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2 years ago

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During a collision, a 63.9-kilogram driver of a car moving at 24 m/s is brought to rest by an inflating airbag in 1.2 seconds. W

hat is the magnitude of the force exerted on the driver by the airbag?

1 answer:

kaheart [24]2 years ago

7 0

Hi there!

We can use impulse for this situation:

I = Δp = mΔv

Impulse = Force × time, so:

I = 63.9(24) = 1533.6 Ns

Find force by dividing by time:

I/t = 1533.6/1.2 = 1278 N

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Consider the f(x) = cos(x-C) function shown in the figure in blue color, where 0 ≤ C ≤ 2π. What is the value of parameter C for

klemol [59]

The value of parameter C for the function in the figure is 2.

<h3>What is amplitude of a wave?</h3>

The amplitude of a wave is the maximum displacement of the wave. It can also be described at the maximum upward displacement of a wave curve.

f(x) = Acos(x - C)

where;

A is amplitude of the wave
C is phase difference of the wave

<h3>What is angular frequency of a wave?</h3>

Angular frequency is the angular displacement of any element of the wave per unit time.

From the blue colored graph; at y = 1, x = -2 cm

1 = cos(2 - C)

(2 - C) = cos^(1)

(2 - C) = 0

C = 2

Thus, the value of parameter C for the function in the figure is 2.

Learn more about phase angle here: brainly.com/question/16222725

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2 years ago

An ideal spring is mounted horizontally, with its left end fixed. The force constant of the spring is 170 N/m. A glider of mass

gizmo_the_mogwai [7]

Answer:

Explanation:

First of all we shall find the velocity at equilibrium point of mass 1.2 kg .

It will be ω A , where ω is angular frequency and A is amplitude .

ω = √ ( k / m )

= √ (170 / 1.2 )

= 11.90 rad /s

amplitude A = .045 m

velocity at middle point ( maximum velocity ) = 11.9 x .045 m /s

= .5355 m /s

At middle point , no force acts so we can apply law of conservation of momentum

m₁ v₁ = ( m₁ + m₂ ) v

1.2 x .5355 = ( 1.2 + .48 ) x v

v = .3825 m /s

= 38.25 cm /s

Let new amplitude be A₁ .

1/2 m v² = 1/2 k A₁²

( 1.2 + .48 ) x v² = 170 x A₁²

( 1.2 + .48 ) x .3825² = 170 x A₁²

A₁ = .0379 m

New amplitude is .0379 m

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3 years ago

MULTIPLE CHOICE QUESTION

Nookie1986 [14]

Which object? More information is needed to answer this question

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3 years ago

Two metal disks, one with radius R1 = 2.45 cm and mass M1 = 0.900 kg and the other with radius R2 = 5.00 cm and mass M2 = 1.60 k

natima [27]

Answer:

part (a) $a_1\ =\ 2.9\ kg$

Part (b) $a_2\ =\ 6.25\ kg$

Explanation:

Given,

mass of the smaller disk = $M_1\ =\ 0.900\ kg$
Radius of the smaller disk = $R_1\ =\ 2.45\ cm\ =\ 0.0245\ m$

mass of the larger disk = $M_2\ =\ 1.6\ kg$
Radius of the larger disk = $R_2\ =\ 5.0\ cm\ =\ 0.05\ m$
mass of the hanging block = m = 1.60 kg

Let I be the moment of inertia of the both disk after the welding, $\therefore I\ =\ I_1\ +\ I_2\\\Rightarrow I\ =\ \dfrac{1}{2}(M_1R_1^2\ +\ M_2R_2^2)\\\Rightarrow I\ =\ 0.5\times (0.9\times 0.0245^2\ +\ 1.6\times 0.05^2)\\\Rightarrow I\ =\ 2.27\times 10^{-3}\ kgm^2$

part (a)

A block of mass m is hanging on the smaller disk,

From the f.b.d. of the block,

Let 'a' be the acceleration of the block and 'T' be the tension in the string.

$mg\ -\ T\ =\ mg\\\Rightarrow T\ =\ mg\ -\ ma\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,eqn (1)$

Net torque on the smaller disk,

$\therefore \tau\ =\ I\alpha\\\Rightarrow TR_1\ =\ \dfrac{Ia}{R_1}\\\Rightarrow T\ =\ \dfrac{Ia}{R_1^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,enq (2)$

From eqn (1) and (2), we get,

$mg\ -\ ma\ =\ \dfrac{Ia}{R_1^2}\\\Rightarrow a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.027^2}\ +\ 1.60}\\\Rightarrow a\ =\ 2.91\ m/s^2$

part (b)

In this case the mass is rapped on the larger disk,

From the above expression of the acceleration of the block, acceleration is only depended on the radius of the rotating disk,

Let ' $a_2$ ' be the acceleration of the block in the second case,

From the above expression,

$\therefore a\ =\ \dfrac{mg}{\dfrac{I}{R_1^2}\ +\ m}\\\Rightarrow a\ =\ \dfrac{1.60\times 9.81}{\dfrac{2.27\times 10^{-3}}{0.05^2}\ +\ 1.60}\\\Rightarrow a\ =\ 6.25\ m/s^2$

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3 years ago

Which of the following is true about acids?

Stells [14]

i would say, are sour and corrode metals

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3 years ago

Read 2 more answers