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The weight of an object on an orbiting space vehicle is measured to be \(42 \mathrm{~N}\) based on an artificial gravitational acceleration of \(6 \mathrm{~m} / \mathrm{s}^{2}\). What is the weight of the object, in \(\mathrm{N}\), on earth, where \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\) ?

Short Answer

Expert verified
Approximately 68.67 N.

Step by step solution

01

- Understanding the Given Values

Identify the given values from the problem.
02

- Calculate the Mass of the Object

Use the formula for weight, which is given by: \[ W = m \times g \] Rearrange to solve for mass: \[ m = \frac{W}{g} \] Substitute the given values of weight (42 N) and artificial gravitational acceleration (6 \mathrm{~m} / \mathrm{s}^{2}): \[ m = \frac{42}{6} = 7 \mathrm{~kg} \]
03

- Calculate the Weight on Earth

Use the calculated mass and Earth's gravitational acceleration to find the weight on Earth. \[ W_{earth} = m \times g_{earth} \] Substitute the mass (7 \mathrm{~kg}) and Earth's gravity (9.81 \mathrm{~m} / \mathrm{s}^{2}): \[ W_{earth} = 7 \times 9.81 \approx 68.67 \mathrm{~N} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

artificial gravitational acceleration
Artificial gravitational acceleration is a concept used in space to create a force similar to gravity, allowing astronauts to stay healthy under long periods in space. This is achieved by rotating the spacecraft or a part of it to simulate gravity. The artificial gravitational accelation in this problem is given as 6 m/s². This force can be adjusted based on the needs of the mission and the health of the astronauts. It's important to understand that artificial gravity is not real gravity; it's a result of rotational forces acting on the mass of objects. It helps in keeping muscles and bones strong, preventing them from becoming weaker over time.
mass calculation
Mass is a fundamental property of an object that does not change, regardless of location. To determine mass when a gravitational force (weight) and gravitational acceleration are known, you can use the equation: \[ W = m \times g \] Rearranging gives: \[ m = \frac{W}{g} \] In this exercise, the weight measured under artificial gravity is 42 N, and the artificial gravitational acceleration is 6 m/s². Using the formula: \[ m = \frac{42}{6} = 7 \text{ kg} \] The calculated mass of the object is 7 kg. Once we know the mass, we can find the weight under different gravitational forces by using the same mass with the new gravitational acceleration.
gravitational force
Gravitational force, often referred to as weight, is the force with which an object is pulled towards the center of a planet, such as Earth, due to gravity. On Earth, gravity accelerates objects downward at 9.81 m/s². To calculate the weight of an object on Earth, you use the object's mass and Earth's gravity: \[ W_{earth} = m \times g_{earth} \] Given the mass calculated earlier as 7 kg, we substitute this value into the equation: \[ W_{earth} = 7 \times 9.81 \approx 68.67 \text{ N} \] The weight of the object on Earth is approximately 68.67 N, demonstrating how the gravitational force changes based on the gravitational acceleration at different locations.

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Most popular questions from this chapter

An object weighs \(25 \mathrm{kN}\) at a location where the acceleration of gravity is \(9.8 \mathrm{~m} / \mathrm{s}^{2}\). Determine its mass, in \(\mathrm{kg}\).

The variation of pressure within the biosphere affects not only living things but also systems such as aircraft and undersea exploration vehicles. (a) Plot the variation of atmospheric pressure, in atm, versus elevation \(z\) above sea level, in \(\mathrm{km}\), ranging from 0 to \(10 \mathrm{~km}\). Assume that the specific volume of the atmosphere, in \(\mathrm{m}^{3} / \mathrm{kg}\), varies with the local pressure \(p\), in \(\mathrm{kPa}\), according to \(v=72.435 / p\). (b) Plot the variation of pressure, in atm, versus depth \(z\) below sea level, in \(\mathrm{km}\), ranging from 0 to \(2 \mathrm{~km}\). Assume that the specific volume of seawater is constant, \(v=0.956 \times\) \(10^{-3} \mathrm{~m}^{3} / \mathrm{kg}\) In each case, \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\) and the pressure at sea level is \(1 \mathrm{~atm}\)

An object whose mass is \(10 \mathrm{~kg}\) weighs \(95 \mathrm{~N}\). Determine (a) the local acceleration of gravity, in \(\mathrm{m} / \mathrm{s}^{2}\). (b) the mass, in \(\mathrm{kg}\), and the weight, in \(\mathrm{N}\), of the object at a location where \(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\)

The following table lists temperatures and specific volumes of water vapor at two pressures: \begin{tabular}{ccccc} \hline\(p=1.0 \mathrm{MPa}\) & & \multicolumn{2}{c}{\(p=1.5 \mathrm{Mpa}\)} \\ \cline { 1 - 2 } \hline\(\left({ }^{\circ} \mathrm{C}\right)\) & \(v\left(\mathrm{~m}^{3} / \mathrm{kg}\right)\) & & \(T\left({ }^{\circ} \mathrm{C}\right)\) & \(v\left(\mathrm{~m}^{3} / \mathrm{kg}\right)\) \\ \hline 200 & \(0.2060\) & 200 & \(0.1325\) \\ 240 & \(0.2275\) & 240 & \(0.1483\) \\ 280 & \(0.2480\) & 280 & \(0.1627\) \\ \hline \end{tabular} Data encountered in solving problems often do not fall exactly on the grid of values provided by property tables, and linear interpolation between adjacent table entries becomes necessary. Using the data provided here, estimate (a) the specific volume at \(T=240^{\circ} \mathrm{C}, p=1.25 \mathrm{MPa}\), in \(\mathrm{m}^{3} / \mathrm{kg}\). (b) the temperature at \(p=1.5 \mathrm{MPa}, v=0.1555 \mathrm{~m}^{3} / \mathrm{kg}\), in \({ }^{\circ} \mathrm{C}\). (c) the specific volume at \(T=220^{\circ} \mathrm{C}, p=1.4 \mathrm{MPa}\), in \(\mathrm{m}^{3} / \mathrm{kg}\).

Determine the total force, in \(\mathrm{kN}\), on the bottom of a \(100 \times 50 \mathrm{~m}\) swimming pool. The depth of the pool varies linearly along its length from \(1 \mathrm{~m}\) to \(4 \mathrm{~m}\). Also, determine the pressure on the floor at the center of the pool, in \(\mathrm{kPa}\). The atmospheric pressure is \(0.98\) bar, the density of the water is \(998.2 \mathrm{~kg} / \mathrm{m}^{3}\), and the local acceleration of gravity is \(9.8 \mathrm{~m} / \mathrm{s}^{2}\).

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