Ask question

Ask question

All categories

3 years ago

14

Calculate the number of vacancies per cubic meter in gold (au) at 900c. the energy for vacancy formation is 0.98 ev/atom. furthe

rmore, the density and atomic weight for au are 18.63 g/cm3 (at 900c) and 196.9 g/mol, respectively.

1 answer:

Sergeeva-Olga [200]3 years ago

8 0

My mom said its f so i think its q

You might be interested in

You have two balls of equal size and smoothness, and you can ignore air resistance. One is heavy, the other is much lighter. You

motikmotik

Answer:

They will hit the ground at the same time.

Explanation:

By ignoring the opposing forces i.e. air resistant, both the heavy and light balls will fall with same acceleration due to gravity (g=9.8 m/s²) and g is independent of mass of the objects. Thus both will hit the ground at the same time.

4 0

3 years ago

How many atoms of each element are in the products?

Kay [80]

The number of atoms in an element must be the same number of atoms of each element in a PRODUCT.

4 0

4 years ago

During a marathon race, a runner’s blood flow increases to 10.0 times her resting rate. Her blood’s viscosity has dropped to 95.

Phoenix [80]

To solve the problem it is necessary to apply the equations related to the Poiseuilles laminar flow law, with which the stationary laminar flow ΦV of an incompressible and uniformly viscous liquid (also called Newtonian fluid) can be determined through a cylindrical tube of constant circular section. Mathematically this can be expressed:

$Q = \frac{\Delta P \pi r^4}{8\eta l}$

Where:

$\eta_i =$ are the viscosities of the concrete before and after the increase

l = Length of the vessel

$r_1, R_2$ = Radio of the vessel before and after the increase

$\Delta P$ = Change in the pressure

$Q_{1,2} =$ The rates of flow before and after he increase

Our values are given as:

$Q_2 = 10Q_1 \rightarrow$ 10 times her resting rate

$\eta_2 = 0.95\eta_1$ 95% of its normal value

$\Delta P_2 = 1.5\Delta P_1$ Increase of 50%

Plugging known information to get

$Q_1 = \frac{\Delta P \pi r^4}{8\eta l}$

$Q_1 8\eta_1 l = \Delta P_1 \pi r_1^4$

$r_1^4 = \frac{Q_1 8\eta_1 l}{\Delta P_1 \pi}$

$r_1 = (\frac{Q_1 8\eta_1 l}{\Delta P_1 \pi})^{1/4}$

$r_2 = (\frac{Q_2 8\eta_2 l}{\Delta P_2 \pi})^{1/4}$

$r_2 = (\frac{10Q_18 \times 0.95\eta_1 l}{1.5\Delta P_1 \pi})^{1/4}$

$r_2 = 1.586r_1$

Therefore the factor of average radio of her blood vessels increased is 1.589 the initial factor after the increase.

7 0

4 years ago

A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a

jeka94

Answer:

a) t_l - t_r = 12.54 us

b) (t_l - t_r) / T = 0.0157

Explanation:

Given:

- Frequency of source f = 1250 Hz

- Distance from source to right ear d_r = 2.6 m

- Distance from source to left ear d_l = ?

- Separation between ears s = 0.15 m

Find:

a. What is the difference in the arrival time of the sound at the left ear and the right ear?

b. What is the ratio of this time difference to the period of the sound wave?

Solution:

- Apply Pythagoras theorem to calculate the distance d_l from source to left ear:

d_l = sqrt ( 2.6^2 + 0.15^2)

d_l = sqrt ( 6.7825 )

d_l = 2.6043 m

- The time deference can be calculated from a simple distance - speed formula:

t_l - t_r = (1 / v) * ( d_l - d_r)

Where, v = 343 m/s speed of sound in air:

t_l - t_r = (1 / 343) * ( 2.6043 - 2.6)

t_l - t_r = ( 0.0043 / 343 )

t_l - t_r = 12.54 us

- Now we compute the Time period of the sound wave:

T = 1 / f

T = 1 / 1250 = 8*10^-4 s

- The ratio of differential time to Time period T is:

(t_l - t_r) / T = 12.54 * 10^-6 / 8*10^-4

(t_l - t_r) / T = 0.0157

3 0

3 years ago

A young's interference experiment is performed with blue-green laser light. the separation between the slits is 0.500 mm, and th

fgiga [73]

What class is this for?

3 0

3 years ago