why does a combination of red, yellow and blue paint look black?​

The magnitude of the maximum possible upwards acceleration of the bucket without the thread breaking is 25m/s^2

The maximum acceleration of the bucket is determined by the maximum tension of the thread. The acceleration of the bucket is determined by Newton's Second Law of Motion, which states that the sum of the forces acting on an object is equal to the mass of the object multiplied by its acceleration.

Therefore, the equation to calculate the maximum acceleration can be expressed as:

F = ma

where F is the maximum tension of the thread (62.5 N), m is the mass of the bucket plus the contents (2.50 kg), and a is the maximum acceleration of the bucket.

Rearranging the equation to solve for a, we get:

a = F/m

Plugging in the values for F and m, we get:

a = 62.5 N/2.50 kg

Simplifying, we get:

a = 25 m/s2

Therefore, the maximum acceleration of the bucket without the thread breaking is 25 m/s2.

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Answer:

To find the magnitude of a vector, you can use the Pythagorean theorem. In this case, the magnitude of the vector is given by:

sqrt((-22.2m)^2 + (12.6m)^2)

This simplifies to:

sqrt(486.24 + 159.96)

Which is equal to:

sqrt(646.2)

Which is approximately equal to:

25.4 m

So the magnitude of the vector (-22.2m, 12.6m) is approximately 25.4 m.

Explanation:

mark brainliest please thanks

Answer:

Explanation:

givens

m = 5 kg

a = 9.8

h = 2 m

Formula

PE = m*g*h

Solution

PE = 5 * 9.8 * 2

PE = 98 This does not take sig digs into account. The answer would be 100 if you have to consider significant digits.

Answer:

μ = 0.41

Explanation:

\(\frac{1}{2} kx^2 = \mu mgd\)

\(\mu = (\frac{1}{2} kx^2)/mgd\)

\(\mu = (\frac{1}{2} (100)(0.2)^2)/(0.5)(9.81)(1)\)

\(\mu = 0.41\)

The coefficient of kinetic friction between the block and the tabletop is 0.41.

The given parameters;

The coefficient of kinetic friction between the block and the tabletop by applying the principle of conservation of energy as shown below.

Fd = Uₓ

μmgd = ¹/₂kx²

where;

The coefficient of kinetic friction between the block and the tabletop;

\(\mu_k = \frac{kx^2}{mgd} \\\\\mu_k = \frac{100\times 0.2^2}{2\times 0.5\times9.8 \times 1 } \\\\\mu_k = 0.41\)

Thus, the coefficient of kinetic friction between the block and the tabletop is 0.41.

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Answer:

Heres on ex.

When the driver of a bus applies brake suddenly, the lower part of the body comes to rest as the bus comes to rest but, the upper part of the body continues to move forward due to inertia of motion.

Explanation:

In exothermic reactions, heat and light are released to the surrounding environment. On the other hand, in an endothermic reaction, heat is required and therefore it can be considered as a reactant.

Learn more in:

Answer:

(Question) Which result occurs during an exothermic reaction?

(Answer) Light or heat are released into the environment.

(Question) Which event is an example of an endothermic reaction?

(Answer) photosynthesis

(Question) How is total reaction energy calculated?

(Answer) reactant bond energy – product bond energy

(Question) When producing hydrogen iodide, the energy of the reactants is 581 kJ/mol, and the energy of the products is 590 kJ/mol. The equation is shown.

H2 + I2 → 2HI

What is the total energy of the reaction? Is this an endothermic or exothermic reaction?

(Answer) –9 kJ/mol, exothermic

(Question) Hydrogen bromide breaks down into diatomic hydrogen and bromine in the reaction shown.

2HBr → H2 + Br2

The energy of the reactant is 732 kJ/mol, and the energy of the products is 630 kJ/mol.

What is the total energy of the reaction? Is this reaction endothermic or exothermic?

(Answer) 102 kJ/mol, endothermic

Explanation:

just finished the quick check UwU

The momentum of the ball in motion is -0.68 kg m/s, and the force between the ball and the wall is -6.8 N.

The state of motion of an object with a certain mass is termed as momentum. It is the product of the mass and velocity of the moving object and is a vector quantity.

