Answer:
The time taken by the object to reach the ground is 0.58 seconds.
Explanation:
Given that,
An object was released from rest at height of 1.65 m with respect to ground. We need to find the time taken by the object to reach the ground. Initial speed of the object is 0 as it is at rest. It will move downward under the action of gravity such that, the distance covered by the object is given by :
t = 0.58 seconds
So, the time taken by the object to reach the ground is 0.58 seconds. Hence, this is the required solution.
The potential energy would be zero. Only kinetic energy is present in this case. To find out what the answer is we do the equation: mv^2/2 soo...
KE =mv^2/2
KE= 1(2^2)/2 which the answer will come up by 2 Joules.
Because when an object is in motion it has kinetic energy
Answer:
a) - 72.5°c
b) pressure = 3625.13 Pa
c) density = 0.063 kg/m^3
d) it is a subsonic aircraft
Explanation:
a) Determine Temperature
Temperature at 19.5 km ( 19500 m )
T = -131 + ( 0.003 * altitude in meters )
= -131 + ( 0.003 * 19500 ) = - 72.5°c
b) Determine pressure and density at 19.5 km altitude
Given :
Po (atmospheric pressure at sea level ) = 101kpa
R ( gas constant of air ) = 0.287 KJ/Kgk
T = -72.5°c ≈ 200.5 k
pressure = 3625.13 Pa
hence density = 0.063 kg/m^3
attached below is the remaining part of the solution
C) determine if the aircraft is subsonic or super sonic
Velocity ( v ) = 
= 
= 283.8 m/s
hence it is a subsonic aircraft
Answer:
Explanation:
ACCORDING TO NEWTONS SECOND LAW;
F = mass * acceleration
F = m(v-u/t)
m is the mass = 0.15kg
v is the final velocity = 11m/s
u is the initial velocity = 0m/s
t is the time = 0.015
Substitute;
F = 0.15(11-0)/0.015
F = 0.15(11)/0.015
F = 1.65/0.015
F = 110N
Hence the net force is 110N
