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Rudik [331]
3 years ago
11

Using the formula for work, Match the Force × Distance problems with their correct "Work Performed" answer. The first example ha

s been completed for you.Force × Distance = Work Performed 10 newtons 6 meters =60 ng Click the item in the left column. Use the plus sign to move it up or the minus sign to move it down until it matches the correct entry in the right column. Lock your answer in place by clicking the square beside the item. (A check mark means it is locked.)5 ft.-lbs 40 pounds 100 feet 62 n-m 2.5 pounds 24 inches 200 ft.-lbs. 62 newtons 1 meter 40,000 ft.-lbs. 1 ton 20 foot 4000 ft.-lbs. 400 pounds 0.5 feet

Physics
1 answer:
Elina [12.6K]3 years ago
3 0

Answer: The answer is in the figure attached.

Explanation:

The Work W done by a Force F refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.  

It should be noted that it is a scalar magnitude, and its unit in the <u>International System of Units</u> is the Joule (J). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter:  

1J=(1N)(1m)=Nm  

However, in the British Engineering and Gravitational Systems its unit is Foot-pound (ft-lb). Where 1ft-lb=1.355818J

Now, when the applied force F) is constant and the direction of the force and the direction of the movement (traveled distance) d  are parallel, the equation to calculate it is:  

W=(F)(d)

So, taking into account the explanation above, the attached table shows the Work done for each situation.

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Answer:

 (7.8) x (9.8 m/s)  =  76.44 m/s

during the time he spent falling.

Since his falling speed was zero when he 'stepped' off of the top,

he hit the ground at  76.44 m/s.

That's about 170 miles per hour.

I'll bet he left one serious crater!

I hope this helps too! :D

Explanation:

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A lunar eclipse occurs only when the Moon is new. A lunar eclipse occurs only when the Moon is new. True False
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A rod extending between x = 0 and x = 13.0 cm has uniform cross-sectional area A = 8.00 cm2. Its density increases steadily betw
mezya [45]

Answer:

The mass is  m  = 3.45408 kg

Explanation:

From the question we are told that

    The extension  of the rod is from , \   x_1 = 0 \to x_2 = 13.0

     The area is  A =  8.0 cm^2

      The density increase as follows from  \  \rho_1 =2.5 g/cm^2 \to  \rho_2= 19.0 g/cm^3

    The equation  \rho =  B + Cx

at  x_1= 0  \rho_1 =2.5 g/cm^2

So

      2.5  =  B  + 0

=>  B =2.5

    So at x_2 = 13.0 ,  \rho_2= 19.0 g/cm^3

So

            19.0 = 2.5 + C(13)

       =>   C = 1.27

Now  

       m  =  8   \int\limits^{13}_{0} {2.5 + 1.27x} \, dx

      m  =  8   [{2.5 +\frac{ 1.27x^2}{2} } ]\left  | 13} \atop {0}} \right.

      m  =  8   [{2.5 +\frac{ 1.27(13)^2}{2} } ]

      m  = 3454.08 g

        m  = 3.45408 kg

         

8 0
3 years ago
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