It is expressed as: p = m × v

First, let's calculate the initial momentum of the ball before it strikes the wall:

p₁ = m₁× v₁ = 40 g × 10 m/s = 0.04 kg ×  10 m/s = 0.4 kg m/s

Where  m₁ is the mass of the ball and v₁ is its initial velocity.

Next, let's calculate the final momentum of the ball after it bounces off the wall:

p₂ = m₁× v₂= 40 g × (-7 m/s) = -0.04 kg × 7 m/s = -0.28 kg m/s

where v₂ is the velocity of the ball after it bounces off the wall.

The change in momentum of the ball is then:

Δp = p₂- p₁ = (-0.28 kg m/s) - (0.4 kg m/s) = -0.68 kg m/s

Note that the change in momentum is negative, indicating that the ball's momentum has decreased in magnitude and changed direction.

To calculate the force between the ball and the wall,

J = Δp

where J is the impulse and Δp is the change in momentum.

The impulse is also equal to the force multiplied by the time during which the force is applied:

J = F × Δt

where F is the force and Δt is the time during which the force is applied.

Setting these two equations equal to each other, we get:

F × Δt = Δp

Solving for the force, we get:

F = Δp / Δt = (-0.68 kg m/s) / (0.1 s) = -6.8 N

The negative sign indicates that the force exerted by the ball on the wall is directed opposite to the direction of the ball's motion, as expected.

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The change in the momentum of the ball will be 0.68kgm/sec. While the force that acts on the body will be 0.68 N.

The measurement of mass in motion is called momentum. Momentum is equal to its mass multiplied by its velocity. Mathematically it is given as;

Momentum = Mass x Velocity

The given data in the problem is

ball mass ( m)= 40 g=0.040 kg

speed of the ball(v) = 10 m/s

rebound speed(V )= -7 m/s

The change in the momentum of the body is given by,

ΔP = mv - mv'

Δp= 0.040 x 10 - 0.057 x (-7)

ΔP = 0.68 kg.m/s

Hence the change in the momentum of the body will be 0.68 kg.m/s.

Impulse is defined as the product of force and time interval of contact.

ΔP = force *time

force = ΔP/time

force = 0.68/1

force = 0.68

Hence the force that acts on the body will be 0.68 N.

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(a) ball velocity = - 2.47 m/s  

(b) block velocity = 1.236 m/s

Two items collide when they briefly come into touch with one another. To put it another way, a collision is a brief reciprocal encounter between two masses in which the momentum and energy of the masses change.

For ball as it is at initially at height of 70.0 cm = 0.7 m

so by energy conservation,

K.E.(i) P.E.(i) + K.E.(f)+P.E.(f)

0 + mg×(0.7)  = 1/2 mv² + 0

v = 3.71 m/s

now, collision b/w ball and block

as the block is in rest initially and there is no external force b/w them

so by momentum conservation,

m₁u₁+ m₂u₂ = m₁v₁+ m₂v₂

or, (0.5 x 3.71) + 0 = (0.5x v₁) + (2.5 x v₂)

3.71 = v₁ + 5v₂  .........(1)

as elastic collision so energy conservation

or, 1/2 m₁ u₁² + 1/2 m₂ u₂² = 1/2 m₁ v₁² + 1/2 m₂ v₂²

or, (1/2 x 0.5 x 3.712) + 0 = (1/2 x 0.5 x v₁²) + (1/2 x 2.5 x v₂²)

or, 3.433 = 0.25xv₁² + 1.25 v₂²      ...............(2)

from eq. 1 and eq. 2

or, 3.433 = 0.25 x (3.71 - 5 v₂)² + 1.25v₂²  

or, v² = 1.236 m/s

and v₁ = 3.71 - 5v₂

or, v₁ = - 2.47 m/s

so after collision

ball velocity = - 2.47 m/s       [(-)ve x direction]

block velocity = 1.236 m/s     [(+)ve x direction]

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Answer:

\(d=2\times 10^{-9}\)

Explanation:

It is given that, DNA's diameter is 0.000000002 meters. We need to convert the diameter into scientific notation.

Any number can be written in the form of scientific notation as follows :

\(N=m\times 10^n\)

Where, m is a real no and n is any integer

We have, d = 0.000000002 m

There are 8 zeroes before 2. We can shift the decimal after 2 so as to write it into scientific notation.

