A 20,000 kg truck traveling at 25 m/s has a head-on inelastic collision with a 1500 kg car traveling at -30 m/s. calculate the i
1 answer:
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Answer:
g'(10) =
Explanation:
Since g is the inverse of f ,
We can write
g(f(x)) = x <em> </em><em>(Identity)</em>
Differentiating both sides of the equation we get,
g'(f(x)).f'(x) = 1
g'(10) = --equation[1] Where f(x) = 10
Now, we have to find x when f(x) = 10
Thus 10 = + 2
= 8
x =
Since f(x) = + 2
f'(x) = -
f'() = -4 × 4 = -16
Putting it in equation 1, we get:
We get g'(10) = -
1) 12 cm
2) 3 N
Explanation:
1)
The relationship between force and elongation in a spring is given by Hooke's law:
where
F is the force applied
k is the spring constant
x is the elongation
For the spring in this problem, at the beginning we have:
So the spring constant is
Later, the force is tripled, so the new force is
Therefore, the new elongation is
2)
In this second problem, we know that the elongation of the spring now is
From part a), we know that the spring constant is
Therefore, we can use the following equation to find the force:
And substituting k and x, we find:
So, the force to produce an elongation of 6 cm must be 3 N.
Hi there!
A.
Since the can was launched from ground level, we know that its trajectory forms a symmetrical, parabolic shape. In other words, the time taken for the can to reach the top is the same as the time it takes to fall down.
Thus, the time to its highest point:
Now, we can determine the velocity at which the can was launched at using the following equation:
In this instance, we are going to look at the VERTICAL component of the velocity, since at the top of the trajectory, the vertical velocity = 0 m/s.
Therefore:
***vsinθ is the vertical component of the velocity.
Solve for 'v':
Now, recall that:
Plug in the expression for velocity:
B.
We can use the same process as above, where T' = 2T and Th = T.
C.
The work done in part B is 4 times greater than the work done in part A.