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1 year ago

A 70kg man spreads his legs as shown calculate the tripping force

Physics

1 answer:

shutvik [7]1 year ago

6 0

This is simply defined as the energy of an object as an attribute of a physical movement

Force

This is simply defined as the energy of an object as an attribute of a physical movement

it is important to note that a tripping for will be a force opposing motion therefore

For a 70kg man trip on a force the for it must me acted upon by a Force

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A small glass bead charged to 5.0 nCnC is in the plane that bisects a thin, uniformly charged, 10-cmcm-long glass rod and is 4.0

GuDViN [60]

Answer:

The total charge on the rod is 47.8 nC.

Explanation:

Given that,

Charge = 5.0 nC

Length of glass rod= 10 cm

Force = 840 μN

Distance = 4.0 cm

The electric field intensity due to a uniformly charged rod of length L at a distance x on its perpendicular bisector

We need to calculate the electric field

Using formula of electric field intensity

$E=\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}$

Where, Q = charge on the rod

The force is on the charged bead of charge q placed in the electric field of field strength E

Using formula of force

$F=qE$

Put the value into the formula

$F=q\times\dfrac{kQ}{x\sqrt{(\dfrac{L}{2})^2+x^2}}$

We need to calculate the total charge on the rod

$Q=\dfrac{Fx\sqrt{(\dfrac{L}{2})^2+x^2}}{kq}$

Put the value into the formula

$Q=\dfrac{840\times10^{-6}\times4.0\times10^{-2}\sqrt{(\dfrac{10.0\times10^{-2}}{2})^2+(4.0\times10^{-2})^2}}{9\times10^{9}\times5.0\times10^{-9}}$

$Q=47.8\times10^{-9}\ C$

$Q=47.8\ nC$

Hence, The total charge on the rod is 47.8 nC.

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1 year ago

What two factors determine the energy production of the Hoover Dam

swat32

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2 years ago

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katen-ka-za [31]

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1 year ago

Two mating steel spur gears are 20 mm wide, and the tooth profiles have radii of curvature at the line of contact of 10 and 15 m

Rom4ik [11]

Answer:

a) The maximum contact pressure is 274.58 MPa and the width of contact is 0.058 mm

b) The maximum shear stress is 82.37 MPa at a distance of 0.023 mm

Explanation:

Given data:

L = 20 mm

F = 250 N

r₁ = 10 mm

r₂ = 15 mm

v = 0.3

E = 2.07x10⁵ MPa

$A=\frac{1-V_{1}^{2} }{E_{1} }-\frac{1-V_{2}^{2} }{E_{2} } =\frac{1-0.3^{2} }{2.07x10^{5} } *2=8.79x10^{-6}$

a) The maximum contact pressure is:

$P=0.564*\sqrt{\frac{F*(\frac{1}{r_{1} }+\frac{1}{r_{2} }) }{LA} } =0.564*\sqrt{\frac{250*(\frac{1}{10} +\frac{1}{15} )}{20*8.79x10^{-6} } } =274.58MPa$

The width of contact is:

$b=1.13*\sqrt{\frac{FA}{L(\frac{1}{r_{1} }+\frac{1}{r_{2} }) } } =1.13*\sqrt{\frac{250*8.79x10^{-6} }{20*(\frac{1}{10} +\frac{1}{15} )} } =0.029mm\\2*b=0.058mm$