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Rus_ich [418]
3 years ago
13

In a typical rear-end collision, the victim's ________ is/are accelerated faster and harder than the torso

Physics
2 answers:
Georgia [21]3 years ago
8 0

The victim's head is accelerated faster and harder than the torso when the victom is involved in a typical rear-end collision.

The traffic accident where a vehicle crashes into another vehicle that is directly in front of it is called a rear-end collision.

 

One of the most common accident in the United States is the rear-end collision, and in a lot of cases, rear-end collisions are prompted by drivers who are inattentive, unfavorable conditions of the road, and poor following distance.

 

<span>An enough room in front of your car so you can stop when the car in front of you stops suddenly is one basic driving rule. The person isn’t driving safely if he / she is behind you and couldn’t stop.</span>

Makovka662 [10]3 years ago
5 0

Answer:

In a typical rear-end collision, the victim's head is accelerated faster and harder than the torso

Explanation:

In a collision the body part most at risk of injury is the head. This is because the victim's torso is secured with the seat belt, so at the moment of collision the torso is held by the belt and prevented from reaching a speed compatible with the collision. However, there is no device that holds the victim's head, so at the moment of collision, the victim's head will be accelerated faster and stronger than the torso.

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When a piano tuner strikes both the A on the piano and a 440 Hz tuning fork, he hears a beat every 2 seconds. The frequency of t
AysviL [449]

Answer:

The  frequency is  f = 439.5 \  Hz        

Explanation:

From the question we are told that

   The frequency of the  tuning fork is  f_t  =  440 \ Hz

   The beat period is  T  =  2 \  s

Generally the beat frequency is mathematically represented as

       f_b  =  \frac{1}{T}

       f_b  =  \frac{1}{2}

      f_b  = 0.5 \  Hz

The  beat frequency is also represented mathematically as

     f_b  =  f_t  -  f

Where  f is the frequency of the piano

 So

       f =  440 -  0.5  

       f = 439.5 \  Hz        

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Determine the magnitude of the resultant force acting on a 1.5 −kg particle at the instant t=2 s, if the particle is moving alon
Phoenix [80]

Answer:

F = 63N

Explanation:

M= 1.5kg , t= 2s, r = (2t + 10)m and

Θ = (1.5t² - 6t).

magnitude of the resultant force acting on 1.5kg = ?

Force acting on the mass =

∑Fr =MAr

Fr = m(∇r² - rθ²) ..........equation (i)

∑Fθ = MAθ = M(d²θ/dr + 2dθ/dr) ......... equation (ii)

The horizontal path is defined as

r = (2t + 10)

dr/dt = 2, d²r/dt² = 0

Angle Θ is defined by

θ = (1.5t² - 6t)

dθ/dt = 3t, d²θ/dt² = 3

at t = 2

r = (2t + 10) = (2*(2) +10) = 14

but dr/dt = 2m/s and d²r/dt² = 0m/s

θ = (1.5(2)² - 6(2) ) = -6rads

dθ/dt =3(2) - 6 = 0rads

d²θ/dt = 3rad/s²

substituting equation i into equation ii,

Fr = M(d²r/dt² + rdθ/dt) = 1.5 (0-0)

∑F = m[rd²θ/dt² + 2dr/dt * dθ/dt]

∑F = 1.5(14*3+0) = 63N

F = √(Fr² +FΘ²) = √(0² + 63²) = 63N

7 0
3 years ago
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