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-Dominant- [34]
3 years ago
10

A worker pushes a large rock to the north while another worker helps by pushing it to the east. If they both exert equal force,

in what direction does the rock move?
Question 2 options:

northeast

east

north

southwest
Physics
1 answer:
alexandr1967 [171]3 years ago
3 0

let the magnitude of force applied by each worker be "F"

consider east-west direction along X-axis and north-south direction along Y-axis

In unit vector form, force vector by worker pushing in east direction is given as

\underset{A}{\rightarrow} = F \hat{i} + 0  \hat{j}


In unit vector form, force vector by worker pushing in north direction is given as

\underset{B}{\rightarrow} = 0 \hat{i} + F  \hat{j}

resultant force is given as the vector sum of two vector forces as

\underset{R}{\rightarrow} = \underset{A}{\rightarrow} + \underset{B}{\rightarrow}

\underset{R}{\rightarrow} = (F \hat{i} + 0  \hat{j} ) + (0 \hat{i} + F  \hat{j} )

\underset{R}{\rightarrow} = F \hat{i} +  F  \hat{j}

direction of the force is hence given as

θ = tan⁻¹(F/F)

θ =  tan⁻¹(1)

θ = 45 degree north of east

hence the direction is north-east


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Explanation:

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F_g = \frac{GmM}{R^2} --- (1)

Where G = Gravitational constant = 6.67408 × 10⁻¹¹ m³ kg⁻¹ s⁻²

m = Mass of the body = 2 kg

M = Mass of the Earth = 5.972 × 10²⁴ kg

R = Distance of the object from the center of the Earth = Radius of the Earth + Object's distance from the surface of the Earth = (6371 * 10³) + 3.0 = 6371003 m

Plug in the values in (1):

(1)=> F_g = \frac{6.67408 * 10^{-11} * 2 * 5.972*10^{24}}{(6371003)^2} = 19.63

Now that we have force strength at the location, we can use:

Force = mass * gravitational-field-strength

Plug in the values:

19.63 = 2.0 * gravitational-field-strength

gravitational-field-strength = 19.63/2 = 9.82 N/kg

Hence the correct answer is Option (3) 9.8 N/kg

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<h3><u>✽</u></h3>

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<u>We need to find time</u>

<h3>We know that ,</h3>

\begin{gathered} {\underline{\boxed{ \rm {\red{Speed \:  =  \:  \frac{Distance}{Time} }}}}}\end{gathered}

<h3>So ,</h3>

\begin{gathered} {\underline{\boxed{ \rm {\orange{Time \:  =  \:  \frac{Distance}{Speed} }}}}}\end{gathered}

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Aquimedes' principle allows to find the result for the apparent weight of the steel sphere is:

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<h3>Archimedes' principle </h3>

The thrust of a fluid on a body is given by Archimedes' principle which states that equal to the weight of the displaced liquid.

              B = \rho \ g \ V_{liquid}  

Where B is the buoyancy, p the density of the liquid, g the acceleration due to gravity and V the volume of the liquid.

 

In the attachment we see a free body diagram of the system, which is a representation of the forces without the details of the bodies.

           

            W_{apparent} = W- B

 

The volume of the liquid is the same as the volume of the sphere.

             V_{liquid} = V_{body} = \frac{4}{3} \pi r^3  

Let's substitute.

            W_{apparent} = m g - \rho_{liquid} \ g \ (\frac{4}{3} \pi r^3 )  

Let's calculate.

          W_{apparent} = 9.8 ( 38 - \frac{4}{3} \pi \ 1000 \ 0.04^3 )  = 9.8 \ 37.73  

         .W_{apparent} = 369.77 N          

In conclusion, using Aquimedes' principle, we can find the result for the apparent weight of the steel sphere:

  • Apparent weight is:  W_{apparent} = 369.77 N  

Learn more about Archimedes' principle here: brainly.com/question/13106989

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