# Look at the attachement and asnwer the questions

1
by rajusetu

2015-05-01T03:11:26+05:30

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8.
Normal force on B by the floor = F
Force exerted by block A on B:  F1
F - m_B g -  F1 = - m_B * 2        =>  F = F1 + m_B g - 2 m_B
F1 - m_A  g = - m_A * 2            =>  F1 = g/2 - 1  = 4  Newtons

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9.  component of weight/acceleration due to gravity along the inclined slope
= g Sin 30 = g/2

Component of weight/acceleration along the groove OA = g/2 Sin 30 = g/4

s = u t + 1/2 a t²    => t² = 2 s / a
t² = 2 * 5 /  (g/4) = 4              =>           t = 2 sec.
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10.      m1 = 100 kg          m2 = 60 kg

a1 = acceleration of 100 kg upwards.  Rope is also moving with the same acceleration.
T - 100 kg * g = 100 kg * a1        => T = 1000+ 100 a1          ---- (1)

Acceleration of the person = a
relative acceleration of the man wrt rope  = a + a1  = 5g/4
=> a = 5g/4 - a1  = 12.5 - a1      upwards
T - 60 kg * g = 60 kg * a
T = 600 + 60 a = 600 + 60 (12.5 - a1) = 1350 - 60 a1      ----- (2)

Solving the two equations          a1 = 350/160  m/sec/sec
T = 1000 + 100 * 35/16 = 1218  N
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11
F = buoyancy force upwards  remains same
F - M g  = M a        => F = M (g+a)
F - (M-m) g = (M-m) a'          =>  a' = [ M g + Ma - Mg + mg ] / (M-m)
a' = (M a + m g)/(M-m)
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12.        T - 5 g = 5 a
10 g - T = 10 a
a = g/3
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13.      F - m g Sin 37  = m a          along the slope
5.5 - 0.50 * g * 0.6  = 0.50 a      =>  a = 5 m/sec/sec

v^2 = 2 a s = 2 * 5 * 10  = 100
v = 10 m/s    at the top of the incline.
angle of projection = 37 deg.
y initial = 10 Sin 37 = 6 meters

Time to reach the top point :    u Sin 37 / g = 10 * 0.6 / 10 = 0.60 sec
height reached = s  =  u sin 37 * 0.60 - 1/2 * 10 * 0.60^2
= 10  * 0.6 * 0.60 - 5 * 0.60^2  = 1.80 meters
total height at the top of the flight = 6 + 1.80 = 7.80 m

time to reach the ground = t  => t^2 = 2s / g =>        t = 2 * 7.80 / 10 = 1.56 sec

total time of the flight = 2.16 sec.
total  horizontal distance travelled after projection from the ramp
=  u cos 37 * 1.56 sec
= 12.48 meters

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