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vekshin1
2 years ago
6

Stress that acts in the plane of a cut section, rather than at right angles to the section is called:_______

Engineering
1 answer:
snow_lady [41]2 years ago
8 0

probably B, because stress and tension are alike.

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Write a system of equations to describe the situation below, solve using any method, and fill in the blanks.
Ann [662]

The number of tubs that each of them sold is; 24 tubs each

The number of days it will take for both of them to sell same amount of tubs is; 4 days

Number of cookies that Nicole had already sold = 4 tubs

Number of cookies sold by Josie before counting = 0 cookies

Nicole now sells 5 tubs per day and

Josie now sells 6 tubs per day.

Let the number of days it will take for them to have sold the same amount of cookies be d.

Thus;

5d + 4 = 6d + 0

6d - 5d = 4

d = 4 days

Thus, total number of cookies for both are;

Total for Nicole = 4 + 5(4) = 24 cookies

Total for Josie = 6(4) = 24 cookies

Read more about proportion at; brainly.com/question/870035

6 0
2 years ago
Plssssssssssssss Alexi is writing a program which prompts users to enter their age. Which function should she use?
aleksandr82 [10.1K]

Answer:

int()

Explanation:

float() is using decimals, so that can't be it, like float(input( "how much does this cost?"))

print() is used to print something, not a user asking, like print("hello")

string() means like a whole, like string( I am good)

By elimination, int() is correct.

Hope this helps!

7 0
2 years ago
Steam enters a steady-flow adiabatic nozzle with a low inlet velocity (assume ~0 m/s) as a saturated vapor at 6 MPa and expands
Sergio [31]
Yea bro I don’t really know
7 0
2 years ago
Kerosene flows through 3/4 standard type K drawn copper tube. The pressure drop measured at two points 50 m apart is 130 kPa. De
jok3333 [9.3K]

Answer:

Q=4.98\times 10^{-3}\ m^3/s.

Explanation:

Given that

L= 50 m

Pressure drop = 130 KPa

copper tube is 3/4 standard type K drawn tube.

From standard chart ,the dimension of 3/4 standard type K copper tube given as

Outside diameter=22.22 mm

Inside diameter=18.92 mm

Dynamic viscosity for kerosene

\mu =0.00164\ Pa.s

We know that

\Delta P=\dfrac{128\mu QL}{\pi d_i^4}

Where Q is volume flow rate

L is length of tube

d_i is inner diameter of tube

ΔP is pressure drop

μ is dynamic viscosity

Now by putting the values

\Delta P=\dfrac{128\mu QL}{\pi d_i^4}

130\times 1000=\dfrac{128\times 0.00164\times 50Q}{\pi \times 0.01892^4}

Q=4.98\times 10^{-3}\ m^3/s

So flow rate is Q=4.98\times 10^{-3}\ m^3/s.

6 0
3 years ago
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
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