PHYSICS COLLEGE

A man with a mass of 65.0 kg skis down a frictionless hill that is 5.00 m high. At the bottom of the hill the terrain levels out. As the man reaches the horizontal section, he grabs a 20.0-kg backpack and skis off a 2.00-m-high ledge. At what horizontal distance from the edge of the ledge does the man land (the man starts at rest)?

Answers

Answer 1

Answer:

Answer:

The horizontal distance is 4.823 m

Solution:

As per the question:

Mass of man, m = 65.0 kg

Height of the hill, H = 5.00 m

Mass of the backpack, m' = 20.0 kg

Height of ledge, h = 2 m

Now,

To calculate the horizontal distance from the edge of the ledge:

Making use of the principle of conservation of energy both at the top and bottom of the hill (frictionless), the total mechanical energy will remain conserved.

Now,

$KE_(initial) + PE_(initial) = KE_(final) + PE_(final)$

where

KE = Kinetic energy

PE = Potential energy

Initially, the man starts, form rest thus the velocity at start will be zero and hence the initial Kinetic energy will also be zero.

Also, the initial potential energy will be converted into the kinetic energy thus the final potential energy will be zero.

Therefore,

$0 + mgH = (1)/(2)mv^(2) + 0$

$2gH = v^(2)$

$v = √(2* 9.8* 5) = 9.89\ m/s$

where

v = velocity at the hill's bottom

Now,

Making use of the principle of conservation of momentum in order to calculate the velocity after the inclusion, v' of the backpack:

$mv = (m + m')v'$

$65.0* 9.89 = (65.0 + 20.0)v'$

$v' = 7.56\ m/s$

Now, time taken for the fall:

$h = (1)/(2)gt^(2)$

$t = \sqrt{(2h)/(g)}$

$t = \sqrt{(2* 2)/(9.8) = 0.638\ s$

Now, the horizontal distance is given by:

x = v't = $7.56* 0.638 = 4.823\ m$

Answer 2

Answer:

Answer

given,

mass of the man = 65 kg

height = 5 m

mass of the back pack = 20 kg

skis off to 2.00 m high ledge

horizontal distance =

speed of the person before they grab back pack is equal to potential and kinetic energy

$mgh= (1)/(2)mv^2$

$v = √(2gh)$

$v = √(2* 9.8 * 5)$

v = 9.89 m/s

now he perform elastic collision

$v = (m_1v_1)/(m_1+m_2)$

$v = (65* 9.89)/(65+20)$

v = 7.57 m/s

time taken by the skies to fall is

$h = (1)/(2)gt^2$

$t = \sqrt{(2h)/(g)}$

$t = \sqrt{(2* 2)/(9.8)}$

t = 0.6388 s

distance

d = v x t

d = 7.57 x 0.6388

d = 4.84 m

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Answers

Answer:

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In this example we will use pendulum motion to actually measure the acceleration of gravity on a different planet. An astronaut on the surface of Mars measures the frequency of oscillation of a simple pendulum consisting of a ball on the end of a string. He finds that the pendulum oscillates with a period of 1.5 s. But the acceleration due to gravity on Mars is less than that on earth, gMars=0.38gearth. Later, during a journey to another planet, the astronaut finds that his simple pendulum oscillates with a period of 0.92 s. What planet is he now on?SOLUTIONSET UP Each planet has a different value of the gravitational acceleration g near its surface. The astronaut can measure g at his location, and from this he can determine what planet he's on. First we use the information about Mars to find the length L of the string that the astronaut is swinging. Then we use that length to find the acceleration due to gravity on the unknown planet.

Answers

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Answers

A weather pattern is defined as a period of time when the weather remains consistent. In the lab, a lot of observation about weather is obtained

What is the definition of a weather pattern?

A weather pattern is defined as a period of time when the weatherremains consistent. Weather changes are crucial to humanexistence.

because they influence our everyday activities and provide moisture for crops.

The rain does not always end within the day, and gloomy days might last just as long as sunny days. Tornadoes and hurricanes, for example, may inflict tremendous damage.

In the lab the following observation about weather is obtained;

1. We will find the graphs and statistics that indicate signs of climate change and engage with an interactivegraphic.

2. You'll also look at and debate maps of global temperature and precipitation patterns that are changing.

3. This lab will teach you about Earth's biomes and the close relationship that exists between them and the climates that serve to define them.

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Explanation:

The question refers to the subjects of meteorology, climatology, and atmospheric science. Meteorology is the study of the atmosphere, atmospheric phenomena, and atmospheric impacts on the Earth's weather. It involves the prediction of weather in the short term based on thousands of measurements of variables such as air pressure and temperature.

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Atmospheric Science is a broad field that encompasses both meteorology and climatology, as well as other disciplines that study the atmosphere. This discipline is typically based heavily on physics and involves the study of weather and climate patterns, predictions of developments in weather and climatic events, and the analysis of the effects of these events on the planet and its inhabitants.

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