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Oduvanchick [21]
3 years ago
9

A dragster starts from rest and travels 1/4 mi in 6.80 s with constant acceleration. What is its velocity when it crosses the fi

nish line?
Physics
1 answer:
Ahat [919]3 years ago
6 0
<h2>Its velocity when it crosses the finish line is 117.65 m/s</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = ?

        Time, t = 6.8 s    

        Displacement, s = 1/4 mi =    400 meters

     Substituting

                      s = ut + 0.5 at²

                      400 = 0 x 6.8 + 0.5 x a x 6.8²

                      a = 17.30 m/s²

Now we have equation of motion v = u + at

     Initial velocity, u = 0 m/s

     Final velocity, v = ?

     Time, t = 6.8 s

      Acceleration, a = 17.30 m/s²

     Substituting

                      v = u + at  

                      v = 0 + 17.30 x 6.8

                      v = 117.65 m/s

Its velocity when it crosses the finish line is 117.65 m/s

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Artyom0805 [142]

If the pulling is done parallel to the floor with constant velocity, then the box is in equilibrium. In particular, the weight and normal force cancel, so that

<em>n</em> = 38 N

The friction force is proportional to the normal force by a factor of 0.27, so that

<em>f</em> = 0.27 (38 N) ≈ 10.3 N

and so the answer is D.

8 0
3 years ago
___ + 3H2O + light —&gt; C3H6O3 + 3O2. What amount and substance balance this reaction?
Alik [6]

Answer:3H2O + light-c3h603+302

Explanation:

5 0
3 years ago
Read 2 more answers
In attempting to pass the puck to a teammate, a hockey player gives it an initial speed of 2.8 m/s. However, this speed is inade
erastova [34]

Answer:

3.95979 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

v^2-u^2=2as\\\Rightarrow as=\frac{v^2-u^2}{2}\\\Rightarrow as=\frac{0^2-2.8^2}{2}\\\Rightarrow as=-3.92

Here s=\frac{1}{2}s

\\\Rightarrow as=-7.84\ m/s^2

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -7.84}\\\Rightarrow u=3.95979\ m/s

Initial velocity of the puck should be 3.95979 m/s

8 0
2 years ago
Your in an escape room. Only its real and your oxygen runs out in 3 minutes unless you solve this problem, The problem :a ball i
alex41 [277]

The sum of the series allows to find the result for the total distance that the ball bounces is:

            total distance = 59.52 in

A series is a set of things or numbers related by a specific operation.

They indicate that the ball falls from an initial height y₀ = 30 in. and it bounces 50% of the height and the process is repeated until it stops, see attached.

Let's build a table to observe the sequence.

drop   height    rebound

   1        30          15

   2       15            7.5

   3        7.5         3.75

If we call the first term y₀  

The first bounce can be found.

                             y₁ = \frac{y_o}{2}

The second bounces.

                             y_2 = \frac{y_1}{2}  \\y_2 = \frac{y_o}{4}

The third bounce.

                             y_3  = \frac{y_2}{2}  \\y_3 = \frac{y_0}{ 8}

By observing this table we can construct a series of the form

 

      Total distance = y_o \ ( 1 + \frac{1}{2} + \frac{1}{4}+  \frac{1}{8} + ... +\frac{1}{2n} )

The sum of the serie has a result of

        sum  = 127/64 = 1,984

Let's calculate

     distance total  = 30 1,984

     Distance total = 59.52 cm

In conclusion, using the sum of the series we can find the result for the total distance that the ball bounces is:

            total distance = 59.52 in

Learn more here: brainly.com/question/8879163

4 0
2 years ago
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The information gathered in an investigation is called
aliya0001 [1]
Data

hoped it helped!!
6 0
3 years ago
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