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Anna35 [415]
4 years ago
14

In a crash test the strapped-in, 75-kg dummy moves a distance of 0.80 m when the test car is slammed straight into a wall at 11.

2 m/s (~25 mph). The average force acting on the dummy during the collision is how many times his own weight?
Physics
1 answer:
Brums [2.3K]4 years ago
3 0

Explanation:

It is given that,

Mass of the dummy, m = 75 kg

Distance, d = 0.8 m

Initial speed of the dummy, u = 0

Final speed of the dummy, v = 11.2 m/s

Firstly finding the acceleration of the test car. Using third equation of motion to find it as :

v^2-u^2=2ad

v^2=2ad

a=\dfrac{v^2}{2d}

a=\dfrac{(11.2)^2}{2\times 0.8}

a=78.4\ m/s^2

Let F is the average force. It is given by the product of mass and acceleration. It is given by :

F=ma

F=75\times 78.4

F = 5880 N

Taking ratios,

\dfrac{F}{W}=\dfrac{ma}{mg}

\dfrac{F}{W}=\dfrac{a}{g}

\dfrac{F}{W}=\dfrac{78.4}{9.8}

\dfrac{F}{W}=8

F=8W

The average force acting on the dummy during the collision is 8 times of his weight. Hence, this is the required solution.

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denpristay [2]
  • Initial velocity=u=15m/s
  • Final velocity=v=10m/s
  • Time=0.08s

\\ \sf\longmapsto Acceleration=\dfrac{v-u}{t}

\\ \sf\longmapsto Acceleration=\dfrac{10-15}{0.08}

\\ \sf\longmapsto Acceleration=\dfrac{-5}{0.08}

\\ \sf\longmapsto Acceleration=-62.5m/s^2

5 0
3 years ago
How do you calculate the volume and density of something that weighs 20 grams?
Nostrana [21]
Mass divided by volume is density, while mass times density is volume. you cannot calculate density without volume and you cannot calculate volume without density.
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6 0
4 years ago
An auto race takes place on a circular track. A car completes one lap in a time of 25.0 s, with an average tangential speed of 5
ludmilkaskok [199]

Answer:

(a) Angular speed will be 0.2512 rad/sec

(b) Radius will be 213.375 m

Explanation:

We have given time to complete 1 lap = 25 sec

We know that 1 lap = 2\pi radian

(a) So angular speed =\frac{2\pi }{25}=\frac{2\times 3.14}{25}=0.2512rad/sec

(B) Tangential velocity is given as v = 53.6 m/sec

We know that tangential velocity is given by

v=\omega r

So radius r=\frac{v}{\omega }=\frac{53.6}{0.2512}=213.375m

4 0
3 years ago
Viewed through a spectroscope, the spectral profile of a yellow street lamp has a narrow line in the yellow region of the visibl
muminat

Answer:

discrete lines are observed by the spectroscope, the emission of the lamp is of the ATOMIC source

Explanation:

Bulbs can emit light in several ways:

* When the emission is carried out by the heating of its filament, the bulb is called incandescent, in general its spectrum is similar to that of a black body, this is a continuous spectrum with a maximum dependent on the fourth power of the temperature of the filament.

* The emission can be by atomic transitions, in this case there is a discrete spectrum formed by the spectral lines of the material that forms the gas of the lamp, in general for the yellow emission the most used materials are mercury and sodium or a mixture of they.

Consequently, as discrete lines are observed by the spectroscope, the emission of the lamp is of the ATOMIC type

6 0
3 years ago
An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 2060 × 103 seconds (about
Fittoniya [83]

Answer:

Acceleration, a=2.22\times 10^{-3}\ m/s^2

Explanation:

It is given that,

Time period of revolution of the moon, T=2060\times 10^3\ s

If the distance from the center of the moon to the surface of the planet is, h=235\times 10^6\ m

The radius of the planet, r=3.9\times 10^6\ m

Let a is the moon's radial acceleration. Mathematically, it is given by :

a=R\times \omega^2, R is the radius of orbit

Since, \omega=\dfrac{2\pi}{T}

The radius of orbit is,

R=r+h

R=3.9\times 10^6\ m+235\times 10^6\ m=238900000\ m

So, a=\dfrac{4\pi^2 R}{T^2}

a=\dfrac{4\pi^2 \times 238900000}{(2060\times 10^3)^2}

a=2.22\times 10^{-3}\ m/s^2

Hence, this is the required solution for the radial acceleration of the moon.

5 0
3 years ago
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