Study the example and choose the influence on perception. Those computer geeks are all the same. They spend all day playing vide
1 answer:
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Answer:
3.07 m/s
Explanation:
We know that from kinematics equation
and here, a=g where v is the final velocity, u is the initial velocity, a is acceleration, s is the distance moved, g is acceleration due to gravity
Making u the subject then
Substituting v for 6.79 m/s, s for 1.87 m and g as 9.81 m/s2 then
Answers:
B.)
C.)
Explanation:
The image attached shows the way the force is acting on the block. Now, if we draw a free body diagram of the situation and write the equations for the Net Force in X and Y, we will have the following:
Net Force in X:
(1)
Where:
is the Normal force
is the magnitude of the force exerted on the block
is the angle
Net Force in Y:
(2)
Where:
is the Friction force (it is expresed with the
sign because this force may be up or down, we cannot know because the block is at rest)
is the gravity force
Rewrittin (1):
(3) This is according to option B
Rewritting (2):
(3) This is according to option C
Answer:
a.
b.
c.
Explanation:
Modeling With Functions
Careful measurements have produced a model of one sprinter's velocity at a given t, and it's is given by
For Carl Lewis's run at the 1987 World Championships, the values of a and b are
Please note we changed the value of b to negative to make the model have sense. Thus, the equation for the velocity is
a. What was Lewis's acceleration at t = 0 s, 2.00 s, and 4.00 s?
To compute the accelerations, we must find the function for a as the derivative of v
For t=0
For t=2
b. Find an expression for the distance traveled at time t.
The distance is the integral of the velocity, thus
To find the value of C, we set X(0)=0, the sprinter starts from the origin of coordinates
Solving for C
Now we complete the equation for the distance
c. Find the time Lewis needed to sprint 100.0 m.
The equation for the distance cannot be solved by algebraic procedures, but we can use approximations until we find a close value.
We are required to find the time at which the distance is 100 m, thus
Rearranging
We define an auxiliary function f(t) to help us find the value of t.
Let's try for t=9 sec
Now with t=9.9 sec
That was a real close guess. One more to be sure for t=10 sec
The change of sign tells us we are close enough to the solution. We choose the time that produces a smaller magnitude for f(t).