The first thing you should know for this case is the definition of distance.
d = v * t
Where,
v = speed
t = time
We have then:
d = v * t
d = 9 * 12 = 108 m
The kinetic energy is:
K = ½mv²
Where,
m: mass
v: speed
K = ½ * 1500 * (18) ² = 2.43 * 10 ^ 5 J
The work due to friction is
w = F * d
Where,
F = Force
d = distance:
w = 400 * 108 = 4.32 * 10 ^ 4
The power will be:
P = (K + work) / t
Where,
t: time
P = 2.86 * 10 ^ 5/12 = 23.9 kW
answer:
the average power developed by the engine is 23.9 kW
Answer:
write the name of any five districts of nepal
In order to solve the problem, it is necessary to apply the concepts related to the conservation of momentum, especially when there is an impact or the throwing of an object.
The equation that defines the linear moment is given by
where,
m=Total mass
Mass of Object
Velocity before throwing
Final Velocity
Velocity of Object
Our values are:
Solving to find the final speed, after throwing the object we have
We have three objects. For each object a launch is made so the final mass (denominator) will begin to be subtracted successively. In addition, during each new launch the initial speed will be given for each object thrown again.
That way during each section the equations should be modified depending on the previous one, let's start:
A) 
B) 
C) 
Therefore the final velocity of astronaut is 3.63m/s
Answer:3.4 seconds
Explanation:
Initial velocity(u)=0
acceleration=34.5m/s^2
Height(h)=200m
Time =t
h=u x t - (gxt^2)/2
200=0xt+(34.5xt^2)/2
200=34.5t^2/2
Cross multiply
200x2=34.5t^2
400=34.5t^2
Divide both sides by 34.5
400/34.5=34.5t^2/34.5
11.59=t^2
t^2=11.59
Take them square root of both sides
t=√(11.59)
t=3.4 seconds