So,

\(d=2\times 10^{-9}\)

Hence, the diameter of the DNA is \(2\times 10^{-9}\).

Temperature of iron (Ti) = 415 °C Temperature of water (Tw) = 15.0 °CTemperature at equilibrium (Te) = 100 °CMass of water (m) = 0.625 kgMass of steam evaporated (ms) = 0.144 kgHeat lost by iron (Q1) = Heat gained by water (Q2) + Heat required to evaporate steam .

Heat lost by iron  = (mass of iron (m) x specific heat capacity of iron (c) x change in temperature of iron (ΔT1))Heat gained by water = (mass of water (m) x specific heat capacity of water (c) x change in temperature of water (ΔT2))Heat required to evaporate steam  = (mass of steam (ms) x specific latent heat of vaporization of water (L))Now, using the above formula we can calculate the mass of the iron block as:

Q3m x c x ΔT1 = m x c x ΔT2 + ms x L

Let's calculate the value of Q1 first.

Q1 = m x c x ΔT1m = Q1 / (c x ΔT1)

We know that

c = 450 J/kg °C and ΔT1 = Ti - Te = 415 - 100 = 315°CQ1 = m x c x ΔT1= m x 450 J/kg

°C x 315°C= 141750 m Jm = Q1 / (c x ΔT1)= 141750 / (450 x 315)= 1.002 kg

Now, let's calculate the value of Q3.Q3 = ms x L= 0.144 kg x 2.26 x 10^6 J/kg= 325440 J

Now, let's calculate the value of Q2

.Q2 = m x c x ΔT2m = (Q2 + Q3) / (c x ΔT2)

We know that ΔT2 = Te - Tw = 100 - 15 = 85°CQ2 = m x c x ΔT2= 0.625 kg x 4186 J/kg °C x 85°C= 276981.25 JNow, let's calculate the mass of the iron block.m =

(Q2 + Q3) / (c x ΔT2)= (276981.25 + 325440) / (450 x 85)= 1.003 kg

Hence, the mass of the iron block is 1.003 kg rounded off to 3 significant figures.

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After pushing the pumpkin hard, you find yourself reversing direction at a speed of 4.0 m/s. 3.0 seconds after being pushed, the pumpkin will slide 12 m. Assume you weigh 60.0 kg.

We can use the conservation of momentum to solve this problem. After the push, the momentum of the system is given by:

p = (449 kg + 60 kg) * v

where v is the speed of the pumpkin and you after the push. Since you end up sliding backward at 4.0 m/s, we have:

v = -4.0 m/s

Substituting this into the expression for momentum, we find:

p = (449 kg + 60 kg) * (-4.0 m/s) = -2036 kg·m/s

The negative sign indicates that the momentum of the system is in the opposite direction of your motion.

During the sliding motion, the net force on the system is given by:

Fnet = (449 kg + 60 kg) * g * sin(θ)

where g is the acceleration due to gravity (9.81 m/s^2) and θ is the angle of the ramp. Since the ramp is smooth and horizontal, θ = 0 and Fnet = 0. Therefore, there is no net force to change the momentum of the system.

Using the equation for motion with constant acceleration, we can find the distance the pumpkin slides in 3.0 seconds:

x = x0 + v0t + (1/2)at²

Since the initial speed of the pumpkin is -4.0 m/s and there is no net force acting on it, its speed remains constant during the slide. Therefore, v0 = -4.0 m/s and a = 0. Substituting these values, we find:

x = x0 + v0t = (-4.0 m/s) * (3.0 s) = -12 m

The negative sign indicates that the pumpkin slides in the opposite direction to your motion. Therefore, the pumpkin slides 12 meters backward (i.e., towards you) in 3.0 seconds after the push.

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Answer:

12.75N

Explanation:

Let's consider the forces acting on the bar:

Weight of the bar (W = 30.0 N): It acts vertically downward, passing through the center of gravity of the bar.

Tension in cord 1 (T1): It acts horizontally and to the left, making an angle of 90° with the bar. Since cord 1 is massless, it does not contribute to the torque.

Tension in cord 2 (T2): It acts at an angle φ = 40.0° with the vertical.

To find the tension in cord 1 and cord 2, we need to set up torque equilibrium equations. The torque of the weight about the point of suspension must be balanced by the torques of the tensions in cords 1 and 2.

