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katovenus [111]
2 years ago
9

During a car crash, why are airbags useful? 0.They decrease the amount of impulse applied to the you.

Physics
2 answers:
Vera_Pavlovna [14]2 years ago
3 0

Answer:

They increase the time it takes to slow you down, which decreases the force applied to you.

Explanation:

Hope it helps :)))

Bumek [7]2 years ago
3 0

Answer:

They increase the time it takes to slow you down, which decreases the force applied to you.

Explanation:

Air bags can only deploy at around 8-14mph

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A protostar reaches equilibrium when gravity is balanced by the outward force of nuclear fusion. 
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I think the Bulb No. 2 will stop emitting light if the bulb No. 1 burns out.

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Problem: A barbell consists of two small balls, each with mass m at the ends of a very low mass rod of length d. The barbell is
zepelin [54]

Answer:

mass of ball 1=m1

mass of ball 2=m2

velocity of ball=r1w1

velocity of ball 2=r2w2

Total angular momentum=m1*v1+m2*v2

but

v1=r1*w1

v2=r2*w2

Substitute values in above equation

Total angular momentum of the system=m1*r1*w1+m2*r2*w2

7 0
3 years ago
Energy is conventionally measured in Calories as well as in joules. One Calorie in nutrition is one kilocalorie, defined as 1 kc
Sergeeva-Olga [200]

Answer:

a) The student must run flight of stairs to lose 1.00 kg of fat 709.5 times.

b) Average power

P(w)= 1062.07 [w]

P(hp)=1.42 [hp]

c) This activity is highly unpractical, because the high amount of repetitions he has to due in order to lose, just 1 Kg of fat.

Explanation:

First, lets consider the required amount of work to move the mass of the student. (considering running stairs just as a vertical movement)

Work:

W= F*d= m*g*d

Where m is the mass of the student, g is gravity (9.8 m/s) and d is the total distance going up the stairs (0.15m *85steps= 12.75m )

W= F*d= m*g*d=85* 9.8*12.75=10620.75 [J]

Converting from Joules to Kcals:

\frac{10620.75}{4186} =2.537 Kcal

Now lets take into account the efficiency of the human body (20%)

2.537 ---> 20%

 x       ---> 100%

x=\frac{2.537*100}{20} =12.685

So the student is consuming 12.685 KCals each time he runs up the stairs.

Now,

1 g --> 9 Kcals

1000 g --> 9000KCals

Burning 1 g of fat, requieres 9 KCals, 1000g burns 9000KCals. So in order to burn a 1Kg of fat:

\frac{9000Kcals}{12.685Kcals} =709.5 times

He must run up the stairs 709.5 times, to burn 1 Kg of fat.

********************

For b) just converting units, taking into account the time lapse. (53103.75 is the 100% of the energy in joules, from converting 12.685Kcals to joules)

Power=\frac{Joules}{Seconds} =\frac{53103.75}{50} =1062.075 [W]\\

P(hp)=\frac{P(w)}{745.7} =\frac{1062.075}{745.7} =1.42[hp]

*****

4 0
3 years ago
Velocity: You leave on a 549-mi trip in order to attend a meeting that will start 10.8 h after you begin your trip. Along the wa
rewona [7]

Answer:

1.55 h

Explanation:

Let the time to spend over dinner be t.

Since I need to spend 10.8 h for the whole trip, then I have 10.8 - t hours left to drive.

Speed = distance/time

v = \dfrac{d}{10.8-t}

d = v(10.8-t)

The distance is 549 and maximum speed is 65 mi/h.

549 = 65(10.8 - t)

10.8-t = 8.45

t = 10.8 - 8.45 = 1.55 \text{ h}

5 0
3 years ago
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