Answer:
1. The electric potential would stay the same.
2. The electric potential energy would double
Explanation:
1. Since the electric potential at the point is V and is not dependent on the charge q, the electric potential at the point still remains the same even if the charge q is doubled to 2q.
So, the electric potential remains the same when the charge is doubled to 2q.
2. Since potential energy U = qV where q = charge and V = electric potential. So when the charge is q and electric potential, V, the electric potential energy at the point is U = qV.
When the charge is doubled to 2q and the electic potential remains the same, the electric potential energy is now U' = 2qV = 2U (since U = qV).
So, the electric potential energy at the point is doubled when the charge is doubled.
Your forehead can withstand a force of about 6.0 kN before it fractures. Your cheekbone on the other hand can only handle about 1.3 kN before fracturing. If a 140 g baseball hits your head at 30.0 m/s and stops in 0.00150 s,
Required:
a. What is the magnitude of the ball's acceleration?
b. What is the magnitude of the force that stops the baseball?
c. What force does the baseball apply to your head? Explain?
d. Are you in danger of a fracture if the ball hits you in the forehead?
Answer:
Explanation:
a)
Final velocity v = 0 ; initial velocity u = 30 m/s , time t = .0015 s
v = u + a t
0 = 30 m/s + a x .0015 s
a = - 30 / .0015
= - 20000 m / s²
b )
Magnitude of force = m x a
= .140 kg x 20,000 m / s²
= 2800 N = 2.8 kN.
c )
The force applied by baseball = 2.8 kN .
d )
Since ball can withstand a force of 1.3 kN so it will break if 2.8 kN force acts on it . SO, head will fracture.
A stone dropped from the top of a 80m high building strikes the ground at 40 m/s after falling for 4 seconds. The stone's potential energy with respect to the ground is equal to its kinetic energy … (use g = 10 m/s 2)
A) at the moment of impact.
B) 2 seconds after the stone is released.
C) after the stone has fallen 40 m.
D) when the stone is moving at 20 m/s.
At the moment of impact both Kinetic Energy and Potential Energy should be 0, right? So it can't be A), right? Or is this wrong? Is it indeed A)? Please show work and explain it well.
Answer:
Explanation:
The answer is C because the building is 80 meters high. Before the stone is dropped, it has ONLY potential energy since kinetic energy involves velocity and a still stone has no velocity. At impact, there is no potential energy because potential energy involves the height of the stone relative to the ground and a stone ON the ground has no height; here there is ONLY kinetic.
From the First Law of Thermodynamics, we know that energy cannot be created or destroyed, it can only change form. Therefore, that means that at the halfway point of 40 meters, half of the stone's potential energy has been lost, and it has been lost to kinetic energy. Here, at 40 meters, there is an equality between PE and KE. It only last for however long the stone is AT 40 meters, which is probably a millisecond of time, but that's where they are equal.
what is the change in energy (in J) for a system that lost J of heat to the surrounding, and did 150J of work
Answer:
Change in energy Δu = 40 J
Explanation:
Given:
System heat ΔQ = 150 J
Missing value of Lost heat ΔW = 110 J
Find:
Change in energy Δu
Computation:
Change in energy Δu = System heat ΔQ - Missing value of Lost heat ΔW
Change in energy Δu = 150 - 110
Change in energy Δu = 40 J
Calculate the minimum area moment of inertia for a rectangular cross-section with side lengths 6 cm and 4 cm.
52 cm4
72 cm4
32 cm4
24 cm4
2 cm4
Answer:
Minimum Area of rectangle = 24 centimeter²
Explanation:
Given:
Length of rectangle = 6 centimeter
Width of rectangle = 4 centimeter
Find:
Minimum Area of rectangle
Computation:
Minimum Area of rectangle = Length of rectangle x Width of rectangle
Minimum Area of rectangle = 6 x 4
Minimum Area of rectangle = 24 centimeter²
A stone is dropped from a bridge. It takes 4s to reach the water below. How high is the bridge above the water?
Answer:
height is 78.4m
Explanation:
h=u.t + 0.5.g.t^2
= 0 + 0.5x9.8x4^2
= 78.4m
At what angle torque is half of the max
If you stand next to a wall on a frictionless skateboard and push the wall with a force of 44 N , how hard does the wall push on you
Answer:
44 N
Explanation:
Given that,
If you stand next to a wall on a frictionless skateboard and push the wall with a force of 44 N, then we need to find the force the wall push on you.
