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xeze [42]
3 years ago
15

A VW Beetle goes from 0 to 60.0 mi/h with an acceleration of +2.35 m/s^2. (a) How much time does it take for the beetle to reach

this speed? (b) A top-fuel dragster can go from 0 to 30 mi/h in 0.600 s. Find the acceleration (in m/s^2) of the dragster?
Physics
1 answer:
Firlakuza [10]3 years ago
8 0
Im going to tell you what to do but not the result. So pay close attention: the first thing you need to do is convert miles/h to m/s. Then for the part a) <span>divide the final velocity by the initial velocity. That will give you the amount of it will take to accelerate to the final velocity.Now for the part b you </span>use the formula v=vo+at. I hope this can help you
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A person throws a ball straight up in the air. The ball rises to a maximum height and then falls back down so that the person ca
Lana71 [14]

Answer:

The acceleration is about 9.8 m/s2 (down) when the ball is falling.

Explanation:

The ball at maximum height has velocity zero

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.8 m/s² (positive downward and negative upward)

v=u+at\\\Rightarrow 0=u-9.8\times t\\\Rightarrow u=9.8t

The accleration 9.8 m/s² will always be acting on the body in opposite direction when the body is going up and in the same direction when the body is going down. The acceleration on the body will never be zero

5 0
4 years ago
What committee decides which bills move to the House floor for debate and sets the terms for the debate?
spayn [35]
 i think the answer is B-House Rules Comittee
6 0
3 years ago
You are working at a company that manufactures electri- cal wire. Gold is the most ductile of all metals: it can be stretched in
Alona [7]

Explanation:

We know that the relation between volume and density is as follows.

      Volume = \frac{\text{mass}}{\text{density}}

So,       V = \frac{10^{-3}}{19.3 \times 10^{3} kg/m^{3}}

               = 5.181 \times 10^{-8} m^{3}

Now, we will calculate the area as follows.

      Area = \frac{\text{volume}}{\text{length}}

               = \frac{5.181 \times 10^{-8} m^{3}}{2.4 \times 10^{3}}

               = 2.15 \times 10^{-11} m^{2}

Formula to calculate the resistance is as follows.

         R = \rho \frac{l}{A}

             = \frac{2.44 \times 10^{-8} \times 2400}{}2.15 \times 10^{-11}}

             = 2.71 \times 10^{6} ohm

Thus, we can conclude that the resistance of given wire is 2.71 \times 10^{6} ohm.

4 0
3 years ago
When astronomers look at distant galaxies, what sort of motion do they see?
arlik [135]
Hello! You can call me Emac or Eric.

I understand your problem, that question is pretty hard. But I found some information that I think you should read. This can get your problem done quickly.

Please hit that thank you button if that helped, I don’t want thank you’s I just want to know that this helped.

Please reply if this doesn’t help, I will try my best to gather more information or a answer.

Here is some good information that could help you out a lot!


Let’s begin by exploring some techniques astronomers use to study how galaxies are born and change over cosmic time. Suppose you wanted to understand how adult humans got to be the way they are. If you were very dedicated and patient, you could actually observe a sample of babies from birth, following them through childhood, adolescence, and into adulthood, and making basic measurements such as their heights, weights, and the proportional sizes of different parts of their bodies to understand how they change over time.

Unfortunately, we have no such possibility for understanding how galaxies grow and change over time: in a human lifetime—or even over the entire history of human civilization—individual galaxies change hardly at all. We need other tools than just patiently observing single galaxies in order to study and understand those long, slow changes.

We do, however, have one remarkable asset in studying galactic evolution. As we have seen, the universe itself is a kind of time machine that permits us to observe remote galaxies as they were long ago. For the closest galaxies, like the Andromeda galaxy, the time the light takes to reach us is on the order of a few hundred thousand to a few million years. Typically not much changes over times that short—individual stars in the galaxy may be born or die, but the overall structure and appearance of the galaxy will remain the same. But we have observed galaxies so far away that we are seeing them as they were when the light left them more than 10 billion years ago.


That is some information, I do have more if you need some! Thanks!

Have a great rest of your day/night! :)


Emacathy,
Brainly Team.


8 0
3 years ago
A boy and a girl are on a spinning merry-go-round. The boy is at a radial distance of 1.2 m from the central axis; the girl is a
Licemer1 [7]

Answer:

E) True. The girl has a larger tangential acceleration than the boy.

Explanation:

In this exercise they do not ask us to say which statement is correct, for this we propose the solution to the problem.

Angular and linear quantities are related

          v = w r

          a = α r

the boy's radius is r₁ = 1.2m the girl's radius is r₂ = 1.8m

as the merry-go-round rotates at a constant angular velocity this is the same for both, but the tangential velocity is different

          v₁ = w 1,2 (boy)

          v₂ = w 1.8 (girl)

whereby

          v₂> v₁

reviewing the claims we have

          a₁ = α 1,2

          a₂ = α 1.8

          a₂> a₁

A) False. Tangential velocity is different from zero

B) False angular acceleration is the same for both

C) False. It is the opposite, according to the previous analysis

D) False. Angular acceleration is equal

E) True. You agree with the analysis above,

8 0
3 years ago
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