Erica (38 kgkg ) and Danny (46 kgkg ) are bouncing on a trampoline. Just as Erica reaches the high point of her bounce, Danny is
1 answer:
Answer:
The speed with which just after he grabs her is 2.68 m/s.
Explanation:
Given that,
Mass of Erica, m = 38 m/s
Mass of Danny, m' = 46 kg
Erica reaches the high point of her bounce, Danny is moving upward past her at 4.9 m/s. At this moment, the initial speed of Erica will be 0. The momentum will remain conserved. Using the conservation of linear momentum as :
So, the speed with which just after he grabs her is 2.68 m/s. Hence, this is the required solution.
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Answer:
The minimum coefficient of friction is 0.27.
Explanation:
To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.
Centripetal force is written as
with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as
with ω denoting the angular velocity, which we are given. With that, the above becomes:
Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?
Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:
and so 45 rev/min = 4.71 rad/s.
A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.
Answer:
Explanation:
Vm = Δs/Δt
700km/h = Δs/1.5h
700 = Δs/1.5
S = 700 x 1.5
S = 1050 Km
*(S = Δs)
Answer: The rocket will have traveled 1050 Km
Hope this help ☺