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Snowcat [4.5K]
3 years ago
6

A car’s brakes decelerate it at a rate of -1.70 m/s2. What is the time, in seconds, required to slow the car from 15 m/s to 9.0

m/s?
Physics
1 answer:
jeka943 years ago
3 0

Answer:

t = 3.52 s

Explanation:

Given that,

The deceleration of a car is, a = -1.7 m/s²

The initial velocity of the car, u = 15 m/s

Final velocity of the car, v = 9 m/s

We need to find the time that is required to slow the car from 15 m/s to 9 m/s. The definition of acceleration is :

a=\dfrac{v-u}{t}\\\\t=\dfrac{v-u}{a}\\\\t=\dfrac{9-15}{-1.7}\\\\t=3.52\ s

So, it will take 3.52 seconds to slow the car from 15 m/s to 9 m/s.

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Explain the difference between an electrically charged and a neutral object
Stella [2.4K]

An electrically charged element is called an "ion". A neutral element is an atom.

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If an object has a constant acceleration, it must be changing velocity at the same rate over and over again..TrueFalse
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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
3 years ago
Read 2 more answers
A horizontal 649 N merry-go-round of radius 1.05 m is started from rest by a constant horizontal force of 61.3 N applied tangent
nexus9112 [7]

Answer:

The kinetic energy of the merry-go-round is 632.82 J

Explanation:

Given;

weight of the merry-go-round, W = 649 N

radius of the merry-go-round, r = 1.05 m

applied horizontal force, F =  61.3 N

acceleration due to gravity, g = 9.8 m/s²

mass of  merry-go-round, m = W/g

                                               = 649/9.8  = 66.225 kg

moment of inertia of merry-go-round, I = ¹/₂mr²

                                                                 = ¹/₂ x 66.225 x (1.05)²

                                                                 = 36.507 kg.m²

Angular acceleration of the merry-go-round, α

τ = Iα = Fr

α = Fr / I

Where;

α is angular acceleration

α = (61.3 x 1.05) / 36.507

α = 1.763 rad/s²

Angular velocity of the merry-go-round, ω

ω = αt

ω = 1.763 x 3.34

ω = 5.888 rad/s

Finally, the kinetic energy of the merry-go-round, K.E

K.E = ¹/₂Iω²

K.E = ¹/₂ x 36.507 x (5.888)²

K.E = 632.82 J

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