Answer:
Group 18
Explanation:
hope this helps, pls mark brainliest :D
By collapsing and crumpling, the force and energy is absorbed by the crumple zone and it is not transferred to the passengers inside. This is a safety feature for the passengers.
Answer:
height is 4 times
Explanation:
speed of ball hit by Luis is u
speed of ball hit by Ron is 2u
Let the maximum height traveled by the ball hit by Luis is h and by the ball hit by Ron is h'
At maximum height, the final velocity of ball is zero in both the cases.
Use third equation of motion, we get
For Luis :
![v^{2}=u^{2}-2gh](https://tex.z-dn.net/?f=v%5E%7B2%7D%3Du%5E%7B2%7D-2gh)
![0^{2}=u^{2}-2gh](https://tex.z-dn.net/?f=0%5E%7B2%7D%3Du%5E%7B2%7D-2gh)
.... (1)
For Ron :
![v^{2}=u^{2}-2gh](https://tex.z-dn.net/?f=v%5E%7B2%7D%3Du%5E%7B2%7D-2gh)
![0^{2}=4u^{2}-2gh'](https://tex.z-dn.net/?f=0%5E%7B2%7D%3D4u%5E%7B2%7D-2gh%27)
.... (2)
Divide equation (2) by equation (1), we get
![\frac{h'}{h}=4](https://tex.z-dn.net/?f=%5Cfrac%7Bh%27%7D%7Bh%7D%3D4)
h' = 4 h
Thus, the statement of third student is wrong.
Answer:
The value is ![A = 39315 \ m^2](https://tex.z-dn.net/?f=A%20%20%20%3D%2039315%20%5C%20%20m%5E2)
Explanation:
From the question we are told that
The velocity which the rover is suppose to land with is
The mass of the rover and the parachute is
The drag coefficient is
The atmospheric density of Earth is ![\rho = 1.2 \ kg/m^3](https://tex.z-dn.net/?f=%5Crho%20%3D%20%201.2%20%5C%20%20kg%2Fm%5E3)
The acceleration due to gravity in Mars is ![g_m = 3.689 \ m/s^2](https://tex.z-dn.net/?f=g_m%20%20%3D%20%203.689%20%5C%20%20m%2Fs%5E2)
Generally the Mars atmosphere density is mathematically represented as
![\rho_m = 0.71 * \rho](https://tex.z-dn.net/?f=%5Crho_m%20%20%3D%20%200.71%20%2A%20%20%5Crho)
=> ![\rho_m = 0.71 * 1.2](https://tex.z-dn.net/?f=%5Crho_m%20%20%3D%20%200.71%20%2A%20%201.2)
=> ![\rho_m = 0.852 \ kg/m^3](https://tex.z-dn.net/?f=%5Crho_m%20%20%3D%200.852%20%5C%20%20kg%2Fm%5E3)
Generally the drag force on the rover and the parachute is mathematically represented as
![F__{D}} = m * g_{m}](https://tex.z-dn.net/?f=F__%7BD%7D%7D%20%3D%20%20m%20%20%2A%20%20g_%7Bm%7D)
=>
=>
Gnerally this drag force is mathematically represented as
![F__{D}} = C__{D}} * A * \frac{\rho_m * v^2 }{2}](https://tex.z-dn.net/?f=F__%7BD%7D%7D%20%3D%20%20%20C__%7BD%7D%7D%20%2A%20%20A%20%2A%20%20%5Cfrac%7B%5Crho_m%20%2A%20v%5E2%20%7D%7B2%7D)
Here A is the frontal area
So
![A = \frac{2 * F__D }{ C__D} * \rho_m * v^2 }](https://tex.z-dn.net/?f=A%20%20%20%3D%20%20%5Cfrac%7B2%20%2A%20%20F__D%20%7D%7B%20C__D%7D%20%20%2A%20%20%5Crho_m%20%20%2A%20v%5E2%20%20%20%7D)
=> ![A = \frac{2 * 8374 }{ 0.5 * 0.852 * 1 ^2 }](https://tex.z-dn.net/?f=A%20%20%20%3D%20%20%5Cfrac%7B2%20%2A%208374%20%7D%7B%200.5%20%2A%20%200.852%20%20%20%20%2A%20%201%20%5E2%20%20%20%7D)
=> ![A = 39315 \ m^2](https://tex.z-dn.net/?f=A%20%20%20%3D%2039315%20%5C%20%20m%5E2)