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3 years ago

although the pivot exerts a force on the ruler it does not exert a torque with respect to the pivot point. why not?

Physics

1 answer:

allsm [11]3 years ago

8 0

Answer:

The situation is not shown, however if one calculates torques about the pivot point, then the pivot point itself exerts no torque because the force applied by the pivot point exerts zero torque because Fp * L = 0 where L = 0 in this case.

If one wants to calculate torques about, say, the end of the ruler then one would need to calculate the torque applied by the pivot point.

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Consider a metal single crystal oriented such that the normal to the slip plane and slip direction are at angles of 60° and 35°,

Vedmedyk [2.9K]

Answer:

19.5324 MPa

Explanation:

Information provided

Angle between the normal to the slip plane with tensile axis, $\alpha=60^{o}
$

Angle by slip direction with tensile axis, $\beta=35^{o}$

Critical resolved shear stress, $\tau_{c}=8 MPa$

Applied stress $\sigma=12 MPa$

Shear stress at slip plane

$\tau=\sigma cos\alpha cos\beta$

$\tau=12cos60^{o}cos35^{o}=4.915 MPa$

$\tau$ hence crystal won’t yield

Applied stress, $\sigma$ for crystal to yield is given by

$\sigma=\frac {\tau_{c}}{cos\alpha cos\beta}$

$\sigma=\frac {8}{cos60cos35}=19.53239342 MPa$

$\sigma=19.5324 MPa
$

7 0

3 years ago

Which event causes the formation of the electric current produced by the

Valentin [98]

Answer:

Explanation:

Light with energy above a certain point can be used to knock electrons loose, freeing them from a solid metal surface, according to Scientific American.

photoelectric effect is the emission of electrons when electromagnetic radiation, such as light, hits a material

6 0

3 years ago

Read 2 more answers

10%) Problem 7: Water flows through a water hose at a rate of Q1 = 620 cm3/s, the diameter of the hose is d1 = 1.99 cm. A nozzle

Semenov [28]

Answer:

a) $A_1 = \frac{\pi d_1^2}{4}$

Explanation:

a) the cross-sectional area of the hose would be the square of radius times pi. And since the sectional radius is half of its diameter d. We can express the cross-sectional area A1 in term of diameter d1

$A_1 = \pi r_1^2 = \pi (d_1/2)^2 = \frac{\pi d_1^2}{4}$

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3 years ago

A TV set is pushed a distance of 2 m with a force of 20 N how much work is done on the set

vladimir2022 [97]

Answer:

40 J

Explanation:

8 0

4 years ago

Derive an expression for the block's centripetal acceleration ac in terms of m , θ , and physical constants, as appropriate

maxonik [38]

The expression for the block's centripetal acceleration is derived as ω²r or v²/r.

<h3>What is centripetal acceleration?</h3>

The centripetal acceleration of an object is the inward or radial acceleration of an object moving in a circular path.

The expression for the block's centripetal acceleration is derived as follows;

ω = dθ/dt

where;

ω is the angular speed
θ is the angular displacement
t is the time of motion

ac = ω²r

where;

r is the radius of the circular path

Also, ω = v/r

ac = (v/r)²r

ac = v²/r

Thus, the expression for the block's centripetal acceleration is derived as ω²r or v²/r.

Learn more about centripetal acceleration here: brainly.com/question/79801

7 0

2 years ago