Taking the left end of the bar as the reference point (pivot), the torque equilibrium equation can be written as:

Torque due to weight = Torque due to T1 + Torque due to T2

The torque due to the weight is calculated as follows:

Torque due to weight = Weight of the bar * Perpendicular distance between the weight and the pivot point

The torque due to T1 is zero since it acts along the line of action passing through the pivot point.

The torque due to T2 can be calculated as follows:

Torque due to T2 = T2 * Perpendicular distance between the cord and the pivot point

Using the given values:

Weight of the bar (W) = 30.0 N

Length of the bar (L) = 3.0 m

Distance of the center of gravity from the left-hand end of the bar (d) = 2.2 m

Angle between cord 2 and the vertical (φ) = 40.0°

We can calculate the perpendicular distance between the weight and the pivot point as:

Perpendicular distance = L/2 - d

Using these values, we can solve for T2:

30.0 N * (L/2 - d) = T2 * L * sin(φ)

Let's substitute the given values and solve for T2:

30.0 N * (3.0/2 - 2.2) = T2 * 3.0 * sin(40.0°)

T2 ≈ 12.75 N

Therefore, the tension in cord 2 (T2) is approximately 12.75 N.

The answer is that the time the diver was in the air is 1.13 seconds.

To determine the time the diver was in the air, we can use the kinematic equation:

Δy = viΔt + 1/2at²,

where Δy is the displacement, vi is the initial velocity, a is the acceleration due to gravity (g), and t is the time.The initial velocity, vi, is given as 5.5 m/s, and since the diver jumps upwards, the displacement, Δy, is equal to the height of the board, which is 3.0 m. The acceleration due to gravity, a, is -9.8 m/s² (negative because it acts downwards).Substituting the known values into the equation:3.0

m = (5.5 m/s)t + 1/2(-9.8 m/s²)t²

Simplifying, we get:

4.9t² + 5.5t - 3.0 = 0

We can solve for t using the quadratic formula:

t = (-5.5 ± √(5.5² - 4(4.9)(-3.0))) / (2(4.9))= (-5.5 ± 1.59) / 9.8= -0.47 s or 1.13 s

Since time cannot be negative, the time the diver was in the air is 1.13 seconds.

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The most appropriate method to study the potential impacts of oil drilling on coastal California would be design a computer simulation of an oil spill in the area to determine the environmental impact design a computer simulation of an oil spill in the area to determine the environmental impact. The correct option to this question is C.

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Answer:

Approaches mathematical learning through inquiry

-Explore real contexts, problems, situations, and models

-Learning through doing shifts the focus on the students

-Problems have multiple entry and exit points

-Links to other disciplines

Explanation:

quizlet

Hope this answer helps you

Answer: the radius of the circular path is approximately 1637.58 meters.

Explanation:

The centripetal force acting on the airplane is provided by the component of the gravitational force that acts towards the center of the circular path. This component is given by:

F_c = m * g * tan(banking angle)

Where:

F_c is the centripetal force

m is the mass of the airplane

g is the acceleration due to gravity

tan(banking angle) is the tangent of the banking angle

Now, the centripetal force is also given by the formula:

F_c = (m * v^2) / r

Where:

v is the speed of the airplane

r is the radius of the circular path

Equating the two expressions for F_c, we get:

(m * g * tan(banking angle)) = (m * v^2) / r

Canceling out the mass (m) on both sides of the equation, we have:

g * tan(banking angle) = v^2 / r

Solving for r, we get:

r = (v^2) / (g * tan(banking angle))

Substituting the given values:

v = 400 km/h = 400,000 m/h

g = 9.8 m/s^2

banking angle = 20°

Converting the speed to m/s:

v = 400,000 m/h * (1/3600) h/s = 111.11 m/s

Converting the banking angle to radians:

banking angle = 20° * (π/180) rad/° = 0.3491 rad

Now, substituting the values into the formula:

r = (111.11^2) / (9.8 * tan(0.3491))

r ≈ 1637.58 meters

Therefore, the radius of the circular path is approximately 1637.58 meters.

Answer:

On Earth: mass = 115 kg; weight = 1127 N

Interspace:  mass = 115 kg; weight = 0 N

Explanation:

On Earth, his mass is 115 kg, and his weight is 115 kg * 9.8 m/s^2 = 1127 N.

