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3 years ago

6

What are valence electrons?

1 answer:

SashulF [63]3 years ago

8 0

Answer:

The correct answer is D. Electrons in an atom that can bond with other atoms.

Explanation:

For those of you that need it still

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4 years ago

Read 2 more answers

A mass resting on a horizontal, frictionless surface is attached to one end of a spring; the other end is fixed to a wall. It ta

Marina CMI [18]

Answer:

(a) The spring constant is 426N/m

(b) The mass is 5.65kg

Explanation:

(a) Energy stored in the spring (E) = 3.6J, extension (e) = 0.13m

E = 1/2Ke^2

3.6 = 1/2 × K × 0.13^2

K (spring constant) = 3.6×2/0.0169 = 426N/m

(b) F = Ke = 426 × 0.13 = 55.38N

Assuming the maximum acceleration (a) is 9.8m/s^2

Mass (m) = F/a = 55.38/9.8 = 5.65kg

7 0

4 years ago

Light of wavelength 650 nm is normally incident on the rear of a grating. The first bright fringe (other than the central one) i

koban [17]

Answer:

A

$N = 1340.86 \ slits / cm$

B

$\theta = 15.7^o$

Explanation:

From the question we are told that

The wavelength is $\lambda = 650 \ nm = 650 *10^{-9} \ m$

The angle of first bright fringe is $\theta = 5^o$

The order of the fringe considered is n =1

Generally the condition for constructive interference is

$dsin (\theta ) = n * \lambda$

=> $d = \frac{1 * 650 *10^{-9 }}{ sin(5)}$

=> $d = 7.458 *10^{-6} \ m$

Converting to cm

$d = 7.458 *10^{-6} \ m = 7.458 *10^{-6} * 100 = 0.0007458 \ cm$

Generally the number of grating pre centimeter is mathematically represented as

$N = \frac{1}{d}$

=> $N = \frac{1}{0.0007458}$

=> $N = 1340.86 \ slits / cm$

Considering question B

From the question we are told that

The first wavelength is $\lambda_1 = 650 \ nm = 650 *10^{-9} \ m$

The second wavelength is $\lambda_2 = 429 \ m = 420 *10^{-9 } \ m$

The order of the fringe is $n = 2$

The grating is $N = 5000 \ slits / cm$

Generally the slit width is mathematically represented as

$d = \frac{1}{N }$

=> $d = \frac{1}{ 5000 }$

=> $d = 0.0002 \ c m = 2.0 *10^{-6} \ m$

Generally the condition for constructive interference for the first ray is mathematically represented as

$d sin(\theta_1) = n * \lambda_1$

=> $\theta_1 = sin^{-1} [\frac{ 2 * \lambda }{d}]$

=> $\theta_1 = sin^{-1} [\frac{ 2 * 650 *10^{-9} }{ 2*10^{-6}}]$

=> $\theta_1 = 40.5 ^o$

Generally the condition for constructive interference for the second ray is mathematically represented as

$d sin(\theta_2) = n * \lambda_2$

=> $\theta_2 = sin^{-1} [\frac{ 2 * \lambda_1 }{d}]$

=> $\theta_2 = sin^{-1} [\frac{ 2 * 420 *10^{-9} }{ 2*10^{-6}}]$

=> $\theta_2 = 24.8 ^o$

Generally the angular separation is mathematically represented as

$\theta = \theta_1 - \theta_1$

=> $\theta = 42.5^o - 24.8^o$

=> $\theta = 15.7^o$

4 0

3 years ago