Question #3bb16

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Answer 1 · 2017-03-31T05:16:22

If you mean the last problem, the answer is 3) 3.92 m/s^2

Forces: $m_B g$
acting on two masses without friction
so
$m_B g = (m_A + m_B)a$
$a = m_B/ (m_A + m_B) g= 3.92$ $m/s^2$

Answer 2 · 2017-03-31T22:02:30

question 3 answer 1

Constant velocity means zero acceleration i.e. sum of forces is $T cos 30 = mu (mg -Tsin 30)$
or $mu = 5 sqrt(3)/2/[(3*9.81) - 2.5] = 0.161$

Answer 3 · 2017-04-01T01:12:59

Q no 11 $->$ option (2)

Initial velocity of the car $u=0$

Applied force $F=400N$

Final velocity of the car $v=14km"/"h=(14xx1000)/3600=35/9m"/"s$

Time taken to change in velocity $t=12s$

Acceleration produced $a=(v-u)/t=35/(12xx9)=35/108m"/"s^2$

So by Newton's law the mass of the car

$M =F/a=400/(35/108)kg=(400xx108)/35kg~~1230kg$ (rounding up to 3 significant figure)

Answer 4 · 2017-04-01T02:24:49

Q No 12 $->$ Option (1) 69.6N

drawn

Considering the equilibrium of forces in the vertical horizontal direction we can write

$T_1sin40+T_2sin45=10xx9.81=98.1N.......[1]$

$T_1cos40=T_2cos45.....[2]$

Adding [1] and [2] we get

$T_1sin40+T_2sin45+T_1cos40=98.1+T_2cos45$

$=>T_1sin40+cancel(T_2/sqrt2)+T_1cos40=98.1+cancel(T_2/sqrt2)$

$=>T_1=9.81/(sn40+cos49)~~69.6N$