It is based on Newton's third law of motion which states that for an action there is an equal and opposite reaction. If the you push the wall with a force of 44 N, the wall push on you is 44 N also as it is based on Newton's third law of motion.
The pan flute is a musical instrument consisting of a number of closed-end tubes of different lengths. When the musician blows over the open ends, each tube plays a different note. The longest pipe is 0.31 m long.
What is the frequency of the note it plays? Assume room temperature of 20∘C.
Answer:
f = 276.6 Hz
Explanation:
This musical instrument can be approximated to a tube system where each tube has one end open and the other closed.
In the closed part there is a node and in the open part a belly or antinode. Therefore the wavelength is
L = λ/ 4
speed is related to wavelength and frequency
v = λ f
λ = v / f
we substitute
L = v / 4f
f = v / 4L
the speed of sound at 20ºC is
v = 343 m / s
let's calculate
f = [tex]\frac{343 }{4 \ 0.31}[/tex]
f = 276.6 Hz
Electric field is always perpendicular to the equipotential surface.
a. True
b. False
Answer:
a: true.
Explanation:
We can define an equipotential surface as a surface where the potential at any point of the surface is constant.
For example, for a punctual charge, the equipotential surfaces are spheres centered at the punctual charge.
Or in the case of an infinite plane of charge, the equipotential surfaces will be planes parallel to our plane of charge.
Now we want to see if the electric field is always perpendicular to these equipotential surfaces.
You can see that in the two previous examples this is true, but let's see for a general case.
Now suppose that you have a given field, and you have a test charge in one equipotential surface.
So, now we can move the charge along the equipotential surface because the potential in the surface is constant, then the potential energy of the charge does not change. And because there is no potential change, then there is no work done by the electric field as the charge moves along the equipotential surface.
But the particle is moving and the electric field is acting on the particle, so the only way that the work can be zero is if the force (the one generated by the electric field, which is parallel to the electric field) and the direction of motion are perpendiculars.
Then we can conclude that the electric field will be always perpendicular to the equipotential surfaces.
The correct option is a.
Charge q is 1 unit of distance away from the source charge S. Charge p is six times further away. The force exerted
between S and and q is __ the force exerted between S and p.
1/6
6 times
1/36
36 times
Answer:
Fq = k q Q / R^2
Fp = k q Q / (6 R)^2
The force exerted between S and p is 1 / 36 of that between S and q
or the force between S and q is 36 times that between S and p.
Two point charges, the first with a charge of 4.47 x 10-6 C and the second with a charge of 1.86 x 10-6 C, are separated by 17.4 mm. What is the magnitude of the electrostatic force experienced by charge 2
Answer: [tex]247.12\ N[/tex]
Explanation:
Given
Magnitude of the charges
[tex]q_1=4.47\times 10^{-6}\ C[/tex]
[tex]q_2=1.86\times 10^{-6}\ C[/tex]
Distance between them [tex]d=17.4\ mm[/tex]
As both charges are of same sign, they must repel each other
Force experienced by second charge is
[tex]\Rightarrow F_{21}=\dfrac{kq_1q_2}{d^2}\\\\\Rightarrow F_{21}=\dfrac{9\times 10^9\times 4.47\times 10^{-6}\times 1.86\times 10^{-6}}{(17.4\times 10^{-3})^2}\\\\\Rightarrow F_{21}=\dfrac{74.82\times 10^{-3}}{302.76\times 10^{-6}}\\\\\Rightarrow F_{21}=0.2471\times 10^3\\\Rightarrow F_{21}=247.12\ N[/tex]
Thus, charge 2 experience a force of [tex]247.12\ N[/tex]
Answer:
The force between the two charges is 247.15 N.
Explanation:
Charge, q = 4.47 x 10^-6 C
charge, q' = 1.86 x 10^-6 C
distance, d = 17.4 mm
Let the force is F.
The force is given by the Coulomb's law:
[tex]F = \frac{K q q'}{r^2}\\\\F =\frac{9\times 10^9\times 4.47\times 10^{-6}\times1.86\times 10^{-6}}{(17.4\times 10^{-3})^2}\\\\F = 247.15 N[/tex]
A 1030 kg car has four 12.0 kg wheels. When the car is moving, what fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles
Answer:
The required fraction is 0.023.
