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Varvara68 [4.7K]
2 years ago
11

if it is known that the toggle clamp is a machine, what assumptions can be made about it? check all that apply. (you must provid

e an answer before moving to the next part.)
Physics
1 answer:
muminat2 years ago
7 0

A toggle clamp is a device that you employ, often but not exclusively as part of a production process, to firmly place components or parts in place.

Explain about toggle clamp.

A toggle clamp's main characteristics are its rapid action and ability to be easily turned on and off by an operator. Toggle clamps also lock in place firmly. Because of this, toggle clamps are frequently employed in production lines where components need to be held firmly and quickly removed during routine manufacturing procedures.

Toggle clamps with standard or light action are often constructed from components of pressed steel that have been zinc-plated.

Depending on the setting in which you work, Stainless Steel is another alternative for Toggle Clamp production. Stainless steel toggle clamps are frequently used in regulated areas where hygiene is crucial, such as the food industry and the pharmaceutical business.

To know more about toggle clamps use link below:

brainly.com/question/20729246

#SPJ4

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What is power?
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We can define power as the rate of doing work, it is the work done in unit time. The SI unit of power is Watt (W) which is joules per second (J/s). Sometimes the power of motor vehicles and other machines are given in terms of Horsepower (hp) which is approximately equal to 745.7 watts.


Power is the rate at which a force is applied to an object for example.current wire
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Read 2 more answers
If f(x) = -x2+x-1 then value of f(f(2)) is
qaws [65]

Answer:

<u>-13</u>

Explanation:

<u>Function</u>

  • f(x) = -x² + x - 1

<u>Solving</u>

  • Substitute 2 in place of x
  • f(2) = -(2)² + 2 - 1
  • f(2) = -4 + 2 - 1
  • f(2) = -3
  • f(f(2)) = f(-3)
  • f(-3) = -(-3)² - 3 - 1
  • f(-3) = -9 - 3 - 1
  • f(f(2)) = <u>-13</u>
6 0
2 years ago
Estimate the electric field at a point 2.40 cm perpendicular to the midpoint of a uniformly charged 2.00-m-long thin wire carryi
nadya68 [22]

Answer:

E = 1.85*10^{12}\frac{N}{C}

Explanation:

Hi!

The perpendicular distance 2.4cm, is much less than the distance to both endpoints of the wire, which is aprox 1m. Then the edge effect is negligible at this field point, and we can aproximate the wire as infinitely long.

The electric filed of an infinitely long wire is easy to calculate. Let's call z the axis along the wire. Because of its simmetry (translational and rotational), the electric field E must point in the radial direction,  and it cannot depende on coordinate z. To calculate the field Gauss law is used, as seen in the image, with a cylindrical gaussian surface. The result is:

E = \frac{\lambda}{2\pi \epsilon_0 r}\\\lambda=\text{charge per unit length}=\frac{4.95 \mu C}{2 m} = 2.475 \frac{C}{m}\\r=\text{perpendicular distance to wire}\\\epsilon_0=8.85*10^{-12}\frac{C^2}{Nm^2}

Then the electric field at the point of interest is estimated as:

E = \frac{\22.475}{2\pi*( 8.85*10^{-12})*(2.4*10^{-2})}\frac{N}{C}=1.85*10^{12}\frac{N}{C}

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4 years ago
A stone weighing 0.7 kilograms rolls down the inclined plane from position B to position A. Position A is located at sea level.
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This is the explanation

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