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3 years ago

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A planet orbits a star, in a year of length 2.35 x 107 s, in a nearly circular orbit of radius 3.49 x 1011 m. With respect to th

e star, determine (a) the angular speed of the planet, (b) the tangential speed of the planet, and (c) the magnitude of the planet's centripetal acceleration.

1 answer:

Naya [18.7K]3 years ago

7 0

Answer:

a) w = 9.599 10⁴ rad / s , b) v = 3.35 10¹⁶ m / s , c) a = 3.22 10²¹ m / s²

Explanation:

For this exercise we must use the relation of angular kinematics

a) angular velocity, the distance remembered in orbit between time (period)

w = 2π r / T

w = 2 π 3.59 10¹¹ / 2.35 10⁷

w = 9.599 10⁴ rad / s

b) linear and angular velocity are related by the equation

v = w r

v = 9,599 10⁴ 3.49 10¹¹

v = 3.35 10¹⁶ m / s

c) the centripetal acceleration is

a = v² / r = w² r

a = (9,599 10⁴)² 3.49 10¹¹

a = 3.22 10²¹ m / s²

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A swimming pool has the shape of a right circular cylinder with radius 21 feet and height 10 feet. Suppose that the pool is full

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Answer:

The water required to pump all the water to a platform 2 feet above the top of the pool is is 6061310.32 foot-pound.

Explanation:

Given that,

Radius = 21 feet

Height = 10 feet

Weighing = 62.5 pounds/cubic

Work = 4329507.37572

Height = 2 feet

Let's look at a horizontal slice of water at a height of h from bottom of pool

We need to calculate the area of slice

Using formula of area

$A=\pi r^2$

Put the value into the formula

$A=\pi\times21^2$

$A=441\pi\ feet^2$

Thickness of slice $t=\Delta h\ ft$

The volume is,

$V=(441\pi\times\Delta h)\ ft^3$

We need to calculate the force

Using formula of force

$F=W\times V$

Where, W = water weight

V = volume

Put the value into the formula

$F=62.5\times(441\pi\times\Delta h)$

$F=27562.5\pi\times\Delta h\ lbs$

We need to calculate the work done

Using formula of work done

$W=F\times d$

Put the value into the formula

$W=27562.5\pi\times\Delta h\times(10-h)\ ft\ lbs$

We do this by integrating from h = 0 to h = 10

We need to find the total work,

Using formula of work done

$W=\int_{0}^{h}{W}$

Put the value into the formula

$W=\int_{0}^{10}{27562.5\pi\\times(10-h)}dh$

$W=27562.5\pi(10h-\dfrac{h^2}{2})_{0}^{10}$

$W=27562.5\pi(10\times10-\dfrac{100}{2}-0)$

$W=4329507.37572$

To pump 2 feet above platform, then each slice has to be lifted an extra 2 feet,

So, the total distance to lift slice is (12-h) instead of of 10-h

We need to calculate the water required to pump all the water to a platform 2 feet above the top of the pool

Using formula of work done

$W=\int_{0}^{h}{W}$

Put the value into the formula

$W=\int_{0}^{10}{27562.5\pi\\times(12-h)}dh$

$W=27562.5\pi(12h-\dfrac{h^2}{2})_{0}^{10}$

$W=27562.5\pi(12\times10-\dfrac{100}{2}-0)$

$W=1929375\pi$

$W=6061310.32\ foot- pound$

Hence, The water required to pump all the water to a platform 2 feet above the top of the pool is is 6061310.32 foot-pound.

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3 years ago

a car headed north at 15 m/s accelerates for 4.25 s to reach a velocity of 28.3 m/s. What is the acceleration of the car?

Liono4ka [1.6K]

<u>Answer:</u>

The acceleration of the car is $3.13 m/s^2$

<u>Explanation:</u>

In the question it is given that car initially heads north with a velocity $15 m/s$ . It then accelerates for $4.25 s$ and in the end its velocity is $28.3 m/s$ .

initial velocity $u = 15 m/s$

time $t=4.25 s$

final velocity $v=28.3 m/s$

The equation of acceleration is

$a= \frac{(v-u)}{t}$

$= \frac{(28.3-15)}{4.25} = \frac {13.3}{4.25} =3.13m/s^2$

The value of acceleration is positive, here since the car is speeding up. If it was slowing down the value of acceleration would be negative.

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