Question

Question #3bb16

Answer 1

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`

Answer 2

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`

Answer 3

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)

Answer 4

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

Explanation:

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`