All categories

2 years ago

With which of the following statements would Whorf and Sapir agree?

2 answers:

6 0

Hi!

The answer is <span>B. Language influences how people understand their world.

Hope this helps!

-Payshence xoxo</span>

krok68 [10]2 years ago

6 0

The answer should be B

You might be interested in

Which statement best describes the energy of activation?

Varvara68 [4.7K]

hi <3

i believe the answer would be D, as when the temperature increases the particles have more energy and can overcome the activation energy more rapidly.

hope this helps :)

4 0

2 years ago

This when a satellite orbits in an oval-shaped path around a central object.

Ilya [14]

Elliptical orbit.<<<<<<<<<<

8 0

3 years ago

A harmonic wave is traveling along a rope. It is observed that the oscillator that generates the wave completes 37.0 vibrations

Goryan [66]

Answer:

0.47 m

Explanation:

$N$ = Number of vibrations = 37

$t$ = total time taken = 33 s

$T$ = time period of each vibration

frequency of vibration is given as

$f = \frac{N}{t} \\f = \frac{37}{33} \\f = 1.12$ Hz

$d$ = distance traveled along the rope = 421 cm = 4.21 m

$t$ = time taken to travel the distance = 8 s

$v$ = speed of the wave

Speed of the wave is given as

$v = \frac{d}{t}\\v = \frac{4.21}{8}\\v = 0.53 ms^{-1}$

$\lambda$ = wavelength of the harmonic wave

wavelength of the harmonic wave is given as

$\lambda = \frac{v}{f} \\\lambda = \frac{0.53}{1.12} \\\lambda = 0.47 m$

8 0

2 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

Vector a has a magnitude of 12.3 units and points due west. vector b points due north. what is the magnitude of b if a - b has a

algol [13]

The magnitude of vector b is 8.58 Unit.

Since both the vectors a and b are perpendicular to each other, so we can apply the Pythagoras theorem to calculate the magnitude of the vector b.

Applying the Pythagoras theorem

(a-b)^2=a^2+b^2

15^2=12.3^2-b^2

b=8.58 unit

Therefor the magnitude of the vector b is 8.58 unit.

8 0

3 years ago