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madam [21]
2 years ago
14

The average radius of Mars is 3,397 km. If Mars completes one rotation in 24.6 hours, what is the tangential speed of objects on

the planet’s surface? Round your answer to the nearest whole number.
Physics
2 answers:
choli [55]2 years ago
7 1

Answer:

25 times the average speed

Allisa [31]2 years ago
6 0

Answer:

Tangential speed of objects on the planet’s surface = 241 m/s

Explanation:

Speed is given by the ratio of distance traveled to the time.

Radius of mars, r = 3397 km = 3.397 x 10⁶ m

Distance traveled = 2πr = 2 x π x 3.397 x 10⁶ = 2.13 x 10⁷ m

Time taken = 24.6 hours = 24.6 x 60 x 60 = 88560 s

\texttt{Tangential speed =}\frac{2.13\times 10^7}{88560}=240.51m/s

Tangential speed of objects on the planet’s surface = 240.51 m/s = 241 m/s

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victus00 [196]

The concept required to perform this exercise is given by the coulomb law.

The force expressed according to this law is given by

F= \frac{kqQ}{r^2}

Where,

k = 8.99 * 10^9 N m^2 / C^2.

q = charges of the objects

r= distance/radius

Our values are previously given, so

q= 2.5*10^{-6}C\\Q= 75*10^{-6}C\\r=0.59

Replacing,

F=\frac{kqQ}{r^2}

F= \frac{(8.99 x 10^9)(2.5*10^{-6})(75*10^{-6})}{0.59^2}

F= 4.8423N

The force acting on the block are given by,

F-mgsin\theta = ma

a = \frac{F-mgsin\theta}{m}

a = \frac{4.8423-(0.445)(9.8)sin(35)}{0.445}a = 10.31m/s^2

Therefore the box is accelerated upward.

3 0
2 years ago
g A cylinder of mass m is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the
sdas [7]

Answer:

The vertical distance is  d = \frac{2}{k} *[mg + f]

Explanation:

From the question we are told that

   The mass of the cylinder is  m

    The kinetic frictional force is  f

Generally from the work energy theorem

    E  =  P +  W_f

Here E the the energy of the spring which is increasing and this is mathematically represented as

       E =  \frac{1}{2} * k  *  d^2

Here k is the spring constant

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     P  = mgd

And

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      W_f  =  f *  d

So

    \frac{1}{2} * k  *  d^2 =  mgd +  f *  d

=>    \frac{1}{2} * k  *  d^2 =  d[mg +  f    ]

=>  \frac{1}{2} * k  *  d =  [mg +  f    ]

=> d = \frac{2}{k} *[mg + f]

5 0
2 years ago
A particular material has an index of refraction of 1.25. What percent of the speed of light in a vacuum is the speed of light i
beks73 [17]

Answer:

80% (Eighty percent)

Explanation:

The material has a refractive index (n) of 1.25

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v = 239 833 966 m/s

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Therefore speed of light in the material (v) is eighty percent of the speed of light in the vacuum (c)

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