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2 years ago

10

Essam is abseiling down a steep cliff. How much gravitational potential energy does he lose for every metre he descends? His mas

s is 72kg and the gravitational field strength is 10N/kg.

2 answers:

murzikaleks [220]2 years ago

8 0

He loses 720 J for every metre he descends

$\texttt{ }$

<h3>Further explanation</h3>

<em>Let's recall </em><em>Kinetic Energy</em><em> Formula as follows:</em>

$\boxed{Ek = \frac{1}{2}mv^2}$

<em>Ek = Kinetic Energy ( Joule )</em>

<em>m = mass of the object ( kg )</em>

<em>v = speed of the object ( m/s )</em>

$\texttt{ }$

$\boxed{Ep = mgh}$

<em>Ep = Potential Energy ( Joule )</em>

<em>m = mass of the object ( kg )</em>

<em>g = gravitational acceleration ( m/s² )</em>

<em>h = height position of the object ( m )</em>

Let us now tackle the problem !

$\texttt{ }$

<u>Given:</u>

mass = m = 72 kg

gravitational field strength = g = 10 N/kg

height = h = 1 m

<u>Asked:</u>

gravitational potential energy = Ep = ?

<u>Solution:</u>

<em>We will solve this problem as follows :</em>

$Ep = m g h$

$Ep = 72 \times 10 \times 1$

$Ep = 720 \texttt{ J}$

$\texttt{ }$

<h3>Conclusion:</h3>

<em>He loses </em><em>720 J</em><em> for every metre he descends</em>

$\texttt{ }$

<h3>Learn more</h3>

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441
Newton's Law of Motion: brainly.com/question/10431582
Example of Newton's Law: brainly.com/question/498822

$\texttt{ }$

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Energy

Dafna11 [192]2 years ago

5 0

Answer:

720 J

Explanation:

The gravitational potential energy that Essam loses for every metre is given by:

$\Delta U=mg \Delta h$

where

m=72 kg is Essam's mass

$g=10 N/kg$ is the gravitational field strength

$\Delta h=1 m$ is the difference in height

By substituting the numbers into the formula, we find

$\Delta U=(72 kg)(10 N/kg)(1 m)=720 J$

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<h3>Kinetic energy</h3>

Kinetic energy is a form of energy. It is defined as the energy associated with bodies that are in motion and this energy depends on the mass and speed of the body.

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1 year ago

The brachialis attaches to the forearm .035 m from the elbow at an angle of 32 deg. If the brachialis produces 750 N of force, w

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From the question we are told that:

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2 years ago

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2 years ago

A swimming pool is 50 ft wide and 100 ft long and its bottom is an inclined plane, the shallow end having a depth of 4 ft and th

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Explanation:

We define force as the product of mass and acceleration.

F = ma

It means that the object has zero net force when it is in rest state or it when it has no acceleration. However in the case of liquids. just like the above mentioned case, the water is at rest but it is still exerting a pressure on the walls of the swimming pool. That pressure exerted by the liquids in their rest state is known as hydro static force.

Given Data:

Width of the pool = w = 50 ft

length of the pool = l= 100 ft

Depth of the shallow end = h(s) = 4 ft

Depth of the deep end = h(d) = 10 ft.

weight density = ρg = 62.5 lb/ft

Solution:

a) Force on a shallow end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(4)}{2}(2(0)+4)$

$F = 25000 lb$

b) Force on deep end:

$F = \frac{pgwh}{2} (2x_{1}+h)$

$F = \frac{(62.5)(50)(10)}{2} (2(0)+10)$

$F = 187500 lb$

c) Force on one of the sides:

As it is mentioned in the question that the bottom of the swimming pool is an inclined plane so sum of the forces on the rectangular part and triangular part will give us the force on one of the sides of the pool.

1) Force on the Rectangular part:

$F = \frac{pg(l.h)}{2}(2(x_{1} )+ h)$

$x_{1} = 0\\h_{s} = 4ft$

$F = \frac{(62.5)(100)(2)}{2}(2(0)+4)$

$F =25000lb$

2) Force on the triangular part:

$F = \frac{pg(l.h)}{6} (3x_{1} +2h)$

here

h = h(d) - h(s)

h = 10-4

h = 6ft

$x_{1} = 4ft\\$

$F = \frac{62.5 (100)(6)}{6} (3(4)+2(6))$

$F = 150000 lb$

now add both of these forces,

F = 25000lb + 150000lb

F = 175000lb

d) Force on the bottom:

$F = \frac{pgw\sqrt{l^{2} + ((h_{d}) - h(s)) } (h_{d}+h_{s}) }{2}$

$F = \frac{62.5(50)\sqrt{100^{2}(10-4) } (10+4) }{2}$

$F = 2187937.5 lb$

7 0

3 years ago