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3 years ago

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A machine with a mechanical advantage of 10 is used to produce an output force of 250 Newton’s. What input force is applied to t

his machine

1 answer:

finlep [7]3 years ago

4 0

Answer:

que tue nesisitas

Explanation:

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A very slow motion of earth's axis that requires 26,000 years to complete is called ________.

agasfer [191]

The answer is axial precession. Axial precession refers to the very slow motion of the Earth’s axis, which almost requires twenty-six thousand (26,000) years to complete a full rotation. This Axial Precession is caused by the effects of gravitational pull from the Sun and the Moon towards the Earth.

5 0

3 years ago

A child on a high dive has a mass of 40 kilograms. If the high dive is 10 meters in the air, what is the potential energy? GPE=m

saw5 [17]

Answer:

Ep = 3924 [J]

Explanation:

To calculate this value we must use the definition of potential energy which tells us that it is the product of mass by the acceleration of gravity by height.

$E_{p}=m*g*h\\$

where:

Ep = potential energy [J] (units of Joules)

m = mass = 40 [kg]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = 10 [m]

$E_{p} =40*9.81*10\\E_{p} = 3924 [J]$

7 0

3 years ago

A motor does 8000j of work in 20 seconds. What is the power of the motor

lozanna [386]

Power = work / time = 8000J / 20s = 400W

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3 years ago

The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This

erica [24]

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

$g = \frac{GM}{(d-R_{CM})^2}$

Where,

G = Gravitational Universal Constant

d = Distance

M = Mass

$R_{CM} =$ Radius earth center of mass

PART B) Using the same expression previously defined we can find the acceleration of the moon on the earth like this,

$g = \frac{GM}{(d-R_{CM})^2}$

$g = \frac{(6.67*10^{-11})(7.35*10^{22})}{(3.84*10^8-4700*10^3)^2}$

$g = 3.4*10^{-5}m/s^2$

PART C) Centripetal acceleration can be found throughout the period and angular velocity, that is

$\omega = \frac{2\pi}{T}$

At the same time we have that centripetal acceleration is given as

$a_c = \omega^2 r$

Replacing

$a_c = (\frac{2\pi}{T})^2 r$

$a_c = (\frac{2\pi}{26.3d(\frac{86400s}{1days})})^2 (4700*10^3m)$

$a_c = 3.34*10^{-5}m/s^2$

3 0

3 years ago

NEED HELP PLEASEEE 15 POINTS

Nezavi [6.7K]

Answer:

c

Explanation:

thamjs

7 0

3 years ago

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