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DaniilM [7]
3 years ago
11

Robert pushes the box to the left at constant speed. In doing so, Robert does ______ work on the box. A. positive B. negative C.

zero
Physics
1 answer:
andrew-mc [135]3 years ago
5 0

Answer:

He does postive work

Explanation:

Because W= F xS sinစ

So since work is done such that Force and distance have the Sam direction

Thus စ=0

Then work is positive in value

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50 POINTS How are force and velocity DIFFERENT?
nirvana33 [79]
Velocity is speed in a specific direction and force is a push or pull that may cuase object to stop,move,or change direction they are always forces acting on it. so the different that in force there are always force acting on it
3 0
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A girl delivering newspapers covers her route by traveling 3.00 blocks west, 4.00 blocks north, and then 6.00 blocks east. What
Harrizon [31]

The concepts used to solve this problem are those related to the Pythagorean theorem for which we will calculate the distance and the pitch.

According to the attached diagram we have that the expression of the resulting displacement is

R = \sqrt{(3b)^2+(4b)^2}

Therefore the resultant displacement of the girl is

R = \sqrt{(9b^2+16b^2)}

R = \sqrt{25b^2}

R = 5b

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4 0
3 years ago
A 55.0-kg skydiver drew falls for a period of time before opening his parachute. what is his kinetic energy when he reaches a ve
Jlenok [28]
Mass (m)=55kg

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final velocity=16m/s This is the velocity (speed) the skydiver reaches

The equation we use is KE=.5mv^2
Kinetic energy=.5 mass x velocity^2

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J=Joules

KE=7040J

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Hope this helps
3 0
4 years ago
The instantaneous velocity of an object is the blank of the object with a blank
777dan777 [17]
The instantaneous velocity of the object is its speed and direction at that instant.
7 0
3 years ago
Read 2 more answers
What is the velocity of the object?
dmitriy555 [2]
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2 /><h2>\huge\boxed{\text{V = 9.5 m/s}}</h2><h2>_____________________________________</h2>

<h2>DATA:</h2>

mass = m = 2kg

Distance = x = 6m

Force = 30N

TO FIND:

Work = W = ?

Velocity = V = ?

<h2>SOLUTION:</h2>

According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.

To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,

{\Longrightarrow}\qquad \qquad \qquad W\ =\ \frac{1}{2}\;(Base)\:(Height)

Base is the x-axis of the graph which is Position i.e. 6m

Height is the y-axis of the graph which is Force i.e. 30N

So,

                           W\ =\ \frac{1}{2}\:6\:x\:30

                           W   =  90 J

The work done is 90 J.

According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad K.E\\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ m\ V^2 \\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ 2\ (V_f-V_i)^2\\\\{V_i\ is\ 0\ because\ the\ object\ was\ initially\ at\ rest}\\\\ {\Longrightarrow}\qquad \qquad \qquad W\quad\ =\ \frac{1}{2}\ x\ 2\ (V_f-0)^2 \\\\{\Longrightarrow}\qquad \qquad \qquad 90\quad = \frac{1}{2}\ x\ 2\ (V_f)^2

\\\\{\Longrightarrow}\qquad \qquad \qquad V_f\quad =\ \sqrt{\frac{2\ (90)\ }{2}}\\\\{\Longrightarrow}\qquad \qquad \qquad \boxed {V_f\quad =\ 9.48\ m/s}

\boxed{The\ Velocity\ of\ the\ Object\ of\ mass\ 2kg\ at\ 6\ meters\ of\ distance\ was\ 9.48\ m/s}

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

8 0
3 years ago
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