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Elza [17]
3 years ago
13

When 10 N force applied at 30 degrees to the end of a 20 cm handle of a wrench, it was just able to loosen the nut. What magnitu

de of the force would require to just loosen the nut, if the force apply perpendicularly at the end of the handle
Physics
1 answer:
fredd [130]3 years ago
3 0

Answer:

<h2>5N</h2>

Explanation:

To get the magnitude of the force would require to just loosen the nut, if the force apply perpendicularly at the end of the handle, we will have to resolve the force perpendicular to the wrench. Torque is the turning effect of a body or force about a point. It is similar to moments.

Torque = Force * radius

Note that the force must be perpendicular to the wrench. On resolving the force perpendicularly to the wrench, we will have to resolve the force to the vertical.

Fy = Fsinθ

Fy = 10sin30°

Fy = 10 * 0.5

Fy = 5N

<em>Torque = Fy * r</em>

<em>Given Fy = 5N and r = 20cm = 0.2m</em>

<em>Torque = 5 * 0.2</em>

<em>Torque = 1Nm</em>

<em />

<em>Hence the magnitude of the force would require to just loosen the nut, if the force apply perpendicularly at the end of the handle is 5N</em>

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Answer:

41.4496148484\ m/s

Explanation:

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\Delta x = 0.57-0.26

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m = Mass of object = 8.3\times 10^{-9}\ kg

Electric field due to a sheet is given by

E=\dfrac{\sigma}{2\epsilon_0}\\\Rightarrow E=\dfrac{5.9\times 10^{-8}}{2\times 8.85\times 10^{-12}}\\\Rightarrow E=3333.33\ V/m

Electric field is given by

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Voltage is given by

V=E\Delta x

Kinetic energy is given by

K=qV

\dfrac{1}{2}mv^2=qE\Delta x\\\Rightarrow v=\sqrt{\dfrac{2qE\Delta x}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 6.9\times 10^{-9}\times 3333.33\times (0.57-0.26)}{8.3\times 10^{-9}}}\\\Rightarrow v=41.4496148484\ m/s

The initial speed of the object is 41.4496148484\ m/s

7 0
3 years ago
How and why does air pressure change with altitude in the atmosphere?
wlad13 [49]
Pressure with Height: pressure decreases with incrementing altitude. The pressure at any caliber in the atmosphere may be interpreted as the total weight of the air above a unit area at any elevation. At higher elevations, there are fewer air molecules above a given surface than a homogeneous surface at lower calibers.
5 0
3 years ago
The membrane of a living cell can be approximated by a parallel-plate capacitor with plates of area 4.50×10−9 m2 , a plate separ
Marizza181 [45]

The energy stored in the membrane is 6.44\cdot 10^{-14} J

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The capacitance of a parallel-plate capacitor is given by

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d is the separation between the plates

For the membrane in this problem, we have

k = 4.6

A=4.50\cdot 10^{-9} m^2

d=8.1\cdot 10^{-9} m

Substituting, we find its capacitance:

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Now we can find the energy stored: for a capacitor, it is given by

U=\frac{1}{2}CV^2

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V=7.55\cdot 10^{-2} V is the potential difference

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U=\frac{1}{2}(2.26\cdot 10^{-11} F)(7.55\cdot 10^{-2})^2=6.44\cdot 10^{-14} J

Learn more about capacitors:

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6 0
3 years ago
A body of mass 25kg, moving at 3 ms per second on a rough horizontal floor brought to rest after sliding through a distance of 2
erastova [34]
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s=2.5
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3 years ago
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See step by step sexplanation

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P₁ * V₁  = P₂ * V₂

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3.-No se puede estar de acuerdo con el criterio del plomero. En su solución no plantea el aumento de la altura del tanque,  para el logro del aumento de la presión que es realmente lo que hay que hacer

4 0
3 years ago
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