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siniylev [52]
3 years ago
6

Use Newton's laws to explain why a falling object dropped from a 57m tower accelerates initially but then reaches constant veloc

ity. Discuss the forces acting on the object
Physics
1 answer:
snow_lady [41]3 years ago
5 0

Answer:

At the point of dropping the object, by Newton's first law due to gravitational force F_g = m × g, accelerates

By Newton's Second law the object reaches impacts on the air with the gravitational force resulting in changing momentum of m×(Final Velocity - Initial Velocity)

As the velocity increases, the rate of change of momentum becomes equivalent to the gravitational force and by Newton's third law, the action action and reaction are equal and opposite hence they cancel each other out

The body then moves at a constant uniform motion down according to Newton's first law

Explanation:

At the point the object of mass, m, is dropped from the height of the tower, the only force acting on the object is the gravitational force such that the object has an acceleration which is the acceleration due to gravity, g, and the gravitational force is therefore = m × g

As the speed of the object increases while the object is falling with the gravitational acceleration the rate at which the object cuts through layers of air which (by Newton's first law of motion, are at rest ) has some buoyancy effect also increases therefore, the object is constantly increasingly changing the momentum of the air which by Newton's second law results, at an high enough velocity, and by Newton's third law, in a force equal to the applied gravitational force

Therefore, the force of the air drag becomes equal to the gravitational force, cancelling each other out and the object then moves according to Newton;s first law, in uniform motion of a constant speed while still falling down.

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Solve this physics for me <br>please with steps<br>​
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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

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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.

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