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Aleksandr [31]
2 years ago
14

During a car accident, a 125 kg driver is moving at 31 m/s and in 1.5 seconds is brought to rest by an inflating air bag. What i

s the magnitude of the change in momentum of the driver?
Physics
2 answers:
MrMuchimi2 years ago
8 0
Momentum = (mass) (speed)

The driver's initial momentum is (125 kg) (31 m/s) = 3,875 kg-m/s

At rest, his momentum is zero.

The change in his momentum is  (0 - 3,875 kg-m/s)  =  - 3,875 kg-m/s .

The average force exerted on him by the airbag is

                          (3,875 kg-m/s) / (1.5 s)

                     =        2,583.333... kg-m/s²

                     =         2,583.333... newtons   (about about 581 pounds)

That's what it takes to stop a 125 kg mass moving at 31m/s in 1.5sec.

The negative acceleration is    31/1.5 m/s² = about

                                                   =  20.333... m/s²  =  about  2.1 'G's.

If he were not wearing his seat belt and the airbag did not deploy,
then he would not be brought to rest ... at least not until he hit the
windshield at 31 m/s ... about  69.3 mph .








              
Vitek1552 [10]2 years ago
5 0

Answer: D

Explanation:

3900kg•m/s

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Answer:

Explanation:

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2 years ago
The electric field at a point 2.8 cm from a small object points toward the object with a strength of 180,000 N/C. What is the ob
givi [52]

Answer:

Charge, Q=1.56\times 10^{-8}\ C

Explanation:

It is given that,

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Distance from a small object, r = 2.8 cm = 0.028 m

Electric field at a point is given by :

E=\dfrac{kQ}{r^2}

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Q=\dfrac{180000\ N/C\times (0.028\ m)^2}{9\times 10^9\ Nm^2/C^2}

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A bee wants to fly to a flower located due North of the hive on a windy day. The wind blows from East to West at speed 6.68 m/s.
Aleonysh [2.5K]

Answer:  53.31\° East of North

Explanation:

We have the following data:

Speed of the wind from East to West: 6.68 m/s

Speed of the bee relative to the air:  8.33 m/s

If we graph these speeds (which in fact are velocities because are vectors) in a vector diagram, we will have a right triangle in which the airspeed of the bee (its speed relative to te air) is the hypotense and the two sides of the triangle will be the <u>Speed of the wind from East to West</u> (in the horintal part) and the <u>speed due North relative to the ground</u> (in the vertical part).

Now, we need to find the direction the bee should fly directly to the flower (due North):

sin \theta=\frac{Windspeed-from-East-to-West}{Speed-bee-relative-to-air}

sin \theta=\frac{6.68 m/s}{8.33 m/s}

Clearing \theta:

\theta=sin^{-1} (\frac{6.68 m/s}{8.33 m/s})

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What is the threshold frequency for sodium metal if a photon with frequency 6.66 × 1014 s−1 ejects a photon with 7.74 × 10−20 J
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Answer:

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Other actions like heating the material,  placing the material in a magnetic field of opposite polarity and hitting the material would lead to demagnetization of the magnetic material.

8 0
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