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Simora [160]
3 years ago
12

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.
Physics
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
AysviL [449]

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.

8 0
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.

7 0
2 years ago
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