Question

1. the skydiver in the figure below falls at a constant speed in the spread - eagle position. after opening the parachute, is the skydiver accelerating? if so, in which direction? explain your answer. 2. compare the force holding a 10.0 - kg rock on earth and on the moon. the gravitational field on the moon is 1.6 n/kg. 3. you are riding in an elevator holding a spring scale with a 1 - kg mass suspended from it. you look at the scale and see that it reads 9.3 n. what does this tell you about the elevators motion? 4. you take a ride in a fast elevator to the top of a tall building and ride back down. compare your apparent and real weights at each part of the journey. sketch free - body diagrams to support your answers. 5. teckle, with a mass of 65.0 kg, is standing on an ice - skating rink. his friend applies a force of 9.0 n to him. what is teckles resulting acceleration? 6. you have a job loading inventory onto trucks at a meat warehouse. each truck has a weight limit of 10,000 n of cargo. you push each crate of meat along a low - resistance roller belt to a scale and weigh it before moving it onto the truck. one night, right after you weigh a 1000 - n crate, the power goes out, leaving you in the dark. describe a way in which you could apply newtons laws to approximate the mass of the remaining crates.

Explanation:

Step1: Identify the weight - force formula

The weight of an object is given by $F = mg$, where $F$ is the weight, $m$ is the mass and $g$ is the gravitational acceleration. On Earth, $g_{Earth}=9.8\ m/s^{2}$, and on the Moon, $g_{Moon} = 1.6\ N/kg$ (which is equivalent to $1.6\ m/s^{2}$).

Step2: Calculate the weight of the rock on Earth

For a rock of mass $m = 10.0\ kg$ on Earth, using $F = mg$, we have $F_{Earth}=m\times g_{Earth}=10.0\ kg\times9.8\ m/s^{2}=98\ N$.

Step3: Calculate the weight of the rock on the Moon

For the same rock of mass $m = 10.0\ kg$ on the Moon, using $F = mg$, we have $F_{Moon}=m\times g_{Moon}=10.0\ kg\times1.6\ m/s^{2}=16\ N$.

Step4: Compare the weights

We can see that $F_{Earth}=98\ N$ and $F_{Moon}=16\ N$. The force holding the 10 - kg rock on Earth is much larger than the force holding the same rock on the Moon. The ratio $\frac{F_{Earth}}{F_{Moon}}=\frac{98\ N}{16\ N}=6.125$. So the force holding the rock on Earth is 6.125 times the force holding it on the Moon.

Answer:

The force holding the 10 - kg rock on Earth ($98\ N$) is 6.125 times the force holding it on the Moon ($16\ N$).