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3 years ago

PLEASE HELPP Look at the picture

Physics

2 answers:

denis23 [38]3 years ago

8 0

Answer: It’s c
explaining: i guessed

sveticcg [70]3 years ago

8 0

It’s either C. Or B. but it’s most probably B. Sorry if I’m incorrect.

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Newton's First and Second Laws of Motion:Question 3Kepler-10b, the first confirmed terrestrial extrasolar planet, is about 564 l

OleMash [197]

Answer:

F = 85696.5 N = 85.69 KN

Explanation:

In this scenario, we apply Newton's Second Law:

$Unbalanced\ Force = ma\\Upthrust - Weight\ of\ Space\ Craft = ma\\F - W = ma\\F - mg = ma\\F = m(g + a)\\$

where,

F = Upthrust = ?

m = mass of space craft = 5000 kg

g = acceleration due to gravity on surface of Kepler-10b = (1.53)(9.81 m/s²)

g = 15.0093 m/s²

a = acceleration required = 2.13 m/s²

Therefore,

$F = (5000\ kg)(15.0093\ m/s^{2} + 2.13\ m/s^{2})\\$

8 0

3 years ago

How are Earth and Venus similar? How is Venus different from Earth? (Provide two similarities and two differences.)

Alika [10]

Venus shares a similar size, surface composition, and has an atmosphere with a complex weather system. Venus is different from Earth because it spins the opposite direction of Earth and it’s rotation is very slow.

8 0

3 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:

$N = 1340.86 \ slits / cm$

$\theta = 15.7^o$

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

I'll mark brainliest

irga5000 [103]

Answer:

A) 35 ft

B) 5 ft

C) Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick

Explanation:

A) Total distance covered by the dog = 20 + 15

= 35 ft

B) Since the other distance covered by the dog before chewing the stick, after the retrieval, was in an opposite direction to the initial direction, then;

total displacement of the dog = 20 - 15

= 5 ft

C) Net displacement = distance covered by the dog to retrieve the stick + distance covered before the dog starts chewing the stick

But, displacement involves a specified direction. The distance covered before the dog starts chewing the stick was in an opposite direction to the initial direction.

Thus,

Net displacement = distance covered by the dog to retrieve the stick - distance covered before the dog starts chewing the stick

7 0

3 years ago

China's GDP passed one trillion USD between Between 2005 and 2010, China's GDP increased by approximately

maksim [4K]

Answer:

1995 and 2000 , 4 trillions

Explanation:

3 0

3 years ago

Read 2 more answers