Answer:
2.5 x 10⁷ J
Explanation:
F = thrust of the engine = 2.3 x 10⁵ N
d = distance traveled = 87 m
Work done by the engine is given as
W = F d = (2.3 x 10⁵) (87) = 200.1 x 10⁵ J
W' = Net work done
W'' = work done by catapult
KE₀ = initial kinetic energy = 0 J
KE = final kinetic energy = 4.5 x 10⁷ J
Net work done is given as
W' = KE - KE₀
W' = 4.5 x 10⁷ J
We know that
W' = W + W''
4.5 x 10⁷ = 2.001 x 10⁷ + W''
W'' = 2.5 x 10⁷ J
Answer:
'A ball is thrown straight up with an initial speed of 12 m/s_ What are the velocity and acceleration when it is at the top of its trajectory? Select all apply. v=12 mls a = 0 v =-12 mls a = 9.8 m/s2 Oa=-9.8 m/s2'
Explanation:
I look it up
Hope this helps
heat engine that uses work to move heat
mark brainliest :)
Answer:
<h2>34.4m/s</h2>
Explanation:
Step one:
given data
initial speed u= 5m/s
time t= 3seconds
acceleration due to gravity g= 9.8m/s^2
Required:
the final velocity v
Step two:
applying the first equation of motion
v=u+gt-----------we used + because the ball is falling with gravity
v=5+9.8*3
v=5+29.4
v=34.4
The final velocity is 34.4m/s
