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gulaghasi [49]
3 years ago
14

Why caring a body and moving with it is not a work done Why :​

Physics
1 answer:
Alecsey [184]3 years ago
8 0

Answer:

Work done on an object is equal to

FDcos(angle).

So, naturally, if you lift a book from the floor on top of the table you do work on it since you are applying a force through a distance.

However, I often see the example of carrying a book through a horizontal distance is not work. The reasoning given is this: The force you apply is in the vertical distance, countering gravity and thus not in the direction of motion.

But surely you must be applying a force (and thus work) in the horizontal direction as the book would stop due to air friction if not for your fingers?

Is applying a force through a distance only work if causes an acceleration? That wouldn't make sense in my mind. If you are dragging a sled through snow, you are still doing work on it, since the force is in the direction of motion. This goes even if velocity is constant due to friction.

Explanation:

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Solve this physics for me <br>please with steps<br>​
Mars2501 [29]

Answer:

The answers are located in each of the explanations showed below

Explanation:

a)

(i) Surface Tension: The tensile force that causes this tension acts parallel to the surface and is due to the forces of attraction between the molecules of the liquid. The magnitude of this force per unit of length is called surface tension.

σ = F/l [N/m]

where:

F = force [N]

l = length [m]

σ = Surface Tension [N/m]

(ii) Frequency is the number of repetitions per unit of time of any periodic event.

f = 1/T [1/s] or [s^-1] or [Hz]

where:

T = period [s] or [seconds]

f = frecuency [Hz] or [hertz]

(iii) Each of the units will be shown for each variable

v = velocity [m/s]

a = accelertion [m/s^2]

s = displacement [m]

[\frac{m}{s} ]^{2} =[\frac{m}{s} ]^{2} + 2* [\frac{m}{s^{2} } ]*[m]\\

[\frac{m^2}{s^2} ] =[\frac{m^2}{s^2} ] +  [\frac{m^{2} }{s^{2} } ]

[\frac{m^2}{s^2} ]

b) To find the velocity we must derivate the function X with respect to t because this derivate will give us the equation for the velocity, it means:

v=\frac{dx}{dt} \\v = 0.75*2*t+5*t

(i) X = 0.75*t^{2} +5*t+1\\X = 0.75*(4)^{2} +5*(4)+1\\X = 33 [m]

ii) replacing in the derivated equation.

v=1.5*(4)+5\\v=11[m/s]

iii) the average velocity is defined by the expresion v = x/t

v = \frac{x-x_{0} }{t-t_{0} } \\

x_{0}=0.75(2)^{2}+5(2)+1 \\ x_{0}=14[m]\\x=0.75(7)^{2}+5(7)+1\\x=72.75[m]\\t = 7 [s]t0= 2[s]Now replacing:[tex]v_{prom} = \frac{72.75-14}{7-2} \\v_{prom} = 11.75 [m/s]

2

a) Pascal's principle or Pascal's law, where the pressure exerted on an incompressible fluid and in balance within a container of indeformable walls is transmitted with equal intensity in all directions and at all points of the fluid.

Therefore:

P1 = pressure at point 1.

P2 = pressure at point 2.

P1 = F1/A1

P2= F2/A2

\frac{F_{1} }{A_{1} }=\frac{F_{2}}{A_{2} }  \\F_{1}=A_{1}*(\frac{F_{2}}{A_{2} })

b) One of the applications of the surface tension is the <u>capillarity</u> this is a property of liquids that depends on their surface tension (which, in turn, depends on the cohesion or intermolecular force of the liquid), which gives them the ability to climb or descend through a capillary tube.

Other examples of surface tension:

The mosquitoes that can sit on the water.

A clip on the water.

Some leaves that remain floating on the surface.

Some soaps and detergents on the water.

5 0
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