Question #3bb16

4 Answers
Mar 31, 2017

Answer:

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

Explanation:

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#

Mar 31, 2017

Answer:

question 3 answer 1

Explanation:

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#

Apr 1, 2017

Answer:

Q no 11 #->#option (2)

Explanation:

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)

Apr 1, 2017

Answer:

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

Explanation:

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#