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yanalaym [24]
3 years ago
9

During an auto accident, the vehicle's air bags deploy and slow down the passengers more gently than if they had hit the windshi

eld or steering wheel. According to safety standards, the bags produce a maximum acceleration of 60 g, but lasting for only 36 ms (or less).
How far (in meters) does a person travel in coming to a complete stop in 36 ms at a constant acceleration of 60g?
Physics
1 answer:
vladimir2022 [97]3 years ago
7 0

Answer:

At a deceleration of 60g, or 60 times the acceleration due to gravity a person will travel a distance of 0.38 m before coing to a complete stop

Explanation:

The maximum acceleration of the airbag = 60 g, and the duration of the acceleration = 36 ms or 36/1000 s or 0.036 s

To find out how far (in meters) does a person travel in coming to a complete stop in 36 ms at a constant acceleration of 60g

we write out the equation of motion thus.

S = ut + 0.5at²

wgere

S = distance to come to complete stop

u = final velocoty = 0 m/s

a = acceleration = 60g = 60 × 9.81

t = time = 36 ms

as can be seen, the above equation calls up the given variable as a function of the required variable thus

S = 0×0.036 + 0.5×60×9.81×0.036² = 0.38 m

At 60g, a person will travel a distance of 0.38 m before coing to a complete stop

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SVEN [57.7K]

You're talking about a grain of sand or a stone or a rock that's drifting in space, and then the Earth happens to get in the way, so the stone falls down to Earth, and it makes a bright streak of light while it's falling through the atmosphere and burning up from the friction.

-- While it's drifting in space, it's a <em>meteoroid</em>.

-- While it's falling through the atmosphere burning up and making a bright streak of light, it's a <em>meteor</em>.

-- If it doesn't completely burn up and there's some of it left to fall on the ground, then the leftover piece on the ground is a <em>meteorite</em>.

4 0
3 years ago
Suppose a certain car supplies a constant deceleration of A meter per second per second. If it is traveling at 90km/hr. When. th
aksik [14]

Answer:

i)-6.25m/s

ii)18 metres

iii)26.5 m/s or 95.4 km/hr

Explanation:

Firstly convert 90km/hr to m/s

90 × 1000/3600 = 25m/s

(i) Apply v^2 = u^2 + 2As...where v(0m/s) is the final speed and u(25m/s) is initial speed and also s is the distance moved through(50 metres)

0 = (25)^2 + 2A(50)

0 = 625 + 100A....then moved the other value to one

-625 = 100A

Hence A = -6.25m/s^2(where the negative just tells us that its deceleration)

(ii) Firstly convert 54km/hr to m/s

In which this is 54 × 1000/3600 = 15m/s

then apply the same formula as that in (i)

0 = (15)^2 + 2(-6.25)s

-225 = -12.5s

Hence the stopping distance = 18metres

(iii) Apply the same formula and always remember that the deceleration values is the same throughout this question

0 = u^2 + 2(-6.25)(56)

u^2 = 700

Hence the speed that the car was travelling at is the,square root of 700 = 26.5m/s

In km/hr....26.5 × 3600/1000 = 95.4 km/hr

3 0
2 years ago
A wave of wavelength 0.3 m travels 900 m in 3.0 s. Calculate its frequency.
zzz [600]

Answer:

1000 Hz

Explanation:

<em>The frequency would be 1000 Hz.</em>

The frequency, wavelength, and speed of a wave are related by the equation:

<em>v = fλ ..................(1)</em>

where v = speed of the wave, f = frequency of the wave, and λ = wavelength of the wave.

Making f the subject of the formula:

<em>f = v/λ.........................(2)</em>

Also, speed (v) = distance/time.

From the question, distance = 900 m, time = 3.0 s

Hence, v = 900/3.0 = 300 m/s

Substitute v = 300 and λ = 0.3  into equation (2):

f = 300/0.3 = 1000 Hz

6 0
3 years ago
How do you calculate the mass of an object accelerating at22.35m/s2 with a force of 120N
Arte-miy333 [17]

Answer:

The mass of object is calculated as 5.36 kg

Explanation:

The known terms to find the mass are:

           acceleration of object (a) = 22.35 m/s^{2}

                        Force exerted (F) = 120N

                        mass of an object (m) = ?

From Newton's second law of motion;

                                   F = ma

                           or, 120 = m × 22.35

                          or, m= \frac{120}{22.35} kg

                           ∴ m = 5.36 kg

3 0
2 years ago
A positively charged metal sphere, A, is held close to but not touching and identical uncharged sphere, sphere B. Sphere A is no
Yuri [45]

Answer:

The sphere C carries no net charge.

Explanation:

  • When brougth close to the charged sphere A, as charges can move freely in  a conductor, a charge equal and opposite to the one on the sphere A, appears on the sphere B surface facing to the sphere A.
  • As sphere B must remain neutral (due to the principle of conservation of charge) an equal charge, but of opposite sign, goes to the surface also, on the opposite part of the sphere.
  • If sphere A is removed, a charge movement happens in the sphere B, in such a way, that no net charge remains on the surface.
  • If in such state, if  the sphere B (assumed again uncharged completely, without any local charges on the surface), is touched by an initially uncharged sphere C, due to the conservation of  charge principle, no net  charge can be built on sphere C.
3 0
3 years ago
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