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sergeinik [125]

2 years ago

5

Discuss why the article says the fact that we’re in our universe complicates our understanding of the expansion of the universe.

2 answers:

disa [49]2 years ago

7 0

It means that that being in just one universe makes it harder to grasp it's expansion

MArishka [77]2 years ago

4 0

<span>Because of our perception of the universe from inside the universe, we are unable to see how and towards what the universe is expanding. Also, our understanding of it is further complicated because we are moving as part of the expansion, thus distorting our perception of it.</span>

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An airplane is flying at an elevation of 5150 ft, directly above a straight highway. two motorists are driving cars on the highw

docker41 [41]

Using trigonometric ratios we can get the distance;
For the first car; The distance from the point on the highway below the plane
tan = opp/adj
tan(36°) = 5150/x
0.727 = 5150/x
0.727x = 5150
x = 7088.37
For the second car we also use tangent; the distance from the point on the highway below the plane will be;
tan(56°) = 5150/y
1.483 = 5150/y
1.483y = 5150
y= 3473.72
The we can add the two distances to get how far apart the cars are;
7088.37 + 3473.72 = 10562.09 feet.
= 10562.09 ft

7 0

2 years ago

If the train can slow down at a rate of 0.625 m/s2, how long, in seconds, does it take to come to a stop from this velocity?

qwelly [4]

Answer:

66.28 s

Explanation:

Step 1:

Acceleration,a= $0.065m/s^2$

Time,t=9.75 min= $9.75\times 60=585s$

1 min=60 s

Initial velocity,u= $3.4m/s$

We have to find the final velocity of train.

We know that

$v=u+at$

Substitute the values

$v=3.4+0.065(585)=41.425m/s$

Step 2:

Now, initial velocity,u=41.425m/s

Deceleration,a= $-0.625m/s^2$

Because the train velocity decreases

Final velocity, v=0

Again, substitute the values in the above formula

$0-41.425=-0.625t$

$-41.425=-0.625t$

$t=\frac{-41.425}{-0.625}$

$t=66.28s$

Hence, the train takes 66.28 s to come to stop from velocity 41.425m/s.

6 0

3 years ago

An artillery shell is fired at an angle of 83.3 ◦ above the horizontal ground with an initial speed of 1600 m/s. the acceleratio

timurjin [86]

T = 2*V₀*sin (α) / g .

t = 2*1600*sin (83.3°) / 9.8 ≈ 2*1600*0.993 / 9.8 ≈ 324 sec or 5.3 min

5 0

3 years ago

Which does not utilize the gyroscope effect? Question 12 options:

garri49 [273]

Answer:

a spinning top

thats the answer and I hope it helps

7 0

2 years ago

Read 2 more answers

Using the image, explain what information you are given and what you could solve for. Derive an equation that solves for the max

mezya [45]

Answer:

h = (v₀²sin²θ)/2g

R = (v₀²sin2θ)/g

Explanation:

general equation is s = s₀ + v₀t + ½at²

if the firing point is origin and UP and RIGHT are positive directions, and if we ignore air resistance.

In the vertical direction, and remembering that gravity opposes the initial vertical velocity, the equation becomes

y = 0 + (v₀sinθ)t - ½(g)t²

y = (v₀sinθ)t - ½(g)t²

at maximum height h, vertical velocity is zero. The initial vertical velocity is reduced to zero in a time of

t = (v₀sinθ)/g

entering this value into our equation for y

h = (v₀sinθ)(v₀sinθ)/g - ½(g)((v₀sinθ)/g)²

h = (v₀²sin²θ)/g - ½(g)(v₀²sin²θ)/g²

h = (v₀²sin²θ)/g - ½(v₀²sin²θ)/g

h = (v₀²sin²θ)/2g

In the horizontal the equation becomes

x = 0 + (v₀cosθ)t + ½(0)t²

x = (v₀cosθ)t

As it will take as long to fall as it took the projectile to rise to the a•pex.

R = (v₀cosθ)(2)((v₀sinθ)/g)

R = 2(v₀²cosθsinθ)/g

trig identity sin2θ = 2cosθsinθ

R = (v₀²sin2θ)/g

The algorithm thinks the word "a•pex" is a swear word so I have to write it like I have... What an a•pex

8 0

3 years ago