<span>let the fsh jump with initial velocity (u) in direction (angle p) with horizontal
it can cross and reach top of trajectory if its top height h = 1.5m
and horizontal distance d = (1/2) Range
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let t be top height time
at top height, vertical component of its velocity =0
vy = 0 = u sin p - gt
t = u sin p/g
h = [u sin p]*t - 0.5 g[t[^2
1.5 = u^2 sin^2 p/g - u^2 sin^2 p/2g
u^2 sin^2 p/2g = 1.5
u^2 sin^2 p = 1.5*2*9.8 = 29.4
u sin p = 5.42 m/s >>>>>>>>>>>>>>> V-component
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t = HALF the time of flight
d = (1/2) Range (R) = (1/2) [2 u^2 sin p cos p/g]
1 = u^2 sin p cos p/g
u sin p * u cos p = 9.8
5.42 * u cos p = 9.8
u cos p = 1.81 m/s >>>>>>>>>>>>> H-component
check>>
u = sqrt[u^2 cos^2 p + u^2 sin^2 p] = 5.71 m/s
u < less than fish's potential jump speed 6.26 m/s
so it will able to cross</span>
Answer:
all I know is C
are there more questions? anyone?
-KARL IS STOOPID
Explanation:
Answer:
W = 0
Explanation:
We are given with, a construction worker is carrying a load of 40 kg over his head and is walking at a constant velocity. He travels a distance of 50 m.
The work done by an object is given by :

F = ma
So,

m is mass
a is acceleration
d is displacement
The worker is moving with constant velocity, its acceleration will be 0. So, the work done by the worker is 0.
Answer:
As the tines of the tuning fork vibrate at their own natural frequency, they created sound waves that impinge upon the opening of the resonance tube. These impinging sound waves produced by the tuning fork force air inside of the resonance tube to vibrate at the same frequency.
Answer:
4.44s
Explanation:
A 34-kg child on an 18-kg swing set swings back and forth through small angles. If the length of the very light supporting cables for the swing is 4.9 m, how long does it take for each complete back-and-forth swing? Assume that the child and swing set are very small compared to the length of the cables
since the mass of the child and that of the swing is negligible, the masses wont be involved in the calculation
T=2π√L/g
g=acceleration due to gravity which is 9.81m/s2
the length of the supporting cable is 4.9m
T the period
period is the time required to make a complete oscillation
T=2*π√4.9/9.81
T=2*π*0.706
T=4.44s
4.44s