On interspace , his mass is still 115 kg,  and his weight is Zero Newtons = 0 N

(a) The mass of space traveler on Earth is 115 kg and its weight is 1127 N.

(b)  The mass of traveler at interplanetary space is 115 kg and its weight is 0 N.

Given data:

The mass of space traveler outside the Earth is, m = 115 kg.

(a)

Since, the mass of any object is always remains constant. In other words, there is no way to destroy mass and hence mass of traveler on Earth will be same as 115 kg.

While the weight will be,

W = mg

W = 115(9.8)

W = 1127 N

Thus, we can conclude that the mass of space traveler on Earth is 115 kg and its weight is 1127 N.

(b)

On interspace , his mass is still 115 kg,  and his weight is Zero Newtons. This is because the gravity at interplanetary space is zero.

That is,

W = mg

W = 115 (0)

W = 0 N

Thus, we can conclude that the mass of traveler at interplanetary space is 115 kg and its weight is 0 N.

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The projectile launch ratios allow finding the result of whether the ball enters the goal is:  

Parameters given

To find

Projectile launching is an application of kinematics where there is not acceleration on the x-axis and there is gravity acceleration on the y-axis.

In this case, to score the goal, the ball must be at maximum h = 2.5 m when it is at x = 30m, let's find the time, see attachment for a schema.

The components of the initial velocity are:

        cos 24 = \(\frac{v_o_x}{v_o}\)  

        sin 24 = \(\frac{v_{oy}}{v_o}\)

        v₀ₓ = v₀ cos 24

        \(v_{oy}\)  = v₀ sin 24

on the x axis there is no acceleration.

        x = v₀ₓ t

        t = \(\frac{x}{v_o \ cos 24y}\)

Let's  calculate

        t = \(\frac{30}{22 \ cos24}\)  

        t = 1.49 s

Now let's find out at what height for this time

       y = y₀ + \(v_{oy}\)  t - ½ g t²

In general for free -kik the ball is on the floor y = 0

      y = v₀ sin 24 t - ½ g t²

Let's calculate

     y = 22 sin 24 1.49 - ½ 9.8 1.49²

     y = 2.45 m

It indicates that the height of the goal crossbar is 2.5 m, therefore when the ball arrives it is below the height and enters scoring the goal.

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Answer:

8.4 N/m

Explanation:

m = Mass of block = 4.63 gm

g = Acceleration due to gravity = \(9.81\ \text{m/s}^2\)

x = Displacement of spring = 0.45 cm

a = Acceleration of subject = 0.832g

k = Spring constant

Force is given by

\(F=ma\)

From Hooke's law

\(F=kx\)

So

\(ma=kx\\\Rightarrow k=\dfrac{ma}{x}\\\Rightarrow k=\dfrac{4.63\times 10^{-3}\times 0.832\times 9.81}{0.45\times 10^{-2}}\\\Rightarrow k=8.4\ \text{N/m}\)

The force constant of the spring is 8.4 N/m.

Answer:

the volocity is 50

Explanation:

Answer:

The slopes for each of the four line segments are \(a_{A} = 6\,\frac{m}{min^{2}}\), \(a_{B} = 0\,\frac{m}{min^{2}}\), \(a_{C} = -4\,\frac{m}{min^{2}}\) and \(a_{D} = 2.667\,\frac{m}{min^{2}}\), respectively.

Explanation:

There are four line segments:

(i) Line A: \(v(0\,min) = 0\,\frac{m}{min}\), \(v(10\,min) = 60\,\frac{m}{min}\)

(ii) Line B: \(v(10\,min) = 60\,\frac{m}{min}\), \(v(15\,min) = 60\,\frac{m}{min}\)

(iii) Line C: \(v(15\,min) = 60\,\frac{m}{min}\), \(v(40\,min) = -40\,\frac{m}{min}\)

(iv) Line D: \(v(40\,min) = -40\,\frac{m}{min}\), \(v(55\,min) = 0\,\frac{m}{min}\)

The slope of each line segment represents the acceleration of the particle, which can calculated by the geometrical concept of secant line. Hence, we proceed to determine the acceleration associated with each line segment:

