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3 years ago

How long does it take a 1.51 × 104 W steam engine to do 8.72 × 106 J of work? Round your answer to three significant figures

Physics

1 answer:

DerKrebs [107]3 years ago

5 0

5.77 x 10^2 So it’s v

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State Pascal's principle of pressure . please help due tomorrow

Romashka [77]

Answer:

Pascal's law says that pressure applied to an enclosed fluid will be transmitted without a change in magnitude to every point of the fluid and to the walls of the container.

Explanation:

The pressure at any point in the fluid is equal in all directions.

3 0

3 years ago

Distance = 6 km south<br><br> 60 minutes<br><br> What was the average velocity

blsea [12.9K]

v = x/t

v = average velocity, x = displacement, t = elapsed time

Given values:

x = 6km south, t = 60min

Plug in and solve for v:

v = 6/60

v = 0.1km/min south

3 0

3 years ago

Read 2 more answers

What is the force per unit area at this point acting normal to the surface with unit nor- Side View √√ mal vector n = (1/ 2)ex +

Mumz [18]

Complete Question:

Given $\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$ at a point. What is the force per unit area at this point acting normal to the surface with $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$ ? Are there any shear stresses acting on this surface?

Answer:

Force per unit area, $\sigma_n = 28 MPa$

There are shear stresses acting on the surface since $\tau \neq 0$

Explanation:

$\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$

equation of the normal, $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$

$\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

Traction vector on n, $T_n = \sigma \b n$

$T_n = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

$T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right]$

$T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z$

To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.

$\sigma_n = T_n . \b n$

$\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa$

If the shear stress, $\tau$ , is calculated and it is not equal to zero, this means there are shear stresses.

$\tau = T_n - \sigma_n \b n$

$\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau = \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z$

$\tau = \sqrt{(-5/\sqrt{2})^2 + (27/\sqrt{2})^2 + (5/\sqrt{2})^2} \\\\ \tau = 19.74 MPa$

Since $\tau \neq 0$ , there are shear stresses acting on the surface.

3 0

3 years ago

If a running system has a total change in heat of 295 joules, and it's running at a temperature of 402 kelvin, what is the entro

Liula [17]

The answer is either 75 or 55

7 0

3 years ago

Calculate the electric potential at point A, the middle of the rectangle, and at point B, the middle of the right-hand side of t

dsp73

Answer:

With the help of formula.

Explanation:

We can calculate the electric potential of any point through the formula of electric potential which is given below.

Electric potential = Coulomb constant x charge/ distance of separation.

Symbolically it can be written as, V = k q/ r where

V = electric potential

k = Coulomb constant

q = charge

r = distance of separation

If we have all these data, we can simply put the data in the formula and we will get the value of electric potential.

5 0

3 years ago