Answer: 12Mg/h
Explanation:
Let the spring is compressed by a distance x,before the lift stops,then
Mg(h+x)= 1/2 kx^2 ............... 1
Kx - Mg = M ( 5g ) ............ 2
Make x the subject in equation 2
Kx = 5Mg + Mg
Kx = 6Mg
x = 6Mg/k ............ 3
Put equation 3 into 1
Mg ( h + x ) = 1/2 kx^2
Mgh + Mgx = 1/2kx^2
Mgh + Mg × 6Mg/k = 1/2k × ( 6Mg/k )^2
Mgh + Mg× 6Mg/k = 1/2k 36M^2g^2/ k^2
h =18Mg/k - 6Mg/h
K = 12Mg/h
If the lightbulb A in the circuit shown in the image burned out, the path for the current to flow is disrupted because one of its terminals is connected direct to the source. So, there will be no current through the lightbulbs B, C, and D, and they will turn off. Similarly it will happen, if the lightbulb D burned out.
If the lightbulb B burned out the current will continue circulating through the lightbulbs A, C, and D, because lightbulb B is connected in parallel. Similarly it will happen, if the lightbulb C burned out.
<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
=====================
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>
