Explanation:
12N by first law of newton is net force after colloision
The only graph that accurately depict the given motion is graph D.
The given parameters;
- initial position of the man = 0
- direction of the man's first displacement = backward
- time of first motion, t₁ = 6 seconds
- velocity of this first displacement = v₁
- time without any motion (<em>zero movement</em>) = 6 seconds
- direction of the second displacement = forward
- velocity of second displacement = 2v₁
Let the acceleration of the first displacement = a
Acceleration of the second displacement = 2a
From the given graphs we can eliminate every graph without initial decrease or motion towards the negative direction.
The only options with initial motion towards the negative direction are;
The difference between graph B and D;
- in graph B there is a uniform motion for 6 seconds
- in graph D there is no motion for 6 seconds (<em>this is obvious as the line fall directly on top of the horizontal axis maintaining a value of zero for 6 seconds</em>).
Thus, the only graph that accurately depict the given motion is graph D.
Learn more here: brainly.com/question/21095906
Given: Heat Qout means useful work = 2800 J
Heat Qin = 8900 J
Required; Efficiency = ?
Formula: Efficiency = Qout/Qin = x 100%
= 2800 J/8900 J = 0.3146 X 100 %
Efficiency = 31.46%
Answer:
Explanation:
There will be reaction force by each vertical post on horizontal plank . Let it be R₁ and R₂ . R₁ is reaction force by the post nearer to woman
Taking torque of all forces about the end far away from the woman
Torque by reaction force = R₁ x 5.5
= 5.5 R₁ upwards
Torque by weight of woman in opposite direction , downwards
= - 804 x ( 5.5 - 1.55 )
= - 3175.8
Torque by weight of the plank in opposite direction , downwards .
= - 27 x 5.5 / 2
= - 74.25
Torque by R₂ will be zero as it passes through the point about which torque is being taken .
Total torque
= 5.5 R₁ - - 3175.8 - - 74.25 = 0 ( For equilibrium )
5.5 R₁ = 3250
R₁ = 590.9 N .
