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.
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The force of the tripped catch exerted on the 2.5 Kg ball moving at 8.5 m/s to the Left is 160 N
<h3>Data obtained from the question </h3>
- Initial velocity (u) = 8.5 m/s
- Final velocity (v) = 7.5 m/s
- Time (t) = 5 ms = 0.25 s
- Mass (m) = 2.5 Kg
- Force (F) = ?
<h3>How to determine the force</h3>
The force exerted on the ball can be obtained as follow:
F = m(v + u) / t
F = [2.5(7.5 + 8.5)]/ 0.25
F = 40 / 0.25
F = 160 N
Thus, the force exerted on the ball is 160 N
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If you only know its speed, that's not enough information to catch it. You could even chase it at DOUBLE that speed, and you'd never catch it if you were chasing in the wrong direction.
You also have to know the DIRECTION the runaway car is going, so that you can chase in the same direction.
Now that you know its speed AND direction, you know its velocity. You need that information to have any chance of catching it.
Derived quantities depend on.( fundamental)..........physical quantity
Are you from Nepal?
Answer:
180 m
Explanation:
The rock follows a free-fall motion - so the vertical distance covered can be found by using the equation
where
g = 10 m/s^2 is the acceleration due to gravity
t = 6.00 s is the time of the fall
Substituting these data, we find the height of the cliff:
