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11 months ago

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

Physics

1 answer:

il63 [147K]11 months 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>

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A 20 cm tall object is placed in front of a concave mirror with a radius of 31 cm. The distance of the object to the mirror is 9

Aloiza [94]

Answer:

The focal length of the concave mirror is -15.5 cm

Explanation:

Given that,

Height of the object, h = 20 cm

Radius of curvature of the mirror, R = -31 cm (direction is opposite)

Object distance, u = -94 cm

We need to find the focal length of the mirror. The relation between the focal length and the radius of curvature of the mirror is as follows :

R = 2f

f is the focal length

$f=\dfrac{R}{2}$

$f=\dfrac{-31}{2}$

f = -15.5 cm

So, the focal length of the concave mirror is -15.5 cm. Hence, this is the required solution.

4 0

2 years ago

2. A common physics experiment involves lowering an open tube into a cylinder of water and moving the tube up and down to adjust

notsponge [240]

Answer:

Explanation:

This question pertains to resonance in air column. It is the case of closed air column in which fundamental note is formed at a length which is as follows

l = λ / 4 where l is length of tube and λ is wave length.

here l = .26 m

λ = .26 x 4 = 1.04 m

frequency of sound = 330 Hz

velocity of sound = frequency x wave length

= 330 x 1.04

= 343.2 m /s

b )

Next overtone will be produced at 3 times the length

so next length of air column = 3 x 26

= 78 cm

c )

If frequency of sound = 256 Hz

wavelength = velocity / frequency

= 343.2 / 256

= 1.34 m

= 134 cm

length of air column for resonance

= wavelength / 4

134/4

= 33.5 cm

7 0

2 years ago

A white blood cell has a diameter of approximately 12 micrometers, or 0.012 um. A model represents its diameter as 24 um. What i

GuDViN [60]

Answer:

2:1

Explanation:

It is given that,

The diameter of a white blood cell is 12 micrometers

The diameter of white blood cell according to a model is 24 micrometers.

We need to find the ratio of model size to actual size. It can be done as follows :

$R=\dfrac{\text{model size}}{\text{actual size}}\\\\R=\dfrac{24\ \mu m}{12\ \mu m}\\\\R=\dfrac{2}{1}$

Hence, the ratio of model size to actual size is 2:1.

8 0

2 years ago

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gladu [14]

Answer:

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11 months ago

PHYSICS HELP<br> PLEASE HELP ITS ABOUT ATWOOD MACHINES

m_a_m_a [10]

Answer:

7.23407 $\frac{m}{s^2}$

Explanation:

(I will not include units in calculations)

I'm assuming FBD's are already drawn, so I will work from there.

Let the 2.2kg block equal $m_2$ , and the 20kg block equal $m_1$ .

Summation equation for $m_2$ : $\sum F_x=F_t_2-(F_f+F_g_x)=m_2a$ , $\sum F_y=F_n-F_g_y=0$

Summation equation for $m_1$ : $\sum F_y=F_g-F_t_1=m_1a$

Torque Summation Equation: $\sum\tau=F_t_1*r-F_t_2*r=I\alpha$

Do some plugging in with the values given: $\sum\tau=F_t_1*r-F_t_2*r=.5Mr^2\alpha$

Replace $\alpha$ with $\frac{a}{r}$ , and cancel out the r's.

$\sum\tau=F_t_1-F_t_2=.5Ma$

This step is important: Rearrange the force summation equation to solve for each tension force.

$F_t_2=m_2a+F_f+F_g_x\\F_t_1=m_1g=m_1a$

Perform Substitution: $\sum\tau=m_1g-m_1a-(m_2a+F_f+F_g_x)=.5Ma$

Now, we need to find the friction force and the horizontal component of the force of gravity.

Note that $F_f=$ μ $F_n$

And based on our earlier summation equation: $F_n=F_g_y$

First, break $F_g$ into x and y components. $F_g_y=F_g\cos(\theta)$ , $F_g_x=F_g\sin(\theta)$

Perform substitution with this and the fact that $F_g=mg$ .

$\sum\tau=m_1g-m_1a-(m_2a+\mu*m_2g\cos(\theta)+m_2g\sin(\theta))=.5Ma$

Solving for a, plugging in numbers yields an answer of 7.23407 $\frac{m}{s^2}$

6 0

2 years ago