Answer:
Energy Lost for group A's car = 0.687 J
Energy Lost for group B's car = 0.55 J
Explanation:
The exact question is as follows :
Given - The energy of an object can be converted to heat due to the friction of the car on the hill. The difference between the potential energy of the car and its kinetic energy at the bottom of the hill equals the energy lost due to friction.
To find - How much energy is lost due to heat for group A's car ?
How much for Group B's car ?
Solution -
We know that,
GPE = 1 Joule (Potential Energy)
Now,
For Group A -
Energy Lost = GPE - KE
= 1 J - 0.313 J
= 0.687 J
So,
Energy Lost for group A's car = 0.687 J
Now,
For Group B -
Energy Lost = GPE - KE
= 1 J - 0.45 J
= 0.55 J
So,
Energy Lost for group B's car = 0.55 J
Answer:
Explanation:
The change in kinetic energy will be simply the difference between the final and initial kinetic energies:
We know that the formula for the kinetic energy for an object is:
where <em>m </em>is the mass of the object and <em>v</em> its velocity.
For our case then we have:
Which for our values is:
Answer:
t ’= , v_r = 1 m/s t ’= 547.19 s
Explanation:
This is a relative velocity exercise in a dimesion, since the river and the boat are going in the same direction.
By the time the boat goes up the river
v_b - v_r = d / t
By the time the boat goes down the river
v_b + v_r = d '/ t'
let's subtract the equations
2 v_r = d ’/ t’ - d / t
d ’/ t’ = 2v_r + d / t
In the exercise they tell us
d = 1.22 +1.45 = 2.67 km= 2.67 10³ m
d ’= 1.45 km= 1.45 1.³ m
at time t = 69.1 min (60 s / 1min) = 4146 s
the speed of river is v_r
t ’=
t ’=
In order to complete the calculation, we must assume a river speed
v_r = 1 m / s
let's calculate
t ’=
t ’= 547.19 s
Answer:
a)6.67 m/s2
b)16.7 rad/s2
c)increasing angular acceleration
Explanation:
a) It's because the system is not just mass of the man, it consists of the man holding a rope wrapped around a cylinder, not just a man free falling. So you would have to consider the rotating cylinder under the torque created by the man gravity force.
Let g = 10m/s2
T = mgd =75*10*0.4 = 300 N.m
The from the mass moments inertial of the solid cylinder:
we can calculate the angular acceleration of the cylinder:
then translate that to acceleration:
c) if the mass of the rope is not neglected, that means the force of gravity increases as the rope unwrapping around the cylinder, so the torque increases. Also the moment of inertial of the rope-cylinder system decreases due to rope unwrapping. In the end, the angular acceleration is no longer constant, but increasing.