All categories

3 years ago

How does inertia affect a person who is not wearing a seatbelt during a collision?

2 answers:

7 0

En caso de una colisión de bajo nivel de gravedad podría golpearse con el tablero del carro o en la parte del sillón (si esta sentado en la parte trasera). Pero si es muy grave la colisión podría salirse del carro por el cristal.

Orlov [11]3 years ago

3 0

Well an object in motion stays in motion. the body would continue to go at the speed the car was going before it was hit, and then the boy would go threw the windshield (if they are in the front seat). Which means inertia wouldn't have any affect on the person until they hit the windshield and then the pavement. Hope this makes sense to you!

You might be interested in

A truck with 0.420-m-radius tires travels at 32.0 m/s. what is the angular velocity of the rotating tires in radians per second?

andrey2020 [161]

Angular velocity of the rotating tires can be calculated using the formula,

v=ω×r

Here, v is the velocity of the tires = 32 m/s

r is the radius of the tires= 0.42 m

ω is the angular velocity

Substituting the values we get,

32=ω×0.42

ω= 32/0.42 = 76.19 rad/s

= 76.19× $\frac{1}{2\pi} *60$ revolution per min

=728 rpm

Angular velocity of the rotating tires is 76.19 rad/s or 728 rpm.

4 0

3 years ago

What distance does light travel in water, glass, and diamond during the time that it travels 1.0 m in vacuum? The refractive ind

lidiya [134]

Answer:

refractive index for water,glass,diamond are 0.752m, 0.667m, 0.413m respectively

Explanation:

refractive index (n) = $\frac{velocity of light in air/vacuum}{velocity of light in substance}$

velocity = $\frac{distance}{time}$

The time for travel is kept constant for all mediums.

refractive index (n) = $\frac{\frac{distance in vacuum}{time} }{\frac{distance in medium}{time} }\\ \\=\frac{distance in vacuum}{distance in medium}$

distance in medium = $\frac{distance in vacuum}{refractive index of medium}$

$S_{medium} = \frac{S_{vacuum} }{n_{medium} }$

For water, n= 1.33

$S_{water} = \frac{1 }{1.33}$

$S_{water} = 0.752m$

For glass, n=1.5

$S_{glass}= \frac{1 }{1.5}$

$S_{glass} = 0.667m$

For diamond, n= 2.42

$S_{diamond} = \frac{1 }{2.42}$

$S_{diamond} = 0.413m$

3 0

3 years ago

A 2.50 kg ball moving at 7.50 m/s is caught by a 70.0 kg man while the man is standing on ice. What is the common velocity of th

yarga [219]

Answer:

V = 2.5*7.0 / ( 2.5 + 70 )

8 0

2 years ago

A 1500 kg car skids to a halt on a wet road where μk = 0.47. You may want to review (Pages 141 - 145) . Part A How fast was the

shusha [124]

The car travels at a speed of 25m/s.

<u>Explanation:</u>

Given-

Mass, m = 1500kg

Coefficient of friction, μk = 0.47

Distance, x = 68m

Speed, s = ?

We know,

$Force, F = ma$

and

F = μ X m X g

Therefore,

μ * m * g = m * a

μ * g = a

Let, g = 9.8m/s²

So,

$a = 0.47 * 9.8 m/s^2$

$a = 4.606m/s^2$

We know,

$v^2 - u^2 = 2as$

where, v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

If the car comes to rest, the final velocity, v becomes 0.

So,

$u^2 = 2 * 4.606 * 68\\\\u^2 = 626.416m/s\\\\u = 25m/s$

The car travels at a speed of 25m/s.

6 0

3 years ago

A sound has a sound level of 30 dB. Its intensity is what multiple of the standard reference level for intensities?

arsen [322]

<h2>Answer:</h2>

1000th multiple of the standard reference level for intensities.

<h2>Explanation:</h2>

The sound intensity level (β), measured in decibels, of a sound with an intensity of I is defined as follows;

β = 10 log (I / I₀) --------------------(i)

Where;

I₀ = reference intensity

Given from the question;

β = sound level = 30dB

Substitute this value into equation (i) as follows;

30 = 10 log (I / I₀)

Divide both sides by 3;

3 = log (I / I₀)

Take antilog of both sides;

10^(3) = (I / I₀)

1000 = I / I₀

Solve for I;

I = 1000I₀

Therefore the intensity of the sound is 1000 times the standard reference level for intensities (I₀)

7 0

3 years ago