Three objects of uniform density—a solid sphere, a solid cylinder, and a hollow cylinder— are placed at the top of an incline (f
ig. cq10.13). they are all released from rest at the same elevation and roll without slipping. (a) which object reaches the bottom first
1 answer:
First you must make a free body diagram of each body to find the acceleration of each one. After you find the general formula of acceleration, you must replace the values of K for each body. (Assuming they all have the same mass). Finally, by replacing K, the body that gets the most acceleration will be the first to reach the bottom.I attached the solution.In this case the solid sphere reaches the bottom first.
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(a). The car's average velocity between t = 1.0s to t = 1.5s will be -
(b). The car's acceleration at t = 1.5s will be -
(c). Car's speed is increasing with time.
We have a a remote controlled toy car that starts from rest and begins to accelerate in a straight line.
We have to determine -
- The car's average velocity (in m/s) in the interval between -
t = 1.0 s to t = 1.5 s.
- The car's acceleration at t = 1.5 s.
- Determining whether car's speed increasing or decreasing with time.
The rate of change of velocity with respect to time is called Acceleration. Mathematically -
According to the question, we have the following data for the Car -
t = 0s → x = 0m
t = 0.5s → x = 0.1m
t = 1.0s → x = 0.4m
t = 1.5s → x = 0.9m
t = 2.0s → x = 1.6m
PART - A
The car's average velocity between t = 1.0s to t = 1.5s will be -
PART - B
Velocity at t = 1.5 s will be -
The car's acceleration at t = 1.5s will be -
PART - C
Since, the acceleration of the car is positive, this means that the car is accelerating in the forward direction. Hence, its speed is increasing with time.
[ The following data was missing in your answer. The complete question would include this data also -
t = 0s → x = 0m
t = 0.5s → x = 0.1m
t = 1.0s → x = 0.4m
t = 1.5s → x = 0.9m
t = 2.0s → x = 1.6m ]
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Answer:
37357 sec
or 622 min
or 10.4 hrs
Explanation:
GIVEN DATA:
Lifting weight 80 kg
1 cal = 4184 J
from information given in question we have
one lb fat consist of 3500 calories = 3500 x 4184 J
= 14.644 x 10^6 J
Energy burns in 1 lift = m g h
= 80 x 9.8 x 1 = 784 J
lifts required
= 18679
from the question,
1 lift in 2 sec.
so, total time = 18679 x 2 = 37357 sec
or 622 min
or 10.4 hrs
gravitational Potential energy is given by
here we have
m = 200 kg
h = 21000 m
now we will have
so GPE of balloon will be 41160000 J above the given height from ground