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dlinn [17]
1 year ago
7

A car was struck broadside, injuring the driver. you would expect that during the collision, the driver's:_________

Physics
1 answer:
il63 [147K]1 year ago
4 0

The driver's Body was pushed laterally while the head remained still.

When two objects collide side-by-side (broadside), the head tends to stay static while the body is forced laterally, injuring the neck. The motorist would presumably be protected by a side-curtain airbag rather than hurt.

<h3>A Broadside Collision: What Is It? </h3>

A broadside collision happens when the front of one car bangs into the side of another, usually at a high rate of speed. Although catastrophic injuries can occur in any automobile accident on the road, broadside collisions are especially dangerous.

<h3>Where is the most likely location for broadside collisions?</h3>

The majority of broadside collisions happen at or near junctions. This is due to the fact that crossroads design makes it easy for a broadside collision to occur, especially when motorists are not paying attention. The likelihood of a broadside collision increases when several vehicles cross paths at an intersection.

learn more about collision here

<u>brainly.com/question/24915434</u>

#SPJ4

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A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s (~ 18 mph). What is the total time it is i
evablogger [386]

Answer:

The total time it is in the air for the ball is 1.6326 s

Given:

Initial velocity = 8 \frac{m}{s}

To find:

the total time it is in the air = ?

Formula used:

t = \frac{v-u}{a}

Where t = time to reach maximum height

v = final velocity of the ball = 0 m/s

u = initial velocity of ball = 8 m/s

a = acceleration due to gravity = -9.8

Acceleration of gravity is taken as negative because ball is moving in opposite direction.

Solution:

A ball is thrown upward at time t=0 from the ground with an initial velocity of 8 m/s.

The time taken by the ball to reach the maximum height is given by,

t = \frac{v-u}{a}

Where t = time to reach maximum height

v = final velocity of the ball = 0 m/s

u = initial velocity of ball = 8 m/s

a = acceleration due to gravity = -9.8

Acceleration of gravity is taken as negative because ball is moving in opposite direction.

t = \frac{0-8}{-9.8}

t = 0.8163 s

Thus, time taken by the ball to reach the ground again = time taken to reach maximum height

So, Total time required for ball to reach ground = 2t = 2 × 0.8163

Total time required for ball to reach ground = 1.6326 s

The total time it is in the air for the ball is 1.6326 s

4 0
3 years ago
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