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Veseljchak [2.6K]

2 years ago

11

What things affect gravity

2 answers:

harina [27]2 years ago

6 0

<span>The force of gravity = GMm / r^2, where G is gravitational constant, M is mass of one object, m is mass of another object, r is distance between them.

So, mass of objects affect gravity as well as the distance between the two objects.</span>

mamaluj [8]2 years ago

5 0

Distance and mass affect gravity. If an object has a greater mass, it's gravitational pull will be stronger. And if an object is further away from the one that is exhorting the gravitational pull, then the gravitational pull on the object will be weaker than on closer objects.

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A standard 1 kilogram weight is a cylinder 54.0 mm in height and 55.0 mm in diameter. what is the density of the material

denis-greek [22]

The radius of the cylinder is equal to half the diameter:

$r=\frac{d}{2}=\frac{55.0 mm}{2}=27.5 mm$

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where h is the heigth of the cylinder. Converting into meters,

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5 0

3 years ago

Mike is standing on the roof of a building looking at the roof of the neighboring building that is 15 meters away and 10 meters

amid [387]

Answer:

Part a)

$t = 1.65 s$

Part b)

$x = 40.4 m$

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

$v = 27.3 m/s$

direction of velocity is given as

$[tex]\theta = 26.35 degree$

Explanation:

Part a)

acceleration due to gravity on this planet is 3/4 times the gravity on earth

So the acceleration due to gravity on this new planet is given as

$a = \frac{3}{4}(9.81)$

$a = 7.36 m/s^2$

now the vertical displacement covered by the canister is given as

$y = 10 m$

now by kinematics we have

$y = \frac{1}{2}gt^2$

$10 = \frac{1}{2}(7.36)t^2$

$t = 1.65 s$

Part b)

Horizontal speed of the canister is given as

$v_x = 24.5 m/s$

now the distance moved by it

$x = v_x t$

$x = 24.5 (1.65)$

$x = 40.4 m$

Since the distance of other building is 15 m so YES it can make it to other building

Part c)

Final velocity in X direction will remains the same

$v_x = 24.5 m/s$

final velocity in Y direction

$v_y = v_i + at$

$v_y = 0 + (7.36)(1.65)$

$v_y = 12.14 m/s$

now magnitude of velocity is given as

$v = \sqrt{v_x^2 + v_y^2}$

$v = \sqrt{24.5^2 + 12.14^2}$

$v = 27.3 m/s$

direction of velocity is given as

$\theta = tan^{-1}\frac{v_y}{v_x}$

$\theta = tan^{-1}\frac{12.14}{24.5}$

$[tex]\theta = 26.35 degree$

6 0

2 years ago