Explanation:
Given that
Mass of a car, m = 1030 kg
Mass of 4 wheels = 12 kg
We need to find the fraction of the total kinetic energy of the car is due to rotation of the wheels about their axles.
The rotational kinetic energy due to four wheel is
[tex]=4\times \dfrac{1}{2}I\omega^2\\\\=4\times \dfrac{1}{2}\times \dfrac{1}{2}mR^2(\dfrac{v}{R})^2\\\\=mv^2[/tex]
Linear kinetic Energy of the car is:
[tex]=\dfrac{1}{2}mv^2\\\\=\dfrac{1}{2}\times Mv^2[/tex]
Fraction,
[tex]f=\dfrac{mv^2}{\dfrac{1}{2}Mv^2}\\\\f=\dfrac{m}{\dfrac{1}{2}M}\\\\f=\dfrac{12}{\dfrac{1}{2}\times 1030}\\\\=0.023[/tex]
So, the required fraction is 0.023.
3. Three blocks of masses m, 2m and 3m are suspended from the ceiling using ropes as shown in diagram. Which of the following correctly describes the tension in the three rope segments?
a. T1< T2 < T3
b. T1< T2 = T3
c. T1 = T2 = T3
d. T1> T2 > T3
please help.show how and which?
see attachment for more detail.
Option d (T₁ > T₂ > T₃) correctly describes the tension in the three rope system.
Let's evaluate each tension.
Case T₃.
[tex] T_{3} - W_{3} = 0 [/tex]
For the system to be in equilibrium, the algebraic sum of the tension force (T) and the weight (W) must be equal to zero. The minus sign of W is because it is in the opposite direction of T.
[tex] T_{3} = W_{3} [/tex]
Since W₃ = mg, where m is for mass and g is for the acceleration due to gravity, we have:
[tex] T_{3} = W_{3} = mg [/tex] (1) Case T₂.
[tex] T_{2} - (T_{3} + W_{2}) = 0 [/tex]
[tex] T_{2} = T_{3} + W_{2} [/tex] (2)
By entering W₂ = 2mg and equation (1) into eq (2) we have:
[tex] T_{2} = T_{3} + W_{2} = mg + 2mg = 3mg [/tex]
Case T₁.
[tex] T_{1} - (T_{2} + W_{1}) = 0 [/tex]
[tex] T_{1} = T_{2} + W_{1} [/tex] (3)
Knowing that W₁ = 3mg and T₂ = 3mg, eq (3) is:
[tex] T_{1} = 3mg + 3mg = 6mg [/tex]
Therefore, the correct option is d: T₁ > T₂ > T₃.
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Correct answer: D. [tex]T_{1} > T_{2} > T_{3}[/tex]
First, we must construct the Equations of Equilibrium for each mass based on Newton's Laws of Motion, then we solve the resulting system for every Tension force:
Mass m:
[tex]\Sigma F = T_{3}-m\cdot g = 0[/tex] (1)
Mass 2m:
[tex]\Sigma F = T_{2}-2\cdot m \cdot g -T_{3} = 0[/tex] (2)
Mass 3m:
[tex]\Sigma F = T_{1}-3\cdot m\cdot g - T_{2} = 0[/tex] (3)
The solution of this system is: [tex]T_{3} = m\cdot g[/tex], [tex]T_{2} = 3\cdot m\cdot g[/tex] and [tex]T_{1} = 6\cdot m\cdot g[/tex], which means that [tex]T_{1} > T_{2} > T_{3}[/tex]. (Correct answer: D.)
A satellite measures a spectral radiance of 8 Watts/m2/um/ster at a wavelength of 10 microns. Assuming a surface emissivity of 0.90, what would be the estimated temperature
If 10 W of power is supplied to 1 kg of water at 100℃, how long will it take to for the water to completely boil away? The time calculated is a little less than actual time of boiling in practice. Why?
