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liraira [26]
3 years ago
15

In order to be considered work, the components that must be present are

Physics
2 answers:
atroni [7]3 years ago
6 0
The components that must be present for work to be considered is a force and a movement in the same direction as the force. In the basic definition of work, a magnitude and displacement that occurs in the same direction is what makes up work. Among the choices, the correct answer is the first one.
Alex Ar [27]3 years ago
5 0

Answer:

A. a force and a movement in the same direction as the force

Explanation:

I did the test and got it.

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Answer:

The resistance is 2 ohms

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Is direction the length of the route between two points ?
zepelin [54]

Answer:

No distance is the length between two routes

Explanation:

Distance is the length of the route between two points. ... Direction is just as important as distance in describing motion. A vector is a quantity that has both size and direction. It can be used to represent the distance and direction of motion.

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2 years ago
When a man travel from hilly to terai what will happen to weight and why?
juin [17]

Answer:

When a man travels from Hilly region to Terai region, his weight gradually increases because the value of g is more at the Terai region than that in hilly region. 3. An object weights 20 N in air and 16 N in liquid, then answer the following questions.

Explanation:

because the value of g is more at the Terai region than that in hilly region. 3. An object weights 20 N in air and 16 N in liquid, then answer the following questions.

6 0
3 years ago
How many significant digits are measurement 0.00210 mg?
jek_recluse [69]
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8 0
3 years ago
Read 2 more answers
A girl weighing 50 kgf wears sandals of pencil heel of area of cross section 1 cm^2, stands on the floor.An elephant weighing 20
Klio2033 [76]

Answer:

\boxed{{\boxed{\blue{ 12.5}}}}

Explanation:

Given, for girl : Weight or force;

\rm \: F_1 = 50 \: kgf

Area of both heels;

\rm \: A_1 =  \; 2 ×1 \;  cm^2 = 2  \: cm^2

\rm \: Pressure \:  P_1  =  \cfrac{F_1}{ A_1 }  =  \dfrac{50 \: kgf}{2 \: cm {}^{2} }  = 25 \: kgf \: cm {}^{ - 1}

For elephant, Weight = Force \rm F_2 = 2000 kg•f

Area of 4 feet;

\rm \: A_2  = \; 4 \times 250 \;  cm^2 = 1000 \:  cm^2

\rm \: Pressure \:  P_2 = {F_2}/{A_2} \;  = \cfrac{2 \cancel{0 00 }\:  kgf}{1 \cancel{000} \: cm^2} =  2 \: kgf \: { \:cm}^{- 1}

Now;

\rm  = \dfrac{Pressure \:  Exerted  \: by  \: the \:  Girl}{Pressure  \: exerted  \: by \:  the  \: elephant}

=  \rm \: P_1/P_2

\implies    \rm\cfrac{25 \: kgf \: \: cm {}^{ - 2} }{2 \: kgf \: cm {}^{ - 2} } =  \rm\cfrac{25 \:  \cancel{kgf \: \: cm {}^{ - 2}} }{2 \: \cancel{ kgf \: cm {}^{ - 2}} } = \boxed{12.5}

Thus, the girl's pointed heel sandals exert 12.5 times more pressure P than the pressure P exerted by the elephant.

I aspire this helps!

3 0
2 years ago
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