Answer:2m/s²
Explanation: Well F=MA so sice F=4N and M=2kg let's plug in the values
4N=2KG*A
A=4N/2KG
A=2m/s²
<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:
please write neater
Explanation:
can you write neater so I can answer th question but also is a equal to b
Answer:
Their efforts would be expressed in units of Joules per second
Explanation:
The unit of their efforts can be derived from the formula of power which is given by the product of mass, acceleration and distance (the product is energy with unit joules) divided by time taken to complete the task (unit is seconds)
Therefore, the unit of their efforts would be joules per second