Answer:
t = 2.26 x 10⁵ s
Explanation:
The energy supplied to the water will be equal to the heat required for the boiling of water:
E = ΔQ
Pt = mL
where,
P = Power = 10 W
t = time = ?
m = mass of water = 1 kg
L = Latent heat of vaporization of water = 2.26 x 10⁶ J/kg
Therefore,
[tex](10\ W)t = (1\ kg)(2.26\ x\ 10^6\ J/kg)\\\\t = \frac{2.26\ x\ 10^6\ J}{10\ W}\\\\[/tex]
t = 2.26 x 10⁵ s
This time will be less than the actual time taken due to some heat loss during the transmission of this heat energy to the container in which water is held.
A 0.4 m long solenoid has a total of 356 turns of wire and carries a current of 79 A. What is the magnitude of the magnetic field at the center of the solenoid?
Answer:
B = 0.088 T
Explanation:
Given that,
The length of a solenoid, l = 0.4 m
No. of turns of wire, N = 356
Current, I = 79 A
We need to find the magnitude of the magnetic field at the center of the solenoid. It is given by the formula.
[tex]B=\mu_onI\\\\B=\mu_o \dfrac{N}{l}\times I\\\\B=4\pi \times 10^{-7}\times \dfrac{356}{0.4}\times 79\\\\B=0.088\ T[/tex]
So, the magnitude of the magnetic field at the center of the solenoid is 0.088 T.
Here we will use a simple example to demonstrate the difference between the rms speed and the average speed. Five ideal-gas molecules chosen at random are found to have speeds of 500, 600, 700, 800, and 900 m/s. Find the rms speed for this collection. Is it the same as the average speed of these molecules
Explanation:
Given that,
Five ideal-gas molecules chosen at random are found to have speeds of 500, 600, 700, 800, and 900 m/s.
The rms speed for this collection is as follows :
[tex]v_{rms}=\sqrt{\dfrac{v_1^2+v_2^2+v_3^2+v_4^2+v_5^2}{5}} \\\\v_{rms}=\sqrt{\dfrac{500^2+600^2+700^2+800^2+900^2}{5}} \\\\v_{rms}=714.14[/tex]
The average speed of these molecules is :
[tex]v_{a}=\dfrac{v_1+v_2+v_3+v_4+v_5}{5}} \\\\v_{a}={\dfrac{500+600+700+800+900}{5}} \\\\v_{a}=700\ m/s[/tex]
So, the rms speed is 714.14 m/s abd the average speed is 700 m/s.
(a) State Hook's law. [2]
(b) The walls of the tyres on a car are made of a rubber
compound. The variation with stress of the strain of a
specimen of this rubber compound is shown in Fig. 1.2.
As the car moves, the walls of the tyres end and straighten
continuously. Use Fig. 1.2 to explain why the walls of the
tyres become warm
[3]
hin
Fig. 1.2.
Hooke's law gives the relationship between the force applied and observed compression and expansion
(a) Hooke's law states that force applied is proportional to the deformation of an object
(b) The tube becomes warm from the excess of the energy absorbed but not given off back as the straightening of the tyre
The reason for the above explanation are as follows;
(a) Hooke's law states that for little deformations, the change in the dimension, Δx, of an extended or compressed object is directly proportional to the applied force, F
Mathematically, we get;
F = k × ΔL
[tex]k = \mathbf{\dfrac{F}{\Delta L}}[/tex]
Given that the material cross sectional area = A, and the original length of the material = L, we get;
F/A = Stress = σ, ΔL/L = strain = ε
[tex]\mathbf{Young's \ Modulus, \ E} = \mathbf{\dfrac{\sigma}{\varepsilon}} = \dfrac{\left(\dfrac{F}{A} \right) }{\left(\dfrac{\Delta L}{L} \right) } = \mathbf{\dfrac{F}{\Delta L } \times \dfrac{L}{A}}[/tex]
Therefore, Hooke's law can be expressed as a form of Young's Modulus for a given length to area ratio of a material
(b) The given graph of stress to strain curve, is attached, from which area under the curve gives the energy absorbed by the the material during deformation
Therefore during bending, the stress is increasing as shown in the top of the two curve, and the energy absorbed is given by the area under the curve
As the tyre straightens, the path of the stress change curve is given by the lower curve
The area under the top (bending) curve is larger than the area under the lower (straightening) curve, therefore, the energy absorbed during bending is larger than the energy given off during straightening
Energy absorbed < Energy released
The balance energy is transformed into other forms of energy, including heat energy, which is observed by the raising in temperature, warming, of the tyre
Energy absorbed = Energy released + Heat energy
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When placed 1.18 m apart, the force each exerts on the other is 11.2 N and is repulsive. What is the charge on each
Answer:
[tex]q=41.62\ \mu C[/tex]
Explanation:
Given that,
Force between two objects, F = 11.2 N
Distance between objects, d = 1.18 m
We need to find the charge on each objects. The force between charges is as follows :
[tex]F=\dfrac{kq^2}{r^2}\\\\q=\sqrt{\dfrac{Fr^2}{k}} \\\\q=\sqrt{\dfrac{11.2\times (1.18)^2}{9\times 10^9}} \\\\q=41.62\ \mu C[/tex]
So, the charge on each sphere is [tex]41.62\ \mu C[/tex].
