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bagirrra123 [75]
3 years ago
15

P7.36 A ship is 125 m long and has a wetted area of 3500 m2. Its propellers can deliver a maximum power of 1.1 MW to seawater at

20C. If all drag is due to friction, estimate the maximum ship speed, in kn.

Physics
1 answer:
Mariana [72]3 years ago
4 0

The solution is in the attachment

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A 1700kg rhino charges at a speed of 50.0km/h. what average force is needed to bring the rhino to a stop in 0.50s?
uranmaximum [27]
From 50km/h to 0km/h in 0.5s we need next acceleration:
First we convert km/h in m/s:
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F=m*a

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How does the table show that the balloon went downwards ?
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What is the orbital period of a spacecraft in a low orbit near the surface of mars? The radius of mars is 3.4×106m.
valkas [14]
<h2>Answer: 56.718 min</h2>

Explanation:

According to the Third Kepler’s Law of Planetary motion<em> </em><em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>

In other words, this law states a relation between the orbital period T of a body (moon, planet, satellite) orbiting a greater body in space with the size a of its orbit.

This Law is originally expressed as follows:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;

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M=6.39(10)^{23}kg is the mass of Mars

a=3.4(10)^{6}m  is the semimajor axis of the orbit the spacecraft describes around Mars (assuming it is a <u>circular orbit </u>and a <u>low orbit near the surface </u>as well, the semimajor axis is equal to the radius of the orbit)

If we want to find the period, we have to express equation (1) as written below and substitute all the values:

T=\sqrt{\frac{4\pi^{2}}{GM}a^{3}}    (2)

T=\sqrt{\frac{4\pi^{2}}{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(6.39(10)^{23}kg)}(3.4(10)^{6}m)^{3}}    (3)

T=\sqrt{11581157.44 s^{2}}    (4)

Finally:

T=3403.1099s=56.718min    This is the orbital period of a spacecraft in a low orbit near the surface of mars

6 0
3 years ago
What is v^2=0.05-4.9 please i need this asap​
Margaret [11]

Answer:

v =2.02

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