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3 years ago

What effect would decreasing the distance between objects have on their gravitational attraction to each other?

2 answers:

8 0

Decreasing the distance between objects will result in a greater gravitational attraction between each other. This is because the distance is inversely related to the gravitational attraction. For instance, in the same way, if the distance between two objects was increased, then their gravitational attraction to each other would decrease.

Lady_Fox [76]3 years ago

5 0

Decreasing the distance between two objects having a considerable mass would increase the attraction on gravitation. The reverse is true that if you separate or inrease the objects distance would substantially decrease their gravitational attraction. Most object in our planet is held by its gravitational force towards it's center.

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Physics help please

zhuklara [117]

Answer: 37.981 m/s

Explanation:

This situation is related to projectile motion or parabolic motion, in which the travel of the ball has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$x=52 m$ is the point where the ball strikes ground horizontally

$V_{o}$ is the ball's initial speed

$\theta=0$ because we are told the ball is thrown horizontally

$t$ is the time since the ball is thrown until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t+\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=120m$ is the initial height of the ball

$y=0$ is the final height of the ball (when it finally hits the ground)

$g=-9.8m/s^{2}$ is the acceleration due gravity

Knowing this, let's start by finding $t$ from (2):

$0=y_{o}+V_{o}sin(0\°) t+\frac{gt^{2}}{2}$ (3)

$0=y_{o}+\frac{gt^{2}}{2}$

$t=\sqrt{\frac{-2 y_{o}}{g}}$ (4)

$t=\sqrt{\frac{-2 (120 m)}{-9.8m/s^{2}}}$ (5)

$t=4.948 s$ (6)

Then, we have to substitute (6) in (1):

$x=V_{o}cos(0\°) t$ (7)

And find $V_{o}$ :

$V_{o}=\frac{x}{t}$ (8)

$V_{o}=\frac{52 m}{4.948 s}$ (9)

$V_{o}=10.509 m/s$ (10)

On the other hand, since we are dealing with constant acceleration (due gravity) we can use the following equation to find the value of the ball's final velocity $V$ :

$V=V_{o} + gt$ (11)

$V=10.509 m/s + (-9.8 m/s^{2})(4.948 s)$ (12)

$V=-37.981 m/s$ (13) This is the ball's final velocity, and the negative sign indicates its direction is downwards.

However, we were asked to find the ball's final speed, which is the module of the ball's final vleocity vector. This module is always positive, hence the speed of the ball just before it strikes the ground is 37.981 m/s (positive).

5 0

2 years ago

Three equal point charges, each with charge 1.15 μCμC , are placed at the vertices of an equilateral triangle whose sides are of

julsineya [31]

Answer:

$U=50.96J$

Explanation:

The electrostatic potential energy for pair of charge is given by

U=1/4π∈₀×(q₁q₂/r)

Hence for a system of three charges the electrostatic potential energy can be found by adding up the potential energy for all possible pairs or charges.For three equal charges on the corners of an equilateral triangle,the electrostatic potential energy is given by:

U=1/4π∈₀×(q²/r)+1/4π∈₀×(q²/r)+1/4π∈₀×(q²/r)

U=3×1/4π∈₀×(q²/r)

Substitute given values

$U=3\frac{1}{4\pi E_{o} }\frac{q^{2} }{r}\\ U=3\frac{1}{4\pi8.85*10^{-12} }\frac{(1.15*10^{-6}C )^{2} }{0.0007m}\\ U=50.96J$

8 0

3 years ago

The balance Lenght of a potent ometer wire for a Cell of emf 1.62v is 90cm. if the Cell is replaced by another one of emf 1.08v.

andrew-mc [135]

Answer:

answer is 3.05v

Explanation:

hope it is helpful and briliant

6 0

2 years ago

N 1800kg car has an of 3.8m/s? What is it on the car? acceleration force acting

nevsk [136]

Answer:

6840 N

Explanation:

The force acting on the car can be found by using Newton's second law:

F = ma

where

F is the net force on the car

m is the mass of the car

a is its acceleration

For the car in this problem,

m = 1800 kg

$a=3.8 m/s^2$

Substituting,

$F=(1800)(3.8)=6840 N$

7 0

3 years ago

What is the difference between a 4x100 and a 4x400 relay

qaws [65]

A 4x100 relay is where 4 people run 100 meters and a 4x400 relay is where 4 people run 400 meters

6 0

3 years ago