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ArbitrLikvidat [17]
3 years ago
11

A car starting from rest accelerates in a straight line at a constant rate of 5.5m/s for 6s.If the car after this acceleration s

lows down uniformly at a rate of 2.4m/s, how long does it take to stop​
Physics
1 answer:
aalyn [17]3 years ago
7 0

Answer:

The time it takes to stop is 13.75 seconds

Explanation:

A body moving with constant acceleration, 'a', for a time, 't', has a final velocity, 'v', given by the following kinematic equation;

v = u + a·t

Where;

v = The final velocity of the body

a = The acceleration of the body

t = The time of acceleration (accelerating period) of the body

u = The initial velocity of the body

The given parameters for the acceleration of the car are;

The initial velocity of the car, u = 0 m/s (a car starting from rest)

The constant acceleration of the car, a =  5.5 m/s²

The acceleration duration, t = 6 s

Therefore, we have;

The final velocity of the car after the acceleration, v = 0 m/s + 5.5 m/s² × 6 s = 33 m/s

The final velocity of the car after the acceleration, v = 33 m/s

When the car slows down uniformly, and comes to a stop (final velocity, v₂ = 0 m/s), it has a constant negative acceleration, (deceleration) '-a₂'

The given parameters when the car slows down  are;

The deceleration, -a₂ = 2.4 m/s²

The final velocity, v₂ = 0 m/s

The initial velocity, u₂ = v = 33 m/s

The time it takes to stop = t₂

-a₂ = 2.4 m/s²

∴ a₂ = -2.4 m/s²

From, v = u + a·t, we have;

v₂ = v + a₂·t₂

By plugging in the values of the variables, we have;

0 m/s = 33 m/s + (-2.4 m/s²) × t₂

∴ 2.4 m/s² × t₂ = 33 m/s

t₂ = 33 m/s/(2.4 m/s²) = 13.75 s

The time it takes to stop, t₂ = 13.75 seconds

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Dmitriy789 [7]

Its 1.6 kg

Remember

E = 1/2mv^2 is the kinetic energy of the ball. The energy is given to you so you need to solve for mass. Rearranging the equation gives you 2E/v^2 = 1.6

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Match each word to its definition. (Picture should be included)
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Answer:

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Why do the lighter isotopes disappear first from the atmosphere? Where do those isotopes go?
denpristay [2]

Lighter molecules move fast and escape from the upper atmosphere relatively quickly.

To find the answer, we have to know more about the lighter isotopes.

<h3>What are lighter isotopes?</h3>
  • Lighter molecules are mobile and soon leave the higher atmosphere.
  • A particular element's stable isotopes have slightly different atomic masses and quantum mechanical energies.
  • The lighter isotope of an element's chemical bonds are more easily broken than the heavier isotope's.
  • As a result, the light isotope typically benefits from chemical reactions.

Thus, we can conclude that, lighter molecules move fast and escape from the upper atmosphere relatively quickly.

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1 year ago
A group of hikers hear an echo 3.3 s after they shout. The temperature is 20◦C.
romanna [79]

Answer:

560 m

Explanation:

The speed of sound in air is approximately:

v ≈ v₀ + 0.6T

where v₀ is the speed of sound at 0°C (273 K) in m/s, and T is the temperature in Celsius.

The speed of sound at 20°C at that altitude is:

v ≈ 327 + 0.6(20)

v ≈ 339 m/s

The sound travels from the hikers to the mountain and back again, so it travels twice the distance.

339 m/s = 2d / 3.3 s

2d = 1118.7 m

d = 559.35 m

Rounding, the mountain is approximately 560 m away.

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3 years ago
A 1.5 kg orange falls from a tree and hits the ground in 0.75s. What is the speed of the orange just before it hits the ground?
Olenka [21]

The final speed of the orange is 7.35 m/s

Explanation:

The motion of the orange is a free fall motion, since there is only the force of gravity acting on it. Therefore, it is a uniformly accelerated motion with constant acceleration g=9.8 m/s^2 towards the ground. So we can use the following suvat equation:

v=u+at

where

v is the  final velocity

u is the initial velocity

a is the acceleration

t is the time elapsed

For the orange in this problem, we have

u = 0 (it is dropped from rest)

a=g=9.8 m/s^2 is the acceleration

Substituting t = 0.75 s, we find the final velocity (and speed) of the orange:

v=0+(9.8)(0.75)=7.35 m/s

Learn more about free fall:

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8 0
2 years ago
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