A metal shear can be used to cut flat stock , round stock , channel iron and which of the following?

The acceleration of a particle is given by a = 2t − 10, where a is in meters per second squared and t is in seconds. Determine t

tensa zangetsu [6.8K]

Answer

given,

a = 2 t - 10

velocity function

we know,

$\dfrac{dv}{dt}=a$

$\dfrac{dv}{dt}=(2t-10)$

integrating both side

$\int dv =\int (2t -10) dt$

v = t² - 10 t + C

at t = 0 v = 3

so, 3 = 0 - 0 + C

C = 3

Velocity function is equal to v = t² - 10 t + 3

Again we know,

$\dfrac{dx}{dt}=v$

$\dfrac{dx}{dt}=(t^2-10t + 3)$

integrating both side

$\int dx =\int (t^2-10t + 3)dt$

$x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t + C$

now, at t= 0 s = -4

$-4 = \dfrac{0^3}{3}- 10\dfrac{0^2}{2} + 0 + C$

C = -4

So,

$x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t-4$

Position function is equal to $x = \dfrac{t^3}{3}- 10\dfrac{t^2}{2} + 3 t-4$

8 0

3 years ago

A silicon carbide plate fractured in bending when a blunt load was applied to the plate center. The distance between the fractur

anyanavicka [17]

Answer:

hello your question has some missing values below are the missing values

Mirror Radius (mm) Bending Failure Stress (MPa)

.603 225

.203 368

.162 442

answer : 191 mPa

Explanation:

Determine the stress present at the time of fracture for the original plate

Bending stress ∝ 1 / ( mirror radius )^n ------ ( 1 )

at 0.603 bending stress = 225

at 0.203 bending stress = 368

at 0.162 bending stress = 442

applying equation 1 determine the value of n for several combinations

( 225 / 368 ) = ( 0.203 / 0.603 )^n

hence : n = 0.452

also

( 368/442 ) = ( 0.162 / 0.203 ) ^n

hence : n = 0.821

also

( 225 / 442 ) = ( 0.162 / 0.603 ) ^n

hence : n = 0.514

Next determine the average value of n

n ( mean value ) = ( 0.452 + 0.821 + 0.514 ) / 3 = 0.596

Calculate estimated stress present at the time of fracture for the original plate

= bending stress at x = 0.796 / bending stress at x = 0.603

= x / 225 = ( 0.603 / 0.796 ) ^ 0.596

therefore X ( stress present at the time of fracture of original plate )

= 225 * 0.84747

= 191 mPa 

3 0

2 years ago

How do you take a picture

Mrrafil [7]

to take a picture just choose “take photo” on the button next to the keypad, if you are on a computer it’s the same thing if your computer has a video screener or whatever they call it but if not I’m not sure, hope this helps!

7 0

2 years ago

Read 2 more answers

As a talented grad of CMPEN 270, your new job as a digital designer is to design a "greedy" vending machine that accepts only ni

Firlakuza [10]

Answer:

See explaination

Explanation:

In the below answer "Z" is the output. please kindly check attachment for further details.

When z=1 , the vending machine gives snacks

z=0 , the vending machine doesn't give snacks.

And z becomes 1( high) only when the machine gets 20 cents or more than 20 cents. Since the machine doesn't give change, after giving snacks it will restart from initial state A (where no coins were put).

3 0

3 years ago

A cylindrical specimen of some metal alloy having an elastic modulus of 108 GPa and an original cross-sectional diameter of 4.5

IgorC [24]

Answer:

The maximum length of the specimen before the deformation was 358 mm or 0.358 m.

Explanation:

The specific deformation ε for the material is:

$\epsilon = \deltaL /L$ (1)

Where δL and L represent the elongation and initial length respectively. From the HOOK's law we also now that for a linear deformation, the deformation and the normal stress applied relation can be written as:

$\sigma = E/ \epsilon$ (2)

Where E represents the elasticity modulus. By combining equations (1) and (2) in the following form:

$L= \delta L/ \epsilon$

$\epsilon =\sigma /E$

So by calculating ε then will be possible to find L. The normal stress σ is computing with the applied force F and the cross-sectional area A:

$\sigma=F/A$