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ser-zykov [4K]
3 years ago
10

When you set something down on the ground wht kind of work are your arms doing?

Physics
2 answers:
umka2103 [35]3 years ago
7 0
When we set something down on the ground, the kind of work that our arms doing is : negative apex
It's happen whenever we do works that are align with the force of Gravity (to the bottom)

hope this helps
Lina20 [59]3 years ago
4 0

Answer:

Negative work.

Explanation:

Physically speaking, the term "work" is used when we refer to the work performed by a force, that is, Mechanical Work. Thus, a force applied in a body accomplishes a work when it produces a displacement in the body.

In this way, we can understand that:

  • When a force has the same direction of movement the work done is positive.
  • When a force has opposite direction to the movement the work done is negative.

In this case we can conclude that the work is negative, because the displacement of the "thing being put on the ground" is down, but the force made by the arms is up.

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A thin, rectangular sheet of metal has mass M and sides of length a and b. Find the moment of inertia of this sheet about an axi
Lubov Fominskaja [6]

Answer:

The moment of inertia is I=\frac{M}{12} a^{2}

Explanation:

The moment of inertia is equal:

I=\int\limits^a_b {r^{2} } \, dm

If r is -\frac{a}{2}

and dm=\frac{M}{a} dr

I=\int\limits^a_b {r^{2}\frac{M}{a}  } \, dr\\a=\frac{a}{2} \\b=-\frac{a}{2}

I=\frac{M}{a} \int\limits^a_b {r^{2}  } \, dr\\\\I=\frac{M}{a} (\frac{M}{3} )_{b}^{a}\\  I=\frac{M}{3a} (\frac{a^{3} }{8} +\frac{a^{3} }{8} )\\I=\frac{M}{12} a^{2}

7 0
3 years ago
A waitperson carrying a tray with a platter on it tips the tray at an angle of 12° below the horizontal. If the gravitational fo
Nadusha1986 [10]

Answer:

The answer is 1.0 N

Explanation:

inclination of tray=12^{\circ}

gravitational Force=5 N

Now this gravitational force has two component i.e.

5\sin \theta  is parallel to the tray =1.039 N

5\cos \theta  is perpendicular to the tray =4.890 N

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3 years ago
A piano tuner sounds two strings simultaneously. One has been previously tuned to vibrate at 293.0 Hz. The tuner hears 3.0 beats
ololo11 [35]

Answer:

Part a)

f_B = 290 Hz

Part B)

percentage increase is

percentage = 1.38%

Explanation:

Part a)

As we know that the beat frequency is

f_A - f_B = 3

after increasing the tension the beat frequency is decreased and hence the tension in string B will increase

So we have

293 - f_B = 3

f_B = 290 Hz

Part B)

percentage increase in the tension of the string will be given as

f_A - f_B' = 1

f_B' = 292 Hz

now we have

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

so we have

T_1 = C (290)^2

T_2 = C(292)^2

so we have

\frac{\Delta T}{T} = \frac{292^2 - 290^2}{290^2}

percentage increase is

percentage = 1.38

4 0
3 years ago
What type of plate boundary decreases the amount of the Earth's crust?
rosijanka [135]

Answer:

Convergent.

Explanation:

Just as oceanic crust is formed at mid-ocean ridges, it is destroyed in subduction zones. Subduction is the important geologic process in which a tectonic plate made of dense lithospheric material melts or falls below a plate made of less-dense lithosphere at a convergent plate boundary.

3 0
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A rigid tank internal energy of fluid 800kJ. Fluid loses 500kJ of heat and padle wheel does 100kJ of work. Find final internal e
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Answer:

 U₂ = 400 KJ      

Explanation:

Given that

Initial energy of the tank ,U₁= 800 KJ

Heat loses by fluid ,Q= - 500 KJ

Work done on the fluid ,W= - 100 KJ

Sign -

1.Heat rejected by system - negative

2.Heat gain by system - Positive

3.Work done by system = Positive

4.Work done on the system-Negative

Lets take final internal energy =U₂

We know that

Q= U₂ - U₁ + W

-500 = U₂ - 800 - 100

U₂ = -500 +900 KJ

U₂ = 400 KJ

Therefore the final internal energy = 400 KJ

6 0
3 years ago
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