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sukhopar [10]
3 years ago
5

A steel ball bearing is released from a height H and

Physics
1 answer:
ArbitrLikvidat [17]3 years ago
4 0

Answer:

ELASTIC collision

kinetic energy is conservate

Explanation:

As the ball bounces to the same height, it can be stated that the impact with the floor is ELASTIC.

As the floor does not move the conservation of the moment

            po = pf

            -mv1 = m v2

- v1 = v2

So the speed with which it descends is equal to the speed with which it rises

Therefore the kinetic energy of the ball before and after the collision is the same

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6 A test of a driver's perception/reaction time is being conducted on a special testing track with level, wet pavement and a dri
mylen [45]

Answer:

a. 10.5 s b. 6.6 s

Explanation:

a. The driver's perception/reaction time before drinking.

To find the driver's perception time before drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (0 m/s)² - (22.35 m/s)²/2(117.35 m)

a =  - 499.52 m²/s²/234.7 m

a = -2.13 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 0 m/s (since he stops), a = deceleration of driver = -2.13 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (0 m/s - 22.35 m/s)/-2.13 m/s²

t = - 22.35 m/s/-2.13 m/s²

t = 10.5 s

b. The driver's perception/reaction time after drinking.

To find the driver's perception time after drinking, we first find his deceleration from

v² = u² + 2as where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver and s = distance moved by driver = 385 ft = 385 × 0.3048 m = 117.35 m

So, a = v² - u²/2s

substituting the values of the variables into the equation, we have

a = v² - u²/2s

a = (13.41 m/s)² - (22.35 m/s)²/2(117.35 m)

a = 179.83 m²/s² - 499.52 m²/s²/234.7 m

a = -319.69 m²/s² ÷ 234.7 m

a = -1.36 m/s²

Using a = (v - u)/t where u = initial speed of driver = 50 mi/h = 50 × 1609 m/3600 s = 22.35 m/s, v = final speed of driver = 30 mi/h = 30 × 1609 m/3600 s = 13.41 m/s, a = deceleration of driver = -1.36 m/s² and t = reaction time

So, t = (v - u)/a

Substituting the values of the variables into the equation, we have

t = (13.41 m/s - 22.35 m/s)/-1.36 m/s²

t = - 8.94 m/s/-1.36 m/s²

t = 6.6 s

4 0
3 years ago
THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THANKS!!! THE RIGHT ANSWER WILL RECEIVE A BRAINLESS AND POINTS AND THAN
timurjin [86]

Answer:

f_{o} = 391.67 Hz

Explanation:

The sound of lowest frequency which is produced by a vibrating sting is called its fundamental frequency (f_{o}).

The For a vibrating string, the fundamental frequency (f_{o}) can be determined by:

f_{o} = \frac{v}{2L}

Where v is the speed of waves of the string, and L is the length of the string.

L = 42.0 cm = 0.42 m

v = 329 m/s

f_{o} = \frac{329}{2*0.42}

   = \frac{329}{0.84}

f_{o} = 391.6667 Hz

The fundamental frequency of the string is 391.67 Hz.

3 0
2 years ago
I NEED THE RIGHT ANSWER ASAP NO LINKS !!!<br> This is a Science question
sdas [7]
3rd number line -5- -4 is -9
also it’s math not physics or science
8 0
2 years ago
Read 2 more answers
Why do outer planets take so long to orbit the sun
Hoochie [10]

Answer:

Because the gravitational pull is weaker, And the outer planets are further away so they have more distance to cover in orbit.

5 0
2 years ago
Read 2 more answers
Which statement explains why a short eyeball causes farsighted
snow_lady [41]
I think the correct answer would be the rays of light focus behind the retina. It is a condition called farsightedness or hyperopia.  Due to his, objects that are close to the eyes seems to be blurry and as it worsens everything would become blurry to the eye.
8 0
3 years ago
Read 2 more answers
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