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Evgesh-ka [11]
4 years ago
11

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

Physics
2 answers:
irina [24]4 years ago
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]4 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!
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Which diagram best represents the field around a positively charged particle?
AveGali [126]
The last one. Opposites attract while similar ones repulse each other so all forces are pushing them apart
6 0
3 years ago
If a car takes a banked curve at less than the ideal speed, friction is needed to keep it from sliding toward the inside of the
PolarNik [594]

Answer:

a. v₁ = 16.2 m/s

b. μ = 0.251

Explanation:

Given:

θ = 15 ° , r = 100 m , v₂ = 15.0 km / h

a.

To determine v₁ to take a 100 m radius curve banked at 15 °

tan θ  = v₁² / r * g

v₁ = √ r * g * tan θ

v₁ = √ 100 m * 9.8 m/s² * tan 15° = 16.2 m/s

b.

To determine μ friction needed for a frightened

v₂ = 15.0 km / h * 1000 m / 1 km * 1h / 60 minute * 1 minute / 60 seg

v₂ = 4.2 m/s

fk = μ * m * g

a₁ = v₁² / r = 16.2 ² / 100 m = 2.63 m/s²

a₂ = v₂² / r = 4.2 ² / 100 m = 0.18 m/s²

F₁ = m * a₁  ,  F₂ = m * a₂

fk = F₁ - F₂   ⇒  μ * m * g = m * ( a₁ - a₂)

μ * g = a₁ - a₂   ⇒  μ = a₁ - a₂ / g

μ = [ 2.63 m/s² - 0.18 m/s² ] / (9.8 m/s²)

μ = 0.251

3 0
3 years ago
If you increase the frequency of a sound wave four times what will happen to its speed
Anika [276]

Answer: The correct answer is "the speed of the wave becomes four times".

Explanation:

The relation between the speed, frequency and the wavelength is as follows:

v=f\lambda

Here, v is the speed of the wave, f is the frequency and \lambda is the wavelength.

The speed of the sound wave is directly proportional to the frequency.

In the given problem, if the speed of the sound wave is increased four times then the speed of the sound becomes four times.

Therefore, the speed of the sound wave becomes four times.

8 0
3 years ago
Read 2 more answers
A sample of metallic frewium weighs 185N on a spring scale in air. When immersed in pure water, the frewium pulls on the scale w
balu736 [363]

Wow !  This one could have some twists and turns in it.
Fasten your seat belt.  It's going to be a boompy ride.

-- The buoyant force is precisely the missing <em>30N</em> .

--  In order to calculate the density of the frewium sample, we need to know
its mass and its volume.  Then, density = mass/volume .

-- From the weight of the sample in air, we can closely calculate its mass.

   Weight = (mass) x (gravity)
   185N = (mass) x (9.81 m/s²)
   Mass = (185N) / (9.81 m/s²) = <u>18.858 kilograms of frewium</u> 

-- For its volume, we need to calculate the volume of the displaced water.

The buoyant force is equal to the weight of displaced water, and the
density of water is about 1 gram per cm³.  So the volume of the
displaced water (in cm³) is the same as the number of grams in it.

The weight of the displaced water is 30N, and weight = (mass) (gravity).

           30N = (mass of the displaced water) x (9.81 m/s²)

           Mass = (30N) / (9.81 m/s²) = 3.058 kilograms

           Volume of displaced water = <u>3,058 cm³</u>

Finally, density of the frewium sample = (mass)/(volume)

      Density = (18,858 grams) / (3,058 cm³) = <em>6.167 gm/cm³</em> (rounded)

================================================

I'm thinking that this must  be the hard way to do it,
because I noticed that

       (weight in air) / (buoyant force) =  185N / 30N = <u>6.1666...</u>

So apparently . . .

        (density of a sample) / (density of water) =

                                  (weight of the sample in air) / (buoyant force in water) .

I never knew that, but it's a good factoid to keep in my tool-box.


3 0
3 years ago
Difference between displacement and vector quantity
daser333 [38]

The only 'difference' is that they are different categories.

It's like asking "What's the difference between Susie and girl ?"

Or "What's the difference between Cadillac and car ?"

Displacement <em>IS</em> a vector quantity.

5 0
3 years ago
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