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brilliants [131]
3 years ago
5

A student at a window on the second floor of a dorm sees her physics professor walking on the sidewalk beside the building. she

drops a water balloon from 18.0 meters above the ground when the professor is 1.00 meter from the point directly beneath the window. if the professor is 1.70 meters tall and walks at a rate of 0.450 m/s, does the balloon hit her? if not, how close does it come? (10 pts)

Physics
1 answer:
klasskru [66]3 years ago
8 0
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.

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All the elements in a particular group will have the same number of valence electrons as follows:

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Your friend decides to generate electrical power by rotating a 100,000 turn coil of wire around an axis in the plane of the coil
MakcuM [25]

Answer:

a) I=35mA

b) P=1.73W

Explanation:

a) The max emf obtained in a rotating coil of N turns is given by:

emf_{max}=NBA\omega

where N is the number of turns in the coil, B is the magnitude of the magnetic field, A is the area and w is the angular velocity of the coil.

By calculating A and replacing in the formula (1G=10^{-4}T) we get:

A=\pi r^2 =\pi(0.23m)^2=0.16m^2

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Finally, the peak current is given by:

I=\frac{emf}{R}=\frac{69.72V}{1400\Omega}=49.8mA

b)

we have that

I_{rms}=\frac{I}{\sqrt{2}}=\frac{0.0498A}{\sqrt{2}}=0.035A

P_{rms}=I^2{rms}R=(0.035A)^2(1400\Omega)=1.73W

hope this helps!!

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