All categories

3 years ago

The achievement of lifting a rocket off the ground and into space can be explained by

Physics

1 answer:

lawyer [7]3 years ago

4 0

Answer:

The achievement of lifting a rocket off the ground and into space can be explained by Newton's third law of motion. What is required for a rocket to lift off into space? Thrust is required for a rocket to lift off into space, ... An object that travels around another object in space is called a satellite.

Explanation:

You might be interested in

Why should you explore the potential roadblocks to your chosen career?

valina [46]

Answer: D

Explanation: D is the most reasonable answer because it's always good to plan ahead for anything, so if you were to plan ahead for future obstacles, then you can overcome them.

6 0

3 years ago

Which of the following temparature is approximately equal to room temperature

Artist 52 [7]

Hello there! :)

$\huge\boxed{\text{C. 293K}}$

Room temperature is approximately 20°C.

We can automatically eliminate choices B and D since they are not equal to 20°C.

Since some choices use the Kelvin scale, we can convert from Celsius to Kelvin using a simple formula:

K = C° + 273

Find room temperature in degrees <u>Kelvin</u>:

K = 20° + 273

K = 293°

Thus, the correct choice would be <u>C. 293K.</u>

6 0

3 years ago

Don’t need to answer a question just say something and you’ll get Marked brainliest (Free points WONT GET BRAINLIEST RIGHT AWAY)

stiks02 [169]

Answer:

Explanation:

7 0

3 years ago

Read 2 more answers

Light of wavelength 630 nm is incident on a long, narrow slit. Determine the angular deflection of the first diffraction minimum

asambeis [7]

Answer:

a) 1.8°

b) 0.18°

c) 0.018°

Explanation:

Wavelength (λ) = 630nm = 630 *10^-9m

The equation that describes the angular deflection of a dark band is

Wsin(βm) = mλ

w = width of the single slit

λ = wavelength of the light

βm = angular deflection of the mth dark band.

a) In order to get the angular deflection of the first dark band for a slit with 0.02mm width, substitute w = 0.02mm = 0.02*10^-3 , m = 1 , λ= 630*10^-9

0.02*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.02*10^-3

Sin(β1) = 0.0315

β1 = Sin^-1(0.0315)

= 1.8°

b) substitute w = 0.2mm = 0.2*10^-3 , m = 1 , λ= 630*10^-9

0.2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 0.2*10^-3

Sin(β1) = 0.00315

β1 = Sin^-1(0.00315)

= 0.18°

c) substitute w = 2mm = 2*10^-3 , m = 1 , λ= 630*10^-9

2*10^-3 sin(β1) = 1 * 630*10^-9

Sin(β1) = 630 * 10^-9 / 2*10^-3

Sin(β1) = 3.15*10^-4

β1 = Sin^-1(3.15*10^-4)

= 0.018°

6 0

3 years ago

Find the current in the 12 ohm resistor.

disa [49]

Answer:

1.5 A

Explanation:

V =I.R

18 = I × 12

I = 18/12

= 3/2 = 1.5 A

3 0

3 years ago