As it is given that the air bag deploy in time

total distance moved by the front face of the bag

Now we will use kinematics to find the acceleration




now as we know that

so we have

so the acceleration is 400g for the front surface of balloon
Answer:
A colorful interference pattern is observed when light is reflected from the top and bottom boundaries of a thin oil film. The different bands form as the film's thickness diminishes from a central run-off-poin
Explanation:
<h3>hope it help brainliest pls</h3>
Answer:
Explanation:
When the central shaft rotates , the seat along with passenger also rotates . Their rotation requires a centripetal force of mw²R where m is mass of the passenger and w is the angular velocity and R is radius of the circle in which the passenger rotates.
This force is provided by a component of T , the tension in the rope from which the passenger hangs . If θ be the angle the rope makes with horizontal ,
T cos θ will provide the centripetal force . So
Tcosθ = mw²R
Tsinθ component will balance the weight .
Tsinθ = mg
Dividing the two equation
Tanθ = 
Hence for a given w , θ depends upon g or weight .
Answer: 37.5 nm
Explanation: speed of light c= 3.00·10^8 m/s.
I use same accuracy to speed of light as it's for frequency.
Frequency f= 8.01·10^15 1/s
Speed c = wavelength · frequency
Wavelength = c/f = 3.745·10^-8 m
The answer is 4.0 kg since the flywheel comes to rest the
kinetic energy of the wheel in motion is spent doing the work. Using the
formula KE = (1/2) I w².
Given the following:
I = the moment of inertia about the
axis passing through the center of the wheel; w = angular velocity ; for the
solid disk as I = mr² / 2 so KE = (1/4) mr²w². Now initially, the wheel is spinning
at 500 rpm so w = 500 * (2*pi / 60) rad / sec = 52.36 rad / sec.
The radius = 1.2 m and KE = 3900 J
3900 J = (1/4) m (1.2)² (52.36)²
m = 3900 J / (0.25) (1.2)² (52.36)²
m = 3.95151 ≈ 4.00 kg