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BlackZzzverrR [31]
3 years ago
15

Two children are riding on a merry-go-round that is rotating with a constant angular speed. Abbie is one meter from the center o

f the merry-go-round, whereas Zak is two meters from the center. Abbie’s acceleration is ___________ Zak’s acceleration. Two children are riding on a merry-go-round that is rotating with a constant angular speed. Abbie is one meter from the center of the merry-go-round, whereas Zak is two meters from the center. Abbie’s acceleration is ___________ Zak’s acceleration. a quarter of twice equal to half of four times
Physics
1 answer:
klio [65]3 years ago
7 0

Answer:

The acceleration of Abbie is half of the Zak's.

Explanation:

The centripetal acceleration of an object on a circular path is given by :

a=r\omega^2

Two children are riding on a merry-go-round that is rotating with a constant angular speed. Let r_1 is distance of Abbie from the merry-go-round and r_2 is distance of Zak's from the merry-go-round. Acceleration of Abbie is :

a_1=r_1\omega^2 ...... (1)

r_1=1\ m

Acceleration of Zak's is :

a_2=r_2\omega^2 .......(2)

r_2=2\ m

Dividing equation (1) and (2) we get :

\dfrac{a_1}{a_2}=\dfrac{r_1}{r_2}\\\\\dfrac{a_1}{a_2}=\dfrac{1}{2}\\\\a_1=\dfrac{a_2}{2}

So, the acceleration of Abbie is half of the Zak's.

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A man-made satellite of mass 6105 kg is in orbit around the earth, making one revolution in 430 minutes. What is the magnitude o
blondinia [14]

Answer:

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

Explanation:

At first we assume that Earth is represented by an uniform sphere, such that the man-made satellite rotates in a circular orbit around the planet. Hence, the following condition must be satisfied:

\left(\frac{4\pi^{2}}{T^{2}} \right)\cdot r = \frac{G\cdot M}{r^{2}} (1)

Where:

T - Period of rotation of the satellite, measured in seconds.

r - Distance of the satellite with respect to the center of the planet, measured in meters.

G - Gravitational constant, measured in newton-square meters per square kilogram.

M - Mass of the Earth, measured in kilograms.

Now we clear the distance of the satellite with respect to the center of the planet:

r^{3} = \frac{G\cdot M\cdot T^{2}}{4\pi^{2}}

r = \sqrt[3]{\frac{G\cdot M\cdot T^{2}}{4\pi^{2}} } (2)

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg and T = 25800\,s, then the distance of the satellite is:

r = \sqrt[3]{\frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6.0\times 10^{24}\,kg)\cdot (25800\,s)^{2}}{4\pi^{2}} }

r \approx 18.897\times 10^{6}\,m

The gravitational force exerted on the satellite by the Earth is determined by the Newton's Law of Gravitation:

F = \frac{G\cdot m\cdot M}{r^{2}} (3)

Where:

m - Mass of the satellite, measured in kilograms.

F - Force exerted on the satellite by the Earth, measured in newtons.

If we know that G = 6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}}, M = 6.0\times 10^{24}\,kg, m = 6105\,kg and r \approx 18.897\times 10^{6}\,m, then the gravitational force is:

F = \frac{\left(6.67\times 10^{-11}\,\frac{N\cdot m^{2}}{kg^{2}} \right)\cdot (6105\,kg)\cdot (6\times 10^{24}\,kg)}{(18.897\times 10^{6}\,m)^{2}}

F = 6841.905\,N

A gravitational force of 6841.905 newtons is exerted on the satellite by the Earth.

4 0
3 years ago
A force of 100 newtons is used to move an object a distance of 15 meters with a power of 25 watts. Find the
valentina_108 [34]


work is distance * force so 15*100=1500

and to find time you know power = diastance * force / time

so 25=15*100/t

25=1500/t

25/1500=t

.016=time


5 0
3 years ago
A 25 kg child is riding on a swing. If the child travels 8.9 m/s at the bottom of their swing, how high into the air is the chil
Setler [38]

Answer:

h = 4.04 m

Explanation:

Given that,

Mass of a child, m = 25 kg

The speed of the child at the bottom of the swing is 8.9 m/s

We need to find the height in the air is the child is able to swing. Let the height is h. Using the conservation of energy such that,

mgh=\dfrac{1}{2}mv^2\\\\h=\dfrac{v^2}{2g}

Put all the values,

h=\dfrac{(8.9)^2}{2\times 9.8}\\\\h=4.04\ m

So, the child is able to go at a height of 4.04 m.

7 0
3 years ago
Kepler's First Law states that the shape of planetary orbits is a/an ________________ ?
harkovskaia [24]

Answer:

elliptical orbit

Explanation:

There are three laws of planetary motion, which are called Kepler's law of planetary motion.

First Law : It states that all the planets revolve around the sun in an elliptical path and the sun is at one focus of that elliptical path.

4 0
3 years ago
HELP DUE TODAY
olga2289 [7]
The answer is B I think sorry if it’s wrong
3 0
3 years ago
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