What is the magnitude of force required to accelerate a car of mass 1.7 * 10 ^ 3 kg by 4.75 m/s
What is the magnitude of force required to accelerate a car of mass 1.7 × 10³ kg by 4.75 m/s²
Answer:
F = 8.075 N
Explanation:
Formula for force is;
F = ma
Where;
m is mass
a is acceleration
F = 1.7 × 10³ × 4.75
F = 8.075 N
How does this work?
Why won't it fall?
Why can't you continue building the tower forever?
From what I've been told, if the center of mass leaves it's support area, the object falls.
But the object on the tops center of mass is certainly outside of its support area (the object at the bottom)
Please explain! Thank you in advance. : )
Answer with explanation:
*Pre-condition: The mass of all blocks are evenly distributed.
As you add more and more blocks, the center of mass of the system changes.
Let the first block be the support. The maximum distance that can be unsupported (hung over) by the second block is equal to half the length of second block.
Why?
Let's take a look at our formula for torque. Torque, [tex]\tau[/tex], is given by:
[tex]\tau=rF\sin \theta[/tex], where [tex]r[/tex] is radius, [tex]F[/tex] is force, and [tex]\theta[/tex] is the angle between the radius and level arm.
[tex]\sin \theta[/tex] is used to calculate the relevant component of a force producing torque. In this case, the only force acting on the object is the force of gravity. Therefore, [tex]\theta =90^{\circ}[/tex] and recall [tex]\sin 90^{\circ}=1[/tex].
Conceptually, it's more important to look at the [tex]r[/tex] term here. From our formula, we can see that if [tex]r=0[/tex], there is no torque.
The point of pivot will be at the center of mass. Here's the important part:
As long as the point of pivot is supported, [tex]r[/tex] will remain zero and no torque will be created from the force of gravity. As you keep stacking blocks, as long as the center of mass of the entire system remains supported from your first support, the tower will not fall.
In the given picture shown, that first block will be your support.
The 6 blocks on top of that first block form a center of mass that is still on that first block, thus allowing the tower to remain standing.
1. A turtle and a rabbit are to have a race. The turtle’s average speed is 0.9 m/s. The rabbit’s average speed is 9 m/s. The distance from the starting line to the finish line is 1500 m. The rabbit decides to let the turtle run before he starts running to give the turtle a head start. If the rabbit started to run 30 minutes after the turtle started, can he win the race? Explain.
Answer:no
Explanation:because 0.9*(30*60)=0.9*1800=1620
The turtle has already won the race
Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours
What will be the speed of the rabbit and the turtle?It is given
[tex]V_{t} = 0.9 \frac{m}{s}[/tex]
[tex]V_{r} = 9 \frac{m}{s}[/tex]
[tex]D=1500 m[/tex]
Time taken by turtle
[tex]T= \dfrac{D}{V_{t} }=\dfrac{1500}{0.9_{} }[/tex]
[tex]T=1666 minutes= 27 hours[/tex]
Time taken by rabbit
[tex]T= \dfrac{D}{V_{r} }=\dfrac{1500}{9_{} }[/tex]
[tex]T=166 minutes[/tex]
since rabbit started 30 minutes after turtle then
[tex]T= 136+30=196 minutes[/tex]
[tex]T= 3.2 hours[/tex]
Hence Yes rabbit will win the race will distance in 3.2 hours and turtle will cover in 27 hours
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find the exit angle relative to the horizontal in an isosceles triangle with 36 °
what
what
what
what
sorry
sorrry
sorry
Write a short note of the following
a) Reflection
b) Refraction
c) Diffraction
Answer:
a) Light that passes through the floor to reveal yourself (not shadow).
b) 2 rays of light that bounce between 2 transparent media.
c) I don't know what is Diffraction?
