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3 years ago

Defining Mechanical Advantage and Efficiency

Physics

2 answers:

Volgvan3 years ago

4 0

Answer:

hola como estas she cartoon Chino

qaws [65]3 years ago

3 0

Answer:

1. B

2.D

Explanation:

I took the Assignment

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What Do You Already Know about Density? Material Design. Number each material and sort the items in order from lowest (1) to hig

Elena-2011 [213]

Answer:

1. Dry Beans - 591.75 kg/m^3

2. Flour - 593 kg/m^3

3. Wax - 900 kg/m^3

4. Wet sand - 2039 kg/m^3

5. Chalk - 2499 kg/m^3

6. Talcum Powder - 2776 kg/m^3

7. Copper - 8960 kg/m^3

Explanation:

Make sure your units are the same

7 0

3 years ago

A meteoroid is in a circular orbit 600 km above the surface of a distant planet. The planet has the same mass as Earth but has a

AVprozaik [17]

<h2>Answer:</h2><h2>The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/ $s^{2}$ </h2>

Explanation:

A meteoroid is in a circular orbit 600 km above the surface of a distant planet.

Mass of the planet = mass of earth = 5.972 x $10^{24}$ Kg

Radius of the earth = 90% of earth radius = 90% 6370 = 5733 km

The acceleration of the meteoroid due to the gravitational force exerted by the planet = ?

By formula, g = $\frac{GM}{r^{2} }$

where g is the acceleration due to the gravity

G is the universal gravitational constant = 6.67 x $10^{-11}$ $m^{3} kg^{-1} s^{-2}$

M is the mass of the planet

r is the radius of the planet

Substituting the values, we get

g = $\frac{(6.67 * 10^{-11}) (5.972 * 10^{24}) }{5733^{2} }$

g = 12.12 m/ $s^{2}$

The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/ $s^{2}$

6 0

3 years ago

Acceleration occurs whenever an ___________________ force acts on an object.

leonid [27]

Acceleration occurs whenever the forces on an object are unbalanced.

It's the group of forces on the object that's either balanced or unbalanced.
There's no such thing as "an unbalanced force".

8 0

3 years ago

Read 2 more answers

If the angular velocity of the pully is -8.4rad/s at a given time, and its anglar acceleration is -2.8rad/s2, what is the angula

snow_lady [41]

Here we know that

$\omega_i = - 8.4 rad/s$

$\alpha = - 2.8 rad/s^2$

$t = 1.5 s$

now from kinematics we have

$\omega_f = \omega_i + \alpha t$

now from above all values we have

$\omega_f = (-8.4 rad/s) + (-2.8 rad/s^2)(1.5)$

$\omega_f = -8.4 + (-4.2)$

$\omega_f = -12.6 rad/s$

so final angular speed is -12.6 rad/s

6 0

3 years ago

E=<br> (500.0lm)<br> 4 (100)

drek231 [11]

Answer:

didn't understand your question

7 0

3 years ago