Line A

\(a_{A} = \frac{v(10\,min)-v(0\,min)}{10\,min-0\,min}\)

\(a_{A} = \frac{60\,\frac{m}{min}-0\,\frac{m}{min}}{10\,min-0\,min}\)

\(a_{A} = 6\,\frac{m}{min^{2}}\)

Line B

\(a_{B} = \frac{v(15\,min)-v(10\,min)}{15\,min-10\,min}\)

\(a_{B} = \frac{60\,\frac{m}{min}-60\,\frac{m}{min} }{15\,min-10\,min}\)

\(a_{B} = 0\,\frac{m}{min^{2}}\)

Line C

\(a_{C} = \frac{v(40\,min)-v(15\,min)}{40\,min-15\,min}\)

\(a_{C} = \frac{-40\,\frac{m}{min}-60\,\frac{m}{min} }{40\min-15\,min}\)

\(a_{C} = -4\,\frac{m}{min^{2}}\)

Line D

\(a_{D} = \frac{v(55\,min)-v(40\,min)}{55\,min-40\,min}\)

\(a_{D} = \frac{0\,\frac{m}{min}-\left(-40\,\frac{m}{min} \right) }{55\,min-40\,min}\)

\(a_{D} = 2.667\,\frac{m}{min^{2}}\)

The slopes for each of the four line segments are \(a_{A} = 6\,\frac{m}{min^{2}}\), \(a_{B} = 0\,\frac{m}{min^{2}}\), \(a_{C} = -4\,\frac{m}{min^{2}}\) and \(a_{D} = 2.667\,\frac{m}{min^{2}}\), respectively.

Wundt studies consciousness using introspection because his subjects examined and shared their own feelings and thoughts. Wundt was the first psychologist.

Introspection refers to the examination of one's own mental and/or emotional mechanisms and processes.

Wilhelm Wundt was a distinguished psychologist who demonstrated that introspection is a highly practiced mechanism of self-examination.

W. Wundt developed the theory of conscious thought by indicating that conscious mental states can be scientifically (empirically) studied by using introspection.

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Answer: B His subjects examined and shared their own feelings and thoughts.

Explanation: Just took the test

Answer:

Explanation:

alot

(a) The induced current in the circuit is clockwise.

b. To find the power (P) using P = Fd/t or P = Fv (since d/t = v). Here, F = ILB (from the Lorentz force), so P = (ILB)v.

(a) The induced current in the circuit is clockwise.

This can be determined using the right-hand rule.

As the metal bar moves to the left through the magnetic field directed out of the plane, the generated force on the electrons (Lorentz force) will push them toward the top rail, creating a clockwise current.

(b) To find the rate at which the applied force is doing work on the bar, first calculate the induced EMF (ε) using Faraday's law:

induced EMF (ε) using Faraday's law:

ε = BLv

= (0.65 T) * (0.36 m) * (5.9 m/s)

= 1.389 Tm²/s

= 1.389 V (since 1 Tm²/s = 1 V)

induced current (I) using Ohm's law:

I = ε/R

= 1.389 V / 45 Ω

= 0.03086 A

force (F) from the Lorentz force law, where F = ILB:

F = ILB

= (0.03086 A) * (0.36 m) * (0.65 T)

= 0.00723 N

Finally, we find the power (P) using P = Fv:

P = Fv

= (0.00723 N) * (5.9 m/s)

= 0.04266 W

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Answer:

Airbags reduce chances of injury by absorbing most of the impact force from the body during a car crash

Explanation:

In a car collision, the speed of the vehicle is suddenly bought to rest. All the kinetic energy is suddenly converted into other forms of energy.

The body of the driver keeps travelling forward under his inertia force due to his mass until he is slammed against the steering wheel. The steering wheel is a very rigid component, and so when the body slams against it, the body takes the deformation, absorbing some of the energy of the moving car. This sudden impact of energy can be fatal enough to gravely injure the driver because the body does not undergo much deformation. When an airbag is used, the crash automatically triggers the release of the airbag. Instead of the body colliding against the rigid steering wheel, it is now collided against the soft air bag. The airbag is very collapsible, and some of the kinetic energy of the car on the driver is converted into the deformation energy used to deform the airbag when they collide. In the process of deformation, the time of impact is extended, reducing the force impacted on the driver, reducing the fatality of the impact.