What advantage do weather do weather satellites have over ground based weather stations

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

The total amount of revenue collected from $1 per unit tariff on this product will be?

Rzqust [24]

The total amount of revenue collected from $1 per unit tariff on this product will be $7.

What is revenue from tariffs?

A tax rate known as a revenue tariff is one that is used to obtain direct income from company revenues. The level of prices is significantly impacted by a revenue tariff.

How do tariffs bring in money?

As imports reach the domestic market, the government will earn more money because a tariff is a tax. Since import prices are artificially inflated, domestic sectors also profit from less competition.

Learn more about the Tarrif with the help of the given link:

brainly.com/question/11612474

#SPJ4

I understand that the question you are looking for is "The total amount of revenue collected from a $1-per-unit tariff on this product will be: $22. $14. $8. $7".

6 0

2 years ago

Is it true or false

tester [92]

Answer:

false

Explanation:

8 0

2 years ago

Erica (37kg ) and Danny (49kg ) are bouncing on a trampoline. Just as Erica reaches the high point of her bounce, Danny is movin

Semmy [17]

Answer:

2.45 m/s

Explanation:

From the law of conservation of momentum,

The total momentum before the grab = Total momentum after the grab.

mu+m'u' = V(m+m')................... Equation 1

Where m = mass of Erica, m' = mass of Danny, u = initial velocity of Erica, u' = initial velocity of Danny, V = common velocity after garb

Make V the subject of the equation

V = (mu+m'u')/(m+m')............. Equation 2

Given: m = 37 kg, m' = 49 kg, u = 0 m/s(at the maximum height), u' = 4.3 m/s

Substitute into equation 2

V = (37×0+49×4.3)/(37+49)

V = 210.7/86

V = 2.45 m/s

Hence their speed just after the grab = 2.45 m/s

8 0

3 years ago

Read 2 more answers

A truck pulls a caravan at 85 km/h [S]. When they hit a bump in the road the caravan detaches! If the caravan decelerates at a r

Brrunno [24]

The caravan has an initial speed of

85 km/h × (1000 m/km) × (1/3600 h/s) ≈ 23.611 m/s

If it slows down at a constant rate, then it will cover a distance x such that

0² - (23.611 m/s)² = 2 (-1.3 m/s²) x

Solve for x :

x = -(23.611 m/s)² / (2 (-1.3 m/s²)) ≈ 214.417 m ≈ 210 m

5 0

3 years ago