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amm1812
3 years ago
10

In addition to 1 m = 39.37 in, the following exact conversion equivalents are

Physics
1 answer:
Norma-Jean [14]3 years ago
5 0

Answer:

3.76 m/s.

Explanation:

Velocity in mile per hour (mph) = 8.4 miles per hour

Velocity in metre per second (m/s) =?

Next, we shall convert 8.4 mph to ft/h. This can be obtained as follow :

1 mph = 5280 ft/h

Therefore,

8.4 mph = 8.4 mph / 1 mph × 5280 ft/h

8.4 mph = 44352 ft/h

Next, we shall convert 44352 ft/h to in/h. This is illustrated below:

1 ft/h = 12 in/h

Therefore,

44352 ft/h = 44352 ft/h / 1 ft/h × 12 in/h

44352 ft/h = 532224 in/h

Next, we shall convert 532224 in/h to m/h. This can be obtained as follow:

39.37 in/h = 1 m/h

Therefore,

532224 in/h = 532224 in/h / 39.37 in/h × 1 m/h

532224 in/h = 13518.51664 m/h

Next, we shall convert 13518.51664 m/h to m/min. This is illustrated below:

1 m/h = 1/60 m/min

Therefore,

13518.51664 m/h = 13518.51664 m/h / 1 m/h × 1/60 m/min

13518.51664 m/h = 225.30861 m/min

Finally, we shall convert 225.30861 m/min to m/s. This is illustrated below:

1 m/min = 1/60 m/s

Therefore,

225.30861 m/min = 225.30861 m/min / 1 m/min × 1/60 m/s

225.30861 m/min = 3.76 m/s

Therefore,

8.4 miles per hour is equivalent to 3.76 m/s.

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Explanation:

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4 0
3 years ago
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;

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

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
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Here's link to the answer:

tinyurl.com/wpazsebu

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2 years ago
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2 years ago
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A 49 kg person is being dragged in their sleeping bag to the lake by a 593 N
crimeas [40]

Answer:

f = 485.62 N

Explanation:

Since, the bag is moving with some acceleration. Hence, the unbalanced force will be given as:

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But, from Newtons Second Law of Motion:

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comparing the equations:

ma = Fx - f

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where,

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Therefore,

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6 0
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