<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>
Given: Mass m = 44 Kg; Velocity v = 10 m/s
Required: Kinetic energy K.E = ?
Formula: K.E = 1/2 mv²
K.E 1/2 (44 Kg)(10 m/s)²
K.E = 2,200 Kg.m²/s²
K.E = 2,200 J Answer is A
Answer:
im pretty sure it is C. Insulation
Explanation:
because it says reduced, and basically ur insulating the fire.
The moon<span> is 1/4 the size of </span>Earth<span>, so the </span>moon's<span> gravity is much less than the </span>earth's gravity, 83.3% (or 5/6) less to be exact. Finally, "weight<span>" is a measure of the gravitational pull between two objects. So of course you would </span>weigh<span> much less on the </span>moon<span>.</span>
Given,
the initial velocity = 0 m /s.
acceleration = 3.20 m / s^2
time = 32.8 s
According to laws of motion.
s = ut + 1/2 at ^2
s = 1/2 at²
s=1/2(3.20)(32.8)²
s= 1721.344 m
the distance traveled before takeoff is 1731.3m