Simple physics question
Answer:
option (b) is the correct answer
How fast should a moving clock travel if it is to be observed by a stationary observer as running at one-half its normal rate?A) 0.50c
B) 0.65c
C) 0.78c
D) 0.87c
Answer:
Option (D) is correct.
Explanation:
Let the speed is v.
[tex]\Delta t = \gamma \Delta t'\\\\\Delta t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\times \frac{\Delta t}{2}\\\\\sqrt{1-\frac{v^2}{c^2}} =\frac{1}{2}\\\\1-\frac{v^2}{c^2}=\frac{1}{4}\\\\\frac{3}{4}c^2 = v^2\\\\v = 0.87 c[/tex]
Option (D) is correct.
A 0.40-kg mass attached to the end of a string swings in a vertical circle having a radius of 1.8 m. At an instant when the string makes an angle of 40 degrees below the horizontal, the speed of the mass is 5.0 m/s. What is the magnitude of the tension in the string at this instant
Answer:
[tex]T=8.1N[/tex]
Explanation:
From the question we are told that:
Mass m=0.40
Radius r=1.8m
Angle Beneath the Horizontal \theta =40 \textdegree
Speed v=5.0m/s
The Tension Angle
[tex]\alpha=90-\theta\\\\\alpha=90-40[/tex]
[tex]\alpha=50 \textdegree[/tex]
Generally the equation for Tension is is mathematically given by
[tex]T=\frac{mv^2}{r}+mgcos \alpha[/tex]
[tex]T=\frac{0.40*5^2}{1.8}+0.40*5cos50[/tex]
[tex]T=8.1N[/tex]
Army is standing still on the ground; Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward. How fast does Bill see Carlos moving and in what direction?
a. 10 mis eastward
b. 5 m/s eastward
c. 15 m/s westward
d. 20 m/s westward
e. 10 m/s westward
Explanation:
Given that,
Bill is riding his bicycle at 5 m/s eastward: and Carlos is driving his car at 15 m/s westward.
Taking eastward as positive direction, we have:
[tex]v_B=+5\ m/s[/tex]is the velocity of Bill with respect to Amy (which is stationary)
[tex]v_c=15\ m/s[/tex] is the velocity of Carlos with respect to Amy.
Bill is moving 5 m/s eastward compared to Amy at rest, so the velocity of Bill's reference frame is
[tex]v_B=+5\ m/s[/tex]
Therefore, Carlos velocity in Bill's reference frame will be
[tex]v_c'=-15\ m/s-(+5\ m/s)\\\\=-20\ m/s[/tex]
So, the magnitude is 20 m/s and the direction is westward (negative sign).
3. The figure below shows the motion of a car. It starts from the origin, O travels 8m
towards the east and then 12m towards the west.
D
8m.
X
X-8
12m.w
()What is the net displacement D from the origin to the final position?
(ii) What is the total distance travelled by the car?
Answer:
i. -4m
ii. 20m
Explanation:
The car travels 8m to the east, then travels 12m to the west which is the opposite of the east. Going west, the car travels 8m back to the origin point and then another 4m due west to make 12m. The displacement from the origin point is -4 (the negative sign shows the direction because displacement is a vector quantity)
Total distance = 8m going east + 8m back to origin + 4m west = 20m
Two identical cars, each traveling at 16 m>s, slam into a concrete wall and come to rest. In car A the air bag does not deploy and the driver hits the steering wheel; in car B the driver contacts the deployed air bag. (a) Is the impulse delivered by the steering wheel to driver A greater than, less than, or equal to the impulse delivered by the air bag to driver B
Answer:
I = - m 16 the two impulses are the same,
Explanation:
The impulse is given by the relationship
I = Δp
I = p_f - p₀
in this case the final velocity is zero therefore p_f = 0
I = -p₀
For driver A the steering wheel impulse is
I = - m v₀
I = - m 16
For driver B, the airbag gives an impulse
I = - m 16
We can see that the two impulses are the same, the difference is that in the air bag more time is used to give this impulse therefore the force on the driver is less
