Answer:
The effective spring constant of the firing mechanism is 1808N/m.
Explanation:
First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:
(This is correct because the horizontal motion has acceleration zero). Then:
Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:
Then, plugging in the given values, we obtain:
Finally, the effective spring constant of the firing mechanism is 1808N/m.
We don't know Carter, and we don't know where he is or what
he's doing, so I'm taking a big chance speculating on an answer.
I'm going to say that if Carter is pretty much just standing there,
or, let's say, lying on the ground taking a nap, then the force of
the ground acting on him is precisely exactly equal to his weight.
Answer:
v=32.9m/s
Explanation:
The acceleration needed to mantain a circular motion of radius r and speed v is given by the equation 
This is the centripetal acceleration. The person will feel what is called a centrifugal acceleration, of the same value, because he is not in an inertial frame (thus subject to fictitious forces, product of inertia).
We want to know the speed of his head when it is subject to 12.5 times the value of the acceleration of gravity while moving on a 8.84m radius circle, so we must do:
Answer:
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
Explanation:
For this problem let's use Newton's second law applied to each body
Body A
X axis
T = m_A a
Axis y
N- W_A = 0
Body B
Vertical axis
W_B - T = m_B a
In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension
We write the equations
T = m_A a
W_B –T = M_B a
We solve this system of equations
m_B g = (m_A + m_B) a
a = m_B / (m_A + m_B) g
In this initial case
m_A = M
m_B = M
a = M / (1 + 1) M g
a = ½ g
Let's find the tension
T = m_A a
T = M ½ g
T = ½ M g
Now we change the mass of the second block
m_B = 2M
a = 2M / (1 + 2) M g
a = 2/3 g
We seek tension for this case
T’= m_A a
T’= M 2/3 g
Let's look for the relationship between the tensions of the two cases
T’/ T = 2/3 M g / (½ M g)
T’/ T = 4/3
T’= 4/3 T
The new tension is 4/3 = 1.33 of the previous tension the answer e
