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3 years ago

10

what acceleration does the force of earth's gravity produceit varies depending on how long it has been falling.it changes direct

ion every 9.8 meters9.8 m/s/sgravity does not produce an acceleration

1 answer:

iren2701 [21]3 years ago

3 0

If gravity is the ONLY force acting on an object, then the force of
Earth's gravity produces acceleration of 9.8 meters per second²
downward. That number doesn't change, and it doesn't depend
on the mass or weight of the object.

If there are any other forces acting on the object at the same time,
like air resistance or some other kind of friction for example, then
all bets are off, and we can't say what the object's acceleration will
be in that case.

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If the action force is 100N what will be the reaction force

erik [133]

Answer:

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Newtons's 3rd Law of Motion

Newton’s 3rd Law of Motion, commonly referred to as the Law of Action and Reaction, describes the phenomena by which all forces come in pairs. If Object 1 exerts a force on Object 2, then Object 2 must exert a force back on Object 1. Moreover, the force of Object 1 on Object 2 is equal in magnitude, or size, but opposite in direction to the force of Object 2 on Object 1. Written mathematically:

Newton's 3rd Law Equation

This has many implications, some of which aren’t immediately obvious. For example, if you punch the wall with your fist with a force of 100N, the wall imparts a force back on your fist of 100N (which is why it hurts!). Or try this. Push on the corner of your desk with your palm for a few seconds. Now look at your palm... see the indentation? That’s because the corner of the desk pushed back on your palm.

running tiger

Although this law surrounds your actions everyday, often times you may not even realize its effects. To run forward, a cat pushes with its legs backward on the ground, and the ground pushes the cat forward. How do you swim? If you want to swim forwards, which way do you push on the water? Backwards, that’s right. As you push backwards on the water, the reactionary force, the water pushing you, propels you forward. How do you jump up in the air? You push down on the ground, and it’s the reactionary force of the ground pushing on you that accelerates you skyward!

As you can see, then, forces always come in pairs. These pairs are known as action-reaction pairs. What are the action-reaction force pairs for a girl kicking a soccer ball? The girl’s foot applies a force on the ball, and the ball applies an equal and opposite force on the girl’s foot.

How does a rocket ship maneuver in space? The rocket propels hot expanding gas particles outward, so the gas particles in return push the rocket forward. Newton’s 3rd Law even applies to gravity. The Earth exerts a gravitational force on you (downward). You, therefore, must apply a gravitational force upward on the Earth!

5 0

3 years ago

A chunk of paraffin (wax) has a mass of 50.4 grams and a volume of 57.9 cm3. What is the density of

ExtremeBDS [4]

$Formula\ for\ density:\\\\
p=\frac{m}{V}\\p-density,\\m-mass,\\V-volume\\\\
Data:\\
m=50,4grams\\
V=57,9cm^3\\\\
p=\frac{50,4g}{57,9cm^3}=0,87\frac{g}{cm^3}\\\\Density\ of\ paraffin\ is\ equal\ to\ 0,87\frac{g}{cm^3}.
$

6 0

4 years ago

Io and Europa are two of Jupiter's many moons. The mean distance of Europa from Jupiter is about twice as far as that for Io and

Anastaziya [24]

The universal gravitation law and Newton's second law allow us to find that the answer for the relation of the rotation periods of the satellites is:

$\frac{T_{Eu}}{T_{Io}}$ = 2.83

The universal gravitation law states that the force between two bodies is proportional to their masses and inversely proportional to their distance squared

$F = G \frac{Mm}{r^2}$

Where G is the universal gravitational constant (G = 6.67 10⁻¹¹ $\frac{N m^2 }{kg^2}$ ), F the force, m and m the masses of the bodies and r the distance between them

Newton's second law states that force is proportional to the mass and acceleration of bodies

F = m a

Where F is the force, m the mass and the acceleration

In this case the body is the satellites of Jupiter and the planet,

$G \frac{Mm}{r^2} = m a$

Suppose the motion of the satellites is circular, then the acceleration is centripetal

a = $\frac{v^2}{r}$ r

Where v is the speed of the satellite and r the distance to the center of the planet

we substitute

$G \frac{Mm}{r^2} = m \frac{v^2}{r} \\G \frac{M}{r} = v^2$

Since the speed is constant, we can use the uniform motion ratio

v = $\frac{\Delta x}{t}$

In the case of a complete orbit, the time is called the period.

The distance traveled is the length of the orbit circle

Δx = 2π r

We substitute

$G \frac{M}{r} = (\frac{2 \pi r}{T} )^2 \\T^2 = (\frac{4 \pi ^2}{GM}) \ r^3$

Let's write this expression for each satellite

Io satellite

Let's call the distance from Jupiter is

r = $r_{Io}$

$T_{Io}^2 = (\frac{4 \pi ^2}{GM} ) \ r_{Io}^3$ TIo² = (4pi² / GM) rIo³

Europe satellite

Distance from Jupiter is

$r_{Eu} = 2 \ r_{Io}$

We calculate

$T_{Eu} = ( \frac{4\pi ^2 }{GM} (2 \ r_{Io})^3\\T_{Eu} = ( \frac{4 \pi ^2 }{GM}) r_{Io} \ 8$

$T_{Eu}^2 = 8 T_{Io}^2$

$\frac{ T_{Eu}}{T_{Io}} = \sqrt{8} = 2.83$

In conclusion, using the universal gravitation law and Newton's second law, we find that the answer for the relationship of the relation periods of the satellites is:

$\frac{T_{Eu}}{T_{Io}} = 2.83$

Learn more about universal gravitation law and Newton's second law here:

brainly.com/question/10693965

6 0

3 years ago

Two particles having charges of 0.440 nC and 11.0 nC are separated by a distance of 1.80 m . 1) At what point along the line con

Harlamova29_29 [7]

Answer:

a) 0.3 m

b) r = 0.45 m

Explanation:

given,

q₁ = 0.44 n C and q₂ = 11.0 n C

assume the distance be r from q₁ where the electric field is zero.

distance of point from q₂ be equal to 1.8 -r

now,

E₁ = E₂

$\dfrac{K q_2}{(1.8-r)^2} = \dfrac{K q_1}{r^2}$

$(\dfrac{1.8-r}{r})^2= \dfrac{q_2}{q_1}$

$\dfrac{1.8-r}{r}= \sqrt{\dfrac{11}{0.44}}$

1.8 = 6 r

r = 0.3 m

<h3>b) zero when one charge is negative.</h3>

let us assume q₁ be negative so, distance from q₁ be r

from charge q₂ the distance of the point be 1.8 +r

now,

E₁ = E₂

$\dfrac{K q_2}{(1.8+r)^2} = \dfrac{K q_1}{r^2}$

$(\dfrac{1.8+r}{r})^2= \dfrac{q_2}{q_1}$

$\dfrac{1.8+r}{r}= \sqrt{\dfrac{11}{0.44}}$

1.8 =4 r

r = 0.45 m

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Ithoshpere is space sheeld. asthenosphere next sheets of sky

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4 years ago