Explain why the top of the loop cannot be the same height as (or higher than) the top of the first hill. Assume the roller coast
er and its surroundings are close to a closed system. Use what you know about energy transformations and the effects of friction within a closed system to write your answer.
Passage
Examine this diagram of a roller-coaster track. The very top of the first hill (point B) is higher than the very top of the loop (point D). After the cars reach point B, this roller coaster needs no additional energy to complete its trip.
Passage
Examine this diagram of a roller-coaster track. The very top of the first hill (point B) is higher than the very top of the loop (point D). After the cars reach point B, this roller coaster needs no additional energy to complete its trip.
2 answers:
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The correct answer is "3. Both increase"
In fact, the frequency of a vibrating string is given by
where L is the length of hte string, T the tension and
the linear mass density.
We can see that string is made shorter, L decreases, so the frequency f increases. And since the pitch has the same behaviour of the frequency, it increases as well.
In fact, the frequency of a vibrating string is given by
where L is the length of hte string, T the tension and
We can see that string is made shorter, L decreases, so the frequency f increases. And since the pitch has the same behaviour of the frequency, it increases as well.
Using Newton's Law of Cooling, the given are:
Temperature of the first bath (Ta1): 25C
a(final temperature)= 50C, b (initial temperature)=100C
x-y is the difference in time. In the first bath, 150s
dT/dt=-k(T-Ta)
ln (T-25) [from a=100 to b=50]= -k*150
ln ((50-25)/(100-25))= -k*150
Solving for k=7.324081X10^-3
For the second bath: find Ta2 with dt=120s
dT/dt=-(7.324081x10^-3)*(T-Ta2)
ln (T-Ta2) [from a=100 to b=50]= -(7.324081x10^-3)*120
ln ((50-Ta2)/(100-Ta2))= -(<span>7.324081x10^-3)</span>*120
Solving for Ta (second bath temperature)= 14.49 degrees Celsius
Temperature of the first bath (Ta1): 25C
a(final temperature)= 50C, b (initial temperature)=100C
x-y is the difference in time. In the first bath, 150s
dT/dt=-k(T-Ta)
ln (T-25) [from a=100 to b=50]= -k*150
ln ((50-25)/(100-25))= -k*150
Solving for k=7.324081X10^-3
For the second bath: find Ta2 with dt=120s
dT/dt=-(7.324081x10^-3)*(T-Ta2)
ln (T-Ta2) [from a=100 to b=50]= -(7.324081x10^-3)*120
ln ((50-Ta2)/(100-Ta2))= -(<span>7.324081x10^-3)</span>*120
Solving for Ta (second bath temperature)= 14.49 degrees Celsius