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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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An object of mass m attached to a spring of force constant k oscillates with simple harmonic motion. The maximum displacement fr
Lady_Fox [76]

An object of mass m attached to a spring of force constant k oscillates with simple harmonic motion. The system's potential energy when kinetic energy of (3/4) E is (1/8) k A².

<h3>What is mechanical energy?</h3>

Mechanical energy is the sum of potential energy and kinetic energy.

Total mechanical energy = P.E max = K.E max

Total mechanical energy = K.E +P.E

Given is the kinetic energy is (3/4)E.

E= (3/4)E + P.E

P.E =  (1/4) E

Maximum potential  energy =E = (1/2) k A²

Here. A is the maximum displacement and k is the spring constant.

The potential energy at kinetic energy of (3/4) E is

P.E = (1/4)E = (1/8) k A²

Therefore, the system's potential energy when kinetic energy of (3/4) E is (1/8) k A².

Learn more about mechanical energy.

brainly.com/question/20904200

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6 0
3 years ago
A train travels 600 km in one hour. What is the trains velocity in meters/second
eimsori [14]

v= s/t

600 ÷ 60

10 m/s

4 0
4 years ago
Read 2 more answers
Martha (50 kg) is attracted to Stewart (70 Kg) who sits 4 m away. What is the gravitational attraction between them? G=6.67 x 10
kodGreya [7K]

Answer:

1.46× 10^-8 N

Explanation:

F=Gm₁m₂/r²

F=(6.67×10⁻¹¹ × 50× 70)/(4)²

F=1.46× 10^-8 N

Hope, this helps you.

8 0
3 years ago
A 65 Kg roller-blade is accelerating at 5 m/s/s across the side walk. What force would be necessary for this acceleration to occ
Neporo4naja [7]
Newton's second law states that the force applied to an object is equal to the product between the mass m of the object and its acceleration a:
F=ma
Using m=65 kg and a=5 m/s^2, we can find the value of the force applied to the roller-blade to obtain this acceleration:
F=(65 kg)(5 m/s^2)=325 N
3 0
3 years ago
3. Suppose that an airplane flying 60 m/s, at a height of 300m, dropped a sack of flour (pere the effect
Arlecino [84]

469.24m. An airplane flying 60m/s at a height of 300m dropped a sack of flour that stack the ground 469.24m from the point of release.

This is a example of horizontal parabolic projectile motion,and we represents this motion in the coordinate axis, which means that the velocity has components in x axis and y axis.

The equation of components on the x axis.

v_{0}x=\frac{x}{t}, where x is the distance and Vox the initial velocity before the drop

The equation of components on the y axis.

y = v_{0}yt+\frac{gt^{2} }{2}, where y is the height, and the velocity in y component before the drop is 0, reducing the equation to y = \frac{gt^{2} }{2}

Clear t from both the equation of components on the x axis and the y axis:

t=\frac{x}{v_{0} x} and t=\sqrt{\frac{2h}{g}}

Equating both equations and clearing the distance x:

\frac{x}{v_{0} x}=\sqrt{\frac{2h}{g}}\\x={v_{0} x}\sqrt{\frac{2h}{g}}

Substituting the values:

x=60\frac{m}{s} \sqrt{\frac{2(300m)}{9.81\frac{m}{s^{2} } }}=469.24m

3 0
3 years ago
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