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Paul [167]
3 years ago
10

An 97 kg climber climbs to the top of Mount Everest, which has a peak height of 8850 m above sea level. What is the climber’s po

tential energy with respect to sea level
Physics
1 answer:
Kisachek [45]3 years ago
4 0
Eg=mgh
Eg=(97)(9.8)(8850)
Eg=8412810J
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Francium-223 has a half-life of 22 minutes. A Geiger counter gives a reading of 58 counts per minute (cpm) from a sample of fran
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Answer:

29 counts per minute (29 cpm)

Explanation:

The half-life of a radioactive isotope is the time needed for the activity of the sample to halve.The half-life of Francium-223 is exactly 22 minutes: this means that its activity after 22 minutes becomes exactly half of its initial value. Since the initial activity of this sample of francium-223 was 58 cpm, the activity after 22 minutes will be

A=\frac{58 cpm}{2}=29 cpm

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Bill Nye- Static Electricity Answer Key?
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What is the relationship between temperature, dew point temperature, and air pressure
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Dew point is the temperature at which the water vapor in the air condenses , then evaporates. The barometric or air pressure is independent from the dew point. 
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3 years ago
Ian walks 2 km to his best friend's house, then walks 0.5 km to the library. He then makes a 2.5 km walk home. The entire walk t
earnstyle [38]

Average speed = (total distance covered) / (time to cover the distance)

Ian's total distance covered = (2km + 0.5km + 2.5km) = 5 km.

His time to cover the distance = 3 hours.

Average speed = (5 km) / (3 hrs)

Average speed = (5/3) (km/hr)

<em>Average speed = 1.67 km/hr</em>

5 0
3 years ago
A pendulum is formed by taking a 2.0 kg mass and hanging it from the ceiling using a steel wire with a diameter of 1.1 mm. it is
Lera25 [3.4K]

Answer: 1.39 s

Explanation:

We can solve this problem with the following equations:

\frac{\Delta l}{l_{o}}=\frac{F}{AY} (1)

T=2 \pi \sqrt{\frac{l_{o}}{g}} (2)

Where:

\Delta l=0.05 mm=5(10)^{-5} m is the length the steel wire streches (taking into account 1mm=0.001 m)

l_{o} is the length of the steel wire before being streched

F=mg=(2 kg)(9.8 m/s^{2})=19.6 N is the force due gravity (the weight) acting on the pendulum with mass m=2 kg

A is the transversal area of the wire

Y=2(10)^{11} Pa is the Young modulus for steel

T is the period of the pendulum

g=9.8 m/s^{2} is the acceleration due gravity

Knowing this, let's begin by finding A:

A=\pi r^{2}=\pi (\frac{d}{2})^{2}=\pi \frac{d^{2}}{4} (3)

Where d=1.1 mm=0.0011 m is the diameter of the wire

A=\pi \frac{(0.0011 m)^{2}}{4} (4)

A=9.5(10)^{-7}m^{2} (5)

Knowing this area we can isolate l_{o} from (1):

l_{o}=\frac{\Delta l AY}{F} (6)

And substitute l_{o} in (2):

T=2 \pi \sqrt{\frac{\frac{\Delta l AY}{F}}{g}} (7)

T=2 \pi \sqrt{\frac{\frac{(5(10)^{-5} m)(9.5(10)^{-7}m^{2})(2(10)^{11} Pa)}{2(10)^{11} Pa}}{9.8 m/s^{2}}} (8)

Finally:

T=1.39 s

3 0
3 years ago
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