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madam [21]
3 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]3 years ago
7 1

Answer:

25 times the average speed

Allisa [31]3 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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A college student is working on her physics homework in her dorm room. Her room contains a total of 6.0 x 10^26 gas molecules. A
IceJOKER [234]

Answer:

Temperature, T = 3.62 kelvin

Explanation:

It is given that,

Total number of gas molecules, N=6\times 10^{26}

Her body is converting chemical energy into thermal energy at a rate of 125 W, P = 125 W

Time taken, t = 6 min = 360 s

Energy of a gas molecules is given by :

\Delta E =\dfrac{3}{2}NkT

T=\dfrac{2E}{3Nk}, k is Boltzmann constant

T=\dfrac{2\times P\times t}{3Nk}

T=\dfrac{2\times 125\times 360}{3\times 6\times 10^{26}\times 1.38\times 10^{-23}}

T = 3.62 K

So, the temperature increases by 3.62 kelvin. Hence, this is the required solution.

4 0
3 years ago
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lyudmila [28]

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7 0
3 years ago
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A boy jumps from rest, straight down from the top of a cliff. He falls halfway down to the water below in 0.866 s. How much time
Iteru [2.4K]
<h2>Entire trip takes 1.22 seconds.</h2>

Explanation:

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 9.81 m/s²  

        Time, t = 0.866 s      

     Substituting

                      s = ut + 0.5 at²

                      s = 0 x 0.866 + 0.5 x 9.81 x 0.866²

                      s = 3.68 m

      Halfway is 3.68 m

Total height = 2 x 3.68 = 7.36 m

We have equation of motion s = ut + 0.5 at²

        Initial velocity, u = 0 m/s

        Acceleration, a = 9.81 m/s²  

        Time, t = ?

        Displacement, s  = 7.36 m

     Substituting

                      s = ut + 0.5 at²

                      7.36 = 0 x t + 0.5 x 9.81 x t²

                      t = 1.22 s

Entire trip takes 1.22 seconds.

7 0
3 years ago
Model rocket engines are sized by thrust, thrust duration, and total impulse, among other characteristics. A size C5 model rocke
umka2103 [35]

Answer:

v_{f} = 115.95 m / s

Explanation:

This is an exercise of a variable mass system, let's form a system formed by the masses of the rocket, the mass of the engines and the masses of the injected gases, in this case the system has a constant mass and can be solved using the conservation the amount of movement. Which can be described by the expressions

        Thrust = v_{e}  \frac{dM}{dt}

        v_{f}-v₀ = v_{e} ln ( \frac{M_{o} }{M_{f}} )

where v_{e} is the velocity of the gases relative to the rocket

let's apply these expressions to our case

the initial mass is the mass of the engines plus the mass of the fuel plus the kill of the rocket, let's work the system in SI units

       M₀ = 25.5 +12.7 + 54.5 = 92.7 g = 0.0927 kg

     

The final mass is the mass of the engines + the mass of the rocket

      M_{f} = 25.5 +54.5 = 80 g = 0.080 kg

thrust and duration of ignition are given

       thrust = 5.26 N

       t = 1.90 s

Let's start by calculating the velocity of the gases relative to the rocket, where we assume that the rate of consumption is linear

          thrust = v_{e} \frac{M_{f} - M_{o}  }{t_{f} - t_{o}  }

          v_{e} = thrust  \frac{\Delta t}{\Delta M}

          v_{e} = 5.26 \frac{1.90}{0.080 -0.0927}

          v_{e} = - 786.93 m / s

the negative sign indicates that the direction of the gases is opposite to the direction of the rocket

now we look for the final speed of the rocket, which as part of rest its initial speed is zero

            v_{f}-0 = v_{e} ln ( \frac{M_{o} }{M_{f} } )

we calculate

            v_{f} = 786.93 ln (0.0927 / 0.080)

            v_{f} = 115.95 m / s

5 0
3 years ago
Which is an example of a vector quantity
swat32

Answer:

The answer to your question is C

Explanation:

acceleration is a vector quantity because it has both magnitude and direction

4 0
3 years ago
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