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2 years ago

How do we use light in space?

Physics

1 answer:

bulgar [2K]2 years ago

5 0

Answer:

They use LED lights.

Explanation:

Hope this helps

-A Helping Friend (mark brainliest pls)

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If the car has a mass of 0.4 kg, the ratio of height to width of the ramp is 19/85, the initial displacement is 1.9 m, and the c

Rasek [7]

Answer:

final displacement lf = 0.39 m

Explanation:

from change in momentum equation:

$\delta p = m \sqrt(2g * y/x)* [\sqrt li + \sqrt lf]$

given: m = 0.4kg, y/x = 19/85, li = 1.9 m,

\delta p = 1.27 kg*m/s.

putting all value to get the final displacement value

$1.27 = 0.4\sqrt(2*9.81 *(19/85))* [\sqrt 1.9 + \sqrt lf]$

final displacement lf = 0.39 m

5 0

2 years ago

It is rate for any motion to

Gnoma [55]

Answer:

a. stay the same for very long

Explanation:

It is rare for any motion to stay the same for a very long time. The force applied on a body causes changes in the magnitude of motion.

For motion to remain constant, there must not be a net force acting on the body
All the forces on the body must be balanced.
This is very hard to come by.
Motion changes very frequently.

3 0

3 years ago

A clam dropped by a seagull takes 3.0 seconds to hit the ground. What is the seagull's approximate height above the ground at th

ankoles [38]

<h2>The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Time, t = 3 s

Substituting

s = ut + 0.5 at²

s = 0 x 3 + 0.5 x 9.81 x 3²

s = 44.145 m

The seagull's approximate height above the ground at the time the clam was dropped is 4 m

4 0

3 years ago

An airplane is flying through a thundercloud at a height of 2000 m (This is a very dangerous thing to do because of updrafts, tu

Doss [256]

Answer:

$400000\ \text{N/C}$

Explanation:

$q_1$ = Charge at 3000 m = 40 C

$q_2$ = Charge at 1000 m = -40 C

$r_1$ = 3000 m

$r_2$ = 1000 m

k = Coulomb constant = $9\times10^9\ \text{Nm}^2/\text{C}^2$

Electric field due to the charge at 3000 m

$E_1=\dfrac{k|q_1|}{r_1^2}\\\Rightarrow E_1=\dfrac{9\times 10^9\times 40}{3000^2}\\\Rightarrow E_1=40000\ \text{N/C}$

Electric field due to the charge at 1000 m

$E_2=\dfrac{k|q_2|}{r_2^2}\\\Rightarrow E_2=\dfrac{9\times 10^9\times 40}{1000^2}\\\Rightarrow E_2=360000\ \text{N/C}$

Electric field at the aircraft is $E_1+E_2=40000+360000=400000\ \text{N/C}$ .

7 0

2 years ago

Peer assessment is a unique educational model. Think back to how you felt about peer assessment at the beginning of the term, an

luda_lava [24]

Answer:

In the beginning, I was not familiar to assess assessments of the other students. Ifelt a little bit weird that is it possible to check assignments while having an instructor.I was also a bit frustrated, to be honest, that why do we have to assess thoseassessments. It was kind of extra burden for me. But after few weeks assessingmore assignments, my feeling had changed because I was learning lots of thingsthat were changing my perspectives. I was gaining extra knowledge from my peersin the form of assessments. Yes, I am comfortable with assessing assessments,because I got to learn many vocabularies and making structures of the sentencecorrectly by improving grammatically as I am not a native English speaker. Thus, inthis way, I was learning something new in each and every assessment.

8 0

3 years ago

Read 2 more answers