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Mademuasel [1]
2 years ago
15

If a sprinter has an average acceleration of 9.5 m/s2, how long does it take for the person to go from rest to 10 m/s

Physics
1 answer:
mojhsa [17]2 years ago
8 0

Answer:bruh Brandon I’m trying to get the answer I look it up on google and I see this I’m dead see you in 4th hour

Explanation:

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Did you diagram all the faults and folds correctly? Which fault and/or fold did you find most difficult to diagram?
butalik [34]

A fold is a stack of planar surfaces that are bent during permanent deformation and this happens when the earth's crust is compressed together.

<h3>What is a Fault?</h3>

This refers to the planar fracture that occurs in a volume of rock that causes significant displacement.

Hence, we can see that there are three types of fold and they are:

  • (1) anticlines,
  • (2) synclines  
  • (3) monoclines.

Please note that your question is incomplete so I gave a general overview of the concept to give you a better understanding.

Read more about faults and folds here:

brainly.com/question/14240712

#SPJ1

8 0
1 year ago
If a dart gun with a 20N/m spring inside of it is compressed a distance of 0.3m, How high into the air can it shoot a 0.15kg dar
n200080 [17]

Answer:

h = 61.16[cm]

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that energy is conserved or equal in two points in space for an instant in time.

In this way we will have the points A & B, the point A for the moment before shooting and the moment B when the Dart is in the highest position.

In this way the energy is:

E_{A}=E_{B}

Now we must identify the energies in the moments A & B. in the instant A we have the spring compressed, in such a way that only elastic energy is stored.

E_{A}=\frac{1}{2} *k*x^{2}

where:

k = spring constant = 20 [N/m]

x = distance = 0.3 [m]

Now, at the moment when the dart is in the highest position (B), it means that it does not go up anymore, that is, its movement is zero, and therefore its kinetic energy is zero, in this way the energy at the highest point corresponds to potential energy.

E_{B}=m*g*h

where:

m = mass = 0.15[kg]

g = gravity acceleration = 9.81 [m/s²]

h = elevation [m]

Now replacing:

\frac{1}{2} *20*(0.3)^{2}=0.15*9.81*h\\0.9=1.4715*h\\h=0.61[m]\\or\\h = 61.16[cm]

6 0
2 years ago
How does space promote science education
Fittoniya [83]
<span>NASA and the Mad Science Group of Montreal, Canada, have teamed in an effort to spark the imagination of children, encouraging more youth to pursue careers in science, technology, engineering and math. The two organizations recently signed a Space Act Agreement, officially launching the development of the Academy of Future Space Explorers.</span>
3 0
2 years ago
An investigation involves determining which metal is better for making pots that will cook food faster. Which is the best hypoth
irina [24]
Because the specific metals aren’t mentioned in this inquiry. The educational guesses that we can propose is that:
<span><span>1.    </span>The hypothetical inquiry: There are existing metals for making pots that will cook food much faster.</span>

<span><span>2.    </span>The one-tailed alternative hypothesis: There are other metals for making pots that will cook food much faster than the other metals.</span> <span><span>
3.    </span>The one-tailed null hypothesis: All metals that are used in making pots will cook food at an equal rate.</span>



8 0
3 years ago
As in the video, we apply a charge +Q to the half-shell that carries the electroscope. This time, we also apply a charge –Q to t
Andreyy89

Answer:

Explanation:

When the positively charged half shell is brought in contact with the electroscope, its needle deflects due to charge present on the shell.

When the negatively charged half shell is brought in contact with the positively charged shell , the positive and negative charge present on each shell neutralises each other  .So both the shells lose their charges .The positive half shell also loses all its charges

When we separate the half shells , there will be no deflection  in the electroscope because both the shell have already lost their charges and they have become neutral bodies . So they will not be able to produce any deflection in the electroscope.

4 0
3 years ago
Read 2 more answers
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