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REY [17]
2 years ago
15

An object traveling at 1.5 rad

Physics
1 answer:
Veronika [31]2 years ago
6 0

The object's final velocity, given the data is 10.5 rad/s

<h3>What is acceleration? </h3>

This is defined as the rate of change of velocity which time. It is expressed as

a = (v – u) / t

Where

  • a is the acceleration
  • v is the final velocity
  • u is the initial velocity
  • t is the time

<h3>How to determine the final velocity</h3>

The following data were obtained from the question

  • Initial velocity (u) = 1.5 rad/s
  • Acceleration (a) = 0.75 rad/s²
  • Time (t) = 12 s
  • Final velocity (v) = ?

The final velocity can be obtained as follow:

a = (v – u) / t

0.75 = (v – 1.5) / 12

Cross multiply

v – 1.5 = 0.75 × 12

v – 1.5 = 9

Collect like terms

v = 9 + 1.5

v = 10.5 rad/s

Thus, the final velocity of the object is 10.5 rad/s

Learn more about acceleration:

brainly.com/question/491732

#SPJ1

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Explanation:

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Which of these equations is dimensionally correct?
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Answer: first and third.

Explanation:

An equation is dimensionally correct if the units are the same in both sides of the equation.

first, let's define the units used:

{m} = kg

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{F} = kg*m/s^2

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{t} = s

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Now, let's analyze each option:

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in the left side the units are:

{m}*{v}/{t} =  kg*(m/s)*(1/s) = kg*m/s^2

And as is written above, these are the units of F, so this is correct.

2) x*v^2 = F*(x^3/x^2)

This is more trivial, in the right side we can see an F, that has mass units (kg)  and in the left side we have x and v, and we know that none of these have mass units, so this expression is not correct.

3) xt= vt^2+at^3

the units in the right side are:

{x}*{t] = m*s

in the right side are:

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3 years ago
A dog runs after a car. The car is travelling at an average speed of 5 m/s.
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Answer:

<h2>The answer is 4 s</h2>

Explanation:

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t =  \frac{d}{v}  \\

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t =  \frac{20}{5}  \\

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<h3>4 s</h3>

Hope this helps you

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Half maximum velocity occurs at the point of half maximum kinetic energy which is exactly halfway down.

<h3>Conservation of energy</h3>

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