Ask question

Ask question

All categories

3 years ago

7

Can someone who knows slot about engineering explain to me by using an example on how to properly read a triangular cone shaped

taper guage used for small gaps? I know how to use it just need to know how to properly read one.

1 answer:

Vlad1618 [11]3 years ago

4 0

srry cant help u there

You might be interested in

A football punter accelerates a .55 kg football

Ronch [10]

Answer:

17.6 N

Explanation:

The force exerted by the punter on the football is equal to the rate of change of momentum of the football:

$F=\frac{\Delta p}{\Delta t}$

where

$\Delta p$ is the change in momentum of the football

$\Delta t=0.25 s$ is the time elapsed

The change in momentum can be written as

$\Delta p = m(v-u)$

where

m = 0.55 kg is the mass of the football

u = 0 is the initial velocity (the ball starts from rest)

v = 8.0 m/s is the final velocity

Combining the two equations and substituting the values, we find the force exerted on the ball:

$F=\frac{m(v-u)}{\Delta t}=\frac{(0.55)(8.0-0)}{0.25}=17.6 N$

5 0

3 years ago

Which statement provides the most complete description of an object's motion?

vivado [14]

Answer:

A. The bird watcher followed the south trail a distance of five kilometers in 45 minutes.

5 0

2 years ago

The maximum wavelength that an electromagnetic wave can have and still eject electrons from a metal surface is 542 nm. What is t

Veseljchak [2.6K]

Answer:

2.295 eV

Explanation:

maximum wavelength, λ = 542 nm = 542 x 10^-9 m

The work function of the metal is defined as the minimum amount of energy falling on the metal so that the photo electrons just ejects the surface of metal.

$W_{o}=\frac{hc}{\lambda }$

where, h is the Plank's constant and c be the speed of light

h = 6.634 x 10^-34 Js

c = 3 x 10^8 m/s

$W_{o}=\frac{6.634\times 10^{-34}\times 3\times 10^{8}}{542\times10^{-9} }$

$W_{o}=3.67\times 10^{-19} J=\frac{3.67\times 10^{-19}}{1.6\times 10^{-19}} eV$

Wo = 2.295 eV

Thus, the work function of this metal is 2.295 eV.

5 0

3 years ago

In a particular experiment to study the photoelectric effect, the frequency of the incident light and the temperature of the met

qwelly [4]

Answer:

B. The number of electrons emitted from the metal per second increases.

Explanation:

Light consists of photons . Energy of each photon depends upon frequency of light . The increase in intensity increases the number of photons . It does not increase energy of photons .

So if a high intensity light falls on a photosensitive plate , each photon ejects one electron . So number of electrons increases if we increase intensity of photon. It does not increase kinetic energy of ejected electrons . Work function depends upon the nature of plate.

6 0

3 years ago

A thin-walled cylindrical pressure vessel is subjected to an internal gauge pressure, p=75 psip=75 psi. It had a wall thickness

Mekhanik [1.2K]

To solve this problem we must apply the concept related to the longitudinal effort and the effort of the hoop. The effort of the hoop is given as

$\sigma_h = \frac{Pd}{2t}$

Here,

P = Pressure

d = Diameter

t = Thickness

At the same time the longitudinal stress is given as,

$\sigma_l = \frac{Pd}{4t}$

The letters have the same meaning as before.

Then he hoop stress would be,

$\sigma_h = \frac{Pd}{2t}$

$\sigma_h = \frac{75 \times 8}{2\times 0.25}$

$\sigma_h = 1200psi$

And the longitudinal stress would be

$\sigma_l = \frac{Pd}{4t}$

$\sigma_l = \frac{75\times 8}{4\times 0.25}$

$\sigma_l = 600Psi$

The Mohr's circle is attached in a image to find the maximum shear stress, which is given as

$\tau_{max} = \frac{\sigma_h}{2}$

$\tau_{max} = \frac{1200}{2}$

$\tau_{max} = 600Psi$

Therefore the maximum shear stress in the pressure vessel when it is subjected to this pressure is 600Psi

6 0

3 years ago