All categories

4 years ago

In a car how does an air bag minimize the force acting on a person during a collision

Physics

2 answers:

Natasha_Volkova [10]4 years ago

6 0

Answer:

It increases the time it takes for the person to stop.

Explanation:

Increasing the time of the impact decreases the force.

Klio2033 [76]4 years ago

4 0

It decreases the momentum in the car.

You might be interested in

QUESTION 9 / 10

schepotkina [342]

Answer:

Explanation:

4 0

3 years ago

Read 2 more answers

Why is the sky blue?

Levart [38]

Answer:

Gases and particles in Earth's atmosphere scatter sunlight in all directions. Blue light is scattered more than other colors because it travels as shorter, smaller waves. This is why we see a blue sky most of the time.

7 0

3 years ago

A point charge (–5.0 µC) is placed on the x axis at x = 4.0 cm, and a second charge (+5.0 µC) is placed on the x axis at x = –4.

AveGali [126]

Answer:

The magnitude of electric force is $7.2\times10^{-3} N$

Explanation:

Coulomb's Law:

The force of attraction or repletion is

directly proportional to the products of charges i.e $F\propto q_1q_2$

inversely proportional to the square of distance i.e $F\propto \frac{1}{r^2}$

$\therefore F\propto \frac{q_1q_2}{r^2}$

$\Rightarrow F=k \frac{q_1q_2}{r^2}$ [ k is proportional constant=9×10⁹N m²/C²]

There are two types of force applied on Q=+2.5 μC=2.5×10⁻⁶ C

Let F₁ force be applied on Q =+2.5 μC by q₁= -5.0 μC = - 5.0×10⁻⁶ C

and F₂ force be applied on Q=+2.5 μC by q₂= 5.0 μC= 5.0×10⁻⁶ C

Since the magnitude of F₁ and F₂ are same. Therefore their y component cancel.

If we draw a line from q₁ to Q .

The it forms a triangle whose base = 4.0 cm and altitude =3.0 cm.

Let hypotenuse = r

Therefore, $r=\sqrt{altitude^2+base^2} =\sqrt{3^2+4^2} =5$

we know,

$cos \theta = \frac{base }{hypotenuse}$

$\Rightarrow cos \theta = \frac{4 }{r}$

Total force $F_Q = 2.F_1 cos\theta \hat{i}$

$=2k\frac{Qq_1}{r^2} cos\theta \hat i$

$=2\ \frac{9\times1 0^9\times2.5 \times 5\times 10^{-12}}{r^2} \frac{4}{r} \hat i$

$=8\ \frac{9\times10^9\times2.5 \times 5\times 10^{-12}}{5^3} \hat i$ [ r=5]

$=7.2\times10^{-3}\hat i$ N

The magnitude of electric force is $7.2\times10^{-3} N$

3 0

3 years ago

Why is the nucleus of an atom dense

Inessa [10]

Explanation:

The nucleus of an atom is dense because it contains more of particles in a very little space.

The nucleus is the center of an atom made up of the protons and neutrons.

Atomic nucleus is very small compared to the size of the atom.

Density is function of mass and volume.

Mass is the amount of matter in a substance.

Volume is the space occupied by a substance.

The more the mass the more the density.

Since protons and neutron are massive bodies occupying a small space, they make the nucleus very dense.

learn more:

Density brainly.com/question/5055270

#learnwithBrainly

7 0

3 years ago

A small fish is dropped by a pelican that is rising steadily at 0.50 m/s.

natima [27]

Answer:

(a) 24.025 m/s. downward.

(b) 31 m

Explanation:

From Newton's equation of motion,

(a)

v = u + gt ................... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, t = time.

Note: Let upward velocity be negative and downward be positive

Given: u = -0.5 m/s (upward), t = 2.5 s

Constant : g = 9.81 m/s²

Substitute into equation 1

v = 0.5+9.81(2.5)

v = -0.5+24.525

v = 24.025 m/s. downward.

(b) using

s₁ = ut + 1/2gt²......................... Equation 2

Where s₁ = distance at which the fish fall after being dropped by the pelican

Given: u = - 0.5 m/s, t = 2.5 s, g = 9.81 m/s²

Substitute into equation 2

s₁ = -0.5(2.5) + 1/2(9.81)(2.5)²

s₁ = -1.25+30.656

s₁ = 29.41 m

also,

s₂ = vt ................ Equation 3

Where s₂ = the distance by which the pelican rise during this time.

Given: v = 0.5 m/s, t= 2.5 s

s₂ = 0.5(2.5)

s₂ = 1.25 m.

Note: Distance between the pelican and fish = s₁ + s₂

Distance between the pelican and fish = 29.41+1.25

Distance between the pelican and fish = 30.66

Distance between the pelican and fish ≈ 31 m

3 0

3 years ago