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Margaret [11]
3 years ago
6

What description refers to fog?

Physics
1 answer:
wolverine [178]3 years ago
4 0
I think the answer is D
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B) ¿Con qué experimentos o instrumentos se calibrarán los patrones distribuidos
Setler [38]
English please !!!!!!!
8 0
3 years ago
Can we use momentum to see how fast the earth is going?
Kisachek [45]

Yes, if we know the Earth's mass

Explanation:

The momentum of an object is a vector quantity given by the equation

p=mv

where

m is the mass of the object

v is its velocity

In this case, we are asked if we can find the velocity of the Earth by starting from its momentum. Indeed, we can. In fact, we can rewrite the equation above as

v=\frac{p}{m}

Therefore, if we know the momentum of the Earth (p) and we know its mass as well (m), we can solve the formula to find the Earth's velocity.

Learn more about momentum:

brainly.com/question/7973509

brainly.com/question/6573742

brainly.com/question/2370982

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#LearnwithBrainly

6 0
4 years ago
The acceleration due to the earth's gravity, in si units, is 9.8 m/s2. in the absence of air friction, a ball is dropped from re
gogolik [260]
We don't know anything about the amount of distance it travels, but that's okay. The only equation we need here is 

velocity(final) = velocity(initial) + acceleration * time
vf = vi + (a * t)

The ball is dropped from rest, so vi = 0 m/s.
We want it so that the ball hits the ground with a final velocity of 60 m/s, so vf = 60 m/s. 
We are given the acceleration due to gravity, a = 9.8 m/s^2.
We are solving for the time, t = ?.

Now we just plug in the values.
vf = vi + (a * t)
60 m/s = 0 m/s + (9.8 m/s^2)*(t)

60 = 9.8t

60 / 9.8 = t

t = 6.122 s

Hopefully this is the right answer.

7 0
4 years ago
What percentage of the takeoff velocity did the plane gain when it reached the midpoint of the runway? a plane accelerates from
ElenaW [278]
When is at the end of the runway the velocity of the plane is given by the equation vf^{2}=0+2*a*s    where s=1800 m is the runway length. Thus
vf^{2}=2*5*1800=18000 (m/s)^{2}      
vf =134.164 (m/s)  

At half runway the velocity of the plane is
v^{2}=2*5* \frac{1800}{2}=9000 ( \frac{m}{s} )^{2}
 
v= \sqrt{9000}=94.87 ( \frac{m}{s})

Therefore at midpoint of runway the percentage of takeoff velocity is
‰P= \frac{v}{vf}=  \frac{94.87}{134.164}=0.707
6 0
3 years ago
Please help answer both of these !! I’m in a rush :(((
Leno4ka [110]

Answer: search it on browser

7 0
3 years ago
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