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1) Answer: C
The speed of boy on landing will be
𝑣 = √2𝑔ℎ
𝑣 = √2(9.81)(3)
𝑣 = 7.67 𝑚/𝑠
Assuming the boy to be having mass m = 60 Kg
Chane in momentum is
∆𝑃 = 𝑚𝑣 − 0
∆𝑃 = 60 (7.67)
∆𝑃 = 4.6 𝑋 102 𝐾𝑔𝑚/𝑠
The closest option is C
2) Answer: C
Distance covered in first 20 s = 3(20) = 60 m
Distance covered in next 20 s = 4(20) = 80 m
Total distance travelled = 60 +80 = 140 m
Average speed is given by
𝑣=
𝑇𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑
𝑇𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
𝑣=
140
40
𝑣 = 3.5 𝑚/𝑠
3) Answer: B
The area under a-t graph gives change in velocity.
4) Answer: C
a
T
Tcosθ
Tsinθ
ma
mg
If we work in the train carriage frame we are in a non-inertial frame of
reference (accelerated frame) so we need to apply a pseudo force ma in a direction opposite to
the acceleration.
Now, balancing the forces
𝑇𝑠𝑖𝑛𝜃 = 𝑚𝑎
𝑇𝑐𝑜𝑠𝜃 = 𝑚𝑔
Dividing the two equations we have
𝑇 𝑠𝑖𝑛𝜃 𝑚𝑎
=
𝑇 𝑐𝑜𝑠𝜃 𝑚𝑔
𝑡𝑎𝑛𝜃 =
𝑎
𝑔
𝑎 = 𝑔 𝑡𝑎𝑛𝜃
5) Answer: A
The force is constant, so the acceleration is also constant. Since the slope of the speedtime graph gives acceleration hence the slope must be constant.
Kinetic energy is square of velocity, so it will be a parabola hence correct answer is A.
6) Answer: D
Initially, the spring force has to balance the weight so spring force is
𝐹 = 3𝑔
When the string is cut the 2 Kg block falls freely so it’s acceleration is g. The block 1
Kg have spring force acting in an upward direction and mg acting downward
𝑚𝑎 = 𝐹 − 𝑚𝑔
1𝑎 = 3𝑔 − 1𝑔
𝑎 = 2𝑔
7) Answer: B
Initial kinetic energy = 100 J
Initial potential energy = mgh
E = 100 + mgh
Now when kinetic energy is doubled, then potential energy will be mgh’
𝐸 = 200 + 𝑚𝑔ℎ′
Total energy at both points is same
100 + 𝑚𝑔ℎ = 200 + 𝑚𝑔ℎ′
𝑚𝑔(ℎ − ℎ′ ) = 100
(ℎ − ℎ′ ) =
(ℎ − ℎ′ ) =
100
𝑚𝑔
100
2(10)
(ℎ − ℎ′ ) = 5 𝑚
8) Answer: D
Power is energy per unit time
𝑚𝑔ℎ
𝑡
600(6)
𝑃=
8
𝑃=
𝑃 = 450 𝑊
9) Answer: D
The heat energy is given by
∆𝑄 = �...