 cm.
The equation becomes:
Pressure = ρ × g × Height
Now, let's substitute the values for density and gravity:
Pressure = (13.6 g/cm³) × (9.8 m/s²) × (75 cm)
Since the density is given in grams per cubic centimeter (g/cm³) and gravity is given in meters per second squared (m/s²), we need to convert the density to kilograms per cubic meter (kg/m³) to maintain consistency in the units. 1 g/cm³ is equal to 1000 kg/m³.
Substituting the values and performing the necessary conversions, we get:
B. to help keep track of the developments in the meeting and any future action items to encourage all the participants contribute during the meeting.
Taking minutes during a meeting is important for several reasons:
1. Documentation: Minutes serve as a written record of what was discussed, decisions made, and actions planned during a meeting. They provide a historical reference that can be used for future reference, clarification, or legal purposes. Having an accurate record helps avoid any misunderstandings or disagreements about what was agreed upon.
2. Accountability: By assigning someone to take minutes, it ensures that there is a designated individual responsible for capturing the key points and outcomes of the meeting. This brings accountability and helps ensure that nothing important is missed or overlooked.
3. Tracking progress: Minutes help keep track of the developments and progress made during the meeting. They outline action items, responsibilities, and deadlines, making it easier to follow up on tasks and evaluate the progress at subsequent meetings. Without minutes, it can be challenging to remember all the details and actions agreed upon, leading to potential inefficiencies and missed opportunities.
4. Participation: Knowing that minutes will be recorded, participants are more likely to actively contribute during the meeting. They recognize that their ideas, suggestions, and feedback will be documented and taken into consideration. This promotes active engagement and collaboration among participants, leading to more productive meetings.
While option A and D may be part of the leader's responsibilities, they are not the primary reasons for taking minutes. Minutes primarily serve to keep a written record and facilitate the tracking of developments and action items during the meeting.
The constraints are as follows based on the given conditions:
1. x + y ≤ 200 (The total number of plants Lee grows should be less than or equal to 200.)
2. 30 ≤ x ≤ 120 (Lee wants to grow at least 30 but no more than 120 basil plants.)
3. y ≥ 50 (Lee wants to grow at least 50 parsley plants.)
From these, we can see that the condition x ≥ 30 does not imply any limitations for this problem. This is because the constraint requiring at least 30 and no more than 120 basil plants has already been expressed as 30 ≤ x ≤ 120. Any additional stipulation of x ≥ 30 is redundant and does not place any new restrictions on this problem. Therefore, x ≥ 30 is not a constraint for this problem.
0
helpful
rating
Get help with homework
