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Lapatulllka [165]
2 years ago
11

After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 54.0 cm. The explorer finds that

the pendulum completes 98.0 full swing cycles in a time of 135s.
Required:
What is the value of g on this planet?
Physics
1 answer:
Natasha2012 [34]2 years ago
3 0

Answer:

g = 11.2 m/s²

Explanation:

First, we will calculate the time period of the pendulum:

T = \frac{t}{n}

where,

T = Time period = ?

t = time taken = 135 s

n = no. of swings in given time = 98

Therefore,

T = \frac{135\ s}{98}

T = 1.38 s

Now, we utilize the second formula for the time period of the simple pendulum, given as follows:

T = 2\pi \sqrt{\frac{l}{g}}

where,

l = length of pendulum = 54 cm = 0.54 m

g = acceleration due to gravity on the planet = ?

Therefore,

(1.38\ s)^2 = 4\pi^2(\frac{0.54\ m}{g} )\\\\g = \frac{4\pi^2(0.54\ m)}{(1.38\ s)^2}

<u>g = 11.2 m/s²</u>

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Copper and aluminum are being considered for a high-voltage transmission line that must carry a current of 51.5 A. The resistanc
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Answer:

a).Jcu= 436.44x10^{3} \frac{A}{m^{2}}

b).λcu=1.05728 \frac{kg}{m}

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d).λcu=0.487 \frac{kg}{m}

Explanation:

a).

ζcu=1.7x10^{-8}Ωm

ζal=2.76x10^{-8}Ωm

A=\frac{w}{R}

wcu=ζcu*l

Acu=\frac{1.7x10^{-8}*1000}{0.144} =1.18x10^{-4} m^{2}

J=\frac{I}{Acu}=\frac{51.5A}{1.18x10^{-4}m^{2}} \\J=436.44x10^{3} \frac{A}{m^{2}}

b).

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λcu=Dcu*Acu

λcu=8960 \frac{kg}{m^{3}}*1.18x10^{-4} m^{2}

λcu=1.05728 \frac{kg}{m}

c).

wal=ζal*l

Aal=\frac{2.7x10^{-8}*1000}{0.144} =0.187x10^{-3} m^{2}

J=\frac{I}{Aal}=\frac{51.5A}{0.187x10^{-3}m^{2}} \\J=274.66x10^{3} \frac{A}{m^{2}}

d).

mass per unit Aluminum

λal=Dal*Aal

λal=2600 \frac{kg}{m^{3}}*0.1875x10^{-3} m^{2}

λcu=0.487 \frac{kg}{m}

3 0
3 years ago
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