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devlian [24]
4 years ago
11

You are exploring a distant planet. When your spaceship is in a circular orbit at a distance of 630 km above the planet's surfac

e, the ship's orbital speed is 5500 m/s . By observing the planet, you determine its radius to be 4.48×106 m. You then land on the surface and, at a place where the ground is level, launch a small projectile with initial speed 14.6 m/s at an angle of 30.8∘ above the horizontal.If resistance due to the planet's atmosphere is negligible, what is the horizontal range of the projectile?
Physics
1 answer:
Anarel [89]4 years ago
3 0

Answer:

R = 24.3 m

Explanation:

As we know that the orbital speed is given as

v = \sqrt{\frac{GM}{R + h}}

here we know that

v = 5500 m/s

R = 4.48 \times 10^6 m

h = 630 km

now we have

5500 = \sqrt{\frac{(6.6 \times 10^{-11})M}{4.48 \times 10^6 + 6.30\times 10^5}}

M = 2.34 \times 10^24 kg

now acceleration due to gravity of planet is given as

a = \frac{GM}{R^2}

a = \frac{(6.6 \times 10^{-11})(2.34 \times 10^{24})}{(4.48\times 10^6)^2}

a = 7.7 m/s^2

now range of the projectile on the surface of planet is given as

R = \frac{v^2 sin2\theta}{g}

R = \frac{14.6^2 sin(2\times 30.8)}{7.7}

R = 24.3 m

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A wave with a period of 1⁄3 second has a frequency of D. 3 Hz. To calculate this we will use the formula that represents the correlation between a frequency (f) and a time period (T): T = 1/f. Or: f = 1/T. The unit for the time period is second "s" while the unit for frequency is Hertz "Hz" (=1/s). We know that T = 1/3 s. That means that f = 1/(1/3s) = 3 1/s = 3 Hz.
3 0
3 years ago
A ladder 10 ft long rests against a vertical wall. If the bottom of the ladder slides away from the wall at a rate of 1.4 ft/s,
Art [367]

Answer:

\dfrac{d\theta}{dt} =-0.233\ rad/s

Explanation:

given,

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let x be the distance of the bottom and y be the distance of the top of ladder.

x² + y² = 100

differentiating with respect to time we get

2 x\dfrac{dx}{dt}+2y\dfrac{dy}{dt} = 0..............(1)

when x = 8 and y = 6 and when \dfrac{dx}{dt} = 1.4ft/s

from equation (1)

now,

16\times 1.4 + 12\dfrac{dy}{dt} = 0

\dfrac{dy}{dt} = -\dfrac{5.6}{3}

let the angle between the ladders be θ

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y = xtan θ

\dfrac{dy}{dt} =\dfrac{dy}{dt} tan\theta + x sec^2\theta\dfrac{d\theta}{dt}

-\dfrac{5.6}{3} =1.4\times \dfrac{6}{8} + 8 (1+\dfrac{9}{16})\dfrac{d\theta}{dt}

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\dfrac{d\theta}{dt} =-0.233\ rad/s

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3 years ago
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Have a nice day!
7 0
3 years ago
find the average velocity of a bicycle that starts 100km south and is 120km south of town after 0.4 hours​
dusya [7]

Answer:

The average velocity is 50 km/h south

Explanation:

The average velocity of an object is its total displacement divided by

the total time taken.

That means it is the rate at which an object changes its position from

one place to another.

Average velocity is a vector quantity.

The SI unit is meters per second.

A bicycle that starts 100 km south and is 120 km south of town after

0.4 hour​.

The displacement = 120 - 100 = 20 km south

The time = 0.4 hour

The average velocity = \frac{D}{T}, where D is the displacement

and t is the time

The average velocity of the bicycle = \frac{20}{0.4}=50 km/h

<em>The average velocity is 50 km/h south</em>

If you want it in meter per second, change the kilometer to meter

and change the hour to seconds

1 km = 1000 m

1 hour = 60 × 60 = 3600 seconds

The average velocity of the bicycle = \frac{50(1000)}{3600}=13.89 m/s south

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