Scientists classify rocks by

IRINA_888 [86]

My best guess would be sun because it is bright but is surrounded by shadows on all sides.

5 0

3 years ago

A charge of 4.5 × 10-5 C is placed in an electric field with a strength of 2.0 × 104 . If the charge is 0.030 m from the source

snow_tiger [21]

Answer:

The electrical potential energy is 0.027 Joules.

Explanation:

The values from the question are

charge (q) = $4.5 \times 10^{-5} C$

Electric Field strength (E) = $2.0 \times 10^{4} N/C$

Distance from source (d) = 0.030 m

Now the formula for the electrical potential energy (U) is given by

$U = q \times E \times d$

So now insert the values to find the answer

$U = 4.5 \times 10^{-5} C \times 2.0 \times 10^{4} N/C \times 0.030 m$

On further solving

$U = 0.027 J$

8 0

3 years ago

As an airplane accelerates in level flight, what parameter must the pilot adjust to keep level and balance Lift = Weight?

Y_Kistochka [10]

Answer:

C. Angle of Attack.

Explanation:

The pilot must adjust the angle of attack parameter. The angle of attack of this plane to get to the desired lift coefficient.

And thus, we have

Lift = Weight

3 0

3 years ago

Some hypothetical alloy is composed of 12.5 wt% of metal A and 87.5 wt% of metal B. If the densities of metals A and B are 4.27

densk [106]

Answer:

The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

Explanation:

Given that,

Weight of metal A = 12.5%

Weight of metal B = 87.5%

Length of unit cell = 0.395 nm

Density of A = 4.27 g/cm³

Density of B= 6.35 g/cm³

Weight of A = 61.4 g/mol

Weight of B = 125.7 g/mol

We need to calculate the density of the alloy

Using formula of density

$\rho=n\times\dfrac{m}{V_{c}\times N_{A}}$

$n=\dfrac{\rho\timesV_{c}\times N}{m}$ ....(I)

Where, n = number of atoms per unit cells

m = Mass of the alloy

V=Volume of the unit cell

N = Avogadro number

We calculate the density of alloy

$\rho=\dfrac{1}{\dfrac{12.5}{4.27}+\dfrac{87.5}{6.35}}\times100$

$\rho=5.98$

We calculate the mass of the alloy

$m=\dfrac{1}{\dfrac{12.5}{61.4}+\dfrac{87.5}{125.7}}\times100$

$m=111.15$

Put the value into the equation (I)

$n=\dfrac{5.9855\times(0.395\times10^{-9}\times10^{2})^3\times6.023\times10^{23}}{111.15}$

$n=1.99\approx 2\ atoms/cell$

Hence, The number of atoms in the unit cell is 2, the crystal structure for the alloy is body centered cubic.

5 0

4 years ago

Light of wavelength 436.1 nm falls on two slits spaced 0.31 mm apart. What is the required distance from the slits to the screen

kakasveta [241]

Answer:

The correct answer is "4.26 m".

Explanation:

Given:

Wavelength,

$\lambda = 436.1 \ nm$

or,

$=436.1\times 10^{-9} \ m$

Distance,

$d = 0.31 \ mm$

or,

$=0.31\times 10^{-3} \ m$

Distance between the 1st and 2nd dark fringes,

$(y_2-y_1) = 6\times 10^{-3} \ m$

As we know,

⇒ $\frac{d}{L} (y_2-y_1) = \lambda$

or,

⇒ $L=\frac{d(y_2-y_1)}{\lambda}$

By substituting the values, we get

$=\frac{0.31\times 6\times 10^{-6}}{436.1\times 10^{-9}}$

$=\frac{0.31\times 6\times 10^3}{436.1}$

$=\frac{1860}{436.1}$

$=4.26 \ m$

3 0

3 years ago