Answer:
A. -5488J
B. 273.8J
C. 372.44N
Explanation:
Given:
m = 40kg
h = 14 m
v= 3.7 m/s
Part(a)
The change in the potential energy of the bear Earth system during the slide
AU = -mgh = -40(9.8) (14) = -5488 J
Part(b)
The kinetic energy of the bear just before hitting the ground is
Ks 1/2 mV^2= (40)(3.7)2 = 547.6 /2 = 273.8J
Part(c)
The change in the thermal energy of the system due to friction is
AEth = fxh=-(AK +AU) = 5488– 273.8 = 5214.2 J
The average frictional force that acts on the sliding bear is
F = Eth / 14= 5214.2/14 =372.44N
Answer:
I=
Explanation:
We are given that
Mass of rod=M
Length of rod=L
Mass of hoop=M
Radius of hoop=R
We have to find the moment of inertia I of the pendulum about pivot depicted at the left end of the slid rod.
Moment of inertia of rod about center of mass=
Moment of inertia of hoop about center of mass=
Moment of inertia of the pendulum about the pivot left end,I=
Moment of inertia of the pendulum about the pivot left end,I=
Moment of inertia of the pendulum about the pivot left end,I=
Moment of inertia of the pendulum about the pivot left end,I=
Moment of inertia of the pendulum about the pivot left end,I=
If the element's atomic mass is accurate enough you should be able to tell from that , however if it isn't you would most likely want the number of neutrons or electrons contained by the atom.
To find the impulse you multiply the mass by the change in velocity (impulse=mass×Δvelocity). So in this case, 3 kg × 12 m/s ("12" because the object went from zero m/s to 12 m/s).
The answer is 36 kg m/s