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3 years ago

What determines gravitational pull

Physics

1 answer:

krok68 [10]3 years ago

6 0

Answer

Remember that the net force on you determines how your motion changes. The strength of the gravitational force between two objects depends on two factors, mass and distance. The Mass of the Objects The more mass two objects have, the greater. the force of gravity the masses exert on each other.

Explanation:

try to look it up on google

I know. I know everything

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What can we learn from space images?

Arte-miy333 [17]

How planets look like,and maybe How the planets were created

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3 years ago

A sliver cylindrical rod has a length of 0.5 m and a radius of 0.4 m, find the density of the rod if it's mass is 2.640 kg

Serga [27]

Answer:

Density=10.50kg/m³

Explanation:

$Solution,\\Height(h)=0.5m\\Radius(r)=0.4m\\Now, \\Volume=\pi r^{2} h\\\\Volume=\pi *(0.4m)^{2} *0.5m\\\\Volume=0.251327m^{3} \\\\Again,\\\\Density=\frac{mass}{volume} \\\\Density=\frac{2.640kg}{0.251327} \\\\Density=10.50kg/m^{3}$

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2 years ago

When using the scientist method, which step comes last?

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3 years ago

A tugboat tows a ship with a constant force of magnitude F1. The increase in the ship's speed during a 10 s interval is 5.0 km/h

Ratling [72]

Answer

given,

time = 10 s

ship's speed = 5 Km/h

F = m a

a is the acceleration and m is mass.

In the first case

F₁=m x a₁

where a₁ = difference in velocity / time

F₁ is constant acceleration is also a constant.

Δv₁ = 5 x 0.278

Δv₁ = 1.39 m/s

$a_1=\dfrac{1.39}{10}$

a₁ = 0.139 m/s²

F₂ =m x a₂

F₃ = F₂ + F₁

Δv₃ = 19 x 0.278

Δv₃ = 5.282 m/s

a₃=Δv₂ / t

$a_3=\dfrac{5.282}{10}$

a₃ = 0.5282 m²/s

m a₃=m a₁ + m a₂

a₃ = a₂ + a₁

0.5282 = a₂ + 0.139

a₂=0.3892 m²/s

F₂ = m x 0.3892...........(1)

F₁ = m x 0.139...............(2)

F₂/F₁

ratio = $\dfrac{0.3892}{0.139}$

ratio = 2.8

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3 years ago

A force vector F1 points due east and has a magnitude of 200N. A second force F2 is added to F1. The resultant of the two vector

PilotLPTM [1.2K]

Answer:

The second vector $\vec{F_2}$ points due West with a magnitude of 600N

Explanation:

The original vector $\vec{F_1}$ points with a magnitude of 200N due east, the Resultant vector $\vec{R}$ points due west (that's how east/west direction can be interpreted, from east to west) with a magnitude of 400N. If we choose East as the positive direction and West as the negative one, we can write the following vectorial equation:

$\vec{F_1}+\vec{F_2}=\vec{R}\implies\vec{F_2}=\vec{R}-\vec{F_1}=-400N-200N=-600N$

With the negative sign signifying that the vector points west.

3 0

3 years ago