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olga2289 [7]
3 years ago
8

How does an airbag help protect a passenger in a car accident?

Physics
2 answers:
Novosadov [1.4K]3 years ago
8 0
Good question. It like a fail safe,if you hit a car tail gate going slightly fast it come out to protect you from hitting your on the the wheel or the mirror. See if you don't have a seat belt or it just don't fully stop you the airbag might help.Like concussions.  I Hope This Help you :)
likoan [24]3 years ago
3 0

Answer:

It shortens the passengers stopping time, and this decreases the force.

I think.

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Difference between freefall and weightlessness.​
Margaret [11]

Answer:

Differences between freefall and weightlessness are as follows:

<h3><u>Freefall</u></h3>
  • When a body falls only under the influence of gravity, it is called free fall.
  • Freefall is not possible in absence of gravity.
  • A body falling in a vacuum is an example of free fall.

<h3><u>Weightlessness</u></h3>
  • Weightlessness is a condition at which the apparent weight of body becomes zero.
  • Weightlessness is possible in absence of gravity.
  • A man in a free falling lift is an example of weightlessness.

Hope this helps....

Good luck on your assignment....

6 0
3 years ago
A cubical Gaussian surface surrounds two positive charges, each has a charge q 1 1 = + 3.90 × 10 − 12 3.90×10−12 C, and three ne
Masteriza [31]

Answer:

The electric flux is zero because charge is zero.

Explanation:

Given that,

Positive charge q_{1}=3.90\times10^{-12}\ C

Negative charge q_{2}=-2.60\times10^{-12}\ C

We need to calculate the total charged

Using formula of charge

Q_{enc}=2q_{1}+3q_{2}

Put the value into the formula

Q_{enc}=2\times3.90\times10^{-12}+3\times(-2.60\times10^{-12})

Q_{enc}=0

We need to calculate the electric flux

Using formula of electric flux

\phi=\dfrac{Q_{enc}}{\epsilon_{0}}

Put the value into the formula

\phi=\dfrac{0}{8.85\times10^{-12}}

Hence, The electric flux is zero because charge is zero.

7 0
3 years ago
A bicycle rider has a speed of 19.0 m/s at a height of 55.0 m above sea level when he begins coasting down hill. The mass of the
lukranit [14]

Answer:

The mechanical energy of the rider at any height will be 6.34 × 10⁴ J.

Explanation:

Hi there!

The mechanical energy of the rider is calculated as the sum of the gravitational potential energy plus the kinetic energy. Since there are no dissipative forces (like friction), the mechanical energy of the rider at a height of 55.0 m above the sea level will be the same at a height of 25.0 m (or at any height), because the loss in potential energy will be compensated by a gain in kinetic energy, according to the law of conservation of energy.

Then, calculating the potential and kinetic energy at 55.0 m and 19 m/s, we can obtain the mechanical energy that will be constant:

Mechanical energy = PE + KE

Where:

PE = potential energy.

KE = kinetic energy.

The potential energy is calculated as follows:

PE = m · g · h

Where:

m = mass of the object.

g = acceleration due to gravity.

h = height.

Then, the potential energy of the rider will be:

PE = 88.0 kg · 9.81 m/s² · 55.0 m = 4.75 × 10⁴ J

The kinetic energy is calculated as follows:

KE = 1/2 · m · v²

Where "m" is the mass of the object and "v" its velocity. Then:

KE = 1/2 · 88.0 kg · (19.0 m/s)²

KE = 1.59 × 10⁴ J

The mechanical energy of the rider will be:

Mechanical energy = PE + KE = 4.75 × 10⁴ J + 1.59 × 10⁴ J = 6.34 × 10⁴ J

This mechanical energy is constant because when the rider coast down the hill, its potential energy is being converted into kinetic energy, so that the sum of potential energy plus kinetic energy remains constant.

5 0
3 years ago
Derived quantities depend on...........physical quantity​
Vesnalui [34]

Derived quantities depend on.( fundamental)..........physical quantity

Are you from Nepal?

7 0
2 years ago
Read 2 more answers
The drawing shows a loudspeaker A and point C, where a listeer is positioned. A second loudspeaker B is located somewhere to the
Liono4ka [1.6K]

Answer: 4.17m

Explanation:

The observer at C will hear a sound on no sound upon whether the interference is constructive or destructive.

If the listeners hears sounds it is caled constructive interference but if he hears no sound its called destructive interference.

So

d2 - d1 = (n *lamba)/ 2

Where n=1,3,5

lamda=v/f =349/62.8

lamda=5.56m

d2= d1 + nlamda/2

d2= 1 + 5.56/2

d2= 3.78m

X'= 1 cos 60= 0.5m

Y= 1 sin60= 0.866m

X"^2 + Y^2 =d2^2

X" =√(y^2 - d2^2)

X"=√(3.78^2 - 0.886^2)

X"= 3.67m

So therefore the closest that speaker A can be to speaker B so the listener does not hear any sound is X' + X"= 0.5 + 3.67

4.17m

3 0
3 years ago
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