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Harlamova29_29 [7]
3 years ago
15

An astronaut on the moon places a package on a scale and finds its weight to be 10. N. () What would the weight of the package

be on the earth?
(b) What is the mass of the package on the moon?
(c) What is the package’s mass on earth?
Physics
1 answer:
jeka943 years ago
8 0

Answer:

(a) 61.25 N

(b) 6.25 kg

(c) 6.25 Kg

Explanation:

Weight on moon = 10 N

Acceleration due to gravity on moon = 1.6 m/s^2

Acceleration due to gravity on earth = 9.8 m/s^2

Let m be the mass of the package.

(a) Weight on earth = mass x acceleration due to gravity on earth

Weight on earth = 6.25 x 9.8 = 61.25 N

(b) Weight on moon = mass x acceleration due to gravity on moon

10 = m x 1.6

m = 6.25 kg

(c) Mass of the package remains same as mass does not change, so the mass of package on earth is 6.25 kg.

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Answer:

Some of the frequency that cannot be produced by the string includes 400Hz, 500Hz 650Hz etc...

Explanation:

Harmonics in strings are defined as the integral multiples of its fundamental frequency. This multiples are in arithmetic progression.

For example if Fo is the fundamental frequency of the string, the harmonics will be 2fo, 3fo, 4fo, 5fo... etc

If the string produces a fundamental frequency of 150Hz, some of the harmonics produced by the string will be 300Hz, 450Hz, 600Hz, 750Hz... etc

Some of the harmonics that cannot be produced include 400Hz, 500Hz 650Hz etc...

6 0
3 years ago
Calculate the number of free electrons per cubic meter for some hypothetical metal, assuming that there are 1.3 free electrons p
boyakko [2]

Answer:

The number of free electrons per cubic meter is 7.61\times 10^{28}\ m^{-3}

Explanation:

It is given that,

The number of free electrons per cubic meter is, 1.3

Electrical conductivity of metal, \sigma=6.8\times 10^7\ \Omega^{-1}m^{-1}

Density of metal, \rho=10.5\ g/cm^3

Atomic weight, A = 107.87 g/mol

Let n is the number of  free electrons per cubic meter such that,

n=1.3\ N

n=1.3(\dfrac{\rho N_A}{A})

Where

\rho is the density of silver atom

N_A is the Avogadro number

A is the atomic weight of silver

n=1.3\times (\dfrac{10.5\ g/cm^3\times 6.02\times 10^{23}\ atoms/mol}{107.87\ g/mol})

n=7.61\times 10^{22}\ cm^{-3}

or

n=7.61\times 10^{28}\ m^{-3}

Hence, this is the required solution.

6 0
3 years ago
Sam is moving house and is carrying a 300N box of books up a flight of steps 5m high, it takes her 30 seconds. Gary follows her
hjlf

Answer:

Sam is providing the biggest power i.e. 50 W

Explanation:

Sam is moving house and is carrying a 300N box of books up a flight of steps 5m high, it takes her 30 seconds.

Sam's power :

P_1=\dfrac{W_1}{t_1}\\\\P_1=\dfrac{F_1d}{t_1}\\\\P_1=\dfrac{300\times 5}{30}\\\\P_1=50\ W

Gary follows her carrying a bag of clothes doing 1000 J of work; it only takes him 25 seconds.

Gary's power :

P_2=\dfrac{W_2}{t_2}\\\\P_2=\dfrac{1000}{25}\\\\P_2=40\ W

So, it is clear that Sam is providing the biggest power.

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