Answer:
Distance travelled is 7 meters and the displacement is 3 meters
Answer:
v = 306.76 Km/h
Explanation:
given,
height of the aircraft = 3000 m
differential pressure reading = 3300 N/m²
density of air = 0.909 Kg/m³
speed of aircraft = ?
Assuming the air flowing above air craft is in-compressible, irrotational and steady so, we can use Bernoulli's equation to solve the problem.
using Bernoulli's equation

where ρ is the density of the air at 3000 m



v = 85.21 m/s

v = 306.76 Km/h
Answer:
blah blah blahblah
Explanation:
because math is math and nobody understands it
Answer:

Now when it will reach at point B then its normal force is just equal to ZERO


Explanation:
Since we need to cross both the loops so least speed at the bottom must be

also by energy conservation this is gained by initial potential energy


so we will have

now we have

here we have
R = 7.5 m
so we have


Now when it will reach at point B then its normal force is just equal to ZERO

now when it reach point C then the speed will be
![mgh - mg(2R_c) = \frac{1}{2]mv_c^2](https://tex.z-dn.net/?f=mgh%20-%20mg%282R_c%29%20%3D%20%5Cfrac%7B1%7D%7B2%5Dmv_c%5E2)


now normal force at point C is given as



As we know that with respect to oxygen atom taken as reference the product of atomic mass and specific heat of a metal will remain constant.
this product is equal to 0.38
so here we will say that let atomic mass of the metal is M
so with respect to oxygen atom its mass is given as

now we will have

now we will have

so atomic mass of the metal is 7 g/mol