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IRINA_888 [86]
3 years ago
6

Who was the first person to walk on the moon ?

Physics
2 answers:
liq [111]3 years ago
8 0

Answer:

god correct correct

Explanation:

inna [77]3 years ago
8 0

Answer:

Neil Armstrong duh

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Traumatic brain injury such as a concussion results when the head undergoes a very large acceleration. Generally an acceleration
eimsori [14]

The complete text of the problem is:

<em>"Traumatic brain injury such as concussion results when the head undergoes a very large acceleration. Generally, an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1000 m/s2 lasting for at least 1 ms will cause injury. Suppose a small child rolls off a bed that is 0.43 m above the floor. If the floor is hardwood, the child's head is brought to rest in approximately 1.8 mm. If the floor is carpeted, this stopping distance is increased to about 1.1 cm. Calculate the magnitude and duration of the deceleration in both cases, to determine the risk of injury. Assume the child remains horizontal during the fall to the floor. Note that a more complicated fall could result in a head velocity greater or less than the speed you calculate. "</em>

<em />

<u>Solution:</u>

1) Acceleration: -2336 m/s^2 on the hardwood floor, -382 m/s^2 on the carpeted floor

First of all, we need to calculate the speed of the child just before he hits the floor. This can be done by using the equation

v^2 - u^2 = 2ad

where

v is the final speed

u = 0 is the initial speed (the child starts from rest)

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

d = 0.43 m is the distance covered by the child as he falls from the bed

Solving for v,

v=\sqrt{2ad}=\sqrt{2(9.8)(0.43)}=2.9 m/s

Now we can analyze the moment of the collision. The child hits the floor with an initial speed of v = 2.9 m/s, and he comes to a stop, so the final speed is v' = 0. If the floor is hardwood, the stopping distance is

d = 1.8 mm = 0.0018 m

So we can find the acceleration by using again the equation

v'^2 - v^2 = 2ad

Solving for a,

a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.0018)}=-2336 m/s^2

For the carpeted floor instead,

d=1.1 cm = 0.011 m

therefore the acceleration is

a=\frac{v'^2 - v^2}{2d}=\frac{0-2.9^2}{2(0.011)}=-382 m/s^2

2) Duration: 1.24 ms for the hardwood floor, 7.59 ms for the carpeted floor

We can find the duration of the collision in both cases by using the equation of the acceleration

a=\frac{v'-v}{t}

where

v' = 0

v = 2.9 m/s

For the hardwood floor,

a=-2336 m/s^2

So the duration of the collision is

t = \frac{v'-v}{a}=\frac{0-2.9}{-2336}=0.00124 s = 1.24 ms

For the carpeted floor,

a=-382 m/s^2

So the duration of the collision is

t = \frac{v'-v}{a}=\frac{0-2.9}{-382}=0.00759 s = 7.59 ms

We can now comment the results using the initial statement of the problem:

"Generally an acceleration less than 800 m/s2 lasting for any length of time will not cause injury, whereas an acceleration greater than 1,000 m/s2 lasting for at least 1ms will cause injury"

Therefore, the fall on the hardwood floor can result in injury (since the acceleration is greater than 1,000 m/s2 for more than 1 ms), while the fall on the carpeted floor is not dangerous (much less than 1000 m/s^2).

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microwave ovens rotate at a rate of about 6.3 rev/min. what is this in revolutions per second? (you do not need to enter any uni
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The microwave ovens rotate at a rate of about 0.105 rev/s.

The microwave rotation is the number of revolutions in a unit of time. To change the unit for angular velocity, assume that the quantity is multiplied by the unit it has. Then change to the desired units. The angular velocity is denoted by ω and has a magnitude of 6.3 rev/min.

ω = 6.3 rev/min

\omega = 6.3 \times \frac{1 \: rev}{1 \: min}

  • 1 minute = 60 seconds
  • The revolution unit didn't change

\omega = 6.3 \times \frac{1 \: rev}{60 \: seconds}

\omega = \frac{6.3 \: rev}{60 \: seconds}

ω = 0.105 rev/s

Learn more about Angular velocity here: brainly.com/question/29344944

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The number of times a machine multiplies the effort force is the
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