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3 years ago

How is Newton’s three laws of motion are used in testing the safety of our automobiles.

Physics

1 answer:

Tanya [424]3 years ago

8 0

Answer

Newton's third law is used for the impact test on the vehicle.

In the frontal collision of the vehicle, the impact of the vehicle is made against the concrete wall then the impact on the passengers and driver is calculated.

In the impact test time of the opening of the airbag is also calculated.

Similar tests also take place when the impact is sideways.

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In a lever, the effort arm is two times as long as the load arm. The resultant force will be

jasenka [17]

Answer is B.

In a lever, the effort arm is 2 times as a long as the load arm. The resultant force will be twice the applied force.

Hope it helped you.

-Charlie

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3 years ago

Read 2 more answers

If a 400-mm diameter pipe with a pipe roughness coefficient of 100 flows full of pressurized water with a head loss of 0.4 ft pe

RoseWind [281]

Answer:

Q = 913.9 gpm

Explanation:

The Hazen Williams equation can be written as follows:

$P = \frac{4.52\ Q^{1.85}}{C^{1.85}d^{4.87}}$

where,

P = Friction Loss per foot of pipe = $\frac{0.4}{1000\ ft}$ = 4 x 10⁻⁴

Q = Flow Rate in gallon/min (gpm) = ?

d = pipe diameter in inches = (400 mm)(0.0393701 in/1 mm) = 15.75 in

C = roughness coefficient = 100

Therefore,

$4\ x \ 10^{-4} = \frac{4.52\ Q^{1.85}}{(100)^{1.85}(15.75)^{4.87}}\\\\Q^{1.85} = \frac{4\ x \ 10^{-4}}{1.33\ x\ 10^{-9}} \\\\Q = (300384.75)^\frac{1}{1.85}$

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3 years ago

A particle with a mass of 0.500 kg is attached to a horizontal spring with a force constant of 50.0 N/m. At the moment t = 0, th

svp [43]

a) $x(t)=2.0 sin (10 t) [m]$

The equation which gives the position of a simple harmonic oscillator is:

$x(t)= A sin (\omega t)$

where

A is the amplitude

$\omega=\sqrt{\frac{k}{m}}$ is the angular frequency, with k being the spring constant and m the mass

t is the time

Let's start by calculating the angular frequency:

$\omega=\sqrt{\frac{k}{m}}=\sqrt{\frac{50.0 N/m}{0.500 kg}}=10 rad/s$

The amplitude, A, can be found from the maximum velocity of the spring:

$v_{max}=\omega A\\A=\frac{v_{max}}{\omega}=\frac{20.0 m/s}{10 rad/s}=2 m$

So, the equation of motion is

$x(t)= 2.0 sin (10 t) [m]$

b) t=0.10 s, t=0.52 s

The potential energy is given by:

$U(x)=\frac{1}{2}kx^2$

While the kinetic energy is given by:

$K=\frac{1}{2}mv^2$

The velocity as a function of time t is:

$v(t)=v_{max} cos(\omega t)$

The problem asks as the time t at which U=3K, so we have:

$\frac{1}{2}kx^2 = \frac{3}{2}mv^2\\kx^2 = 3mv^2\\k (A sin (\omega t))^2 = 3m (\omega A cos(\omega t))^2\\(tan(\omega t))^2=\frac{3m\omega^2}{k}$

However, $\frac{m}{k}=\frac{1}{\omega^2}$ , so we have

$(tan(\omega t))^2=\frac{3\omega^2}{\omega^2}=3\\tan(\omega t)=\pm \sqrt{3}\\$

with two solutions:

$\omega t= \frac{\pi}{3}\\t=\frac{\pi}{3\omega}=\frac{\pi}{3(10 rad/s)}=0.10 s$

$\omega t= \frac{5\pi}{3}\\t=\frac{5\pi}{3\omega}=\frac{5\pi}{3(10 rad/s)}=0.52 s$

c) 3 seconds.

When x=0, the equation of motion is:

$0=A sin (\omega t)$

so, t=0.

When x=1.00 m, the equation of motion is:

$1=A sin(\omega t)\\sin(\omega t)=\frac{1}{A}=\frac{1}{2}\\\omega t= 30\\t=\frac{30}{\omega}=\frac{30}{10 rad/s}=3 s$

So, the time needed is 3 seconds.

d) 0.097 m

The period of the oscillator in this problem is:

$T=\frac{2\pi}{\omega}=\frac{2\pi}{10 rad/s}=0.628 s$

The period of a pendulum is:

$T=2 \pi \sqrt{\frac{L}{g}}$

where L is the length of the pendulum. By using T=0.628 s, we find

$L=\frac{T^2g}{(2\pi)^2}=\frac{(0.628 s)^2(9.8 m/s^2)}{(2\pi)^2}=0.097 m$

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3 years ago

Imagine you are on the top of a mountain with four basketballs. The mass of the four basketballs are 100 kilograms, 200 kilogram

pentagon [3]

Answer:

400 kilogram

Explanation:

Force exerted is directly proportional to the mass of an object.

F=ma where m is mass and a is acceleration. Taking uniform acceleration on all objects then the larger the mass the higher the force and vice versa. Theredore, among the masses given, 400 kilograms is the largest hence it exerts the largest force at the bottom

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3 years ago

Static cling is an example of

cestrela7 [59]

I would say c hope it helps:)

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4 years ago