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3 years ago

The force component along the displacement varies with the magnitude of the displacement, as shown in the graph. (a) 0 to 1.0 m,

1 answer:

5 0

Work is force*displacement if the force and displacement is parallel.

a. You can average the force over the distance so W = Fave*d

b The force part of that multiplication is zero. 

c. You can form the average force for the interval from 2 to 3 and find the work for that section and then consider the interval from 3 to 4, find the work and add the 2 work results.

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!

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When atoms _____, they break down into atoms of different elements and release energy.

SCORPION-xisa [38]

Answer:

It may be combine?

Do you have multiple choice i can see?

Explanation:

5 0

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

Nina [5.8K]

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

What are the rules for setting up an integral of rotation?

Angelina_Jolie [31]

Setting up an integral of rotation is used as a method of of calculating the volume of a 3D object formed by a rotated area of a 2D space. Finding the volume is similar to finding the area, but there is one additional component of rotating the area around a line of symmetry.

First the solid of revolution should be defined. The general function is y=f(x), on an interval [a,b].

Then the curve is rotated about a given axis to get the surface of the solid of revolution. That is the integral of the function.

It all depends of the function f(x), which must be known in order to calculate the integral.

3 0

2 years ago

An element’s atomic number is 35. how many portions would an atom this element have

kotykmax [81]

35 protons are present in an element whose atomic number is 35.

Explanation:

In an atom of an element, the number of protons = atomic number

Number of protons = number of electrons

Mass Number = number of protons + number of neutrons

Hence, an atom with atomic number 35 will have 35 protons.

4 0

3 years ago

A father is pulling his child on a sled in the snow. According to Newton’s Third Law, the force the father exerts on the sled is

viktelen [127]

That is because there are other forces like the friction forces that apply differently on both of them. The frictional forces applied to the sled are smaller than they are on the father, for example, so it's possible for him to pull it.

5 0

3 years ago