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3 years ago

In a car, how does an air bag minimize the force acting on a person during a collision?

Physics

2 answers:

Olenka [21]3 years ago

7 0

Answer:

C: It increases the time it takes for the person to stop.

Explanation:

on edge! hope this helps!!~ o(〃＾▽＾〃)o

Vlad [161]3 years ago

6 0

Answer:

It increases the time it takes for the person to stop.

Explanation:

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Mama L [17]

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3 years ago

Where is most of Earth’s freshwater found?

STALIN [3.7K]

In lakes and streams

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3 years ago

What is the magnitude of your displacement when you follow directions that tell you to walk 225 m in one direction, make a 90° t

Gekata [30.6K]

Start by facing East. Your first displacement is the vector

d₁ = (225 m) i

Turning 90º to the left makes you face North, and walking 350 m in this direction gives the second displacement,

d₂ = (350 m) j

Turning 30º to the right would have you making an angle of 60º North of East, so that walking 125 m gives the third displacement,

d₃ = (125 m) (cos(60º) i + sin(60º) j )

d₃ ≈ (62.5 m) i + (108.25 m) j

The net displacement is

d = d₁ + d₂ + d₃

d ≈ (287.5 m) i + (458.25 m) j

and its magnitude is

|| d || = √[ (287.5 m)² + (458.25 m)² ] ≈ 540.973 m ≈ 541 m

7 0

3 years ago

A particle with charge − 2.74 × 10 − 6 C −2.74×10−6 C is released at rest in a region of constant, uniform electric field. Assum

s2008m [1.1K]

Answer:

241.7 s

Explanation:

We are given that

Charge of particle= $q=-2.74\times 10^{-6} C$

Kinetic energy of particle= $K_E=6.65\times 10^{-10} J$

Initial time= $t_1=6.36 s$

Final potential difference= $V_2=0.351 V$

We have to find the time t after that the particle is released and traveled through a potential difference 0.351 V.

We know that

$qV=K.E$

Using the formula

$2.74\times 10^{-6}V_1=6.65\times 10^{-10} J$

$V_1=\frac{6.65\times 10^{-10}}{2.74\times 10^{-6}}=2.43\times 10^{-4} V$

Initial voltage= $V_1=2.43\times 10^{-4} V$

$\frac{\initial\;voltage}{final\;voltage}=(\frac{initial\;time}{final\;time})^2$

Using the formula

$\frac{V_1}{V_2}=(\frac{6.36}{t})^2$

$\frac{2.43\times 10^{-4}}{0.351}=\frac{(6.36)^2}{t^2}$

$t^2=\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}$

$t=\sqrt{\frac{(6.36)^2\times 0.351}{2.43\times 10^{-4}}}$

$t=241.7 s$

Hence, after 241.7 s the particle is released has it traveled through a potential difference of 0.351 V.

6 0

3 years ago

a block (mass 0.6kg) is released from rest at point A at the top of a ramp inclined at 36.9 degress above the horizontal. The bl

bija089 [108]

14.136 J as shown on the photo with two thought processes but overall same calculation

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3 years ago