Magnitude is used to show the brightness of one object compared with the brightness of another object with the _____ magnitude o
1 answer:
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Answer:
A) and B) are correct.
Explanation:
Let's take a look at the attached picture. Now
The total voltage across both capacitors is the same as the sum of the voltage from each device, that statement is true for any electrical device connected in series. So a) is TRUE
The equivalent capacitance is going to be:
And that value can be mathematically proven that is always less than any of the values of each capacitor. So b is TRUE
And through both capacitors flow the same current, but the amount of charge depends on the value of the capacitors, so only could be the same if the capacitors are the same value. Otherwise, don't. C) not always, so FALSE
Refer to the diagram shown below.
In order for the balloon to strike the professor's head, th balloon should drop by 18 - 1.7 = 16.3 m in the time at the professor takes to walk 1 m.
The time for the professor to walk 1 m is
t = (1 m)/(0.45 m/s) = 2.2222 s
The initial vertical velocity of the balloon is zero.
The vertical drop of the balloon in 2.2222 s is
h = (1/2)*(9.8 m/s²)*(2.2222 s)² = 24.197 m
Because 24.97 > 16.3, the balloon lands in front of the professor, and does not hit the professor.
The time for the balloon to hit the ground is
(1/2)*(9.8)*t² = 18
t = 1.9166 s
The time difference is 2.2222 - 1.9166 = 0.3056 s
Within this time interval, the professor travels 0.45*0.3056 = 0.175 m
Therefore the balloon falls 0.175 m in front of the professor.
Answer:
The balloon misses the professor, and falls 0.175 m in front of the professor.
In order for the balloon to strike the professor's head, th balloon should drop by 18 - 1.7 = 16.3 m in the time at the professor takes to walk 1 m.
The time for the professor to walk 1 m is
t = (1 m)/(0.45 m/s) = 2.2222 s
The initial vertical velocity of the balloon is zero.
The vertical drop of the balloon in 2.2222 s is
h = (1/2)*(9.8 m/s²)*(2.2222 s)² = 24.197 m
Because 24.97 > 16.3, the balloon lands in front of the professor, and does not hit the professor.
The time for the balloon to hit the ground is
(1/2)*(9.8)*t² = 18
t = 1.9166 s
The time difference is 2.2222 - 1.9166 = 0.3056 s
Within this time interval, the professor travels 0.45*0.3056 = 0.175 m
Therefore the balloon falls 0.175 m in front of the professor.
Answer:
The balloon misses the professor, and falls 0.175 m in front of the professor.